Physics 132: What is an Electron? What is Light?

Physics 132: What is an Electron? What is Light?

Roger Hinrichs

Paul Peter Urone

Paul Flowers

Edward J. Neth

William R. Robinson

Klaus Theopold

Richard Langley

Julianne Zedalis

John Eggebrecht

E.F. Redish

University of Massachusetts Amherst Libraries

Amherst, MA

Physics 132: What is an Electron? What is Light?

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Physics 132: What is an Electron? What is Light? by Roger Hinrichs, Paul Peter Urone, Paul Flowers, Edward J. Neth, William R. Robinson, Klaus Theopold, Richard Langley, Julianne Zedalis, John Eggebrecht, and E.F. Redish is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

Welcome to Physics 132 - Introduction to the Course

1

Hello, welcome to Physics 132 at University of Massachusetts, Amherst! This course is where we get to use the ideas from Physics 131 (forces, energy, etc.) to really understand two fundamental objects: electrons and light. These two fundamental objects are all around you. You can see this page due to light. How many electronic devices are you carrying right now? Moreover, understanding these two objects is key for understanding the physical original of biological processes. One cannot hope to understand molecular pathways within cells such as photosynthesis and neural activation without talking about electrons and light, but what are electrons? What is light? The goal of this course is to help you develop your own understanding of these questions.

How do you define what something is? Especially, as is the case for light and electrons, when the object you are trying to define is subatomic and so very far removed from our everyday experience? These are not scientific questions: we cannot design an experiment to test their answers. Thus, this physics course must, right out of the gate, go beyond physics to philosophy. Specifically, we must venture into metaphysics: a branch of philosophy that explores the nature of being, existence, and reality. The word physics actually comes from a Greek word ΦIΣIK meaning “nature.” META is a Greek word meaning “beyond.” So metaphysics literally means beyond nature. In particular, to answer the question of “how do you define what something is?” we need a branch of metaphysics called ontology.

So how will we define things like electrons and light? In philosophy, we would ask, “what will be our ontological framework?”

We will construct our definitions of light and electrons in this course by:

  • Listing what characteristics objects have
  • Listing how objects interact with other constructs

Thus, our definition of an electron will be a list of its properties and interactions. In defining light and electrons in this way, we will see that we must actually look at two other objects: electric and magnetic fields to complete our picture. Does listing properties and interactions really entirely define what it means to be an electron or light? Probably not! There are certainly other possible ontological frameworks, but those are topics for a philosophy class. Science is a powerful way to understand the world, but its requirement of experimental falsifiability does have limitations which is why general education courses are so important for scientists!

How to Use This Book

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This book has been specifically designed for this course out of free-and-open resources such as the OpenStax College Physics textbookhttps://openstax.org/details/college-physics, University of Maryland’s UMD NEXUS Wikibookhttp://umdberg.pbworks.com/w/page/90716129/Working%20content%20I%20(2015) and http://umdberg.pbworks.com/w/page/104048687/Working%20content%20II%20(2016), as well as other resources from around the internet. While the text has all the information you need, some sections are also available as videos on our course YouTube page. These sections will have the link at the beginning with the section below.

Many students often complain that physics lectures, “spend all of their time on the easy problems and never have time to work through harder ones.” In this course, we want to make sure that you have time to work through harder ideas and problems in-class where you can get help from the TAs and instructor. To make sure we have time to explore these concepts, you will need to have some familiarity with basic facts before coming to class. We will expect you to learn the relevant formulas, symbols, and terminology as well as review relevant material from 131 before coming to class using the resources in this book. We will NOT expect you to learn how to solve new types of advanced problems. This style of class, with the expectation of completing readings first is called a flipped classroom and is based on researchStoltzfus, Matthew W. “Active Learning in the Flipped Classroom: Lessons Learned and Best Practices To Increase Student Engagement.” In The Flipped Classroom Volume 1: Background and Challenges, 1223:105–22. ACS Symposium Series 1223. American Chemical Society, 2016.https://doi.org/10.1021/bk-2016-1223.ch008..

This book, like the course, is divided into units. Each unit will begin with a Unit on a Page which summarizes the key points of the unit, both prep and in-class, on a single page. Due to the constraint of fitting on a single page, these pages are necessarily dense. The goal is to help you focus on the underlying themes, see what is really important. I would recommend you print these pages and keep them with you both during your prep and in-class to help you see the big picture.

After the Unit on a Page, you will find a short description of the context that we will be using to explore these physical concepts. We want physics to be relevant to you. Thus, each unit has one (or more!) biological applications that we will use as common themes throughout the unit. To make sure everyone is on the same page, we will explain the needed biology.

After the biology, there will be the physics material that you need to master before coming to class: readings, videos, flashcards to help you remember, and the occasional simulation for you to play with. To help you ensure that you have understood the material, there are homework problems linked from the text that will appear in special colored boxes that look like the box below to help you notice them. These problems are interspersed throughout the text, including on the biology material, to make it easy for you to know where to look for the information for each problem. To make sure you don’t miss any homework problems, we will also make a list of all the problems at the end of the chapter. The homework problems are hosted in the Edfinity homework system. Follow the procedures outlined by your instructor on how to get an account, and check the syllabus for the grading policies.

Homework Problem:

Your homework problems will appear as links in boxes like this one. Click the link to complete the homework in the Edfinity homework system.

They are placed in-line with the text so that you know where to look for the information to solve each problem.

To further ensure that everyone is ready for what will be discussed in class, there will be quizzes on the material as outlined in your syllabus. We WANT you to be prepared for these quizzes; the entire purpose of this book is to prepare you. As such, there will be UMass-Amherst Instructor’s Notes like the one below which will tell you exactly what you need to focus on to be ready for the quizzes.

 

Instructor’s Notes

 

Be sure to pay attention to the text in this box! This is the material that will be on your quiz

 

We encourage your feedback on this book. If you have suggestions for comments or would like to report an error, please go to https://goo.gl/forms/VQaiFUtKD1PxJfH83 and complete the form.

We hope this book is helpful!

Editors:

Brokk Toggerson

Emily Hansen

Goals for The Course

3

The big questions: What is an electron? What is light?

These are the big questions that I am hoping to provide at least some answers to  over the course of this semester. Light and electrons are two of the most fundamental building-blocks of our Universe (as far as we know they have no substructure!). Understanding these two basic elements and how they interact with each other will help you better understand many other fields of science and technology from chemistry to electronics.

Below are the fundamental goals that I would hope that you will take away from the “lecture” portion of this course. These are things that I hope that you will remember many years from now when you have forgotten all of the details. They are divided into two categories: Physics Goals and Skills Goals. There are a completely separate set of goals for the laboratory portion of the course which you will see in lab.

Physics Goals

These are the basic goals of any introductory physics course and are deeply connected to the material we will be covering. While these goals are generally similar to those from P131, I will be expecting that you will be developing a greater proficiency in these goals. Moreover, there are some changes for you to note.

  1. Physics is a set of principles and the fundamental ideas that relate them, NOT a list of equations… This is quite possibly the most common misconception that people have about physics: they tend to think of physics as a list of formulae to be memorized and that all solving physics problems entails is finding the correct formula to get from where you are to where you want to be. This could not be further from the truth. Physics is a list of conceptual ideas expressed mathematically. These conceptual ideas form the “rules” that all of the other sciences, biology, chemistry, etc., have to follow. Thus, knowing the basic principles of physics is beneficial to any scientist!
  2. (More Emphasis!) These principles of physics can be expressed in multiple different ways… When people think of physics they tend to think of equations. However, the ideas of physics can be represented in words (as was done in Europe in the days before Isaac Newton!) pictures, and graphs. As a budding scientist, it is important for you to be able to think of ideas in multiple formats and to be able to decide which format is the best for a given situation. Given the increased remoteness of the material in this course from your everyday experience, I am going to increase the emphasis on this goal relative to P131; being able to write clearly and succinctly about physics concepts using words will be important!
  3. Appreciate the value of the problem solving method used by the discipline of physics… This is the goal most commonly given by non-physics faculty as to why they want their students to take physics, “problem solving.” However, the problem solving method used by physicists is a bit different than that to which you may be accustomed. In physics, we reason from the fundamental principles at play in a given situation. This requires you to look beyond the surface features in a given problem to see how a person throwing a ball, a block sliding down a ramp, and a building standing still are all similar in that they all rely on the same fundamental principle of . As a consequence of     starting from fundamental principles, we physicists like to employ what is known as a “reductionist” approach and think about an idealized world first. This idealized world can seem quite bizarre to people new to physics as it is populated with point masses moving across frictionless surfaces. The goal of this approach is to move to a problem where the fundamental principles are easier to see and understand. The complications that are present in the real world are then added back in later.
  4. (New relative to Physics 131) The fundamental principles of nature do not need to conform with “common sense”… You already saw some of this in P131 where some results can seem counter-intuitive. In this course, however, we will encounter many more phenomena that may seem as though they do not make sense. Some of this will be due to the fact that you do not have as much experience with the world of electrons as you do with the world of friction and springs. Some of these topics’ seemingly nonsensical nature, on the other hand, will be due to the fact that the topics represent fundamentally new concepts without any corresponding analog in your experience. For these ideas any analogy will inherently be imperfect and beginning from first principles (see Physics Goal #3) will be even more critical!
  5. Learn how to use fundamental principles to generalize from one specific situation to a class of similar ones… Often, we will study a particular principle or idea within the context of a specific situation. The beauty of the laws of physics, however, is that these laws can be applied anywhere and the results of one analysis can often be applied to other related problems. For example, the motion of ANY object near the surface of the earth shares certain features. Thus, by studying the motion of a basketball, you can infer something about a skydiver. The trick is knowing what aspects of a given problem transfer to a new situation and which do not – your guide here are the fundamental principles (see Physics Goal #1).
  6. (More Emphasis!) Understand that the physics we study is connected to your everyday experience and the material in your other courses… We want you to see the applications of physics all around you and to connect to your other courses. As such, we may ask you to pull information from your everyday experience or other knowledge to solve problems. Our hope is that physics will provide a new perspective on the material in your other classes. In this class, we will, in particular, be making a lot of connections to the subject of chemistry.

Skills Goals

In addition to physics content, there is a certain set of skills that I want you to take away from this class. These Skills Goals, however, are a bit different and more advanced than for P131 – as benefits a second semester course. These skills are just for the “lecture” portion of the course; the lab portion has additional goals emphasizing data analysis that will be discussed in your first lab sections.

  1. (More Emphasis!) Becoming more comfortable working in symbols… This is a very important skill as every field becomes ever more quantitative; you need to be able to work in a wholly symbolic fashion and be able to read and interpret what the symbols in an equation  mean. You began developing this skill in P131, but we will focus on it more heavily in this class.
  2. (New!) Be able to combine different ideas into a single analysis… Most of the situations we looked at in P131 were single principle situations; we used either Newton’s Laws or conservation of energy for our analysis. We almost never used more than one idea. However, almost all of modern science is what is called interdisciplinary: the result of combining ideas from multiple places. The trend of holistic health is a good example as it looks at not only biological factors but also sociological and psychological. The study of light and electrons is a great place to practice this skill as they are interesting multifaceted objects and we will need to be able to put multiple principles together to understand them.
  3. (New!) Interpretation of mathematical results… In P131, when problems required a number or formula, you would solve it out and be done. P132 will force you to extend your problem solving skills beyond P131 to include one more step critical to more advanced analysis: interpretation of your result. In this class, formulas will sometimes give you special results that require additional consideration. Other times, the result of a calculation may even be nonsense: infinity or imaginary! One of the amazing things about physics is that even these seemingly meaningless answers can tell you about how the world works. In this class, you will expected to develop the skills to interpret these results and learn what they can tell you.

Teamwork goals

I firmly believe in the critical role of teamwork to the success of the scientific enterprise. My own personal experience has confirmed this role many times and research shows that people learn better when they engage with ideas in conjunction with others. Throughout the course, regardless of if you are on a formal team or not, you will be furthering the scientific skills you developed in P131:

  1. Appreciate that the “solitary genius” image of a scientist which is so pervasive in our culture no longer exists (if they ever did)… Science, all science, is now done in teams ranging in size from small teams of three to massive collaborations with memberships in the thousands. These skills are something you can put on your CV/Resume as having developed in this class!
  2. Appreciate that the work done by a team is usually better than the product of even its strongest member…. You will see over the course of the semester that this is true!. However, the environment in which we find ourselves is not as conducive to formalized teamwork as P131. However, I would strongly encourage you to work with other people both in-class on activities and out-of-class on homework and exam preparation.

Biology, Chemistry, Physics, and Mathematics

4

Editors’ note: This section is based upon work fromE.F. Redish, “The disciplines: Physics, Biology, Chemistry, and Math,” in Introductory Physics for the Life Sciences I - NEXUS Physics, University of Maryland, College Park, 2013.

To become a biologist or health-care professional, you have to study a variety of scientific disciplines — biology, chemistry, physics, and math. You might have noted that the world doesn’t actually divide itself in this way. Rather, the disciplines historically have been a way of choosing a sub-class of the phenomena that occur in the world and looking at a particular aspect of them with a particular purpose in mind. Different disciplines have different sets of tools and ways of knowing. Looking at something from different disciplinary perspectives adds a richness and depth to our understanding — like taking two 2-D pictures and merging them into a 3-D image.

Your introductory science and math classes often provide you with some basics — tools, concepts, and vocabulary — but may not give you a perspective on what each discipline adds to what you are learning and how they all fit together. Each discipline has its own orientation and perspective towards the development of a professional scientist. Here’s a brief (and oversimplified) overview of the different disciplines that you encounter in studying biology.

Biology

Biology, as you well know, is the study of living organisms. The approach taken by biology is guided by and constrained by the fact that the subject is about living organisms.

Chemistry

Chemistry starts with the idea that all matter is made up of certain fundamental pieces – atoms of about 100 different kinds (elements) – and is about the ways those elements combine to form more complex structures – molecules. But chemistry is not just about building molecules. It’s about what you can do with that knowledge in our macroscopic world.

For a chemist, most of what happens in biology is “macroscopic” – there are lots and lots of atoms involved – even though you might need a microscope to study it. In introductory chemistry you often assume that reactions are taking place at standard temperature and pressure (300 K and 1 atm).

Physics

The goal of physics is to find the fundamental laws and principles that govern all matter — including biological organisms. Those laws and principles can lead to many types of complex and apparently different phenomena. Physics as traditionally taught at the introductory level tends to explicitly introduce four scientific skills that may seem different to what you see in introductory biology and chemistry classes, but these four skills will prove valuable for your career.

This way of doing science is a bit different from the way biology is often done — but elements of this approach and the constraints imposed on biology by the laws of physics are becoming increasingly important both for research biologists and health-care professionals. For more discussion, see the page, What Physics Can do for Biologists.

Math

Math is a bit different from the sciences. In its essence math is about abstract relationships. Since math is about abstract relationships and how they behave, it’s not “about” anything in the physical world. But it turns out that a lot of relationships in science can be modeled by relations that obey mathematical rules, often very accurately. (If you think this is surprising or strange, you aren’t alone. For fun, take a look at the interesting article by the Nobel Prize-winning nuclear and mathematical physicist, Eugene Wigner, entitled, “The Unreasonable Effectiveness of Mathematics in the Natural Sciences.“)

Math as taught in math classes often is primarily about the abstract relationships — learning how to use the tools of math. Making the transition to using math in real-world situations may be quite jarring as there are now additional things to pay attention to other than the math itself — such as figuring out how the elements of the real-world system get translated into a mathematical model and worrying about whether the mathematical model is good enough or not. I like to think of it this way: in math class you learn the “grammar” of the language of science. Here, and in your other science courses, you need to start learning “vocabulary.” With grammar and vocabulary together, you can begin to describe the Universe.

Bringing these disciplines together

Bringing these all together to permit coherent and productive thinking is a challenge! In this class we expect and encourage you to bring to bear knowledge you have from your other science classes — to try to see how they fit together, support each other, and to learn to identify when a particular disciplinary approach might be most appropriate and useful.

While these different scientific disciplines are all ultimately working to the same end: understanding the Universe. They did evolve semi-independently historically. As such, there are cultural differences between the sciences just as there are cultural differences between countries (driving on the right or left, for example). These cultural differences are not about “right” or “wrong” ways of doing things. They are just different. In fact, these differences in perspective are a strength! The different viewpoints between disciplines have often led to many important discoveries throughout history and are still where many of the most exciting advancements are being made. They can, however, be confusing. We have, therefore, worked with biology, chemistry, and mathematics instructors here at University of Massachusetts – Amherst. These discussions have resulted in some common language used in this book, which might, therefore, be different than in other physics text you may look at. Even so, there are sometimes places where we need to leverage the strength of a different viewpoint. We will point out these differences using boxes like the one below.

A Venn-diagram showing the different sciences

We will use boxes like this to point out important differences between disciplines. Again, these differences are not “right” or “wrong” ways of doing things. They are simply artifacts of the different sciences evolving independently for centuries with influences from different cultures from across the globe.

Unit I

I

Unit I On-a-Page

Principles and Definitions

If you have had Dr. Toggerson or Dr. Bourgeois for Physics 131, you are familiar with a distinction made  between principles and definitions. The principles are the fundamental rules of the Universe that describe how things work. Concepts which are definitions, on the other hand, simply describe a quantity. For example,

\vec{p} = m \vec{v}p⃗=mv⃗p⃗=mv⃗widevec {p} =m widevec {v}">

is the definition of momentum for a massive particle; this equation offers no deep foundational insights on how the universe works. We physicists simply noted that the quantity m\vec{v} came up a lot and we gave it a name \vec{p}. In order to describe how the Universe works, principles will often involve multiple definitions. Note, sometimes a principle or definition has an equation, other times it is just stated in words! This connects to Physics Goals 1 and 2 for this course.

To help get those of you who may not be used to this distinction acquainted and to help organize the huge amount of factual information in this particular unit, I will list the principles for this unit. You can quickly see how short this list is.

Principles for this Unit

Basic Properties of Light

  • Light in a vacuum always travels at the speed of light c=299792458 \, \frac{\mathrm{m}}{\mathrm{s}}

Basic Properties of Waves

  • Fundamental connection between wavelength \lambda, frequency \nu, and wave speed v: v = \lambda \nu
  • The amplitude A is independent of frequency \nu

Basics of Energy

  • Energy is conserved. The change in energy \Delta E is caused by the exchange of energy through heat Q and work W: \Delta E = Q + W.

Wave Particle Duality

  • You can convert from the wave picture to the particle picture through the de Broglie relation: p = \frac{h}{\lambda} where p is the momentum of the particle.
A Venn-diagram showing the different sciencesIn chemistry, you probably saw the conversion between energy and wavelength for photons done through the equation E = \frac{hc}{\lambda}. We will NOT be considering that as a principle to begin analyses / solving problems. The reason is that, in chemistry, you only considered converting between energy and wavelength for photons. We want to think about BOTH photons AND electrons. It turns out, that E = \frac{hc}{\lambda} ONLY works for photons, while p = h / \lambda works for both photons and electrons! Thus we consider p = h/\lambda to be our principle. Many students get tripped up by applying E = \frac{hc}{\lambda} to electrons. Don’t fall into this trap!
  • The probability of finding a particle in a given location is proportional to the square of the amplitude P \propto A^2.
    • Increase amplitude by 3, probability goes up by 9.
    • For light, we represent the amplitude not by A but by E. This is NOT the energy (confusing I know, but it is what it is). We will see why we use E later in the semester.

Standing Waves

  • The wave must “fit” in the box or on the ring. For a box, this means that the box must be an integer number of 1/2 wavelengths. For a ring, an integer number of waves must fit around the ring.

Instructor’s Note

 

The ideas in this unit can be connected in many different ways. One possible useful way to organize such information is in a “concept map” like the one shown below. The map is also available at this link.

In this map:

  • UMass maroon bubbles are big ideas
  • Yellow bubbles apply to massless particles like light
  • Green bubbles apply to massive particles like electrons

I would recommend printing a copy for use in class!

A concept map of all of the ideas in Unit 1
Unit 1 on a page!

Unit I - Introduction and Context for the Unit

1

Interdisciplinary questions we want to answer in this unit

A Venn-diagram showing the different sciences

The motivation for this unit is pretty straightforward: you know from Chemistry electrons and photons can behave as both particles and wave a property known as wave-particle duality.

  • What does this wave-particle duality actually mean?
  • WHY does what you learned in chemistry actually work the way it does?
    • How does this wave-particle duality result in some of the most fundamental properties of chemistry such as discrete energy levels in atoms?
    • Why is the wavelength of a photon emitted from an atom \Delta E = \frac{hc}{\lambda}?

Introduction to the Unit

In this unit, we will follow our ontological framework and begin exploring what light and electrons are by listing some of their basic properties. One of the most famous properties of both light and electrons is known as wave-particle duality: sometimes they behave like particles and sometimes they behave like waves. However, electrons and light are neither particles nor waves: they are a completely new type of object with properties of both. This duality is a reflection of the fact that both light and electrons do not obey the laws of Classical Physics that you learned in Physics 131, they are too small. Instead, electrons and light obey Quantum Mechanics. The way I would recommend that you think about the relationship between classical physics and quantum mechanics is in the paradigm of physics striving for ever-more-accurate approximations to reality. Classical mechanics is a good enough approximation to get people to the moon (they did it!). When things get small, you need a better approximation: quantum mechanics. There is a principle, the correspondence principle, which states the quantum mechanics must reproduce classical mechanics for large objects.

In this unit we will explore both the wave and particle natures of light and electrons. First however, we must define to ourselves what waves and particles are! Following our ontological framework, we will therefore need to look at some of the basic properties that characterize waves and particles in general. Thus, we will begin with some review of particles from physics 131 and then a discussion of waves, which may be familiar to some of you. Once we have defined waves and particles through listing their properties we will explore how these properties manifest for electrons and light.

As you read, you MUST keep in mind that light and electrons are neither particles nor waves. They are something completely new (quantum mechanical objects) that you have zero previous experience with. Particles and waves are simply ways of visualizing these objects in ways our brains can understand. Neither picture is 100% correct. The correct approach is to jump back-and-forth between these two pictures

What we will do in class

This idea of electrons and light being neither particles no waves but having properties of both is a hard one to get used to. In class, we will spend a lot of time practicing jumping between the wave and particle pictures: seeing the benefits that each picture can bring in various situations. The goal is that, through practice, you become more comfortable with bouncing back and forth between pictures as the situation demands.

After a bit of practice with moving between the wave picture and particle pictures of light and electrons, we will combine this understanding with one of the most important ideas in physics: conservation of energy. Thinking about this fundamental principle in conjunction with the fundamentally quantum nature of light and electrons will allow us to understand many different phenomena such as, “Why do electrons in atoms have defined energy levels?” You know from chemistry courses that they do. Our goal is to explain why!

What you should get out of this prep

In order to explore these ideas in class, you need to have a grasp on the basic terminology of waves and particles. You need to know that particles are characterized by their energy, momentum, and number while waves are described by their wavelength and frequency/period, and amplitude. Both waves and particles can be characterized by a speed: a critical fact for converting between the wave and particle pictures. The following chapters will refresh energy and momentum from 131 and introduce the needed concepts for waves: amplitude, frequency, and wavelength. You need to know what all these terms mean, the basic formulas for them, and how they are connected

Also in this unit, some of the basic properties of light and electrons. We will establish electrons via a tour of the atom (probably review for most of you). We will also introduce anti-matter: a type of matter identical to the normal matter with which you are familiar in every respect with two exceptions. First, anti-matter has the opposite charge from normal matter (anti-electrons have positive charge). Second, when matter and anti-matter collide, the result is light. While there are other interesting questions about anti-matter (why isn’t it everywhere?) those two points are all you need to know from the prep. With regards to light, we will introduce the different kinds of light (radio, infrared, …) and the particle of light: the photon. Some of the basic properties of the photon will be introduced. The most important of which you need for the prep are the fact that the mass of the photon is zero and the fact that it always travels at the speed of light c. Finally, we will also explore de Broglie’s relationship on how to connect wave and particle picture

The last topic in this unit’s preparation you need to be familiar with is the idea of intensity: energy per area per time. The text will introduce the idea of power (energy per time) and its unit the Watt. This concept will also be important for converting between the wave and particle pictures.

Instructor’s Notes

 

This is a lot of information. Remember, we are not expecting mastery of it all and we are certainly not expecting you to have a complete picture of how everything Just make sure you know the definitions, formulas, and units for everything. To help keep you focused on what is important, there is a summary tables below (only focusing on the facts for your quiz). Keep these tables handy as you go through the reading: the information will all be explained as you go.

Waves and Particles

This information applies to both electrons and light.

Intensity, detailed later, is the energy per time per area (or power P = E/t):

I = \frac{P}{A} = \frac{E}{t\cdot A}

Properties Convert between them
Waves
  • Amplitude A, or E for light.
  • Wavelength \lambda, measured in meters.
  • Period T, measured in seconds.
  • Frequency \nu = \frac{1}{T}, measured in 1/s or Hz.
  • Speed v = \lambda \nu measured in m/s.
  • Intensity: I \propto A^2
  • de Broglie relation: p = \frac{h}{\lambda}
    • Do NOT use E = \frac{hc}{\lambda} as that only works for light!
  • Amplitude: A^2 \propto P where P is the probability of finding a particle.
  • Intensity: A^2 \propto \frac{(\mathrm{Number \, hitting \, per \, second})\cdot(\mathrm{Energy \, of \, each})}{\mathrm{Area}}
Particles
  • Energy E
  • Momentum \vec{p}
  • Speed v
  • Number N
  • Intensity: I = \frac{(\mathrm{Number \, hitting \, per \, second})\cdot(\mathrm{Energy \, of \, each})}{\mathrm{Area}}

Electrons and Photons

Note, in general, if you see a c or an \epsilon_0 in an equation, it applies to light only! The wave and particle information in the table still applies to both electrons and photons!

Waves Particles
Light
  • Travel at c
  • Different frequencies → different kinds of radiation
  • Amplitude E (which, again, is NOT energy) is explicitly related to intensity: I = \frac{1}{2} \epsilon_0 E^2.
  • Called photons \gamma
  • Mass: Zero
  • Electric charge: Zero
  • Speed: Always travel at c
  • Energy E and momentum p are related by E = pc
  • Different energies → different types of radiation
Electrons
  • Wavelength determined, as always, by de Broglie relation p = h/\lambda.
  • Mass: m_e = 9.11 \times 10^{-31} \, \mathrm{kg}
  • Electric charge |q_e| = 1.602 \times 10^{-19} \, \mathrm{C}
  • Speed: Less than the speed of light
  • Momentum: \vec{p} = m \vec{v}
  • Kinetic energy: E = \frac{1}{2} mv^2 = \frac{p^2}{2m}
  • There is also a particle called anti-electron or positron with the same mass and charge except it has positive charge while the familiar electron has negative charge. When the two meet, they destroy each other.

Basics of Matter

2

Static Electricity and Charge

Editors’ note: This section is derived from Derived from 18.2 Static Electricity and Charge: Conservation of Charge by OpenStax, Bobby Bailey

This piece of gold-colored amber from Malaysia has been rubbed and polished to a smooth, rounded shape.
Figure 1. Borneo amber was mined in Sabah, Malaysia, from shale-sandstone-mudstone veins. When a piece of amber is rubbed with a piece of silk, the amber gains more electrons, giving it a net negative charge. At the same time, the silk, having lost electrons, becomes positively charged. (credit: Sebakoamber, Wikimedia Commons)

What makes plastic wrap cling? Static electricity. Not only are applications of static electricity common these days, its existence has been known since ancient times. The first record of its effects dates to ancient Greeks who noted more than 500 years B.C. that polishing amber temporarily enabled it to attract bits of straw (see Figure 1). The very word electric derives from the Greek word for amber (electron).

Many of the characteristics of static electricity can be explored by rubbing things together. Rubbing creates the spark you get from walking across a wool carpet, for example. Static cling generated in a clothes dryer and the attraction of straw to recently polished amber also result from rubbing. Similarly, lightning results from air movements under certain weather conditions. You can also rub a balloon on your hair, and the static electricity created can then make the balloon cling to a wall. We also have to be cautious of static electricity, especially in dry climates. When we pump gasoline, we are warned to discharge ourselves (after sliding across the seat) on a metal surface before grabbing the gas nozzle. Attendants in hospital operating rooms must wear booties with aluminum foil on the bottoms to avoid creating sparks which may ignite the oxygen being used.

How do we know there are two types of electric charge? When various materials are rubbed together in controlled ways, certain combinations of materials always produce one type of charge on one material and the opposite type on the other. By convention, we call one type of charge “positive”, and the other type “negative.” For example, when glass is rubbed with silk, the glass becomes positively charged and the silk negatively charged. Since the glass and silk have opposite charges, they attract one another like clothes that have rubbed together in a dryer. Two glass rods rubbed with silk in this manner will repel one another, since each rod has positive charge on it. Similarly, two silk cloths so rubbed will repel, since both cloths have negative charge. Figure 2 shows how these simple materials can be used to explore the nature of the force between charges.

 

(a) Negatively charged cloth is attracted to the positively charged glass rod which is hanging by the thread. (b) A positively charged glass rod is hanging with a thread. When another positively charged glass rod brought close to the first rod it deflects due to the repulsive force. (c) Two negatively charged silk cloth brought close to each other repel each other.
Figure 2: A glass rod becomes positively charged when rubbed with silk, while the silk becomes negatively charged. (a) The glass rod is attracted to the silk because their charges are opposite. (b) Two similarly charged glass rods repel. (c) Two similarly charged silk cloths repel.

More sophisticated questions arise. Where do these charges come from? Can you create or destroy charge? Is there a smallest unit of charge? Exactly how does the force depend on the amount of charge and the distance between charges? Such questions obviously occurred to Benjamin Franklin and other early researchers, and they interest us even today.

Instructor’s Note

 

The structure of the atom will be discussed in more detail in a later section.

Charge Carried by Electrons and Protons

Franklin wrote in his letters and books that he could see the effects of electric charge but did not understand what caused the phenomenon. Today we have the advantage of knowing that normal matter is made of atoms, and that atoms contain positive and negative charges, usually in equal amounts.

Figure 3 shows a simple model of an atom with negative electrons orbiting its positive nucleus. The nucleus is positive due to the presence of positively charged protons. Nearly all charge in nature is due to electrons and protons, which are two of the three building blocks of most matter. (The third is the neutron, which is neutral, carrying no charge.) Other charge-carrying particles are observed in cosmic rays and nuclear decay, and are created in particle accelerators. All but the electron and proton survive only a short time and are quite rare by comparison.

Three electrons are shown moving in different direction around the nucleus and their motion is similar to planetary motion.
Figure 3: This simplified (and not to scale) view of an atom is called the planetary model of the atom. Negative electrons orbit a much heavier positive nucleus, as the planets orbit the much heavier sun. There the similarity ends, because forces in the atom are electromagnetic, whereas those in the planetary system are gravitational. Normal macroscopic amounts of matter contain immense numbers of atoms and molecules and, hence, even greater numbers of individual negative and positive charges.

The charges of electrons and protons are identical in magnitude but opposite in sign. Furthermore, all charged objects in nature are integral multiples of this basic quantity of charge, meaning that all charges are made of combinations of a basic unit of charge. Usually, charges are formed by combinations of electrons and protons. The magnitude of this basic charge is

|q_e| = 1.602\times 10^{-19} \mathrm{C}

The symbol q is commonly used for charge and the subscript e indicates the charge of a single electron (or proton).

The SI unit of charge is the coulomb (C). The number of protons needed to make a charge of 1.00 C is

1.00 \, \mathrm{C} \times \frac{1 \, proton}{1.602 \times 10^{-19} \, \mathrm{C}} = 6.25 \times 10^{18} \, \mathrm{protons}

Similarly, 6.25 \times 10^{18} electrons have a combined charge of −1.00 coulomb. Just as there is a smallest bit of an element (an atom), there is a smallest bit of charge. There is no directly observed charge smaller than |q_e|, and all observed charges are integral multiples of |q_e|.

Figure 4 shows a person touching a Van de Graaff generator and receiving excess positive charge. The expanded view of a hair shows the existence of both types of charges but an excess of positive. The repulsion of these positive like charges causes the strands of hair to repel other strands of hair and to stand up. The further blowup shows an artist’s conception of an electron and a proton perhaps found in an atom in a strand of hair.

A girl is touching a Van de Graaff generator with her hair standing up. A magnified view of her single hair is shown which is filled with electrons and protons.
Figure 4: When this person touches a Van de Graaff generator, she receives an excess of positive charge, causing her hair to stand on end. The charges in one hair are shown. An artist’s conception of an electron and a proton illustrate the particles carrying the negative and positive charges. We cannot really see these particles with visible light because they are so small (the electron seems to be an infinitesimal point), but we know a great deal about their measurable properties, such as the charges they carry.

Key Takeaways: Things Great and Small: The Submicroscopic Origin of Charge

With the exception of exotic, short-lived particles, all charge in nature is carried by electrons and protons. Electrons carry the charge we have named negative. Protons carry an equal-magnitude charge that we call positive. (See Figure 4.) Electron and proton charges are considered fundamental building blocks, since all other charges are integral multiples of those carried by electrons and protons. Electrons and protons are also two of the three fundamental building blocks of ordinary matter. The neutron is the third and has zero total charge.

Instructor’s Note

 

This fact that there are no observed free particles with less than |q_e| of charge is important and will be used in some of your homework problems.

Separation of Charge in Atoms

Charges in atoms and molecules can be separated—for example, by rubbing materials together. Some atoms and molecules have a greater affinity for electrons than others and will become negatively charged by close contact in rubbing, leaving the other material positively charged. (See Figure 5.) Positive charge can similarly be induced by rubbing. Methods other than rubbing can also separate charges. Batteries, for example, use combinations of substances that interact in such a way as to separate charges. Chemical interactions may transfer negative charge from one substance to the other, making one battery terminal negative and leaving the first one positive.

When materials are rubbed together, charges can be separated, particularly if one material has a greater affinity for electrons than another. (a) Both the amber and cloth are originally neutral, with equal positive and negative charges. Only a tiny fraction of the charges are involved, and only a few of them are shown here. (b) When rubbed together, some negative charge is transferred to the amber, leaving the cloth with a net positive charge. (c) When separated, the amber and cloth now have net charges, but the absolute value of the net positive and negative charges will be equal.
Figure 5: When materials are rubbed together, charges can be separated, particularly if one material has a greater affinity for electrons than another. (a) Both the amber and cloth are originally neutral, with equal positive and negative charges. Only a tiny fraction of the charges are involved, and only a few of them are shown here. (b) When rubbed together, some negative charge is transferred to the amber, leaving the cloth with a net positive charge. (c) When separated, the amber and cloth now have net charges, but the absolute value of the net positive and negative charges will be equal.

No charge is actually created or destroyed when charges are separated as we have been discussing. Rather, existing charges are moved about. In fact, in all situations the total amount of charge is always constant. This universally obeyed law of nature is called the law of conservation of charge.

Play with the Simulation

Below is a simulation of a balloon and a sweater. As you probably know, if you rub a balloon on a sweater, it will stick to a wall.

A few things to note:

  • The total number of charges is conserved – electrons move from the sweater to the balloon.
  • If you have two balloons with negative charge, they will repel, just like in real life (check it for real if you don’t believe us!)
  • When you bring the balloon near the wall, what happens to the electrons in the wall?

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Law of Conservation of Charge

The total charge is constant in any process. This law is truly universal and of such importance that we will revisit it again in a later section.

In more exotic situations, such as in particle accelerators, mass, \Delta m, can be created from energy in the amount using Einstein’s famous relation:

\Delta m = \frac{E}{c^2}.

Sometimes, the created mass is charged, such as when an electron is created. Whenever a charged particle is created, another having an opposite charge is always created along with it, so that the total charge created is zero. Usually, the two particles are “matter-antimatter” counterparts. For example, an anti-electron would usually be created at the same time as an electron. The anti-electron has a positive charge (it is called a positron), and so the total charge created is zero. (See Figure 6.) All particles have antimatter counterparts with opposite signs. When matter and antimatter counterparts are brought together, they completely annihilate one another. By annihilate, we mean that the mass of the two particles is converted to energy E, again obeying the relationship

\Delta m = \frac{E}{c^2}.

Since the two particles have equal and opposite charge, the total charge is zero before and after the annihilation; thus, total charge is conserved.

Instructor’s Note

 

We will be occasionally dealing with antimatter in this class. You need to know that matter and antimatter are identical in mass, but opposite in charge: an anti-electron has a positive charge. You also need to know that when matter and anti-matter come together the result is pure energy.

Here energy is shown by a vector. Initially electrostatic charge q tot is equal to zero. Now energy gets converted into matter and creates one electron and antielectron pair but final electrostatic charge is equal to zero so change in mass delta m is equal to two m e, which is equal to E divided by c square. (b) In this figure, Electron and antielectron are colliding with each other. The electrostatic charge q tot before collision is zero and after collision it will remain zero.
Figure 6: a) When enough energy is present, it can be converted into matter. Here the matter created is an electron–antielectron pair. (me is the electron’s mass.) The total charge before and after this event is zero. (b) When matter and antimatter collide, they annihilate each other; the total charge is conserved at zero before and after the annihilation.

 

Making Connections: Conservation Laws

Only a limited number of physical quantities are universally conserved. Charge is one—energy, momentum, and angular momentum are others. Because they are conserved, these physical quantities are used to explain more phenomena and form more connections than other, less basic quantities. We find that conserved quantities give us great insight into the rules followed by nature and hints to the organization of nature. Discoveries of conservation laws have led to further discoveries, such as the weak nuclear force and the quark substructure of protons and other particles.

The law of conservation of charge is absolute—it has never been observed to be violated. Charge, then, is a special physical quantity, joining a very short list of other quantities in nature that are always conserved. Other conserved quantities include energy, momentum, and angular momentum.

Section Summary

  • There are only two types of charge, which we call positive and negative.
  • Like charges repel, unlike charges attract, and the force between charges decreases with the square of the distance.
  • The vast majority of positive charge in nature is carried by protons, while the vast majority of negative charge is carried by electrons.
  • The electric charge of one electron is equal in magnitude and opposite in sign to the charge of one proton.
  • An ion is an atom or molecule that has nonzero total charge due to having unequal numbers of electrons and protons.
  • The SI unit for charge is the coulomb (C), with protons and electrons having charges of opposite sign but equal magnitude; the magnitude of this basic charge |q_e| = 1.602 \times 10^{-19} \, \mathrm{C}
  • Whenever charge is created or destroyed, equal amounts of positive and negative are involved.
  • Most often, existing charges are separated from neutral objects to obtain some net charge.
  • Both positive and negative charges exist in neutral objects and can be separated by rubbing one object with another. For macroscopic objects, negatively charged means an excess of electrons and positively charged means a depletion of electrons.
  • The law of conservation of charge ensures that whenever a charge is created, an equal charge of the opposite sign is created at the same time.

 

A Deeper Structure of the Atom

This section is available both as a video and as text. Below, you see the video as well as a text transcript. The content is the same: read or watch as is your preference.

Instructor’s Note

 

The things you need to know for your homework and quizzes are:

  • Electrons and protons are charged, neutrons are not; the size of the charge on the electron and proton is the same, but the signs are different, so same magnitude different sign
  • Opposites attract and that is what holds the atom together
  • Protons and neutrons have the same mass, and electrons are way lighter
  • Protons and neutrons made of stuff, electrons are fundamental
  • The nucleus is super tiny relative to atom

 

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You should be familiar with the basic structure of the atom, but as a review, in the middle of the atom the positively-charged protons and neutrons are huddled together in the nucleus.

The basic structure of the Atom.
The basic structure of the Atom.

However, you might not be familiar with the related symbols. This symbol,

p+p+{p} ^ {+}means proton, p for proton, and then plus to remind us that it has a positive electrical charge (If you’re wondering, can you have a negatively charged proton? Yes, it’s called an antimatter proton.) You also have neutrons. Neutron has zero charge, so it’s symbol is

n0n0{n} ^ {0}Those are huddled together in the nucleus, and then surrounding the nucleus is a big cloud of negatively charged electrons, so we will use the symbol

e-e-{e} ^ {-}for electron. It’s the attraction between the positively charged protons and the negatively charged electrons that sort of hold the entire atom together, and how that all works will be the emphasis of Unit III.

Electrons are a big focus of this course, so it is worth discussing what they are made of. To our best of our knowledge they are not made up of anything. They are fundamental, we have been trying to smash them apart, but no luck. Maybe it’s possible, but no one’s been able to do it. If it is possible, we haven’t hit it hard enough. That’s very much the particle physics approach to everything- hit it harder and see if it breaks. So, electrons are fundamental building blocks- as far as we know they’re not made up of anything.

Protons and neutrons on the other hand, are a lot more fun, because they are made up of smaller pieces called quarks. There are six kinds of quarks, we have ‘up’, ‘down’, ‘strange’, ‘charm’, ‘top’, and ‘bottom’. Those are their official scientific names, I kid you not. In Figure 7 below, the sizing of the circles shows you the masses, how heavy this stuff is (but not their sizes – as far as we know the quarks have zero size!). The ‘top’ quark, the heaviest of the known quarks, actually has about the same mass as an entire atom. It’s quite a heavy little thing. Three of these quarks, ‘top’, ‘bottom’, and ‘charm’, are actually heavier than protons.

 

Quark Masses as balls
Figure 7: The masses of the quarks: u for up, d for down, c for charm, s for strange, b for bottom, and t for top. Again, the size represents the mass, NOT the size; as far as we know all of these quarks have zero size! The grey ball in the lower left is a proton for scale. The small red dot inside the grey ball is an electron. (Credit: Incnis Mrsi [CC BY-SA (https://creativecommons.org/licenses/by-sa/3.0)])

But, if charms, bottoms, and tops are all heavier than protons and neutrons, so what makes up a proton in a neutron? Protons and neutrons are made up of just these two ‘ups’ and ‘downs’. So, a proton is made up of two ‘up’ quarks and one ‘down’ quark, a neutron on the other hand is two ‘down’ quarks and an ‘up’ quark as shown below in Figure 8.

Quarks
Quarks make up both protons and neutrons. Here you can see the smaller-and-smaller steps all the way down to the two ups and a down which make up the proton. (Credit: Finches & quarks [CC BY-SA (https://creativecommons.org/licenses/by-sa/4.0)])

Now you start doing math, so you need the proton to have +1 charge, the neutron to have 0 charge, you have two ‘ups’ and a ‘down’, and two ‘downs’ and an ‘up’. If you play with those numbers, what do you get? You have that ‘up’ quarks have a charge of

2323{2} over {3}that of the proton, and down quarks are

−13−13- {1} over {3}that of the proton. And this works out: think ‘up’ ‘up’ ‘down’ so that’s

23+23−13=+123+23−13=+1{2} over {3} + {2} over {3} – {1} over {3} =+1the charge of a proton, and it works out.

Similarly, think,

−13−13+23=0−13−13+23=0-{1} over {3}- {1} over {3} + {2} over {3} =0the charge of a neutron, they work out.

Instructor’s Note

 

What’s your big takeaway for this? Electrons are fundamental to our knowledge and cannot be broken apart. Protons and neutrons are made up of smaller stuff.

The other key things to know about atoms: protons and neutrons are very very close to the same mass, but neutrons are a tiny bit heavier, but not by much. Electrons on the other hand are way lighter than protons or neutrons. In fact, the electron is the lightest known particle to have electric charge with a mass of 9.11 \times 10^{-31} \, \mathrm{kg}. Protons on the other hand, are much bigger, 1.67 \times 10^{-27} \, \mathrm{kg}.

Instructor’s Note

 

What should you take away from this? Protons are way bigger than electrons, roughly 2,000 times (1836 times to be specific).

If you prefer to think about atoms instead of protons and neutrons, you can think about a Helium atom, you know there are two protons, two neutrons, and two electrons. The electrons make up 0.03% of the mass of helium. Electrons don’t weigh squat. They don’t really matter as far as mass goes. While the nucleus has most of the mass, it doesn’t take up a lot of space. The standard analogy that people make is if you blow up the atom to the size of a large college football stadium, bigger than ours, the nucleus is roughly the size of a marble. Atoms are a whole bunch of empty nothing. The nucleus is about the size of a pea but it is 99.97% of the mass is in that marble comparatively speaking.

More on Conservation of Electric Charge

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We are going to do examples using conservation of electric charge.

Let’s say I rub a plastic rod with some fur, basic friction, and I move about 10^6 electrons from the fur to the rod. What’s the charge of everything when we’re done?

Well we have 10^6 electrons going from the fur to the rod. Now charge has to be conserved. The rod is clearly going to end up with a negative charge from the added electrons, but since charge has to be conserved, and I took those electrons from the fur, my fur has to have an equal positive charge. Charges had to come from somewhere, and they came from the fur.

Now let’s talk about how much charge. Now let’s talk about how much charge. We know it’s 10^6 electrons worth. We know it’s 1.602 \times 10^{−19} \, \mathrm{C} for each electron. So, the rod will have a charge of:

\left( 10^{6} \, \mathrm{electrons} \right) \times \left( \frac{-1.602 \times 10^{-19} \, \mathrm{C}}{\mathrm{electron}} \right) = -1.602 \times 10^{−13} \, \mathrm{C}

The rod will thus have a total charge of q_{\mathrm{rod}} = -1.602 \times 10^{−13} \, \mathrm{C} (remembering that q is the symbol we use for charge).

Since charge is conserved, the fur will have the exact same positive charge.

 

Basics of Particles

3

What is a Particle?

What is a particle? The simplest image of a particle is probably just a ball. What properties apply to all particles? We talked about particles a lot in Physics 131 in the point mass approximation, but it’s probably best that we flush out our definitions.

In its most generic sense, a particle is a chunk of stuff. It exists in a particular place and at a particular time and a particle doesn’t go around corners. If I throw a ball at a door, it’ll either go through the door or bounce back, it won’t curve around it. Particles can, but do not necessarily have to, have mass, we will talk about a massless particle in a later section. But all particles can be thought of as having momentum, that quantity from 131 of mass times velocity. Particles can also be thought of as having energy.

Instructor’s Note

 

In summary, you need to know that particles can be thought of as balls with defined position and speed and are characterized by:

  • Their energy E
  • Their momentum \vec{p}
  • How many of them there are N

Linear Momentum and Force (Review from Physics 131)

This material is review from physics 131, but we will use these ideas in this unit, so here is a short refresher.

Instructor’s Note

 

You quiz will cover:

  • Calculate the momentum for any object
  • Recall that momentum is a vector
  • From the change in momentum, compute the average force

The scientific definition of linear momentum is consistent with most people’s intuitive understanding of momentum: a large, fast-moving object has greater momentum than a smaller, slower object. Linear momentum is defined as the product of a system’s mass multiplied by its velocity. In symbols, linear momentum is expressed as

\vec{p}=m\vec{v}.

Momentum is directly proportional to the object’s mass and also its velocity. Thus the greater an object’s mass or the greater its velocity, the greater its momentum. Momentum p is a vector having the same direction as the velocity v. The SI unit for momentum is kg⋅m/s.

Example Calculating Momentum: A Football Player and a Football

(a) Calculate the momentum of a 110-kg football player running at 8.00 m/s.

(b) Compare the player’s momentum with the momentum of a hard-thrown 0.410-kg football that has a speed of 25.0 m/s.

Strategy

No information is given regarding direction, and so we can calculate only the magnitude of the momentum, p. In both parts of this example, the magnitude of momentum can be calculated directly from the definition of momentum given in the equation, which becomes

p=mv

when only magnitudes are considered.

Solution for (a)

To determine the momentum of the player, substitute the known values for the player’s mass and speed into the equation.

p_{\mathrm{player}} = (110 \, \mathrm{kg}) (8.00 \, \mathrm{m/s}) =880 \, \mathrm{kg\cdot m/s}

Solution for (b)

To determine the momentum of the ball, substitute the known values for the ball’s mass and speed into the equation.

p_{\mathrm{ball}} = (0.410 \, \mathrm{kg})(25.0 \, \mathrm{m/s}) = 10.3 \, \mathrm{kg \cdot m/s}

The ratio of the player’s momentum to that of the ball is

\frac{p_{\mathrm{player}}}{p_{\mathrm{ball}}} = \frac{880}{10.3} = 85.9.

Discussion

Although the ball has greater velocity, the player has a much greater mass. Thus the momentum of the player is much greater than the momentum of the football, as you might guess. As a result, the player’s motion is only slightly affected if he catches the ball.

Instructor’s Note

 

The example above is representative of what you will be asked to do on your homework and quizzes.

 

Momentum and Newton’s 2nd Law (Optional)

All you need to know from this section is the definition of momentum. The following connection to Newton’s 2nd Law is just to help you put this info into context.

The importance of momentum, unlike the importance of energy, was recognized early in the development of classical physics. Momentum was deemed so important that it was called the “quantity of motion.” Newton actually stated his second law of motion in terms of momentum: The net external force equals the change in momentum of a system divided by the time over which it changes. Using symbols, this law is

\sum \vec{F} = \frac{\Delta \vec{p}}{ \Delta t},

where\sum \vec{F} is the net external force,  \Delta \vec{p} is the change in momentum, and  \Delta t is the change in time.

Newton’s 2nd Law in Terms of Momentum

The net external force equals the change in momentum of a system divided by the time over which it changes.

\sum \vec{F} = \frac{\Delta \vec{p}}{ \Delta t},

Making Connections: Force and Momentum

Force and momentum are intimately related. Force acting over time can change momentum, and Newton’s second law of motion, can be stated in its most broadly applicable form in terms of momentum. Momentum continues to be a key concept in the study of atomic and subatomic particles in quantum mechanics.

This statement of Newton’s second law of motion includes the more familiar \sum \vec{F} = m \vec{a} as a special case. We can derive this form as follows. First, note that the change in momentum \Delta p is given by

\Delta p= \Delta (mv).

If the mass of the system is constant, then

\Delta (mv)=m \Delta v.

So that for constant mass, Newton’s second law of motion becomes

\sum \vec{F} = \frac{\Delta \vec{p}}{\Delta t} =m \frac{\Delta \vec{v}}{ \Delta t}.

Because

\frac{\Delta \vec{v}}{ \Delta t} = \vec{a},

we get the familiar equation

\sum \vec{F} = m \vec{a}

when the mass of the system is constant.

Newton’s second law of motion stated in terms of momentum is more generally applicable because it can be applied to systems where the mass is changing, such as rockets, as well as to systems of constant mass. We will consider systems with varying mass in some detail; however, the relationship between momentum and force remains useful when mass is constant, such as in the following example.

Example Calculating Force: Venus Williams’ Racquet

During the 2007 French Open, Venus Williams hit the fastest recorded serve in a premier women’s match, reaching a speed of 58 m/s (209 km/h). What is the average force exerted on the 0.057-kg tennis ball by Venus Williams’ racquet, assuming that the ball’s speed just after impact is 58 m/s, that the initial horizontal component of the velocity before impact is negligible, and that the ball remained in contact with the racquet for 5.0 ms (milliseconds)?

Strategy

This problem involves only one dimension because the ball starts from having no horizontal velocity component before impact. Newton’s second law stated in terms of momentum is then written as

\sum F = \frac{ \Delta p}{ \Delta t }.

As noted above, when mass is constant, the change in momentum is given by

\Delta p=m \Delta v=m(v_f−v_i).

In this example, the velocity just after impact and the change in time are given; thus, once \Delta p is calculated,

\sum F= \frac{\Delta p}{ \Delta t}

can be used to find the force.

Solution

To determine the change in momentum, substitute the values for the initial and final velocities into the equation above.

\Delta p = m(v_f-v_i)(0.057 \, \mathrm{kg})(58 \, \mathrm{m/s} - 0 \, \mathrm{m/s}) = 3.306 \, \mathrm{kg \cdot m/s} \approx 3.3 \, \mathrm{kg \cdot m/s}

Now the magnitude of the net external force can determined by using \sum F= \frac{\Delta p}{ \Delta t}:

\sum F= \frac{\Delta p}{ \Delta t} = \frac{3.306 \, \mathrm{kg \cdot m/s}}{5.0 \times 10^{−3} \, \mathrm{s}} = 661 \, \mathrm{N} \approx 660 \, \mathrm{N},

where we have retained only two significant figures in the final step.

Discussion

This quantity was the average force exerted by Venus Williams’ racquet on the tennis ball during its brief impact (note that the ball also experienced the 0.56-N force of gravity, but that force was not due to the racquet). This problem could also be solved by first finding the acceleration and then using \sum \vec{F} =m \vec{a}, but one additional step would be required compared with the strategy used in this example.

Chapter Summary

Review of Conservation of Energy

4

Instructor’s Note

 

This unit, in fact this entire course, will spend a lot of time talking about energy: a topic covered extensively in Physics 131 as well as in your Biology and Chemistry courses. This chapter is therefore a bit different: we provide links to the relevant sections on energy from the Physics 131 textbook Forces, Energy, Entropy for your reference, with the key takeaways from each section. Just review what you need.

There are also a few homework problems at the end, just to make sure everyone is on the same page.

Relevant parts from Physics 131: Forces, Energy, Entropy:

A Video Reviewing Problem Solving with Conservation of Energy

This example can be either watched or read

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With what minimum speed must you toss a 140 \, \mathrm{g} ball straight up to hit the 14 \, \mathrm{m} meter high roof of a gymnasium if you release the ball 1.3 \, \mathrm{m}  above the ground? With what speed does the ball hit the ground?

You can use conservation of energy to solve this problem.

What is the initial energy state of the ball? We have some kinetic energy and some potential energy, so we have both. How do we know we have kinetic energy? Because we throw the ball, if the ball has no initial Kinetic energy which means it’s not moving that means it doesn’t go up, it had to have had some kinetic energy for it to actually go up and had to have some initial velocity when we threw it. Does it have any initial potential energy? The ball starts 1.3 \, \mathrm{m} above the ground initially, this tells me it started out with some potential energy, it’s already above the ground.

What is its final energy state in the perfect world in physics land? Does it have Kinetic energy at the roof? No, we’re assuming it just touches the roof and has zero velocity at the roof for that moment in time, so its kinetic final energy is actually zero. All we have left is potential final energy.

E_i = E_f

K_i + U_i = K_f + U_f

\frac{1}{2} m v^2 + mgh_i = 0 + mgh_f

\frac{1}{2}v^2 + gh_i = gh_f

What speed does it hit the ground? Energy initial equals energy final, what’s the initial energy state? My initial is the ball at the top of the ceiling. My final is just before it hits the ground. How fast does it hit the ground? It started from the roof, falls down. What is the energy state at the roof? It’s all potential. What’s the energy state the moment before it hits the ground? It’s lost all its potential energy, and its converted into kinetic energy.

E_i = E_f

U_i = K_f

mgh_i = \frac{1}{2} mv^2_f

gh_i = \frac{1}{2} v^2_f

v_f = \sqrt{2 g h_i}

v_f = \sqrt{2 \cdot (9.8 \, \mathrm{m/s^2}) \cdot (14 \, \mathrm{m}) } = 16.6 \, \mathrm{m/s}

Your friends Frisbee has become stuck 26 meters above the ground in a tree. You want to dislodge the Frisbee by throwing a rock at it. The Frisbee is stuck pretty tight, so you figure the rock needs to be traveling at least 5.4m/s when it hits the Frisbee. If you release the rock 1.6 meters above the ground, with what minimum speed must you throw it?

Energy initial has to equal energy final, what is my initial state of affairs? When I’m throwing the rock, that’s my initial state of affairs. Do I have kinetic energy in the beginning? I must have it. How do I know I must have kinetic energy? Because I’m throwing the rock, so the rock has to have some initial velocity. Do I have any initial potential energy? Yes, because I started 1.6 meters above the ground. What’s my final state of affairs? Do I have any kinetic energy at the end? When the rocks up there at the frisbee, does it have any kinetic energy? I know that it had to have a velocity, 5.4m/s, I know that the moment before I hit the frisbee I had to have this velocity. Therefore, I know I had some kinetic energy up there. Do I have any final potential energy? Yes, because it is up in the tree.

E_i = E_f

K_i + U_i = K_f + U_f

\frac{1}{2} mv^2_i + mgh_i = \frac{1}{2} mv^2_f + mgh_f

\frac{1}{2} v^2_i + gh_i = \frac{1}{2} v^2_f + gh_f

v_i = \sqrt{2 \left( \frac{1}{2} v_f^2 + gh_f - gh_i \right) }

Homework

 

 

Some Energy-Related Ideas that Might be New or are Particularly Important

5

Power

Power—the word conjures up many images: a professional football player muscling aside his opponent, a dragster roaring away from the starting line, a volcano blowing its lava into the atmosphere, or a rocket blasting off, as in Figure 1.

Space shuttle launch
Figure 1: This powerful rocket on the Space Shuttle Endeavor did work and consumed energy at a very high rate. (credit: NASA)

These images of power have in common the rapid performance of work, consistent with the scientific definition of power (P) as the rate at which work is done or energy is converted.

Power

Power is the rate at which work is done.

P=W/ t

The SI unit for power is the watt (W), where 1 watt equals 1 joule/second (1 W=1 J/s).

Because work is energy transfer, power is also the rate at which energy is expended. A 60-W light bulb, for example, expends 60 J of energy per second. Great power means a large amount of work or energy developed in a short time. For example, when a powerful car accelerates rapidly, it does a large amount of work and consumes a large amount of fuel in a short time.

Calculating Power from Energy: Calculating the Power to Climb Stairs

What is the power output for a 60.0-kg woman who runs up a 3.00 m high flight of stairs in 3.50 s, starting from rest but having a final speed of 2.00 m/s? (See Figure 2.)

Woman runnnig up stairs
Figure 2: When this woman runs upstairs starting from rest, she converts the chemical energy originally from food into kinetic energy and gravitational potential energy. Her power output depends on how fast she does this.

 

Strategy and Concept

The work going into mechanical energy is

\Delta E = Q + W

There is no heat transfer in this situation, so Q = 0.

\Delta E = W

E_f - E_i = W

(U_f + K_f) - (U_i + K_i) = W

At the bottom of the stairs, we take both K = 0 and U_g = 0; thus,

(0) - (K_f + U_f) = W

-W = \frac{1}{2} mv^2 + mgh_f

where h is the vertical height of the stairs and the minus sign means the energy is leaving her body.Because all terms are given, we can calculate work and then divide it by time to get power.

Solution

Substituting the expression for W in the previous equation, P=W/t yields

P = W / t

P = \frac{ \frac{1}{2}mv_f^2 + mgh}{t}

Entering known values yields

P = \frac{ 0.5(60.0 \, \mathrm{kg})(2.00 \, \mathrm{m/s})^2 + (60.0 \, \mathrm{kg})(9.80 \, \mathrm{m/s^2})(3.00 \, \mathrm{m})}{3.50 \, \mathrm{s}} = \frac{120 \, \mathrm{J} + 1764 \, \mathrm{J}}{3.50 \, \mathrm{s}} = 538 \, \mathrm{W}

Discussion

The woman does 1764 J of work to move up the stairs compared with only 120 J to increase her kinetic energy; thus, most of her power output is required for climbing rather than accelerating.

It is impressive that this woman’s useful power output is slightly less than 1 horsepower (1 hp=746 W)! People can generate more than a horsepower with their leg muscles for short periods of time by rapidly converting available blood sugar and oxygen into work output. (A horse can put out 1 hp for hours on end.) Once oxygen is depleted, power output decreases and the person begins to breathe rapidly to obtain oxygen to metabolize more food—this is known as the aerobic stage of exercise. If the woman climbed the stairs slowly, then her power output would be much less, although the amount of work done would be the same.

Making Connections: Take-Home Investigation—Measure Your Power Rating

Determine your own power rating by measuring the time it takes you to climb a flight of stairs. We will ignore the gain in kinetic energy, as the above example showed that it was a small portion of the energy gain. Don’t expect that your output will be more than about 0.5 hp.

Examples of Power

Examples of power are limited only by the imagination, because there are as many types as there are forms of work and energy. (See Table 1 for some examples.) Sunlight reaching Earth’s surface carries a maximum power of about 1.3 kilowatts per square meter (kW/m2). This quantity of power per area is called intensity, and will be explored more in the chapter on Basics of Waves.

A tiny fraction of this is retained by Earth over the long term. Our consumption rate of fossil fuels is far greater than the rate at which they are stored, so it is inevitable that they will be depleted. Power implies that energy is transferred, perhaps changing form. It is never possible to change one form completely into another without losing some of it as thermal energy. For example, a 60-W incandescent bulb converts only 5 W of electrical power to light, with 55 W dissipating into thermal energy. Furthermore, the typical electric power plant converts only 35 to 40% of its fuel into electricity. The remainder becomes a huge amount of thermal energy that must be dispersed as heat transfer, as rapidly as it is created. A coal-fired power plant may produce 1000 megawatts; 1 megawatt (MW) is 106 W of electric power. But the power plant consumes chemical energy at a rate of about 2500 MW, creating heat transfer to the surroundings at a rate of 1500 MW. (See Figure 3.)

A chinese power station
Figure 3: Tremendous amounts of electric power are generated by coal-fired power plants such as this one in China, but an even larger amount of power goes into heat transfer to the surroundings. The large cooling towers here are needed to transfer heat as rapidly as it is produced. The transfer of heat is not unique to coal plants but is an unavoidable consequence of generating electric power from any fuel—nuclear, coal, oil, natural gas, or the like. (credit: Kleinolive, Wikimedia Commons)

 

Table 1: Power Output or Consumption
Object or Phenomenon Power in Watts
Milky Way galaxy 1037
The Sun 4×1026
Volcanic eruption (maximum) 4×1015
Lightning bolt 2×1012
Nuclear power plant (total electric and heat transfer) 3×109
Aircraft carrier (total useful and heat transfer) 108
Dragster (total useful and heat transfer) 2×106
Car (total useful and heat transfer) 8×104
Football player (total useful and heat transfer) 5×103
Clothes dryer 4×103
Person at rest (all heat transfer) 100
Typical incandescent light bulb (total useful and heat transfer) 60
Heart, person at rest (total useful and heat transfer) 8
Electric clock 33 size 12{3} {}
Pocket calculator 10−3

Power and Energy Consumption

We usually have to pay for the energy we use. It is interesting and easy to estimate the cost of energy for an electrical appliance if its power consumption rate and time used are known. The higher the power consumption rate and the longer the appliance is used, the greater the cost of that appliance. The power consumption rate is P=W/t=E/t, where E is the energy supplied by the electricity company. So the energy consumed over a time t is

E=Pt

Electricity bills state the energy used in units of kilowatt-hours (kW⋅h) which is the product of power in kilowatts and time in hours. This unit is convenient because electrical power consumption at the kilowatt level for hours at a time is typical.

Calculating Energy Costs

What is the cost of running a 0.200-kW computer 6.00 h per day for 30 days if the cost of electricity is $0.120 per kW⋅hkW⋅h?

Strategy

Cost is based on energy consumed; thus, we must find from [latex] E=Pt and then calculate the cost. Because electrical energy is expressed in kW⋅h, at the start of a problem such as this it is convenient to convert the units into kW and hours.

Solution

The energy consumed in kW⋅h is

E = Pt

E = (0.200 \, \mathrm{kW})(6.00 \, \mathrm{h/d})(30 \, \mathrm{d})

E = 36.0 \, \mathrm{kW \cdot h}

and the cost is simply given by

cost=(36.0 kW⋅h)($0.120 per kW⋅h)=$4.32 per month

Discussion

The cost of using the computer in this example is neither exorbitant nor negligible. It is clear that the cost is a combination of power and time. When both are high, such as for an air conditioner in the summer, the cost is high.

The motivation to save energy has become more compelling with its ever-increasing price. Armed with the knowledge that energy consumed is the product of power and time, you can estimate costs for yourself and make the necessary value judgments about where to save energy. Either power or time must be reduced. It is most cost-effective to limit the use of high-power devices that normally operate for long periods of time, such as water heaters and air conditioners. This would not include relatively high power devices like toasters, because they are on only a few minutes per day. It would also not include electric clocks, in spite of their 24-hour-per-day usage, because they are very low power devices. It is sometimes possible to use devices that have greater efficiencies—that is, devices that consume less power to accomplish the same task. One example is the compact fluorescent light bulb, which produces over four times more light per watt of power consumed than its incandescent cousin.

Modern civilization depends on energy, but current levels of energy consumption and production are not sustainable. The likelihood of a link between global warming and fossil fuel use (with its concomitant production of carbon dioxide), has made reduction in energy use as well as a shift to non-fossil fuels of the utmost importance. Even though energy in an isolated system is a conserved quantity, the final result of most energy transformations is waste heat transfer to the environment, which is no longer useful for doing work.

Units of Energy

 

Instructor’s Notes

 

While this is covered in the 131 material linked to in the previous chapter, it is of such importance to this class that we are including it again.
Your Quiz will Cover
  • Converting between the different units of energy

In this course, we will be using Joules, electron-Volts exclusively, and kW-hrs exclusively. We are including these other units for your reference.

Note that the eV is significantly smaller than the joule; eV will generally be the smallest unit of energy used in this course.

If energy is defined as the ability to do work, then energy and work must have the same units. Thus, the SI unit of the energy is the Joule (recall 1J=1Nm=1kgms2). Energy, however, is one quantity where there are many other units in common use in scientific literature including electron-Volts (eV), kilowatt-hours (kW∙hr), calories, and Calories.

Electron-Volts

A common quantity in chemistry is the electron-Volt or eV. One electron-Volt is the amount of energy gained by an electron as it travels between the two ends of a 1 Volt battery (a concept that will be discussed in more detail when you study electricity). Numerically, 1eV = 1.602×10-19J. The reason this unit is common in chemistry is that the energies of atomic bonds are typically about 1eV as shown in the table below. The bond-dissociation energy is the energy released when the bond is formed.

From “Bond-Dissociation Energy – Wikipedia.” Accessed August 1, 2017.
https://en.wikipedia.org/wiki/Bond-dissociation_energy.
Bond Bond-dissociation energy at 298K (eV/Bond) Comment
C-C 3.60-3.69 Strong, but weaker than C–H bonds
Cl-Cl 2.51 Indicated by the yellowish colour of this gas
H-H 4.52 Strong, nonpolarizable bondCleaved only by metals and by strong oxidants
O-H 4.77 Slightly stronger than C–H bonds
OH-H 2.78 Far weaker than C–H bonds
C-O 11.16 Far stronger than C–H bonds
O-CO 5.51 Slightly stronger than C–H bonds
O=O 5.15 Stronger than single bondsWeaker than many other double bonds
N=N 9.79 One of the strongest bondsLarge activation energy in production of ammonia
H3C-H 4.550 One of the strongest aliphatic C–H bonds

Kilowatt Hours and Calories

Ad described above, when you buy electricity from the power company, the bill says how many kilowatt hours you have purchased. A Watt is a unit of a quantity called power and 1 Watt is equal to 1 Joule/second: 1W = 1 J/s. Thus, a kilowatt hour is therefore:

(1kW∗hr)(1000WkW)(1J/sW)(3600shr)=3.6106J=3.6MJs(1kW∗hr)(1000WkW)(1J/sW)(3600shr)=3.6106J=3.6MJsleft (1 kW * hr right ) left ({1000 W} over {kW} right ) left ({1 J/s} over {W} right ) left ({3600 s} over {hr} right ) =3.6106 J=3.6 MJ s">(1kWhr)(1000WkW)(1J/sW)(3600shr)=3.6106J=3.6MJs

The calorie is an imperial unit of energy that is still in common use in the nutritional sciences in the United States. One calorie (lowercase c) is the amount of energy needed to raise 1g of water 1oC or 1 cal = 4.814J. On food labels, you will see energy listed in Calories (capital C). One Calorie is equal to 1kilocalorie; in other words, 1 Cal = 1000 cal. Thus, one 1 Cal = 4814 J. In other countries, you will see food labels in both Calories and Joules like the one shown in Figure 1.

A food label from the UK showing the energy of the food in both Joules and kcal (or Calories).
A food label from the UK showing the energy of the food in both Joules and kcal (or Calories).

The Potential Energy of Electrons in Atoms and Molecules

For this first unit, the primary source of microscopic potential energy with which we shall concern ourselves is the potential energy stored be electrons in atoms and in chemical bonds: U_{\mathrm{chem}}. This potential energy is a result of the force of electrical attraction between different atoms (recall electricity and magnetism was one of our fundamental forces) or between an electron and its nucleus.

The strength of chemical bonds is typically quoted in one of two ways: either the energy in the bond, called the bond dissociation energy, is quoted directly (typically in eV) or the enthalpy per mole will be quoted. For example, the Cl-Cl bond has a bond dissociation energy of 2.51 eV/bond or a bond dissociation enthalpy per mole of ΔH=242kJ/molΔH=242kJ/molΔH=242 kJ/mol">ΔH=242kJ/mol. For atoms, the relevant energy is the ionization energy, which is the amount of energy needed to remove an electron: quoted directly in eV or, occasionally in kJ/mol. How do we interpret these numbers in terms of potential energy? We use the same freedom to choose the zero of potential energy discussed in Unit IV – Chapter 3.3: Macroscopic Potential Energy from the Physics 131 textbook which discusses potential energy at the macroscopic scale.

Thinking about gravity, we tend to put the zero of potential energy at ground level; objects above the ground then have positive potential energy while objects underground have negative potential energy. This use of negative potential energy makes sense, an object at ground level will fall to below ground level if allowed to do so and lose energy in the process.

Where to put the zero of gravitational potential energy? The top of a building? The ground? In the subway below? The choice is arbitrary.
Where to put the zero of gravitational potential energy? The top of a building? The ground? In the subway below? The choice is arbitrary.

For atoms and molecules, we have a similar freedom to choose where to put zero potential energy. The standard convention is to say that free atoms that are far apart have zero potential energy. Atoms in most bonds have lower potential energy than free atoms (that is why the bonds form!). Therefore the potential energy of the atoms is less than zero: the potential energy of atoms in bonds is negative. This may seem like a weird choice for the zero of potential energy, but it is the convention and it makes sense when you think about it!

Two Cl atoms separated by a great distance have zero potential energy while two bonded Cl atoms have a potential energy of -2.51 eV.
Two Cl atoms separated by a great distance have zero potential energy while two bonded Cl atoms have a potential energy of -2.51 eV. Remember, potential energy is the due to the relative position of two objects, so it does not make sense to ask which atom in the bonded pair has the potential energy. The potential energy is due to the two of them!

Let’s return to the quoted Cl-Cl bond with dissociation energy of 2.51 eV/bond. What does this value mean? It means that two Cl atoms bonded together have a potential energy of 2.51 eV less than if they were free. Said another way, the potential energy of Cl atoms in Cl2 is -2.51 eV, while the potential energy of free Cl atoms is 0 eV. This is consistent with what you probably already know about Chlorine: Cl2 is the lower energy state than free Cl atoms. I would get 2.51 eV of energy for every Cl-Cl bond that is formed, as the atoms move from zero potential energy to -2.51 eV. Similarly, I would need to add 2.51 eV of energy to break a Cl-Cl bond and move the two atoms up to zero potential energy. The same idea holds for electrons in atoms. If you look up the ionization energy of hydrogen, you will see 13.6 eV. This value meas that the electron has 13.6 eV lesspotential energy than if it were free. Thus, a more accurate way to write this energy would be U = -13.6 eV.

The Connection Between Kinetic Energy and Momentum

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We will now develop a useful relationship between momentum and kinetic energy. This is a useful relationship that we will use throughout this course.

By now you should have refreshed your memory and know that for a standard particle with mass, such as an electron, the momentum of the particle is

\vec{p} = m \vec{v}

and the kinetic energy of the particle is

K = \frac{1}{2} mv^2.

If you look at these two expressions, they are fairly similar, both involve the mass of the particle m and its velocity v.

Now there are some important differences. The momentum is a vector, including the direction of motion, whereas the kinetic energy is a scalar and is independent of the particle’s direction of motion. However, there’s a useful way to relate these two.

Begin with the magnitude of the momentum, removing the vectors, in which case is just

p=mv.

Square both sides of this expression so you have

p^2=m^2v^2.

Now divide both sides of the expression by 2m, so now you have

\frac{p^2}{2m} = \frac{1}{2}mv^2,

Which is the kinetic energy.

The big punch line is that the kinetic energy of a particle with mass is

K= \frac{p^2}{2m}.

This is a useful expression that we’ll be using throughout this course.

 

 

Basics of Waves

6

What is a Wave?

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Now, we have talked about particles. What about waves? Before we start talking about waves, it’s probably best to give a few different examples of waves. If I ask you to think of a wave the first thing that probably would come to most your minds is a water wave, but we could also have waves on a string, or even sound waves. The most generic picture that a lot of you have, is probably some sort of sine or cosine shape traveling along, but this is not representative of all waves and we want our definition to be in terms of properties that apply to every possible wave that we can think of.

Let’s go through a few questions and develop a definition of a wave.

Does the wave actually have to go anywhere, does a wave travel? No. Sure, most waves go somewhere, water waves travel across an ocean for example, but think of a guitar string, when you pluck it, certainly the string waves back and forth but the string doesn’t go anywhere, the wave stands on the string. This is called a standing wave. So, traveling cannot be part of our definition of a wave.

Does a wave have to be a repeating pattern? Again, not really. While this might be the image that a lot of you have in mind when I say the word wave, remember we can have just a single pulse going back and forth on a string.

Does the wave have to have up and down motion? Well again, no. The standard picture of a wave that you have in your head might look like Figure 1, but I could also send a compression wave down the slinky like in figure 2 where the links of the slinky move back and forth in the same direction as the waves motion.

A transverse wave moving up and down perpendicular to the wave's motion on the slinky.
Figure 1: A wave can go up and down – a transverse wave. This is probably what you are picturing when I say wave.
A compression or longitudinal wave down a slinky
Figure 2: A wave can go down the slinky as a compression – or longitudinal – wave.

Now we need a little bit of terminology. Waves that do wiggle perpendicular to the direction of motion of the wave are called transverse waves. These are the waves that you probably have in mind and these are the ones that were mostly going to be interested in.

The basic terminology of transverse waves, we’ll introduce some more later, are that waves have a peak and a trough, and then the distance from the zero line to either a peak or a trough is called the amplitude. This amplitude is labelled y in Figure 3.

A picture of a wave showing all the quantities.
Figure 3: Parts of a wave (Credit: Krishnavedala)

What other properties of a wave could we perhaps use? Can a wave bend around corners? We know that particles don’t bend around corners, what about waves? Well, it turns out that waves do bend around corners. Think of a water wave, it spreads out, bending around the corner. In the video on the next slide we’ll see some other important properties of waves that all waves do share.

I go and just hit the water with a single stick, we see we get waves coming out in all directions, radiating away from the spot where the ball hits the water.

A ball hitting water makes circular waves

Things get a little bit more interesting, however, if we have two sources of waves going at the same time next to each other like so.

Two balls hitting water next to each other, each making waves which interact.

So now we get two waves, each radiating out from its source. In some places the waves line up peak to peak, or trough to trough, and add up, resulting in a larger wave at that point.

Constructive interference

In other places, the peak of one wave meets the trough of the other, resulting in some cancellation.

Desctructive interference.

This phenomenon of waves adding in some places and cancelling in others is known as interference and is a characteristic property of waves.

So, what is a wave? Well a wave is a disturbance that can, but doesn’t necessarily have to, travel or it can just store energy and momentum. A traveling wave will carry energy from one position to another, think of the water wave that carries energy as it moves across the ocean and also momentum, as that wave hits you, you feel the momentum of the wave. For a standing wave, that energy is just being stored. When I pluck a guitar string, the energy is just being stored in the string and then ultimately releases this sound that we hear. A wave need not necessarily repeat, we can have simple pulse waves. But a wave can bend around corners, and waves of the same kind can interact with each other or with themselves, adding in some places and canceling in other places, through this idea of interference. These are the fundamental characteristics of waves. They don’t exist at a particular place, they sort of spread out over a couple of different places and they can carry energy and momentum while bending around corners and interacting with themselves or other waves of the same kind.

Instructor’s Notes

 

To Summarize:

  • Particles are localized in space, they don’t bend around corners, but can carry energy and momentum.
  • Waves on the other hand, are spread out in space, they are some kind of disturbance that can transfer or store energy and momentum.
  • However, waves, unlike particles, can bend around corners and also waves can interact with themselves or other waves of the same kind through this phenomenon of interference.

Period and Frequency in Oscillations

Close-up of a vibrating guitar string
Figure 2: The strings on this guitar vibrate at regular time intervals. (credit: JAR)

When you pluck a guitar string, the resulting sound has a steady tone and lasts a long time. Each successive vibration of the string takes the same time as the previous one. We define periodic motion to be a motion that repeats itself at regular time intervals, such as exhibited by the guitar string or by an object on a spring moving up and down. The time to complete one oscillation remains constant and is called the period T. Its units are usually seconds, but may be any convenient unit of time. The word period refers to the time for some event whether repetitive or not; but we shall be primarily interested in periodic motion, which is by definition repetitive.

A concept closely related to period is the frequency of an event. For example, if you get a paycheck twice a month, the frequency of payment is two per month and the period between checks is half a month. Frequency \nu is defined to be the number of events per unit time. For periodic motion, frequency is the number of oscillations per unit time. The relationship between frequency and period is

\nu = \frac{1}{T}.

 

The SI unit for frequency is the cycle per second, which is defined to be a hertz (Hz):

1 \, \mathrm{Hz} = \frac{1 \, \mathrm{cycle}}{\mathrm{s}} or 1 \, \mathrm{Hz} = \frac{1}{s}

A cycle is one complete oscillation. Note that a vibration can be a single or multiple event, whereas oscillations are usually repetitive for a significant number of cycles.

A Venn-diagram showing the different sciencesThe different sciences use different symbols for frequency. If you have seen waves in a physics course before, they probably used the symbol f. However, in your chemistry classes, you probably used \nu. Both are in use and mean the same quantity. We will, in general, stick to using the \nu you used in chemistry, but don’t worry if you see an f used somewhere in your homework or in the text, it means the same quantity.

Fun with the history of science, different disciplines discovered the same quantity and gave it different names!

Determine the Frequency of Two Oscillations: Medical Ultrasound and the Period of Middle C

We can use the formulas presented in this module to determine both the frequency based on known oscillations and the oscillation based on a known frequency. Let’s try one example of each.

(a) A medical imaging device produces ultrasound by oscillating with a period of 0.400 µs. What is the frequency of this oscillation?

(b) The frequency of middle C on a typical musical instrument is 264 Hz. What is the time for one complete oscillation?

Strategy

Both questions (a) and (b) can be answered using the relationship between period and frequency. In question (a), the period T is given and we are asked to find frequency\nu. In question (b), the frequency\nuis given and we are asked to find the period T.

Solution (a):

Substitute 0.400μs for T in \nu = \frac{1}{T}:

\nu = \frac{1}{T} =\frac{1}{0.400×10^{−6} \, \mathrm{s}}.

Solve to find

\nu=2.50×10^6 \, \mathrm{Hz}

Discussion (a):

The frequency of sound found in (a) is much higher than the highest frequency that humans can hear and, therefore, is called ultrasound. Appropriate oscillations at this frequency generate ultrasound used for noninvasive medical diagnoses, such as observations of a fetus in the womb.

 

Solution (b):

Identify the known values:

The time for one complete oscillation is the period T:

\nu = \frac{1}{T}.

Solve for T:

T=\frac{1}{\nu}.

Substitute the given value for the frequency into the resulting expression:

T= \frac{1}{264\, \mathrm{Hz}}

T =3.79×10^{−3} \, \mathrm{s}=3.79 \, \mathrm{ms}.

Discussion (b)

The period found in (b) is the time per cycle, but this value is often quoted as simply the time in convenient units (ms or milliseconds in this case).

Everyday Periods and Frequencies

Identify an event in your life (such as receiving a paycheck) that occurs regularly. Identify both the period and frequency of this event.

Solution

I visit my parents for dinner every other Sunday. The frequency of my visits is 26 per calendar year. The period is two weeks.

Section Summary

  • Periodic motion is a repetitious oscillation.
  • The time for one oscillation is the period T
  • The number of oscillations per unit time is the frequency \nu (or sometimes f).
  • These quantities are related by \nu = \frac{1}{T}.

Homework

Problem 10: What is the frequency of a stroboscope?

If you are curious about what a stroboscope is, check out The Stroboscope Wikipedia page.

Detailed description of a wave

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Let’s begin by thinking about the one thing we have already figured out, which is that light has some wave-like properties. In the figure below we have a wave, and we can talk about the wavelength, \lambda, which is the distance from one point on the wave through the same point as shown in Figure 4. Wavelength could be from peak to peak or from zero to zero or from trough to trough. All those distances are the same, and the wavelength is measured in meters.

A picture of a wave showing all the quantities.
Figure 4: Parts of a wave (Credit: Krishnavedala

We also have the idea of the amplitude y in the figure. The difference from either peak to the middle average or from trough to the middle average is what we call the amplitude.

In addition to wavelength and amplitude, we can also talk about how long it takes for a point to go up and down. Think about one point on the wave, bouncing up and down. We can talk about how long it takes for that point on the wave to go from trough to peak to trough, the amount of time that takes is called the period, T, and, as discussed in the last section, will be measured in seconds.

For this class, we will work in SI, International System of Units, we will not work in, what I call barbaric units, there shall be no inches. Meters, kilograms, and seconds will be the norm. On the Moodle page there’s a document of math I expect you to know, it’s things like trigonometry and area of a circle, but I also expect you to know the SI prefixes Nano – Giga. I will not give you these, and on an exam if you come to the TA’s and ask how big a micrometer is, we’re going to have to say tough cookie, that’s something you were supposed to know.

The period is the time it takes for a point on the wave to go up and down, measured in seconds, but I can also count how many that point oscillates in one second. I could say how many oscillations per second this point on this wave make? That quantity, again looking back to the last section, is known as the frequency \nu and its value is

\nu = \frac{1}{T},

the unit of frequency is how many per second or one over seconds, which is called Hertz: Hz.

We will use these basic terms for all of the waves we discuss, electrons and light. We know that wavelength is measured in meters and we know that frequency is in Hertz, or 1 over seconds, so \lambda \nu will be meters over seconds which is a velocity. What’s the only velocity that we could have? The speed of the wave, it’s the only speed we could talk about, so we have the speed of the wave is going to be

v = \lambda \nu

Exploring v = λν

If we know that the speed of a wave is fixed, for example, light travels at a fixed speed, then if the frequency goes up, what’s the wavelength going to do?

Solution:

We know that

v = \lambda \nu.

The speed is fixed, if the frequency goes up, and \lambda times \nu has to be the same thing, then that is going to tell me that the wavelength must go down.

Discussion:

This question is based on mathematical reasoning using symbols and not numbers. Remember one of the goals for this course was working in symbols and not numbers, so here’s an example of that.

Energy in Waves: Intensity

Instructors Notes

 

The key takeaways that you will be potentially quizzed on are:

  • Intensity is power per area: I = \frac{P}{A} with units W/m2
  • The intensity is related to the square of the amplitude I \propto A^2

The energy effects of a wave depend on time as well as amplitude. For example, the longer deep-heat ultrasound is applied, the more energy it transfers. Waves can also be concentrated or spread out. Sunlight, for example, can be focused to burn wood. Earthquakes spread out, so they do less damage the farther they get from the source. In both cases, changing the area the waves cover has important effects. All these pertinent factors are included in the definition of intensity

IIsize 12{I} {}  s

as power per unit area:

I=PAI=PAsize 12{I= { {P} over {A} } } {}

where

PPsize 12{P} {}

is the power carried by the wave through area

AAsize 12{A} {}

. The definition of intensity is valid for any energy in transit, including that carried by waves. The SI unit for intensity is watts per square meter (

W/m2W/m2size 12{“W/m” rSup { size 8{2} } } {

). For example, infrared and visible energy from the Sun impinge on Earth at an intensity of

1300W/m21300W/m2size 12{“1300″`”W/m” rSup { size 8{2} } } {

just above the atmosphere.

Calculating intensity and power: How much energy is in a ray of sunlight?

The average intensity of sunlight on Earth’s surface is about

700W/m2700W/m2size 12{7″00″`”W/m” rSup { size 8{2} } } {}.

(a) Calculate the amount of energy that falls on a solar collector having an area of

0.500m20.500m2size 12{0 “.” “500”`”m” rSup { size 8{2} } } {}in

4.00h4.00hsize 12{4 “.” “00”`”h”} {}.

(b) What intensity would such sunlight have if concentrated by a magnifying glass onto an area 200 times smaller than its own?

Strategy a

Because power is energy per unit time or

P=EtP=Etsize 12{P= { {E} over {t} } } {}the definition of intensity can be written as

I=PA=E/tAI=PA=E/tAsize 12{I= { {P} over {A} } = { { {E} slash {t} } over {A} } } {}and this equation can be solved for E with the given information.

Solution a

  1. Begin with the equation that states the definition of intensity:
    I=PA.I=PA.size 12{I= { {P} over {A} } } {}
  2. Replace
    Pwith its equivalentE/t:E/tsize 12{E/t} {}:
    I=E/tA.I=E/tA.size 12{I= { { {E} slash {t} } over {A} } } {}
  3. Solve for
    E:Esize 12{P} {}:
    E=IAt.E=IAt.size 12{E= ital “IAt”} {}
  4. Substitute known values into the equation:
    E=700W/m20.500m24.00h3600s/h.E=700W/m20.500m24.00h3600s/h.size 12{E= left (“700″” W/m” rSup { size 8{2} } right ) left (0 “.” “500”” m” rSup { size 8{2} } right ) left [ left (4 “.” “00”” h” right ) left (“3600″” s/h” right ) right ]} {}
  5. Calculate to find
    E
    and convert units:
    5.04×106J,5.04×106J,size 12{5 “.” “04” times “10” rSup { size 8{6} } “J”} {}

Discussion a

The energy falling on the solar collector in 4 hours is enough to be useful—for example, for heating a significant amount of water.

Strategy b

Taking a ratio of new intensity to old intensity and using primes for the new quantities, we will find that it depends on the ratio of the areas. All other quantities will cancel.

Solution b

  1. Take the ratio of intensities, which yields:
    I′I=P′/A′P/A=AA′(The powers cancel becauseP′=P).I′I=P′/A′P/A=AA′(The powers cancel becauseP′=P).
  2. Identify the knowns:
    A=200A′,A=200A′,size 12{A=”200″A’} {}
    I′I=200.I′I=200.size 12{ { {I rSup { size 8{‘} } } over {I} } =”200″} {}
  3. Substitute known quantities:
    I′=200I=200700W/m2.I′=200I=200700W/m2.size 12{I’=”200″I=”200″ left (“700″`”W/m” rSup { size 8{2} } right )} {}
  4. Calculate to find
    I′:I′size 12{I’} {}:
    I′=1.40×105W/m2.I′=1.40×105W/m2.size 12{ { {I}} sup { ‘ }=1 “.” “40” times “10” rSup { size 8{5} } `”W/m” rSup { size 8{2} } } {}

Discussion b

Decreasing the area increases the intensity considerably. The intensity of the concentrated sunlight could even start a fire.

Determine the combined intensity of two waves: Perfect constructive interference

If two identical waves, each having an intensity of

1.00W/m21.00W/m2size 12{1 “.” “00”`”W/m” rSup { size 8{2} } } {}, line up perfectly peak-to-peak, what is the intensity of the resulting wave?

Strategy

If two waves, which have equal amplitudes

XXsize 12{X} {}, line up exactly, the resulting wave has an amplitude of

2X2Xsize 12{2X} {}. Because a wave’s intensity is proportional to amplitude squared, the intensity of the resulting wave is four times as great as in the individual waves.

Solution

  1. Recall that intensity is proportional to amplitude squared.
  2. Calculate the new amplitude:
    I′∝X′2=2X2=4X2.I′∝X′2=2X2=4X2.size 12{I rSup { size 8{‘} } prop left (X rSup { size 8{‘} } right ) rSup { size 8{2} } = left (2X right ) rSup { size 8{2} } =4X rSup { size 8{2} } } {}
  3. Recall that the intensity of the old amplitude was: I \propto X^2
  4. Take the ratio of new intensity to the old intensity. This gives:
    I′I=4.I′I=4.size 12{ { {I} over {I rSup { size 8{‘} } } } =4} {}
  5. Calculate to find
    I′:I′size 12{I’} {}:
    I′=4I=4.00W/m2.I′=4I=4.00W/m2.size 12{I’=4I=4 “.” “00”`”W/m” rSup { size 8{2} } } {}

Discussion

The intensity goes up by a factor of 4 when the amplitude doubles. This answer is a little disquieting. The two individual waves each have intensities of

1.00W/m21.00W/m2size 12{1 “.” “00”`”W/m” rSup { size 8{2} } } {}, yet their sum has an intensity of

4.00W/m24.00W/m2size 12{4 “.” “00”`”W/m” rSup { size 8{2} } } {}, which may appear to violate conservation of energy. This violation, of course, cannot happen. What does happen is intriguing. The area over which the intensity is

4.00W/m24.00W/m2size 12{4 “.” “00”`”W/m” rSup { size 8{2} } } {}is much less than the area covered by the two waves before they interfered. For each spot where the waves line up peak-to-peak, there is somewhere else where they line up peak-to-trough and cancel. For example, if we have two stereo speakers putting out

1.00W/m21.00W/m2size 12{1 “.” “00”`”W/m” rSup { size 8{2} } } {}each, there will be places in the room where the intensity is

4.00W/m24.00W/m2size 12{4 “.” “00”`”W/m” rSup { size 8{2} } } {}, other places where the intensity is zero, and others in between. Figure 5 shows what this interference might look like.

Two speakers are shown at the top of the figure at left and right side. Rarefactions are shown as dotted curves and compression as dark curves. The interference of the sound waves from these two speakers is shown. There are some red spots, showing constructive interference, are shown on the interfering waves.
Figure 5: These stereo speakers produce both constructive interference and destructive interference in the room, a property common to the superposition of all types of waves. The shading is proportional to intensity.

 

 

Basics of Light

7

Where Does Light Come From?

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Light is generated any time a charge undergoes acceleration; this is a connection to an idea from Physics 131. Just like in Physics 131 it’s not the motion of the charge that matters, but its acceleration. Moving charges don’t generate light only accelerating ones do. To expand upon this connection to 131 a little bit more, if a charge accelerates by slowing down, it is still accelerating then from Newton’s second law,

\sum \vec{F} = m \vec{a}

, we know that a force has acted upon it. If it takes some distance for this slowing down to occur, then the force must have been applied for some distance and we know that work was therefore done on the charged particle. By the statement of conservation of energy, or equivalently the first law of thermodynamics, if work is done on a particle then the particles energy must change, that energy must go somewhere and where does it often go? It goes into light.

Here’s an example with which you might be familiar from your chemistry class. An electron in an outer energy level of an atom falls to a lower energy level. There’s a change in energy as the electron falls, that energy has to go somewhere. It goes into the release of light.

Electrons changing energy levels, however, is not the only way to produce light. Think about an old-school incandescent lamp with the filament in it that get hot as you turn them on, to understand why these incandescent lights give off light we have to understand a little bit about what temperature is.

Recall from Physics 131 that temperature is related to the average kinetic energies of particles moving around randomly on the atomic and subatomic scales. As these particles are bouncing around randomly, they’re changing directions. From 131 we know that acceleration is a vector, so because velocity changes direction, then we know that there is acceleration. So once again, even any object with temperature will emit light due to the accelerating charges bouncing around on the atomic and subatomic scale.

Instructor’s Notes

 

In summary

  • Light is generated by charges accelerating.
  • Every object with a temperature (i.e. everything) will emit some amount of light of some type.
  • Our eyes, however, are only sensitive to certain kinds of light and we therefore cannot see this light from everyday objects such as you and I. We don’t see light coming off of us because our eyes are not sensitive to the kind of light that we emit due to our temperature.
  • However, we can build devices that can see the light given off by more everyday objects such as people by using technologies such as infrared cameras.

Properties of Light

Instructor’s Notes

 

The video for this section uses  f for frequency. The text, on the other hand uses \nu. This is a good example of the fact that you need to get used to the idea that different disciplines use different letters for the same quantity!

On your equation sheet, in class, an on exams, we will use \nu to be consistent with what you have used in chemistry.

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Like all waves, light waves are characterized by a wavelength, a frequency, a speed, which follows the usual relationship of v = \lambda \nu, and an amplitude. However, there are some important unique characteristics of light waves. For light, the wave in the vacuum speed is always the same, c = 3 \times 10^8 \, \mathrm{m}/\mathrm{s}. In a vacuum v = \lambda \nu, turns into c = \lambda \nu, because all light waves, regardless of their wavelength or frequency or amplitude, travel at this same fundamental speed.

For the amplitude of the light wave we will not use the symbol A we will instead use the symbol E and the amplitude of a light wave has the units of Newton’s per Coulomb \mathrm{N}/\mathrm{C}, Newton’s are the unit of force and Coulomb, as you’ve already discussed elsewhere in your prep, is the unit of a charge. The amplitude of a light wave is a Newton per Coulomb. We will see why this is the unit of a light waves amplitude later in this particular course, but for right now you just need to know that those are the units.

There are many different kinds of light. Where do these different kinds of light come from? Well different wavelengths or frequencies represent different kinds of light. Light is also sometimes called electromagnetic radiation, and so the kinds of light are called the E/M spectrum. You’ll see the terms ‘electromagnetic spectrum’ or ‘E/M spectrum’ used, which just means the kinds of light. You’ll explore more of the different kinds of light in the next section.

But this is giving you a bit of a hint on where this whole course is going and how light, electricity, and magnetism are all going to be deeply connected in some fundamental way, which will come to by the end of this course.

We’ve now seen that the frequency or wavelength of a light wave tells us what kind of light we are going to have. What does the amplitude of the light wave correspond to?

The amplitude, remember we’re using E for the amplitude, is related to the intensity of the light, as in the watts per square meter, by this expression

I = \frac{1}{2} c \epsilon_0 E^2

where c is the usual speed of light c = 3 \times 10^8 \, \mathrm{m}/\mathrm{s} and \epsilon_0 is a property of just empty space. You might not think of empty space as having properties, but it does! The quantity \epsilon_0 is a property of empty space called the permittivity of free space, and it has this value \epsilon_0 = 8.85 \times 10^{-12} \, \frac{\mathrm{C}^2}{\mathrm{J} \cdot \mathrm{m}}. We will talk more about this number throughout this course, for now, you just need to know it’s a property of empty space.

A Venn-diagram showing the different sciences

 

 

An example converting between wavelength and frequency for light (from Chemistry – so this should be familiar!)

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Let’s do some examples what is what is the frequency of light that has 396.15 nanometer as wavelength?

Solution:

Wavelength equals c over frequency: \lambda = c / \nu, meaning that frequency equals c over lambda \nu = c / \lambda. The speed of light in vacuum is given: 2.998 \times 10^8 \, \mathrm{m/s}. For this question, the wavelength is in nanometers while the unit of the speed of light is in meters, so I know that I have to change the nanometer:

\nu = \frac{c}{\lambda}

\nu = \frac{2.998 \times 10^8 \, \mathrm{m/s}}{\left( 396.15 \, \mathrm{nm} \right) \left(\frac{10^{-9} \, \mathrm{m}}{\mathrm{nm}} \right)}

\nu = 7.568 \times 10^{14} \, \mathrm{s}^{-1} = 7.568 \times 10^{14} \, \mathrm{Hz}

Discussion:

That means 7.568 \times 10^{14} is how many waves will pass per one second.

 

Instructor’s Notes

 

In summary

  • Light is a wave with a: wavelength, frequency, speed, and amplitude.
  • The speed of all light waves in vacuum is the same c = 3 \times 10^8 \, \mathrm{m}/\mathrm{s}
  • The units of the amplitude of a light wave are Newtons/Coulomb
  • We will use E for the amplitude of a light wave instead of A.
    • Keep in mind this is NOT the energy!
    • The amplitude has units Newtons/Coulomb
    • Newtons/Coulomb are not the same unit as the Joules we use for energy!
    • I know it is confusing, but we are running out of letters and there is a good reason for E which we will see later in the course
  • While, in general, we know that intensity is proportional to amplitude squared I \propto A^2, for light we have exact equation:

I = \frac{1}{2} c \epsilon_0 E^2

  • \epsilon_0 = 8.85 \times 10^{-12} \, \frac{\mathrm{C}^2}{\mathrm{J} \cdot \mathrm{m}} is a constant of the Universe, just like the speed of light. We will revisit this constant later.

The Main Parts of the Electromagnetic Spectrum

Instructor’s Notes

 

As scientifically trained people, you should have a basic familiarity with the electromagnetic spectrum. Thus, while this course is generally not about memorization, I will ask you to memorize the large basic divisions of the electromagnetic spectrum: radio, microwaves, infrared, visible, ultraviolet, x-rays, and gamma rays. You need to know that radio represents the longest wavelength and gamma rays represent the shortest wavelength. You should also know that, within visible light, red is the longest going through the rainbow to violet. You do NOT need to know the frequencies or wavelengths corresponding to each range. The only exception to this rule is that I do expect you to know that red is about 700nm wavelength while violet is about 350nm. The different types of radiation come up so frequently in scientific discussion that it is important to know some basic facts.

 

A Venn-diagram showing the different sciences

 

 

 

 

 

 

Below, you can find a video that summarizes the parts of the electromagnetic spectrum taken from General Chemistry I (Chem 111 at UMass-Amherst), prepared by Dr. Al-Hariri. Please use it to familiarize yourself with the parts of the spectrum if needed.

An additional graphic can be found below the video and its transcript.

 

 

 

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Everyday we’re bombarded with different types of radiation like the radio radiation from radio tower close to us, to microwave radiation,  to the light radiation and so on; and if you look at the different wavelengths displayed in this picture you can see that the difference between between them is the length of that wave

Now, here are a couple of different types  of electromagnetic radiation and the difference way and the different wavelengths of each:

Different electromagnetic radiations have different wavelengths.

 

A diagram of the Milton spectrum, showing the type, wavelength (with examples), frequency, the black body emission temperature.
Figure 1: A diagram of the EM spectrum, showing the type, wavelength (with examples), frequency, the black body emission temperature. (Inductiveload, NASA [CC BY-SA (http://creativecommons.org/licenses/by-sa/3.0/)])

Introduction to the Photon

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We’ve talked about light as a wave, we’ve talked about its frequency, its wavelength, its speed, its amplitude. We’ve talked about the wave properties of light, now we’re going to move and think about the particle properties of light. What happens when we think of light as a particle as opposed to as a wave?

Let’s say we have a laser, can I keep making this laser dot dimmer and dimmer and dimmer forever? This may seem like a very abstract philosophical question. I’m going to flip it on its head for you. Can I take a sample of water and keep reducing its amount forever? No, eventually I get down to one water molecule, and I’m done. This was the basis for the atomic theory. You can’t separate matter forever. I’m just asking you the exact same question for a dot of light, can I keep having it forever? And it turns out the answer is no, I can’t. At some point I reach the bottom, there’s a smallest dimness, just like there’s a smallest amount of water you can have, there’s a smallest amount of light you can have, and we call this smallest amount of light we say it’s a particle of light, and we call it a photon, and we are going to use this symbol \gamma, the Greek letter gamma, for photon.

We can think of this laser as a light wave, where I change the amplitude to make it brighter or darker, or we can flip that on its head and say it’s a bunch of photons flying along together and to make it brighter or darker I changed the number of photons. Already we’re sort of bouncing back and forth between thinking of things as waves and particles. This photon image is really good when we think about light being absorbed by materials or emitted from materials; that’s when thinking in terms of particles tends to be a good picture. Waves on the other hand tend to do really well when we’re thinking about light flying through space.

Let’s go through the properties of the photon. We are now imagining light to be made up of little balls, but we are imagining them to be made up of little massless spheres. Little massless particles that travel at the speed of light, c. But even though they are massless they still carry energy and momentum.

Thinking about detecting/absorbing light? Think particles!

Almost all detection systems talked about thus far—eyes, photographic plates, photomultiplier tubes in microscopes, and CCD cameras—rely on particle-like properties of photons interacting with a sensitive area. A change is caused and either the change is cascaded or zillions of points are recorded to form an image we detect. These detectors are used in biomedical imaging systems, and there is ongoing research into improving the efficiency of receiving photons, particularly by cooling detection systems and reducing thermal effects.

Photon Momentum – Relationship to Wavelength

A Venn-diagram showing the different sciences
In this part, we are explicitly trying to delve deeper into an equation you saw in Chemistry: E = \frac{hc}{\lambda}. We will see that this equation, while fine for chemistry, is NOT a fundamental principle and thus will NOT be a starting point for us in this class. If you wish to review the chemistry perspective, watch the video below. The video has captions. I did not include the transcript as this video is simply provided to review the chemistry perspective, not as a main focus for our course.
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The quantum of EM radiation we call a photon has properties analogous to those of particles we can see, such as grains of sand. A photon interacts as a unit in collisions or when absorbed, rather than as an extensive wave. Massive quanta, like electrons, also act like macroscopic particles—something we expect, because they are the smallest units of matter. Particles carry momentum as well as energy. Despite photons having no mass, there has long been evidence that EM radiation carries momentum. (Maxwell and others who studied EM waves predicted that they would carry momentum.) It is now a well-established fact that photons do have momentum. In fact, photon momentum is suggested by the photoelectric effect, where photons knock electrons out of a substance. Figure 2 shows macroscopic evidence of photon momentum.

(a) Trajectory of a comet with a nucleus and tail as it passes by the Sun is shown as a partial parabolic path with Sun near the vertex of the parabolic path. (b) The photograph of a moving Hale Bopp comet in space is shown as bright lighted object.
Figure 2: The tails of the Hale-Bopp comet point away from the Sun, evidence that light has momentum. Dust emanating from the body of the comet forms this tail. Particles of dust are pushed away from the Sun by light reflecting from them. The blue ionized gas tail is also produced by photons interacting with atoms in the comet material. (credit: Geoff Chester, U.S. Navy, via Wikimedia Commons).

Figure 2. shows a comet with two prominent tails. What most people do not know about the tails is that they always point away from the Sun rather than trailing behind the comet (like the tail of Bo Peep’s sheep). Comet tails are composed of gases and dust evaporated from the body of the comet and ionized gas. The dust particles recoil away from the Sun when photons scatter from them. Evidently, photons carry momentum in the direction of their motion (away from the Sun), and some of this momentum is transferred to dust particles in collisions. Gas atoms and molecules in the blue tail are most affected by other particles of radiation, such as protons and electrons emanating from the Sun, rather than by the momentum of photons.

Conservation of Momentum

Not only is momentum conserved in all realms of physics, but all types of particles are found to have momentum. We expect particles with mass to have momentum, but now we see that massless particles including photons also carry momentum.

Some of the earliest direct experimental evidence of photon momentum came from scattering of X-ray photons by electrons in substances, named Compton scattering after the American physicist Arthur H. Compton (1892–1962). Around 1923, Compton observed that X-rays scattered from materials had a decreased energy and correctly analyzed this as being due to the scattering of photons from electrons. This phenomenon could be handled as a collision between two particles—a photon and an electron at rest in the material. Energy and momentum are conserved in the collision. (See Figure) He won a Nobel Prize in 1929 for the discovery of this scattering, now called the Compton effect, because it helped prove that photon momentum is given by the de Broglie relation

p = \frac{h}{\lambda}

where h is Planck’s constant: a fundamental constant of the Universe (just like the speed of light c or \epsilon_0). The value for Planck’s constant is h = 6.626 \times 10^{-34} \, \mathrm{J \cdot s}, or in terms of electron volts eV (described in the review of energy) h = 4.135 \times 10^{-15} \, \mathrm{eV \cdot s}. This constant, like all constants, is provided on your equation sheet.

Instructor’s Note

 

We will see in a later chapter on matter waves, that this same relation works for electrons as well. Thus, the de Broglie relation

p = \frac{h}{\lambda}

is one of the fundamental principles for this unit! It connects the particle nature of matter (p is a particle property) and matter’s wave nature (\lambda is a wave property).

The Compton effect is the name given to the scattering of a photon by an electron shown in Figure 3. Energy and momentum are conserved, resulting in a reduction of both for the scattered photon. Studying this effect, Compton verified that photons have momentum. We can see that photon momentum is small, since p=h/λ, and h is very small. It is for this reason that we do not ordinarily observe photon momentum. Our mirrors do not recoil when light reflects from them (except perhaps in cartoons). Compton saw the effects of photon momentum because he was observing x rays, which have a small wavelength and a relatively large momentum, interacting with the lightest of particles, the electron. We will explore this particular phenomenon more in class.

Collision of an electron with a photon of energy E equal to h f is shown. The electron is represented as a spherical ball and the photon as an ellipse enclosing a wave. After collision the energy of the photon becomes E prime equal to h f prime and the final energy of an electron K E sub e is equal to E minus E prime. The direction of electron and photon before and after collision is represented by arrows.
Figure 3: The Compton effect is the name given to the scattering of a photon by an electron. Energy and momentum are conserved, resulting in a reduction of both for the scattered photon. Studying this effect, Compton verified that photons have momentum.

 

Electron and Photon Momentum Compared

(a) Calculate the momentum of a visible photon that has a wavelength of 500 nm. (b) Find the velocity of an electron having the same momentum.

Strategy

Finding the photon momentum is a straightforward application of its definition: p = h/\lambda. Then, we use the formulas we know from 131 to find the electron’s momentum and velocity.

Solution for (a)

Photon momentum is given by the equation:

p = h/\lambda.

Entering the given photon wavelength yields

p = \frac{6.63 \times 10^{-34} \, \mathrm{J \cdot s}}{500 \times 10^{-9} \, \mathrm{m}} = 1.33 \times 10^{-27} \, \mathrm{kg \cdot m/s}.

Solution for (b)

Since this momentum is indeed small, we will use the classical expression p=mv to find the velocity of an electron with this momentum. Solving for v and using the known value for the mass of an electron gives

v = \frac{p}{m} = \frac{1.33 \times 10^{-27} \, \mathrm{kg \cdot m/s}}{9.11 \times 10^{-31} \, \mathrm{kg}} = 1460 \, \mathrm{m/s}.

Discussion

Photon momentum is indeed small. Even if we have huge numbers of them, the total momentum they carry is small. An electron with the same momentum has a 1460 m/s velocity, which is clearly nonrelativistic. A more massive particle with the same momentum would have an even smaller velocity. This is borne out by the fact that it takes far less energy to give an electron the same momentum as a photon. But on a quantum-mechanical scale, especially for high-energy photons interacting with small masses, photon momentum is significant. Even on a large scale, photon momentum can have an effect if there are enough of them and if there is nothing to prevent the slow recoil of matter. Comet tails are one example, but there are also proposals to build space sails that use huge low-mass mirrors (made of aluminized Mylar) to reflect sunlight. In the vacuum of space, the mirrors would gradually recoil and could actually take spacecraft from place to place in the solar system. (See Figure 4.)

(a) A payload having an umbrella-shaped solar sail attached to it is shown. The direction of movement of payload and direction of incident photons are shown using arrows. (b) A photograph of the top view of a silvery space sail.
Figure 4: (a) Space sails have been proposed that use the momentum of sunlight reflecting from gigantic low-mass sails to propel spacecraft about the solar system. A Russian test model of this (the Cosmos 1) was launched in 2005, but did not make it into orbit due to a rocket failure. (b) A U.S. version of this, labeled LightSail-1, is scheduled for trial launches in the first part of this decade. It will have a 40-m2 sail. (credit: Kim Newton/NASA)

Photon Momentum – Relationship to Energy

Photons, in addition to having energy, also have momentum. This is the part that tends to get folks, because in 131. we told you that momentum was mass times velocity which is mostly true. It’s true as long as you’re not going too fast, once you start getting close to the speed of light this will actually break down on you. You need a new expression. But as long as you’re going slow, this is fine. But clearly this does not work for photons because for photons mass is zero. Special relativity has an answer, it’s the momentum of a photon is the energy divided by the speed of light,

p=\frac{E_\gamma}{c} or E_\gamma = pc.

 

Instructor’s Note

 

If you look at the Unit I On-a-Page, you will see that this is one of the fundamental definitions of this unit: the definition of a photon’s momentum in terms of its energy

p=\frac{E_\gamma}{c}.

A way to help keep all of these formula straight: if the formula contains a c then it only applies to light, if the formula contains an m, then it only applies to particles with mass (like electrons)!

From the fundamental principle of this unit, the de Broglie relation p = \frac{h}{\lambda}, and this definition of a photon’s momentum in terms of its energy p=\frac{E}{c}, we can derive a formula that was given to you in your chemistry classes. While I, in general, try to avoid derivations, I think this one is useful as it is short and shows you why what you learned in chemistry is the way it is. That is, after all, one of the motivations for this unit: why does chemistry work?

So we know, p = \frac{h}{\lambda} and p=\frac{E}{c}. Since both equations are equal to p, we can set them equal to each other:

\frac{h}{\lambda} = \frac{E_\gamma}{c}

which, after some rearranging (move the c over) we get the familiar

E = \frac{hc}{\lambda}.

A Venn-diagram showing the different sciences

 

 

 

 

 

 

 

You can start with this equation that you know from chemistry. However, keep in mind that it is NOT a fundamental relationship: it comes from combining:

  • The fundamental principle of the de Broglie relation: p = h/\lambda that connects the wave and particle natures for all matter.
  • The definition of a photon’s momentum in terms of its energy: E_\gamma = pc, which is only specific to photons.

Therefore, the relationship E = \frac{hc}{\lambda} only applies to photons. I have seen many students make mistakes of trying to apply it to electrons!

Compare the Energies of the Photon and Electron from the Last Example

What is the energy of the 500 nm photon, and how does it compare with the energy of the electron with the same momentum?

The electron:

There are two ways of approaching this problem.

1. Use the fact that we know the electron’s velocity to be 1460 \, \mathrm{m/s}, and the expression for kinetic energy from Physics 131: K = \frac{1}{2} m v^2:

K = \frac{1}{2} (9.11 \times 10^{-31} \, \mathrm{kg})(1460 \, \mathrm{m/s})^2

K = 9.64 \times 10^{-25} \, \mathrm{J} = 6.02 \times 10^{-6} \, \mathrm{eV}

2. Directly use the fact that we already know the electron’s momentum from the previous problem p = 1.33 \times 10^{-27} \, \mathrm{kg \cdot m/s}. Combine this knowledge and the  idea of converting directly from momentum to energy for particles with mass using the formula derived in Some Energy-Related Ideas that Might be New: The Connection between Energy and Momentum:

K = \frac{p^2}{2m}

K = \frac{(1.33 \times 10^{-27} \, \mathrm{kg \cdot m/s})^2}{2(9.11 \times 10^{-31} \, \mathrm{kg})} = 9.64 \times 10^{-25} \, \mathrm{J} = 6.02 \times 10^{-6} \, \mathrm{eV}

Clearly, both approaches give the same response as they must.

The photon:

Again, there are two approaches:

1. Use the momentum of the photon to get the energy using E_\gamma = pc:

E_\gamma = pc

E_\gamma = ((1.33 \times 10^{-27} \, \mathrm{kg \cdot m/s})(3 \times 10^8 \, \mathrm{m/s})

E_\gamma = 3.99 \times 10^{-19} \, \mathrm{J} = 2.49 \, \mathrm{eV}.

Where eV are electron volts discussed in Units of Energy.

2. The second approach, is to use the wavelength, coupled with the expression we just derived / you learned in chemistry:

E = \frac{hc}{\lambda}

E = \frac{(6.626 \times 10^{-34} \, \mathrm{J \cdot s})(3 \times 10^8 \, \mathrm{m/s})}{500 \times 10^{-9} \, \mathrm{m}}

E = 3.99 \times 10^{-19} \, \mathrm{J} = 2.49 \, \mathrm{eV}.

Again, both approaches give the same value, as they must.

 

Photon Energies and the Electromagnetic Spectrum

A photon is a quantum of EM radiation whose momentum is related to its wavelength by p = \frac{h}{\lambda}. Combined with the connection between a photon’s energy and momentum E_\gamma = pc yields the energy-wavelength relationship E = \frac{hc}{\lambda}.

All EM radiation is composed of photons. Figure 5 shows various divisions of the EM spectrum plotted against wavelength, frequency, and photon energy. Previously in this book, photon characteristics were alluded to in the discussion of some of the characteristics of UV, x rays, and γ-rays, the first of which start with frequencies just above violet in the visible spectrum. It was noted that these types of EM radiation have characteristics much different than visible light. We can now see that such properties arise because photon energy is larger at high frequencies.

The electromagnetic spectrum including energy
Figure 5: The EM spectrum, showing major categories as a function of photon energy in eV, as well as wavelength and frequency. Certain characteristics of EM radiation are directly attributable to photon energy alone.

Photons act as individual quanta and interact with individual electrons, atoms, molecules, and so on. The energy a photon carries is, thus, crucial to the effects it has. Table 1 lists representative submicroscopic energies in eV. When we compare photon energies from the EM spectrum in Figure 5 with energies in the table, we can see how effects vary with the type of EM radiation.

Table 1

Representative Energies for Submicroscopic Effects
(Order of Magnitude Only)

Energy between outer electron shells in atoms 1 eV
Binding energy of a weakly bound molecule 1 eV
Energy of red light 2 eV
Binding energy of a tightly bound molecule 10 eV
Energy to ionize atom or molecule 10 to 1000 eV

Ionizing Radiation

Gamma rays

A form of nuclear and cosmic EM radiation, can have the highest frequencies and, hence, the highest photon energies in the EM spectrum. For example, a γ-ray photon with \nu = 10^{21} \, \mathrm{Hz} has an energy E=h \nu = 6.63 \times 10^{-13} \, \mathrm{J} = 4.14 \, \mathrm{MeV}. This is sufficient energy to ionize thousands of atoms and molecules, since only 10 to 1000 eV are needed per ionization. In fact, γ rays are one type of ionizing radiation, as are x rays and UV, because they produce ionization in materials that absorb them. Because so much ionization can be produced, a single γ-ray photon can cause significant damage to biological tissue, killing cells or damaging their ability to properly reproduce. When cell reproduction is disrupted, the result can be cancer, one of the known effects of exposure to ionizing radiation. Since cancer cells are rapidly reproducing, they are exceptionally sensitive to the disruption produced by ionizing radiation. This means that ionizing radiation has positive uses in cancer treatment as well as risks in producing cancer. However, the high photon energy also enables γ rays to penetrate materials, since a collision with a single atom or molecule is unlikely to absorb all the γ ray’s energy. This can make γ rays useful as a probe, and they are sometimes used in medical imaging.

X-rays

X-rays, as you can see in Figure 5, overlap with the low-frequency end of the γ ray range. Since x rays have energies of keV and up, individual x-ray photons also can produce large amounts of ionization. At lower photon energies, x rays are not as penetrating as γ rays and are slightly less hazardous. X-rays are ideal for medical imaging, their most common use, and a fact that was recognized immediately upon their discovery in 1895 by the German physicist W. C. Roentgen (1845–1923). (See Figure 6.) Within one year of their discovery, x rays (for a time called Roentgen rays) were used for medical diagnostics. Roentgen received the 1901 Nobel Prize for the discovery of x rays.

One of the first x-ray images, taken by Röentgen himself. The hand belongs to Bertha Röentgen, his wife.
Figure 6: One of the first x-ray images, taken by Röentgen himself. The hand belongs to Bertha Röentgen, his wife. (credit: Wilhelm Conrad Röntgen, via Wikimedia Commons)

While γ rays originate in nuclear decay, x rays are produced by the process shown in Figure 7. Electrons ejected by thermal agitation from a hot filament in a vacuum tube are accelerated through a high voltage, gaining kinetic energy from the electrical potential energy. When they strike the anode, the electrons convert their kinetic energy to a variety of forms, including thermal energy. But since an accelerated charge radiates EM waves, and since the electrons act individually, photons are also produced. Some of these x-ray photons obtain the kinetic energy of the electron. The accelerated electrons originate at the cathode, so such a tube is called a cathode ray tube (CRT), and various versions of them are found in older TV and computer screens as well as in x-ray machines.

X rays are produced when energetic electrons strike the copper anode of this cathode ray tube (CRT). Electrons (shown here as separate particles) interact individually with the material they strike, sometimes producing photons of EM radiation.
Figure 7: X rays are produced when energetic electrons strike the copper anode of this cathode ray tube (CRT). Electrons (shown here as separate particles) interact individually with the material they strike, sometimes producing photons of EM radiation.

Figure 8 shows the spectrum of x rays obtained from an x-ray tube. There are two distinct features to the spectrum. First, the smooth distribution results from electrons being decelerated in the anode material. A curve like this is obtained by detecting many photons, and it is apparent that the maximum energy is unlikely. This decelerating process produces radiation that is called bremsstrahlung (German for braking radiation). The second feature is the existence of sharp peaks in the spectrum; these are called characteristic x rays, since they are characteristic of the anode material. Characteristic x rays come from atomic excitations unique to a given type of anode material. They are akin to lines in atomic spectra, implying the energy levels of atoms are quantized.

X-ray spectrum obtained when energetic electrons strike a material. The smooth part of the spectrum is bremsstrahlung, while the peaks are characteristic of the anode material. Both are atomic processes that produce energetic photons known as x-ray photons.
Figure 8: X-ray spectrum obtained when energetic electrons strike a material. The smooth part of the spectrum is bremsstrahlung, while the peaks are characteristic of the anode material. Both are atomic processes that produce energetic photons known as x-ray photons.

Connections: Conservation of Energy

Once again, we find that conservation of energy allows us to consider the initial and final forms that energy takes, without having to make detailed calculations of the intermediate steps.

Find the minimum wavelength of an x-ray photon produced by electrons accelerated through a potential energy difference of 50.0 keV in a CRT like the one in Figure 7.

Strategy

Electrons can give all of their kinetic energy to a single photon when they strike the anode of a CRT. The kinetic energy of the electron comes from electrical potential energy. Thus we can simply equate the maximum photon energy to the electrical potential energy

Solution

\Delta E = Q + W

E_f - E_i = Q + W

In the initial state, we have an electron with 50 keV of potential energy and no kinetic energy: E_i = U_i = 50 \, \mathrm{keV}. At the end, all that energy is in the photon: E_f = E_\gamma. No other energy enters or leaves the system (the photon and electron are everything we care about!), so Q = W = 0

E_\gamma - U_i = 0

E_\gamma = U_i

\frac{hc}{\lambda} = U_i

\frac{1}{\lambda} = \frac{U_i}{hc}

\lambda = \frac{hc}{U_i}

\lambda = \frac{(6.626 \times 10^{-34} \, \mathrm{J \cdot s})(3 \times 10^8 \, \mathrm{m/s})}{(50 \times 10^3 \, \mathrm{eV}) \left( \frac{1.602 \times 10^{-19} \, \mathrm{J}}{\mathrm{eV}} \right)}

\lambda = 2.48 \times 10^{-11} \, \mathrm{m} = 0.025 \, \mathrm{nm}

Ultraviolet Radiation

Ultraviolet radiation (approximately 4 eV to 300 eV) overlaps with the low end of the energy range of x rays, but UV is typically lower in energy. UV comes from the de-excitation of atoms that may be part of a hot solid or gas. These atoms can be given energy that they later release as UV by numerous processes, including electric discharge, nuclear explosion, thermal agitation, and exposure to x rays. A UV photon has sufficient energy to ionize atoms and molecules, which makes its effects different from those of visible light. UV thus has some of the same biological effects as γ-rays and x-rays. For example, it can cause skin cancer and is used as a sterilizer. The major difference is that several UV photons are required to disrupt cell reproduction or kill a bacterium, whereas single γ-ray and x-ray photons can do the same damage. But since UV does have the energy to alter molecules, it can do what visible light cannot. One of the beneficial aspects of UV is that it triggers the production of vitamin D in the skin, whereas visible light has insufficient energy per photon to alter the molecules that trigger this production. Infantile jaundice is treated by exposing the baby to UV (with eye protection), called phototherapy, the beneficial effects of which are thought to be related to its ability to help prevent the buildup of potentially toxic bilirubin in the blood.

Photon Energy and Effects for UV

Short-wavelength UV is sometimes called vacuum UV, because it is strongly absorbed by air and must be studied in a vacuum. Calculate the photon energy in eV for 100-nm vacuum UV, and estimate the number of molecules it could ionize or break apart.

Strategy

Using the equation E= \frac{hc}{\lambda} and appropriate constants, we can find the photon energy and compare it with energy information in Table 1.

Solution

The energy of a photon is given by

E= \frac{hc}{\lambda}

Using hc=1240 eV⋅nm,

we find that

E=hc/λ=(1240 eV⋅nm)/100 nm=12.4 eV.

Discussion

According to Table 1, this photon energy might be able to ionize an atom or molecule, and it is about what is needed to break up a tightly bound molecule, since they are bound by approximately 10 eV. This photon energy could destroy about a dozen weakly bound molecules. Because of its high photon energy, UV disrupts atoms and molecules it interacts with. One good consequence is that all but the longest-wavelength UV is strongly absorbed and is easily blocked by sunglasses. In fact, most of the Sun’s UV is absorbed by a thin layer of ozone in the upper atmosphere, protecting sensitive organisms on Earth. Damage to our ozone layer by the addition of such chemicals as CFC’s has reduced this protection for us.

Visible Light

The range of photon energies for visible light from red to violet is 1.63 to 3.26 eV, respectively. These energies are on the order of those between outer electron shells in atoms and molecules. This means that these photons can be absorbed by atoms and molecules. A single photon can actually stimulate the retina, for example, by altering a receptor molecule that then triggers a nerve impulse. As reviewed from chemistry in a future chapter, photons can be absorbed or emitted only by atoms and molecules that have precisely the correct quantized energy step to do so. For example, if a red photon of frequency \nu encounters a molecule that has an energy step, \Delta E = h \nu, then the photon can be absorbed. Violet flowers absorb red and reflect violet; this implies there is no energy step between levels in the receptor molecule equal to the violet photon’s energy, but there is an energy step for the red.

There are some noticeable differences in the characteristics of light between the two ends of the visible spectrum that are due to photon energies. Red light has insufficient photon energy to expose most black-and-white film, and it is thus used to illuminate darkrooms where such film is developed. Since violet light has a higher photon energy, dyes that absorb violet tend to fade more quickly than those that do not. (See Figure 9.) Take a look at some faded color posters in a storefront some time, and you will notice that the blues and violets are the last to fade. This is because other dyes, such as red and green dyes, absorb blue and violet photons, the higher energies of which break up their weakly bound molecules. (Complex molecules such as those in dyes and DNA tend to be weakly bound.) Blue and violet dyes reflect those colors and, therefore, do not absorb these more energetic photons, thus suffering less molecular damage.

Photograph of a worn-out movie advertisement poster on a wall.
Figure 9: Why do the reds, yellows, and greens fade before the blues and violets when exposed to the Sun, as with this poster? The answer is related to photon energy. (credit: Deb Collins, Flickr)

 

Transparent materials, such as some glasses, do not absorb any visible light, because there is no energy step in the atoms or molecules that could absorb the light. Since individual photons interact with individual atoms, it is nearly impossible to have two photons absorbed simultaneously to reach a large energy step. Because of its lower photon energy, visible light can sometimes pass through many kilometers of a substance, while higher frequencies like UV, x-ray, and γγ size 12{γ} {}">γ-rays are absorbed, because they have sufficient photon energy to ionize the material.

How Many Photons per Second Does a Typical Light Bulb Produce?

Assuming that 10.0% of a 100-W light bulb’s energy output is in the visible range (typical for incandescent bulbs) with an average wavelength of 580 nm, calculate the number of visible photons emitted per second.

Strategy

Power is energy per unit time, and so if we can find the energy per photon, we can determine the number of photons per second. This will best be done in Joules, since power is given in Watts, which are Joules per second.

Solution

The power in visible light production is 10.0% of 100 W, or 10.0 J/s. The energy of the average visible photon is found by substituting the given average wavelength into the formula

E = \frac{hc}{\lambda}

This produces

E = \frac{ (6.63 \times 10^{-34} \, \mathrm{J \cdot s})(3.00 \times 10^8 \mathrm{m/s})}{580 \times 10^{-9} \, \mathrm{m}} = 3.43 \times 10^{-19} \, \mathrm{J}.

The number of visible photons per second is thus

\frac{\mathrm{photon}}{\mathrm{s}} = \frac{10.0 \, \mathrm{J/s}}{3.43 \times 10^{-19} \, \mathrm{J/photon}}= 2.92 \times 10^{19} \, \mathrm{photon/s}.

Discussion

This incredible number of photons per second is verification that individual photons are insignificant in ordinary human experience. It is also a verification of the correspondence principle—on the macroscopic scale, quantization becomes essentially continuous or classical. Finally, there are so many photons emitted by a 100-W lightbulb that it can be seen by the unaided eye many kilometers away.

Lower Energy Photons

Infrared Radiation (IR)

Infrared radiation (IR) has even lower photon energies than visible light and cannot significantly alter atoms and molecules. IR can be absorbed and emitted by atoms and molecules, particularly between closely spaced states. IR is extremely strongly absorbed by water, for example, because water molecules have many states separated by energies on the order of 10–5eV 10–5 eV size 12{" 10" rSup { size 8{"–5"} } " eV "} {}">10–5eV to 10–2eV, 10–2 eV, size 12{" 10" rSup { size 8{"–2"} } " eV "} {}">10–2eV, well within the IR and microwave energy ranges. This is why in the IR range, skin is almost jet black, with an emissivity near 1—there are many states in water molecules in the skin that can absorb a large range of IR photon energies. Not all molecules have this property. Air, for example, is nearly transparent to many IR frequencies.

Microwaves

Microwaves are the highest frequencies that can be produced by electronic circuits, although they are also produced naturally. Thus microwaves are similar to IR but do not extend to as high frequencies. There are states in water and other molecules that have the same frequency and energy as microwaves, typically about 10–5eV. 10–5 eV. size 12{" 10" rSup { size 8{"–5"} } " eV "} {}">10–5eV. This is one reason why food absorbs microwaves more strongly than many other materials, making microwave ovens an efficient way of putting energy directly into food.

Photon energies for both IR and microwaves are so low that huge numbers of photons are involved in any significant energy transfer by IR or microwaves (such as warming yourself with a heat lamp or cooking pizza in the microwave). Visible light, IR, microwaves, and all lower frequencies cannot produce ionization with single photons and do not ordinarily have the hazards of higher frequencies. When visible, IR, or microwave radiation is hazardous, such as the inducement of cataracts by microwaves, the hazard is due to huge numbers of photons acting together (not to an accumulation of photons, such as sterilization by weak UV). The negative effects of visible, IR, or microwave radiation can be thermal effects, which could be produced by any heat source. But one difference is that at very high intensity, strong electric and magnetic fields can be produced by photons acting together. Such electromagnetic fields (EMF) can actually ionize materials.

Misconception Alert: High-Voltage Power Lines

Although some people think that living near high-voltage power lines is hazardous to one’s health, ongoing studies of the transient field effects produced by these lines show their strengths to be insufficient to cause damage. Demographic studies also fail to show significant correlation of ill effects with high-voltage power lines. The American Physical Society issued a report over 10 years ago on power-line fields, which concluded that the scientific literature and reviews of panels show no consistent, significant link between cancer and power-line fields. They also felt that the “diversion of resources to eliminate a threat which has no persuasive scientific basis is disturbing.”

Lower Energy than Microwaves

It is virtually impossible to detect individual photons having frequencies below microwave frequencies, because of their low photon energy. But the photons are there. A continuous EM wave can be modeled as photons. At low frequencies, EM waves are generally treated as time- and position-varying electric and magnetic fields with no discernible quantization. This is another example of the correspondence principle in situations involving huge numbers of photons.

 

Review from Chemistry of Application of Conservation of Energy to Photons and Atoms

8

A Venn-diagram showing the different sciencesIn your general chemistry courses, you all ready did some of what will be a big part of this unit: namely using the ideas of wave-particle duality in conjunction with conservation of energy. In chemistry, you did this in the context of looking at atomic transitions. In this chapter, you will review the ideas from chemistry, and then be exposed to some differences in how we will treat this same situation in a physics course. The reasons for the differences are three-fold. First, as described elsewhere in this book, each scientific discipline grew with its own history and conventions. The second reason for a different perspective is that the view taken by chemistry, while perfectly fine for all situations you encountered in that course, will break down for some of the situations we want to analyze in this course. The third, and arguably most important reason, mirrors the motivation for a variety of cultural perspectives in a humanities course: by exploring the same processes from different perspectives you gain a deeper and more holistic understanding of the material. Just as your understanding of history is incomplete if you only consider white men, so your understanding of conservation of energy is incomplete if you only look at it from a biology, chemistry, or physics perspective.

Review of Connecting Conservation of Energy to the Wave and Particle Natures of Light in the Context of the Hydrogen Atom from ChemistryPaul Flowers et al. Chemistry: Atoms First 2e. Open Stax, 2014.

Instructor’s Notes

 

NOTE: This is review. If you are familiar with this material, feel free to skip to the problems at the end!

What we expect you to know from this review:

  • Electrons in atoms have discrete energy levels.
  • In order for an electron to transition from one level to another it must absorb or emit a photon with the exact amount of energy that corresponds to the energy difference. If the energy does not exactly match, then the transition will not occur.
  • If I know the energy difference, I can solve for the wavelength of the emitted photon.
  • Be familiar with the idea of energy level diagrams such as the one shown below
A diagram showing the different electron energy levels as horizontal lines with vertical position representing energy.
  • The difference in energy levels in a hydrogen atom is

\Delta E = - Rhc \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)

where R is the Rydberg constant R = 1.097 \times 10^7 \, \mathrm{m}^{-1}.

Following the work of Ernest Rutherford and his colleagues in the early twentieth century, the picture of atoms consisting of tiny dense nuclei surrounded by lighter and even tinier electrons continually moving about the nucleus was well established. This picture was called the planetary model, since it pictured the atom as a miniature “solar system” with the electrons orbiting the nucleus like planets orbiting the sun. The simplest atom is hydrogen, consisting of a single proton as the nucleus about which a single electron moves. The electrostatic force attracting the electron to the proton depends only on the distance between the two particles. This classical mechanics description of the atom is incomplete, however, since as described in Basics of Light: Where does Light Come From? an electron moving in an orbit would be accelerating (by changing direction) and, according to classical electromagnetism, it should continuously emit electromagnetic radiation. This loss in orbital energy should result in the electron’s orbit getting continually smaller until it spirals into the nucleus, implying that atoms are inherently unstable.

In 1913, Niels Bohr attempted to resolve the atomic paradox by ignoring classical electromagnetism’s prediction that the orbiting electron in hydrogen would continuously emit light. Instead, he incorporated into the classical mechanics description of the atom Planck’s ideas of quantization and Einstein’s finding that light consists of photons whose energy is proportional to their frequency (Basics of Light: Introduction to the Photon). Bohr assumed that the electron orbiting the nucleus would not normally emit any radiation (the stationary state hypothesis), but it would emit or absorb a photon if it moved to a different orbit. The energy absorbed or emitted would reflect differences in the orbital energies according to this equation:

\left| \Delta E \right| = \left| E_f - E_i \right| = \frac{hc}{\lambda}

In this equation, h is Planck’s constant and Ei and Ef are the initial and final orbital energies, respectively. The absolute value of the energy difference is used, since frequencies and wavelengths are always positive. Instead of allowing for continuous values of energy, Bohr assumed the energies of these electron orbitals were quantized:

E_n = -\frac{Rhc}{n^2}

Where R is the so-called Rydberg constant that had been previously determined by experimental analysis of hydrogen spectra R = 1.097 \times 10^7 \, \mathrm{m}^{-1}. Inserting this expression into the equation for \Delta E gives

\Delta E = -Rhc \left( \frac{1}{n_f}^2 - \frac{1}{n_i}^2 \right) = \frac{hc}{\lambda}.

The lowest few energy levels are shown in Figure 1. One of the fundamental laws of physics is that matter is most stable with the lowest possible energy. Thus, the electron in a hydrogen atom usually moves in the n = 1 orbit, the orbit in which it has the lowest energy. When the electron is in this lowest energy orbit, the atom is said to be in its ground electronic state (or simply ground state). If the atom receives energy from an outside source, it is possible for the electron to move to an orbit with a higher n value and the atom is now in an excited electronic state (or simply an excited state) with a higher energy. When an electron transitions from an excited state (higher energy orbit) to a less excited state, or ground state, the difference in energy is emitted as a photon. Similarly, if a photon is absorbed by an atom, the energy of the photon moves an electron from a lower energy orbit up to a more excited one. We can relate the energy of electrons in atoms to what we learned previously about energy. The law of conservation of energy says that we can neither create nor destroy energy. Thus, if a certain amount of external energy is required to excite an electron from one energy level to another, that same amount of energy will be liberated when the electron returns to its initial state (Figure 2).

The energy levels of the hydrogen atom in the Bohr model
Figure 1: Quantum numbers and energy levels in a hydrogen atom. The more negative the calculated value, the lower the energy.

 

A diagram showing the different electron energy levels as horizontal lines with vertical position representing energy.
Figure 2: An electron in hydrogen absorbing and then re-emitting a photon with an energy corresponding to the exact difference between two energy levels.

Calculating the Energy and Wavelength of Electron Transitions in Hydrogen Using Bohr’s Formula

What is the energy (in joules) and the wavelength (in meters) of the line in the spectrum of hydrogen that represents the movement of an electron from Bohr orbit with n = 6 to the orbit with n = 2? In what part of the electromagnetic spectrum do we find this radiation?

Solution

In this case, the electron starts out with n = 6, so ni = 6. It comes to rest in the n = 2 orbit, so nf = 2. The difference in energy between the two states is given by this expression:

\Delta E = E_f - E_i = Rhc \left( \frac{1}{n_f} - \frac{1}{n_i} \right)

\Delta E = (1.097 \times 10^7 \, \mathrm{m}^{-1})(6.626 \times 10^{-34} \, \mathrm{J \cdot s})(3.00 \times 10^8 \, \mathrm{m/s}) \left( \frac{1}{2^2} - \frac{1}{6^2} \right)

\Delta E = -4.85 \times 10^{-19} \, \mathrm{J}

This energy difference is negative, indicating a photon leaves the system (is emitted) as the electron falls from the n = 6 orbit to the n = 2 orbit.

The wavelength of a photon with this energy is found by the expression E=hc/λ.

Rearrangement gives:

\lambda = \frac{hc}{E}

\lambda = \frac{ (6.626\times 10^{-34} \, \mathrm{J \cdot s})(3.00 \times 10^8 \, \mathrm{m/s})}{4.85 \times 10^{-19} \, \mathrm{J}} = 4.10 \times 10^{-7} \, \mathrm{m}

From the illustration of the electromagnetic spectrum in Basics of Light: The Main Parts of the Electromagnetic Spectrum, we can see that this wavelength is found in the violet portion of the electromagnetic spectrum.

Thinking about Atomic Transitions from a Physics Perspective

In chemistry, the starting point for the analysis was generally

\Delta E = \frac{hc}{\lambda}

or if, you were looking at the hydrogen atom specifically,

\Delta E = - Rhc \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right).

Moreover, when solving the problem, you would probably just consider the absolute value of the change in energy and then figure out emission or absorption.

How would we look at this same problem in physics? In physics, we like to start with fundamental principles of the Universe and then apply definitions as described in Unit I On-A-Page: Principles and Definitions. In this case, the fundamental principle is Conservation of Energy: \Delta E = Q + W; this basic idea will, therefore be our starting point. To see how this works in practice, let’s look at an example, the n = 6 \rightarrow n = 2 transition discussed in the video.

The n = 6 \rightarrow n = 2 hydrogen transition from a physics perspective

Problem: What wavelength of light is does an electron in the n = 6 state of the hydrogen atom emit if it falls to the n = 2 state?

Solution – Starting with Principles:

First we identify the relevant fundamental principle of the Universe. In this situation, we know the relevant principle will be conservation of energy as the problem describes an electron losing energy as it moves from one state to another. Thus, we begin with

\Delta E = Q + W.

Expanding out the \Delta,

E_f - E_i = Q + W.

Now we need to consider the physics of the situation, or as I often call it, “what is the story that describes our phenomenon?” In this case, the story is:

I begin with an electron in a high energy state (n = 6). This electron then falls to a lower energy state (n=2). Since energy is conserved, that means that the energy lost by the electron has to go somewhere. In this case, the energy leaves the atom as light.

Having this story helps us fill in the parts of our conservation of energy equation:

  • E_f: is the energy of the electron at the end. In our case, the energy of the n = 2 state: E_{(n = 2)}.
  • E_i: is the energy of the electron at the beginning. In our case, the energy of the n = 6 state: E_{(n = 6)}.
  • Energy is leaving the atom through microscopic interactions, i.e. the emission of a photon. Thus, Q is the energy of the photon E_\gamma.
  • There are no forces being applied to the atom. Since work is a force applied for some distance, W = Fd \cos \theta,  we know that the work in this case is zero: W = 0.

Substituting our story into our equation (E_f \rightarrow E_{(n = 2)}, E_i \rightarrow E_{(n = 6)}, Q \rightarrow E_\gamma, and W = 0) then yields:

E_{(n = 2)} - E_{(n = 6)} = E_\gamma + 0.

Now add definitions:

From what we have covered, thus far, we can now put in definitions for the various quantities:

  • We know, from reviewing chemistry in the section above, that the energy of an electron in the hydrogen atom is given by E_n = \frac{-Rhc}{n^2}.

We can now put these in definitions into our equation:

\left( \frac{-Rhc}{2^2} \right) - \left( \frac{-Rhc}{6^2} \right) = E_\gamma.

Notice, as always, we are working in symbols which is one of our goals of the course. Now, we have a math problem! We are looking for \lambda, which we can get from E_\gamma, and h, c, and R are all just constants of nature. We begin by noticing that we can factor out the -Rhc on the left hand side:

-Rhc \left( \frac{1}{2^2} - \frac{1}{6^2} \right) = \frac{1}{\lambda}.

(This should now be starting to look like what you reviewed from chemistry!) We, of course, know 22 and 62, converting these and substituting in the values for R = 1.097 \times 10^7 \, \mathrm{m}^{-1}, c = 3 \times 10^8 \, \mathrm{m/s}, and h = 6.626 \times 10^{-34} \, \mathrm{J \cdot s} we have:

\left( 1.097 \times 10^7 \, \mathrm{m}^{-1} \right)\left( 6.626 \times 10^{-34} \, \mathrm{J \cdot s} \right) \left( c = 3 \times 10^8 \, \mathrm{m/s} \right) \times \left( \frac{1}{4} - \frac{1}{36} \right) = E_\gamma

E_\gamma = -4.85 \times 10^{-19} \, \mathrm{J} = -3.02 \, \mathrm{eV}

Notice, we never took any absolute values, and the result was negative. This negative sign has meaning! It tells us that the energy was lost to the atom! Of course, we already knew that, but this is a way to check our answers: we know energy is lost and we get a negative answer. The math takes care of itself!

Now we know from Basics of Light: Photon Momentum – Relationship to Energy that the energy of a photon is E_\gamma = \frac{hc}{\lambda}, and can get a wavelength:

E_\gamma = \frac{hc}{\lambda}

Here we will take an absolute value, because negative wavelengths, unlike negative energies, don’t make sense. So we have

\lambda = \frac{hc}{E_\gamma}

\lambda = \frac{ \left( 6.626 \times 10^{-34} \, \mathrm{J \cdot s} \right) \left( c = 3 \times 10^8 \, \mathrm{m/s} \right) }{ 4.85 \times 10^{-19} \, \mathrm{J} }

\lambda = 4.10 \times 10^{-7} \, \mathrm{m} = 410 \, \mathrm{nm}

Exactly the blue line described in the video.

 

Why we do it this way

You may be saying to yourself, that the physics method seems a lot longer with a lot more steps. Why not just follow chemistry and start with

\frac{hc}{\lambda} = -Rhc \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)?

The reason for the physics approach is that it is a lot more versatile: it can be applied to loads of situations from x-ray emissions, to photo electric effects, to LEDs, to gravitational redshifts, to particle/anti-particle annihilations, to…, you get the idea. In contrast,

\frac{hc}{\lambda} = -Rhc \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)

works ONLY for the wavelengths of photons and ONLY for hydrogen atoms. Learning to think in the physics perspective is one of the goals for this course!

 

Matter as a Wave

9

De Broglie Wavelength

In 1923 a French physics graduate student named Prince Louis-Victor de Broglie (1892–1987) made a radical proposal based on the hope that nature is symmetric. If EM radiation has both particle and wave properties, then nature would be symmetric if matter also had both particle and wave properties. If what we once thought of as an unequivocal wave (EM radiation) is also a particle, then what we think of as an unequivocal particle (matter) may also be a wave. De Broglie’s suggestion, made as part of his doctoral thesis, was so radical that it was greeted with some skepticism. A copy of his thesis was sent to Einstein, who said it was not only probably correct, but that it might be of fundamental importance. With the support of Einstein and a few other prominent physicists, de Broglie was awarded his doctorate.

De Broglie took both relativity and quantum mechanics into account to develop the proposal that all particles have a wavelength, given by

p = \frac{h}{\lambda} (matter and photons)λ=hp(matter and photons),λ=hp(matter and photons), size 12{λ = { {h} over {p} } } {}">,

where hh size 12{h} {}">h is Planck’s constant and pp size 12{p} {}">p is momentum. This is defined to be the de Broglie wavelength. (Note that we already have this for photons, from the equation p=h/λp=h/λ size 12{p = h/λ} {}">p=h/λ.) The hallmark of a wave is interference. If matter is a wave, then it must exhibit constructive and destructive interference. Why isn’t this ordinarily observed? The answer is that in order to see significant interference effects, a wave must interact with an object about the same size as its wavelength. Since hh size 12{h} {}">h is very small, λλ size 12{λ} {}">λ is also small, especially for macroscopic objects. A 3-kg bowling ball moving at 10 m/s, for example, has

p = \frac{h}{\lambda}

\lambda = \frac{h}{p}

\lambda = \frac{h}{mv}

\lambda = \frac{6.626 \times 10^{-34} \, \mathrm{J \cdot s}}{(3 \, \mathrm{kg})(10 \, \mathrm{m/s})} = 2 \times 10^{-35} \, \mathrm {m}λ=h/p=(6.63×10–34J·s)/[(3 kg)(10 m/s)]=2×10–35m.λ=h/p= (6.63 × 10–34 J·s)/[(3 kg)(10 m/s)] = 2 × 10–35 m. size 12{λ = h/p"= " \( 6 "." "63 " times " 10" rSup { size 8{"–34"} } " J·s" \) / \[ \( "3kg" \) \( "10 m/s" \) " = 2 " times " 10" rSup { size 8{"–35"} } " m"} {}">.

This means that to see its wave characteristics, the bowling ball would have to interact with something about 10–35m 10–35 m size 12{" 10" rSup { size 8{"–35"} } " m"} {}">10–35m in size—far smaller than anything known. When waves interact with objects much larger than their wavelength, they show negligible interference effects and move in straight lines (such as light rays in geometric optics). To get easily observed interference effects from particles of matter, the longest wavelength and hence smallest mass possible would be useful. Therefore, this effect was first observed with electrons.

Connections: Waves

All microscopic particles, whether massless, like photons, or having mass, like electrons, have wave properties. The relationship between momentum and wavelength is fundamental for all particles.

American physicists Clinton J. Davisson and Lester H. Germer in 1925 and, independently, British physicist G. P. Thomson (son of J. J. Thomson, discoverer of the electron) in 1926 scattered electrons from crystals and found diffraction patterns. These patterns are exactly consistent with interference of electrons having the de Broglie wavelength and are somewhat analogous to light interacting with a diffraction grating. (See Figure 1.)

Diffraction pattern obtained for electrons diffracted by crystalline silicon is shown. The diffraction pattern has a bright spot at the center of a circle with brighter and darker regions occurring in a symmetric manner.
Figure 1: This diffraction pattern was obtained for electrons diffracted by crystalline silicon. Bright regions are those of constructive interference, while dark regions are those of destructive interference. (credit: Ndthe, Wikimedia Commons)

De Broglie’s proposal of a wave nature for all particles initiated a remarkably productive era in which the foundations for quantum mechanics were laid. In 1926, the Austrian physicist Erwin Schrödinger (1887–1961) published four papers in which the wave nature of particles was treated explicitly with wave equations. At the same time, many others began important work. Among them was German physicist Werner Heisenberg (1901–1976) who, among many other contributions to quantum mechanics, formulated a mathematical treatment of the wave nature of matter that used matrices rather than wave equations. We will deal with some specifics in later sections, but it is worth noting that de Broglie’s work was a watershed for the development of quantum mechanics. De Broglie was awarded the Nobel Prize in 1929 for his vision, as were Davisson and G. P. Thomson in 1937 for their experimental verification of de Broglie’s hypothesis.

Electron Wavelength versus Velocity and Energy

For an electron having a de Broglie wavelength of 0.167 nm (appropriate for interacting with crystal lattice structures that are about this size): (a) Calculate the electron’s velocity. (b) Calculate the electron’s kinetic energy in eV.

Strategy

For part (a), since the de Broglie wavelength is given, the electron’s velocity can be obtained from λ=h/p

by using the nonrelativistic formula for momentum, p=mv. For part (b), once v is obtained (and it has been verified that v is nonrelativistic), the classical kinetic energy is simply (1/2)mv2.

Solution for (a)

Substituting the formula for momentum (p=mv) into the de Broglie wavelength gives

p=h/λ
mv=h/λ

Solving for v gives

v=h/mλ

Substituting known values yields
v = \frac{6.63 \times 10^{-34} \, \mathrm{J \cdot s}}{(9.11 \times 10^{-31} \, \mathrm{kg})(0.167 \times 10^{-9} \, \mathrm{m})} = 4.36 \times 10^{6} \, \mathrm{m/s}.

Solution for (b)
While fast compared with a car, this electron’s speed is not close to the speed of light, and so we can comfortably use the classical formula to find the electron’s kinetic energy and convert it to eV as requested.

K = \frac{1}{2} mv^2
K = \frac{1}{2}(9.11 \times 10^{-31} \, \mathrm{kg})(4.36 \times 10^6 \, \mathrm{m/s})^2
K = 86.4×10^{-18} \, \mathrm{J} = 54.0 \, \mathrm{eV}

Electron Microscopes

One consequence or use of the wave nature of matter is found in the electron microscope. As we have discussed, there is a limit to the detail observed with any probe having a wavelength. Resolution, or observable detail, is limited to about one wavelength. Since a potential of only 54 V can produce electrons with sub-nanometer wavelengths, it is easy to get electrons with much smaller wavelengths than those of visible light (hundreds of nanometers). Electron microscopes can, thus, be constructed to detect much smaller details than optical microscopes. (See Figure 2.)

Figure a shows a schematic of an electron microscope. Figure b shows an image of a shark tooth.
Figure 2: Schematic of a scanning electron microscope (SEM) (a) used to observe small details, such as those seen in this image of a tooth of a Himipristis, a type of shark (b). (credit: Dallas Krentzel, Flickr)

There are basically two types of electron microscopes. The transmission electron microscope (TEM) accelerates electrons that are emitted from a hot filament (the cathode). The beam is broadened and then passes through the sample. A magnetic lens focuses the beam image onto a fluorescent screen, a photographic plate, or (most probably) a CCD (light sensitive camera), from which it is transferred to a computer. The TEM is similar to the optical microscope, but it requires a thin sample examined in a vacuum. However it can resolve details as small as 0.1 nm (10−10m10−10m size 12{"10" rSup { size 8{ - "10"} } `m} {}">1010m), providing magnifications of 100 million times the size of the original object. The TEM has allowed us to see individual atoms and structure of cell nuclei.

The scanning electron microscope (SEM) provides images by using secondary electrons produced by the primary beam interacting with the surface of the sample (see Figure 2). The SEM also uses magnetic lenses to focus the beam onto the sample. However, it moves the beam around electrically to “scan” the sample in the x and y directions. A CCD detector is used to process the data for each electron position, producing images like the one at the beginning of this chapter. The SEM has the advantage of not requiring a thin sample and of providing a 3-D view. However, its resolution is about ten times less than a TEM.

Electrons were the first particles with mass to be directly confirmed to have the wavelength proposed by de Broglie. Subsequently, protons, helium nuclei, neutrons, and many others have been observed to exhibit interference when they interact with objects having sizes similar to their de Broglie wavelength. The de Broglie wavelength for massless particles was well established in the 1920s for photons, and it has since been observed that all massless particles have a de Broglie wavelength pλ=h/p.λ=h/p. size 12{λ = h/p} {}">=h/λ.

The wave nature of all particles is a universal characteristic of nature. We shall see in following sections that implications of the de Broglie wavelength include the quantization of energy in atoms and molecules, and an alteration of our basic view of nature on the microscopic scale. The next section, for example, shows that there are limits to the precision with which we may make predictions, regardless of how hard we try. There are even limits to the precision with which we may measure an object’s location or energy.

Making Connections: A Submicroscopic Diffraction Grating

The wave nature of matter allows it to exhibit all the characteristics of other, more familiar, waves. Diffraction gratings, for example, produce diffraction patterns for light that depend on grating spacing and the wavelength of the light. This effect, as with most wave phenomena, is most pronounced when the wave interacts with objects having a size similar to its wavelength. For gratings, this is the spacing between multiple slits.) When electrons interact with a system having a spacing similar to the electron wavelength, they show the same types of interference patterns as light does for diffraction gratings, as shown at top left in Figure 3.

Atoms are spaced at regular intervals in a crystal as parallel planes, as shown in the bottom part of Figure 3. The spacings between these planes act like the openings in a diffraction grating. At certain incident angles, the paths of electrons scattering from successive planes differ by one wavelength and, thus, interfere constructively. At other angles, the path length differences are not an integral wavelength, and there is partial to total destructive interference. This type of scattering from a large crystal with well-defined lattice planes can produce dramatic interference patterns. It is called Bragg reflection, for the father-and-son team who first explored and analyzed it in some detail. The expanded view also shows the path-length differences and indicates how these depend on incident angle θθ size 12{θ} {}">θ in a manner similar to the diffraction patterns for x rays reflecting from a crystal.

An electron beam is striking at an angle theta on a crystal and the reflected rays are detected by a detector. A magnified view of the crystal is also shown with two rays of electrons striking the various layers of crystal at an angle theta and reflected back. A graph is shown of intensity variation versus theta. Intensity is along the y axis and theta is along the x axis.The shape of the curve is like a wave and each subsequent peak diminishes as we move out the x axis.
Figure 3: The diffraction pattern at top left is produced by scattering electrons from a crystal and is graphed as a function of incident angle relative to the regular array of atoms in a crystal, as shown at bottom. Electrons scattering from the second layer of atoms travel farther than those scattered from the top layer. If the path length difference (PLD) is an integral wavelength, there is constructive interference.

 

Section Summary

  • Particles of matter also have a wavelength, called the de Broglie wavelength, given by pλ=hpλ=hp size 12{λ = { {h} over {p} } } {}">=h/λ, where pp size 12{p} {}">p is momentum.

 

 

All Homework Problems

10

  1. Given the amount of charge on a speck of dust and the number of protons, how many electrons are there?
  2. Common static electricity ranges from nanocoulombs to microcoulombs. How many electrons is this?
  3. Compare the momenta of elephants, humans, and tranquilizer darts!
  4. Assuming negligible air resistance, what is the final speed of a rock thrown from a bridge?
  5. How many DNA molecules can a single electron from an old-fashioned TV break?
  6. How long can you play tennis off a candy bar?
  7. How long for a car of fixed power to get up to speed?
  8. Which element has an outer electron with the lowest potential energy?
  9. Exploring the relationship between momentum and kinetic energy.
  10. What is the frequency of a stroboscope?
  11. Label the parts of a wave.
  12. If the frequency of a wave is changed, which of the other properties must also change assuming the speed of the wave remains fixed?
  13. Speed, wavelength, and frequency for sound.
  14. How long to collect a certain amount of sunlight?
  15. What is the power output of an ultrasound machine for a given intensity and area?
  16. Which situations will create electromagnetic radiation?
  17. Speed dependencies for electromagnetic waves.
  18. What is the frequency of a radio station given the wavelength?
  19. Rank the types of waves in the EM spectrum by wavelength.
  20. Find the momentum of a microwave photon.
  21. From momentum, calculate the wavelength and energy of a photon.
  22. An AM radio transmitter radiates some power at a given frequency. How many photons per second does the emitter emit?
  23. If the brightness of a beam of light is increased, the ________ of the _____________ will also increase.
  24. Determine the wavelength of the third Balmer line (transition from n = 5 to n = 2 ).
  25. Find the wavelength of a golf ball.
  26. Given an electron’s wavelength, what is its speed?

Unit II

II

Unit II On-a-Page

Terminology

Instructor’s Note

 

This Unit is very heavy on vocabulary, there is a set on flashcards on Quizlet to help you.

Principles for Unit II

This unit in particular has a lot of terminology. To help you stay focused, the principles (where we will begin analyzing situations) are:
How light interacts with surfaces and materials
  • Law of reflection which is best understood in the particle picture.
  • Law of refraction which is best understood in the wave picture.
  • Light slows when it enters a medium, but the energy of a photon cannot change. So the wavelength must!
Optical elements and ray diagrams
  • The position of the image formed by an optical system is located at \frac{1}{i} + \frac{1}{o} = \frac{1}{f} where:
    • o is the image distance, using the sign conventions used above.
    • i is the image distance, using the sign conventions used above.
    • f is the focal length with its correct sign.
  • For ray diagrams: one ray in parallel and out through focal point, one ray in through focal point and out parallel, one ray using the center of the system.
  • And point where photons seem to emerge can be used as an object. Such a point could be a real source of photons (an object) or an image from another optical element.

Motivating Context for Unit II

11

Homework Assignment

The full homework assignment for this unit can be found at this link. The problems here are numbered the same as in that assignment.

 

 

A Venn-diagram showing the different sciences

This unit will focus on how light (and in some cases electrons) travel through both empty space and matter as well as how those interactions can be used to manipulate the paths and make images. The most familiar optical system to most of you is the one you are probably using to read these very words: your eyes! We will therefore be looking at the eye quite a bit throughout this unit, both human eyes and simpler eyes in the animal kingdom. Another common biological application of optics is the microscope. To make sure that everyone is on the same page, you will find below some information about the anatomy of the human eye as well as some basic information about microscopes. Please be familiar with this terminology as we will use it in class.

The Human Eye. Derived from 36.5 Vision by OpenStax Biology

Vision is the ability to detect light patterns from the outside environment and interpret them into images. Animals are bombarded with sensory information, and the sheer volume of visual information can be problematic. Fortunately, the visual systems of species have evolved to attend to the most-important stimuli. The importance of vision to humans is further substantiated by the fact that about one-third of the human cerebral cortex is dedicated to analyzing and perceiving visual information.

Anatomy of the Eye

The photoreceptive cells of the eye, where the conversion of light to nervous impulses occurs, are located in the retina (shown in Figure 2) on the inner surface of the back of the eye. But light does not impinge on the retina unaltered. It passes through other layers that process it so that it can be interpreted by the retina (Figure 2b). The cornea, the front transparent layer of the eye, and the crystalline lens, a transparent convex structure behind the cornea, both refract (bend) light to focus the image on the retina. The iris, which is conspicuous as the colored part of the eye, is a circular muscular ring lying between the lens and cornea that regulates the amount of light entering the eye. In conditions of high ambient light, the iris contracts, reducing the size of the pupil at its center. In conditions of low light, the iris relaxes and the pupil enlarges.
Cross section of the human eye.
Figure 2. (a) The human eye is shown in cross section. (b) A blowup shows the layers of the retina.

Changes in material are crucial to image formation using lenses. Each material has an index of refraction which will be discussed in The Ray Aspect of Light: The Speed of Light in Materials. The biggest change in material, and consequently the biggest bending of rays, actually occurs at the cornea rather than the lens. The cornea provides about two-thirds of the power of the eye, owing to the fact that speed of light changes considerably while traveling from air into cornea. The lens provides the remaining power needed to produce an image on the retina. The cornea and lens can be treated as a single thin lens, even though the light rays pass through several layers of material (such as cornea, aqueous humor, several layers in the lens, and vitreous humor), changing direction at each interface. The image formed is much like the one produced by a single convex lens.

The goal of the cornea and lens is to focus light on the retina and, in particular, the fovea centralis. The fovea centralis is a small, central pit composed of closelpacked cones in the eye. The fovea is responsible for sharp central vision, which is necessary in humans for activities for which visual detail is of primary importance, such as reading and driving. The lens is dynamic, focusing and re-focusing light as the eye rests on near and far objects in the visual field. The lens is operated by muscles that stretch it flat or allow it to thicken, changing the focal length of light coming through it to focus it sharply on the retina. With age comes the loss of the flexibility of the lens, and a form of farsightedness called presbyopia results. Presbyopia occurs because the image focuses behind the retina. Presbyopia is a deficit similar to a different type of farsightedness called hyperopia caused by an eyeball that is too short. For both defects, images in the distance are clear but images nearby are blurry. Myopia (nearsightedness) occurs when an eyeball is elongated and the image focus falls in front of the retina. In this case, images in the distance are blurry but images nearby are clear.

There are two types of photoreceptors in the retina: rods and cones, named for their general appearance as illustrated in . Rods are strongly photosensitive and are located in the outer edges of the retina. They detect dim light and are used primarily for peripheral and nighttime vision. Cones are weakly photosensitive and are located near the center of the retina. They respond to bright light, and their primary role is in daytime, color vision.

Rods and Cones of the eye.
Figure 3. Rods and cones are photoreceptors in the retina. Rods respond in low light and can detect only shades of gray. Cones respond in intense light and are responsible for color vision. (credit: modification of work by Piotr Sliwa)

The fovea is the region in the center back of the eye that is responsible for acute vision. The fovea has a high density of cones. When you bring your gaze to an object to examine it intently in bright light, the eyes orient so that the object’s image falls on the fovea. However, when looking at a star in the night sky or other object in dim light, the object can be better viewed by the peripheral vision because it is the rods at the edges of the retina, rather than the cones at the center, that operate better in low light. In humans, cones far outnumber rods in the fovea.

 

Transduction of Light

The rods and cones are the site of transduction of light to a neural signal. Both rods and cones contain photopigments. In vertebrates, the main photopigment, rhodopsin, has two main parts (Figure 4) an opsin, which is a membrane protein (in the form of a cluster of α-helices that span the membrane), and retinal—a molecule that absorbs light. When light hits a photoreceptor, it causes a shape change in the retinal, altering its structure from a bent (cis) form of the molecule to its linear (trans) isomer. This isomerization of retinal activates the rhodopsin, starting a cascade of events that ends with the closing of Na+ channels in the membrane of the photoreceptor. Thus, unlike most other sensory neurons (which become depolarized by exposure to a stimulus) visual receptors become hyperpolarized and thus driven away from threshold (Figure 5).

Changes of proteins in the eye.
Figure 4. (a) Rhodopsin, the photoreceptor in vertebrates, has two parts: the trans-membrane protein opsin, and retinal. When light strikes retinal, it changes shape from (b) a cis to a trans form. The signal is passed to a G-protein called transducin, triggering a series of downstream events.

 

Protein in the eye
Figure 5. When light strikes rhodopsin, the G-protein transducin is activated, which in turn activates phosphodiesterase. Phosphodiesterase converts cGMP to GMP, thereby closing sodium channels. As a result, the membrane becomes hyperpolarized. The hyperpolarized membrane does not release glutamate to the bipolar cell.

Trichromatic Coding

There are three types of cones (with different photopsins), and they differ in the wavelength to which they are most responsive, as shown in Figure 6 . Some cones are maximally responsive to short light waves of 420 nm, so they are called S cones (“S” for “short”); others respond maximally to waves of 530 nm (M cones, for “medium”); a third group responds maximally to light of longer wavelengths, at 560 nm (L, or “long” cones). With only one type of cone, color vision would not be possible, and a two-cone (dichromatic) system has limitations. Primates use a three-cone (trichromatic) system, resulting in full color vision. The color we perceive is a result of the ratio of activity of our three types of cones. The colors of the visual spectrum, running from long-wavelength light to short, are red (700 nm), orange (600 nm), yellow (565 nm), green (497 nm), blue (470 nm), indigo (450 nm), and violet (425 nm). Humans have very sensitive perception of color and can distinguish about 500 levels of brightness, 200 different hues, and 20 steps of saturation, or about 2 million distinct colors.

Absorbance of eye cones.
Figure 6. Human rod cells and the different types of cone cells each have an optimal wavelength. However, there is considerable overlap in the wavelengths of light detected.

Retinal Processing

Visual signals leave the cones and rods, travel to the bipolar cells, and then to ganglion cells. A large degree of processing of visual information occurs in the retina itself, before visual information is sent to the brain.

Photoreceptors in the retina continuously undergo tonic activity. That is, they are always slightly active even when not stimulated by light. In neurons that exhibit tonic activity, the absence of stimuli maintains a firing rate at a baseline; while some stimuli increase firing rate from the baseline, and other stimuli decrease firing rate. In the absence of light, the bipolar neurons that connect rods and cones to ganglion cells are continuously and actively inhibited by the rods and cones. Exposure of the retina to light hyperpolarizes the rods and cones and removes their inhibition of bipolar cells. The now active bipolar cells in turn stimulate the ganglion cells, which send action potentials along their axons (which leave the eye as the optic nerve). Thus, the visual system relies on change in retinal activity, rather than the absence or presence of activity, to encode visual signals for the brain. Sometimes horizontal cells carry signals from one rod or cone to other photoreceptors and to several bipolar cells. When a rod or cone stimulates a horizontal cell, the horizontal cell inhibits more distant photoreceptors and bipolar cells, creating lateral inhibition. This inhibition sharpens edges and enhances contrast in the images by making regions receiving light appear lighter and dark surroundings appear darker. Amacrine cells can distribute information from one bipolar cell to many ganglion cells.

You can demonstrate this using an easy demonstration to “trick” your retina and brain about the colors you are observing in your visual field. Look fixedly at Figure 7 for about 45 seconds. Then quickly shift your gaze to a sheet of blank white paper or a white wall. You should see an afterimage of the Norwegian flag in its correct colors. At this point, close your eyes for a moment, then reopen them, looking again at the white paper or wall; the afterimage of the flag should continue to appear as red, white, and blue. What causes this? According to an explanation called opponent process theory, as you gazed fixedly at the green, black, and yellow flag, your retinal ganglion cells that respond positively to green, black, and yellow increased their firing dramatically. When you shifted your gaze to the neutral white ground, these ganglion cells abruptly decreased their activity and the brain interpreted this abrupt downshift as if the ganglion cells were responding now to their “opponent” colors: red, white, and blue, respectively, in the visual field. Once the ganglion cells return to their baseline activity state, the false perception of color will disappear.

 

Optical illusion.
Figure 7. View this flag to understand how retinal processing works. Stare at the center of the flag (indicated by the white dot) for 45 seconds, and then quickly look at a white background, noticing how colors appear.

Introduction to Geometric Optics

12

The Ray Aspect of Light

Instructor’s Note

 

In Unit I we talked about how light and electrons have both a wave and particle nature. In this particular unit, we will be using that fact extensively: sometimes thinking about waves and sometimes thinking about particles. the ‘ray’ aspect of light being described here is most easily pictured as the path of the individual particles, either electrons or photons, traveling in a straight line.

There are three ways in which light can travel from a source to another location. (See Figure 1.) It can come directly from the source through empty space, such as from the Sun to Earth. Or light can travel through various media, such as air and glass, to the person. Light can also arrive after being reflected, such as by a mirror. In all of these cases, light is modeled as traveling in straight lines called rays. Light may change direction when it encounters objects (such as a mirror) or in passing from one material to another (such as in passing from air to glass), but it then continues in a straight line or as a ray. The word ray comes from mathematics and here means a straight line that originates at some point. It is acceptable to visualize light rays as laser rays (or even science fiction depictions of ray guns).

Light rays through a window
Figure 1. Three methods for light to travel from a source to another location. (a) Light reaches the upper atmosphere of Earth traveling through empty space directly from the source. (b) Light can reach a person in one of two ways. It can travel through media like air and glass. It can also reflect from an object like a mirror. In the situations shown here, light interacts with objects large enough that it travels in straight lines, like a ray.

Experiments, as well as our own experiences, show that when light interacts with objects several times as large as its wavelength, it travels in straight lines and acts like a ray. Its wave characteristics are not pronounced in such situations. Since the wavelength of light is less than a micron (a micrometer μm or a thousandth of a millimeter), it acts like a ray in the many common situations in which it encounters objects larger than a micron. For example, when light encounters anything we can observe with unaided eyes, such as a mirror, it acts like a ray, with only subtle wave characteristics. We will concentrate on the ray characteristics in this chapter. Since light moves in straight lines, changing directions when it interacts with materials, it is described by geometry and simple trigonometry. This part of optics, where the ray aspect of light dominates, is therefore called geometric optics. There are two laws that govern how light changes direction when it interacts with matter. These are the law of reflection, for situations in which light bounces off matter, and the law of refraction, for situations in which light passes through matter.

Section Summary

  • A straight line that originates at some point is called a ray.
  • The part of optics dealing with the ray aspect of light is called geometric optics.
  • Light can travel in three ways from a source to another location: (1) directly from the source through empty space; (2) through various media; (3) after being reflected from a mirror.

The Law of Reflection

Whenever we look into a mirror, or squint at sunlight glinting from a lake, we are seeing a reflection. When you look at the page of a printed book, you are also seeing light reflected from it. Large telescopes use reflection to form an image of stars and other astronomical objects.

The law of reflection is illustrated in Figure 1, which also shows how the angles are measured relative to the perpendicular to the surface at the point where the light ray strikes. We expect to see reflections from smooth surfaces, but Figure 1 illustrates how a rough surface reflects light. Since the light strikes different parts of the surface at different angles, it is reflected in many different directions, or diffused. Diffused light is what allows us to see a sheet of paper from any angle, as illustrated in Figure 1. Many objects, such as people, clothing, leaves, and walls, have rough surfaces and can be seen from all sides. A mirror, on the other hand, has a smooth surface (compared with the wavelength of light) and reflects light at specific angles, as illustrated in Figure 1. When the moon reflects from a lake, as shown in Figure 1, a combination of these effects takes place.

Reflected Rays
Figure 1. The law of reflection states that the angle of reflection equals the angle of incidence - \theta_r = \theta_i. The angles are measured relative to the perpendicular to the surface at the point where the ray strikes the surface.
Light reflected from a rough surface.
Figure 2. Light is diffused when it reflects from a rough surface. Here many parallel rays are incident, but they are reflected at many different angles since the surface is rough.
Light rays reflecting from paper.
Figure 3. When a sheet of paper is illuminated with many parallel incident rays, it can be seen at many different angles, because its surface is rough and diffuses the light.
Light rays reflecting form a mirror.
Figure 4. A mirror illuminated by many parallel rays reflects them in only one direction, since its surface is very smooth. Only the observer at a particular angle will see the reflected light.
Moonlight reflected by a lake.
Figure 5. Moonlight is spread out when it is reflected by the lake, since the surface is shiny but uneven. (credit: Diego Torres Silvestre, Flickr)

The law of reflection is very simple: The angle of reflection equals the angle of incidence.

THE LAW OF REFLECTION

The angle of reflection equals the angle of incidence.

Instructor’s Note

 

This is important! You should ALWAYS measure your angles with respect to a line perpendicular to the surface. This line perpendicular to the surface is called a ‘normal’ line.

When we see ourselves in a mirror, it appears that our image is actually behind the mirror. This is illustrated in Figure 6. We see the light coming from a direction determined by the law of reflection. The angles are such that our image is exactly the same distance behind the mirror as we stand away from the mirror. If the mirror is on the wall of a room, the images in it are all behind the mirror, which can make the room seem bigger. Although these mirror images make objects appear to be where they cannot be (like behind a solid wall), the images are not figments of our imagination. Mirror images can be photographed and videotaped by instruments and look just as they do with our eyes (optical instruments themselves). The precise manner in which images are formed by mirrors and lenses will be treated in Applications of Geometric Optics: Terminology of Images.

A woman looking at her reflection in the mirror.
Figure 6. Our image in a mirror is behind the mirror. The two rays shown are those that strike the mirror at just the correct angles to be reflected into the eyes of the person. The image appears to be in the direction the rays are coming from when they enter the eyes.

TAKE-HOME EXPERIMENT: LAW OF REFLECTION

Take a piece of paper and shine a flashlight at an angle at the paper, as shown in Figure 3. Now shine the flashlight at a mirror at an angle. Do your observations confirm the predictions in Figure 3 and Figure 4? Shine the flashlight on various surfaces and determine whether the reflected light is diffuse or not. You can choose a shiny metallic lid of a pot or your skin. Using the mirror and flashlight, can you confirm the law of reflection? You will need to draw lines on a piece of paper showing the incident and reflected rays. (This part works even better if you use a laser pencil.)

Section Summary

Law of Reflection in Terms of the Particle Picture of Light

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We’re going to begin with the basics of reflection. The Law of Reflection is simply, \theta_i = \theta_f , the incident angle equals the final angle. Reflection makes the most sense in terms of the particle picture of light. It’s easiest to see if we just imagine light, a photon, as a ball. When a ball is bounced, the incident angle and the final angle are the same before and after it bounces. You can see that relative to the normal, which is how you should always measure your angles, the incident angle and the final angle are the same.

Angle of Incidence and Angle of Reflection
Figure 1. The Law of Reflection (Credit: Wikipedia, C M Vidyashree)

Speed of Light in Materials

It is easy to notice some odd things when looking into a fish tank. For example, you may see the same fish appearing to be in two different places. (See Figure 1.) This is because light coming from the fish to us changes direction when it leaves the tank, and in this case, it can travel two different paths to get to our eyes. The changing of a light ray’s direction (loosely called bending) when it passes through variations in matter is called refraction. Refraction is responsible for a tremendous range of optical phenomena, from the action of lenses to voice transmission through optical fibers.

REFRACTION

The changing of a light ray’s direction (loosely called bending) when it passes through variations in matter is called refraction.

Fish tank
Figure 1. Looking at the fish tank as shown, we can see the same fish in two different locations, because light changes directions when it passes from water to air. In this case, the light can reach the observer by two different paths, and so the fish seems to be in two different places. This bending of light is called refraction and is responsible for many optical phenomena.

Why does light change direction when passing from one material (medium) to another? It is because light changes speed when going from one material to another. So before we study the law of refraction, it is useful to discuss the speed of light and how it varies in different media.

The Speed of Light

Early attempts to measure the speed of light, such as those made by Galileo, determined that light moved extremely fast, perhaps instantaneously. The first real evidence that light traveled at a finite speed came from the Danish astronomer Ole Roemer in the late 17th century. Roemer had noted that the average orbital period of one of Jupiter’s moons, as measured from Earth, varied depending on whether Earth was moving toward or away from Jupiter. He correctly concluded that the apparent change in period was due to the change in distance between Earth and Jupiter and the time it took light to travel this distance. From his 1676 data, a value of the speed of light was calculated to be 2.26 \times 10^{8} m/s (only 25% different than today’s accepted value). In more recent times, physicists have measured the speed of light in numerous ways and with increasing accuracy. One particularly direct method, used in 1887 by the American physicist Albert Michelson (1852–1931), is illustrated in Figure 2. Light reflected from a rotating set of mirrors was reflected from a stationary mirror 35 km away and returned to the rotating mirrors. The time for the light to travel can be determined by how fast the mirrors must rotate for the light to be returned to the observer’s eye.

Optics experiment
Figure 2. A schematic of early apparatus used by Michelson and others to determine the speed of light. As the mirrors rotate, the reflected ray is only briefly directed at the stationary mirror. The returning ray will be reflected into the observer’s eye only if the next mirror has rotated into the correct position just as the ray returns. By measuring the correct rotation rate, the time for the round trip can be measured and the speed of light calculated. Michelson’s calculated value of the speed of light was only 0.04% different from the value used today.

The speed of light is now known to great precision. In fact, the speed of light in a vacuum, c, is so important that it is accepted as one of the basic physical quantities and has the fixed value.

2.99792458 \times 10^{8} m/s \approx 3.0 \times 10^{8} m/s,

where the approximate value of 3.0 \times 10^{8} m/s is used whenever three-digit accuracy is sufficient. The speed of light through matter is less than it is in a vacuum, because light interacts with atoms in a material. The speed of light depends strongly on the type of material, since its interaction with different atoms, crystal lattices, and other substructures varies. We define theindex of refraction n of a material to be

n = \frac {c}{v},

wherev is the observed speed of light in the material. Since the speed of light is always less thanc in matter and equals c only in a vacuum, the index of refraction is always greater than or equal to one.

 

VALUE OF THE SPEED OF LIGHT

2.99792458 \times 10^{8} m/s \approx 3.0 \times 10^{8} m/s

INDEX OF REFRACTION

n = \frac {c}{v}

That is, n \ge 1 Table 1 gives the indices of refraction for some representative substances. The values are listed for a particular wavelength of light, because they vary slightly with wavelength. (This can have important effects, such as colors produced by a prism.) Note that for gases, n is close to 1.0. This seems reasonable, since atoms in gases are widely separated and light travels atc in the vacuum between atoms. It is common to taken = 1for gases unless great precision is needed. Although the speed of light v in a medium varies considerably from its valuecin a vacuum, it is still a large speed.

 

 

Index of Refraction in Various Media
Index of refraction in Various Media
Medium n
Gases at 0^{\circ} C , 1 atm
Air 1.000293
Carbon Dioxide 1.00045
Hydrogen 1.000139
Oxygen 1.000271
Liquids at 20^{\circ} C
Benzene 1.501
Carbon disulfide 1.628
Carbon tetrachloride 1.461
Ethanol 1.361
Glycerine 1.473
Water, fresh 1.333
Solids at 20^{\circ} C
Diamond 2.419
Fluorite 1.434
Glass, crown 1.52
Glass, flint 1.66
Ice at 20 ^{\circ}C 1.309
Polystyrene 1.49
Plexiglass 1.51
Quartz, crystalline 1.544
Quartz, fused 1.458
Sodium chloride 1.544
Zircon 1.923

 

Speed of Light in Matter

Calculate the speed of light in zircon, a material used in jewelry to imitate diamond.

Strategy

The speed of light in a material, v, can be calculated from the index of refraction n of the material using the equation n = \frac {c}{v}

Solution

The equation for index of refraction states that n = \frac {c}{v}, Rearranging this to determine v gives

v = \frac {c}{n}

The index of refraction for zircon is given as 1.923 in Table 1, and c is given in the equation for speed of light. Entering these values in the last expression gives

v = \frac { 3.0 \times 10^{8} m/s}{1.923}
= 1.56 \times 10^{8} m/s

Discussion

This speed is slightly larger than half the speed of light in a vacuum and is still high compared with speeds we normally experience. The only substance listed in Table 1 that has a greater index of refraction than zircon is diamond. We shall see later that the large index of refraction for zircon makes it sparkle more than glass, but less than diamond.

Section Summary

 

Why Light Bends

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So let’s talk about why light bends. So this is what makes starting with wave particle duality so useful: to understand why light bends, a wave picture is very useful, to understand which properties of light change it is useful to think about the particle nature.
Let’s watch, now we’re thinking about light as a wave. So my wave is coming in from the upper left and you can see that this part of the wave hits the interface first and
slows down, while this part of the wave, keeps going. That’s what makes it bend, different parts of the wave hit it at different times. The right edge of this wave hits the interface first and slows down, while the left edge keeps traveling at the faster speed and as a consequence the light as a whole bends. You can see in this video they even got the fact that the wavelength shrinks, they even got that right.
A light ray bends because one side hits the interface first and slows while the rest continues at the faster speed. The wavelength in the slower material is smaller.
A light ray bends because one side hits the interface first and slows while the rest continues at the faster speed. The wavelength in the slower material is smaller.

A wave is a wave is a wave

We have seen in our first unit that electrons and photons are similar in many ways; “a wave is a wave is a wave.” With that in mind, consider the following situation. An electron is traveling in some region when it enters another region where it travels more slowly (perhaps because of more potential energy and thus a decrease in kinetic energy). Which path of the electron is qualitatively correct?

An electron moves from one region where it is fast to another where it is slower. How does it travel in the new material: A bend towards normal, B go straight, or C bend away from the normal.

Solution:

A is correct.

 

Just like a light wave, if I can get the electron to slow down I can get it
to bend towards the normal just like a light wave.
I can make it bend, I can build a lens.
If I can build a lens, I can build a microscope.
This is why electron microscopes work, because I can bend electrons by speeding them up or slowing them down.

Wave hitting interface perpendicular

If the wave comes in straight perpendicular, does it bend?

Solution:

Remember that the bending came from the fact that different parts of the wave slowed down at different times, if we come straight at it, that doesn’t happen and so they all slow down together, and you don’t get a bend: the wave goes straight.

Digging More into Wave-Particle Duality and RefractionA note to more advanced readers - the following derivation of why the wavelength changes and not the frequency is not 100% correct, there are more complex effects at play due to Einstein's Theories of Relativity. However, the essence of the argument depending on energy conservation is correct and so is the result.

Now, let’s think about some of the other properties of the light wave, beyond speed, and how they might change as we go from one material to another. Starting with wavelength.

Wavelength

We know from Unit I that light is made of photons and that these photons have energy

E_\gamma = \frac{hc}{\lambda}.

The c in this equation, however, is trying to tell us something. The value c = 3 \times 10^8 \, \mathrm{m}{s} is the speed of light in vacuum. We know now that, in a material, light will slow to some v < c, and our resulting expression will now be

E_\gamma^{\mathrm{material}} = \frac{hv}{\lambda}

where v is the speed of the light in the material. Because of conservation of energy, the energy of the photon cannot change. Thus, according to our equation, if the speed goes down, the wavelength must also decrease by the same factor.

Example: Reduction of wavelength in materials

Say we have a light source in a vacuum that emits light with a wavelength of \lambda_0 = 6 \, \mathrm{nm} = 6 \times 10^{-9} \, \mathrm{m}. The light then enters a material where the speed of light is only 2 \times 10^{8} \, \mathrm{m}/\mathrm{s}. What is the wavelength \lambda_n in this new material?

Solution:

Given that

E_\gamma^{\mathrm{vacuum}} = \frac{hc}{\lambda_0}

and

E_\gamma^{\mathrm{material}} = \frac{hv}{\lambda_n}

and given he energy of the photon cannot change due to conservation of energy:

E_\gamma^{\mathrm{vacuum}} = E_\gamma^{\mathrm{material}}

we can set the two expressions equal to another:

\frac{hc}{\lambda_0} = \frac{hv}{\lambda_n}

\frac{c}{\lambda_0} = \frac{v}{\lambda_n}

\frac{c}{v} = \frac{\lambda_0}{\lambda_n}.

We recognize the quantity c/v as the index of refraction n = c/v, which in this case is

n = \frac{c}{v} = \frac{3 \times 10^8 \, \mathrm{m/s}}{2 \times 10^8 \, \mathrm{m/s}} = 1.5

Thus, we have

n = \frac{\lambda_0}{\lambda_n}

\lambda_n = \frac{\lambda_0}{n}

Substituting in our values, we have:

\lambda_n = \frac{6 \, \mathrm{nm}}{1.5} = 4 \, \mathrm{nm}.

Discussion:

The speed of light went down by a factor of 2/3 and so did the wavelength!

Frequency

What happens to the frequency of a light wave in matter? Well the fundamental relationship for all waves v = \lambda \nu must still be obeyed. As a light wave goes from a vacuum into a material, the speed changes c \rightarrow v = c/n, and so does the wavelength \lambda_0 \rightarrow \lambda_n = \lambda_0/n. What happens to the frequency?

v = \lambda_n \nu

\Downarrow

\frac{c}{n} = \frac{\lambda_0}{n} \nu

cancel the n and we are left with the true statement c = \lambda_0 \nu. What to make of this? It means that the frequency of the light wave does not change!

Amplitude / Number of Photons

As you know from Unit I, the amplitude of the light wave and the number of photons are both related to the light’s intensity. Thus, these quantities are more about how much light is absorbed by the material than its index of refraction. Glass and air absorb very little light in the visible range, meaning that the amplitude and number of photons is not very much reduced in these materials. Water on the other hand, is very effective at absorbing visible light photons. As shown in the Figure, at a depth of 200m, almost no visible light penetrates resulting in creatures with special adaptations to live in complete darkness.

 

The amount of light decreases with ocean depth.
Light penetration as a function of color and depth: NOAA – National Oceanic and Atmospheric Administration [Public domain]

The Law of Refraction

Figure 1 shows how a ray of light changes direction when it passes from one medium to another. As before, the angles are measured relative to a perpendicular to the surface at the point where the light ray crosses it. (Some of the incident light will be reflected from the surface, but for now we will concentrate on the light that is transmitted.) The change in direction of the light ray depends on how the speed of light changes. The change in the speed of light is related to the indices of refraction of the media involved. In the situations shown in Figure 1, medium 2 has a greater index of refraction than medium 1. This means that the speed of light is less in medium 2 than in medium 1. Note that as shown in Figure 1 (a), the direction of the ray moves closer to the perpendicular when it slows down. Conversely, as shown in Figure 1 (b), the direction of the ray moves away from the perpendicular when it speeds up. The path is exactly reversible. In both cases, you can imagine what happens by thinking about pushing a lawn mower from a footpath onto grass, and vice versa. Going from the footpath to grass, the front wheels are slowed and pulled to the side as shown. This is the same change in direction as for light when it goes from a fast medium to a slow one. When going from the grass to the footpath, the front wheels can move faster and the mower changes direction as shown. This, too, is the same change in direction as for light going from slow to fast.

The law of refraction
Figure 1. The change in direction of a light ray depends on how the speed of light changes when it crosses from one medium to another. The speed of light is greater in medium 1 than in medium 2 in the situations shown here. (a) A ray of light moves closer to the perpendicular when it slows down. This is analogous to what happens when a lawn mower goes from a footpath to grass. (b) A ray of light moves away from the perpendicular when it speeds up. This is analogous to what happens when a lawn mower goes from grass to footpath. The paths are exactly reversible.

The amount that a light ray changes its direction depends both on the incident angle and the amount that the speed changes. For a ray at a given incident angle, a large change in speed causes a large change in direction, and thus a large change in angle. The exact mathematical relationship is the law of reflection, or “Snell’s Law,” which is stated in equation form

n_1 \sin \theta_1 = n_2 \sin \theta_2

Here n_1 and n_2 are the indices of refraction for medium 1 and 2, and \theta_1 and \theta_2 are the angles between the rays and the perpendicular in medium 1 and 2, as shown in Figure 1. The incoming ray is called the incident ray and the outgoing ray the refracted ray, and the associated angles the incident angle and the refracted angle. The law of refraction is also called Snell’s law after the Dutch mathematician Willebrord Snell (1591–1626), who discovered it in 1621. Snell’s experiments showed that the law of refraction was obeyed and that a characteristic index of refraction n could be assigned to a given medium. Snell was not aware that the speed of light varied in different media, but through experiments he was able to determine indices of refraction from the way light rays changed direction. Below is a simulation where you can shine light through different materials and see how it bends. I encourage you to play with it to get a feel for refraction. You can even add a protractor and see that the simulation obeys Snell’s Law.

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THE LAW OF REFRACTION

n_1 \sin \theta_1 = n_2 \sin \theta_2

TAKE-HOME EXPERIMENT: A BROKEN PENCIL

A classic observation of refraction occurs when a pencil is placed in a glass half filled with water. Do this and observe the shape of the pencil when you look at the pencil sideways, that is, through air, glass, water. Explain your observations. Draw ray diagrams for the situation.

Determine the Index of Refraction form Refraction Data

Find the index of refraction for medium 2 in Figure 1 (a), assuming medium 1 is air and given the incident angle is 30^{\circ} and the angle of refraction is 22^{\circ}

Strategy 

The index of reflection for air is taken to be 1 in most cases (and up to four significant figures, it is 1.00). Thus n_1 = 1.00. Form the given information, \theta_1 = 30^{\circ} and \theta_2 = 22^{\circ}. With this information, the only unknown in Snell’s law is n_2, so it can be used to find this unknown.

Solution

Snell’s law is

n_1 \sin \theta_1 = n_2 \sin \theta_2

Rearranging to isolate n_2 gives

n_2 = n_1 \frac{\sin \theta_1} {\sin \theta_2}

Entering known values,

n_2 = 1.00 \frac{\sin 30.0^{\circ}} {\sin 22 .0^{\circ}} = \frac{0.500}{0.375}

n_2 = 1.33

Discussion

This is the index of refraction for water, and Snell could have determined it by measuring the angles and performing this calculation. He would then have found 1.33 to be the appropriate index of refraction for water in all other situations, such as when a ray passes from water to glass. Today we can verify that the index of refraction is related to the speed of light in a medium by measuring that speed directly.

A Larger Change in Direction

Find the index of refraction for medium 2 in Figure 1 (a), assuming medium 1 is air and given the incident angle is 30^{\circ} and the angle of refraction is 22^{\circ}

Strategy 

Again the index of refraction for air is taken to be n_1 = 1.00 and we are given \theta_1 = 30^{\circ}. We can look up the index of refraction for a diamond in Table 1 (Speed of Light in Materials), finding n_2 = 2.419. The only unknown in Snell’s law is \theta_2, which we wish to determine.

Solution

Solving Snell’s law for \ sin \theta_2 yields

\ sin \theta_2 = \ sin \theta_1 \frac{n_1} { n_2}

Entering known values,

\ sin \theta_2 = \ sin 30^{\circ} \frac{1.00} {2.419} = (0.413) (0.500) = 0.207

And angle is thus

\theta_2 = \sin^{-1} 0.207 = 11.9^{\circ}

Discussion

This is the index of refraction for water, and Snell could have determined it by measuring the angles and performing this calculation. He would then have found 1.33 to be the appropriate index of refraction for water in all other situations, such as when a ray passes from water to glass. Today we can verify that the index of refraction is related to the speed of light in a medium by measuring that speed directly.

For the same 30^{\circ} angle of incidence, the angle of refraction in diamond is significantly smaller than in water (11.9^{\circ} rather than 22^{\circ}—see the preceding example). This means there is a larger change in direction in diamond. The cause of a large change in direction is a large change in the index of refraction (or speed). In general, the larger the change in speed, the greater the effect on the direction of the ray.

Section Summary

Producing Images with Geometric Optics

13

Terminology of Images and Optical Elements

Instructor’s Note

 

By the end of this section you should:

  • Be able to define several terms associated with images and optics.
  • Know and apply the sign conventions associated with objects and images in optics.
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An optical element is any lens or mirror. A few examples include the lens in your eye that you’re using to read this or the aforementioned shaving or makeup mirror that many people have to help them get ready in the morning, it makes your face look a little bigger than it is. The lens or the mirror are examples of optical elements. Anything that’s a lens or a mirror is an optical element.

Optical Element
Figure 1.

Optical elements or combinations of them can be used to make images. An image is the apparent reproduction of an object formed by an optical element, or collection of them, through the reflection and/or refraction of light. In these two examples, the person is our object and the images are the image on the back of your retina formed by your eyeball, or the image of your reflection in the mirror.

Optical Element
Figure 2.

Images can either be erect, with the same orientation as the object, or inverted, upside down with respect to the object. In the makeup mirror your face is right-side up, and so the image is erect. The images on your retina are actually upside down your brain corrects them to put them right side up, and this is an example of an inverted image.

Optical Element
Figure 3.

Before we move on to talk about image and object distances, we need to introduce two more terms: the optical axis and the vertex.

The optical axis is an imaginary line that passes through the optical element in a way that’s perpendicular to it. Below, we have a converging lens at the top, with a diverging lens below that, with a converging mirror below that, and a diverging mirror at the bottom. This dashed line that always meets the lens or mirror perpendicular is what we call the optical axis.

Lens
Figure 4.

Looking at a few examples, we can see that the optical axis meets the lens perpendicular here in the middle, same for the diverging lens. Moving on to the mirrors we see the optical axis meets the two mirrors in a way that’s perpendicular. The point where the optical axis meets the optical element is called the vertex.

Optical axis and Vertex of lenses
Figure 5.

The reason we needed these terms is because the image distance i and object distance o are measured along the optical axis from the vertex and these distances have signs that can be positive or negative and the sides are relative to the path of the light if the object is on the same side as the incoming light, then the object distance will be positive, otherwise the object distance is negative. If the image is on the same side as the outgoing light, Then the image distance is positive otherwise the image distance is negative. Note that for our lens the incoming and outgoing sides are different, light passes through a lens, so the light comes in one side and goes out the other. For a mirror on the other hand, the incoming and outgoing sides are the same, light bounces off of a mirror. These may seem like a rather convoluted set of rules, but it turns out that this is actually the simplest set of rules that works for all lenses and mirrors, so any other set of rules you might try to come up with will necessarily be more complicated.

Let’s employ these sign conventions in the terms of two examples, one with a lens and one with a mirror.

To begin with a lens, let’s say you’re looking at a person about 10 meters away. Your eye produces an image on the back of your retina, which is about an inch behind the lens of your eye or 2.5 centimeters, what are the image and object distances including the signs?

Eye
Figure 6.

First, we define our optical axis, passing through the lens you’ll notice that the light is bending here, so that’s what we’re actually defining as our lens. The point where the optical axis meets the lens is the vertex, this is the point from which we are measuring our image and object distances. Now the person of course doesn’t shine off by their own light, but from light bouncing off of them, which means the light is coming from the left.

Since the person is on the same side as where the light is coming from the object distance is going to be positive. Now we also know the person is 10 meters away, so we would say the object distance is 10 meters. For the image on the back of your retina, the image is 0.25 centimeters past the vertex on the side where the light is going out, which means the image distance is also positive leading to an image distance of 2.5 centimeters.

Now let’s look at an example with a mirror. A can of shaving cream sits 30 centimeters in front of a flat plane mirror, like you have in your bathroom. You see the image of the shaving can apparently 30 centimeters behind the mirror. What are the image distances, i, and the object distance, o?

Mirror and eye
Figure 7.

Once again, we define our optical axis so that it meets the mirror perpendicular. The can does not shine by its own light but from light bouncing off of it, so the light is coming from the left, which means the object is on the same side as the incoming light, which means the object distance is positive and 30 centimeters, so o is 30 centimeters.

The image on the other hand is not on the side of the outgoing light because the light bounces off the mirror and back the way it came from. The image is on the side opposite the outgoing and so the image distance, i, is actually – 30 centimeters.

One more practice – Looking into a Spoon

Go get a metal spoon and look into each side. Pictures from Dr. Toggerson are below, but this really works better if you do it yourself. If you look in the back, as in Figure 8a, your image seems to be behind the spoon. In contrast, if you look carefully in the inside, as in Figure 8b, your image seems to hover in front of the spoon (take a piece of paper or a toothpick and try to poke it and you will see what I mean).

What are the sign conventions for o and i in these cases?

Figure 8: Dr. Toggerson’s reflections in a spoon.
Dr. Toggerson in the back of a spoon. His image is behind the spoon and definitely smaller than he is!
Figure 8a: Dr. Toggerson in the back of a spoon. His image is behind the spoon and definitely smaller than he is!
Dr. Toggerson's reflection on the inside of a spoon. Again, the image is smaller than the object. This time, the image is also inverted. Finally, if you look closely, the image appears to hover in front of the spoon!
Figure 8b: Dr. Toggerson’s reflection on the inside of a spoon. Again, the image is smaller than the object. This time, the image is also inverted. Finally, if you look closely, the image appears to hover in front of the spoon!

Solution – back of spoon:

For the back of the spoon:

  • The light is coming off your face and hitting the spoon, thus the object is on the same side as the incoming light: image 0 " title="Rendered by QuickLaTeX.com" height="12" width="38" style="vertical-align: 0px;">.
  • The image is on the opposite side as the outgoing light: the light bounces back towards you and does not go behind the spoon. Thus i < 0
Shows reflection of a face in a spoon with light coming from nose and bouncing to eye, yeilding reflection behind the spoon.
The light comes from the face, bounces off the spoon and into the eye. The object (face) is on the same side as the incoming light so has a positive object distance. The image is on the opposite side as the outgoing light so has a negative image distance.

 

Solution – Front of spoon

If you are looking in a spoon carefully, you will see that the image appears to hover in front of it. As such:

  • The object is still on the same side as the incoming light: image 0 " title="Rendered by QuickLaTeX.com" height="12" width="38" style="vertical-align: 0px;">
  • The image is now on the same side as the outgoing light: image 0 " title="Rendered by QuickLaTeX.com" height="12" width="36" style="vertical-align: 0px;">
Showing the reflection of a face slightly in front of the inside of a spoon with the light ray from the nose going to the eye.
The light comes from the face, bounces off the spoon and into the eye. The object (face) is still on the same side as the incoming light so has a positive object distance. The image, however, is now on the same side as the outgoing light so also has a positive image distance.

Discussion:

I know that these rules seem odd and obtuse and that it will be tempting to make up your own rules: DON’T. These rules work for all situations. You just need to think about the perspective of where the light is traveling.

Section Summary

 

 

Magnification of Images

 

Instructors Note

 

 

Your quiz will cover:

  • Given two of magnification, image height, and object height, find the third
  • Know that the magnification of an inverted image is negative

 

When you look in a flat bathroom mirror, your image is the same height as you are. Look at the picture of Dr. Toggerson below in Figure 1. Figure 1a is taken 20cm in front of a mirror (you can see the camera). The image seems to be exactly the same distance, 20cm, behind the mirror as Dr. Toggerson is in front of it. This is shown diagrammatically in Figure 2 where the woman is the same distance in front of the mirror as her reflection is behind it; also she and her reflection are the same height. This is confirmed with Figure 1b, which is taken with the camera at 40cm from Dr. Toggerson’s face as the mirror. The height of Dr. Toggerson’s face is the same in Figures 1a and 1b. In contrast, look at Figure 3a below which is taken in the back of a spoon, Dr. Toggerson’s face is much smaller, about 2.5cm (1 in, it fits on the spoon!). Similarly, in Figure 3b looking at the inside of a spoon, his image, while larger than in the back of the spoon, is again smaller than his face. Moreover, the image on the inside of the spoon is upside down! The flat mirror and the back of the spoon have erect images, while the image formed by the inside of the spoon is inverted. The different heights are covered by the concept of magnification. Magnification is simply the ratio of image height to object height:

m = \frac{h_i}{h_o}

where h_i is the height of the image and h_o is the height of the object. As a ratio of heights, magnification is, itself, unit-less and just represents how many times bigger or smaller the image is than the object. One last note: if the image is inverted, we say that its height is negative. Thus inverted images have negative magnification. Hopefully, this choice seems reasonable.

Figure 1: Two pictures of Dr. Toggerson. Note that he seems to be the same height in each. The image in a flat mirror is the same height as the object.
A selfie of Dr. Toggerson taken 20cm in front of a mirror
Figure 1a: A selfie of Dr. Toggerson taken 20cm in front of a mirror
A selfie of Dr. Toggerson taken from 16 inches away.
Figure 1b: A selfie of Dr. Toggerson taken from 40cm away. Twice the face-mirror distance
A woman looking at her reflection in the mirror.
Figure 2: The image in a mirror is the same height as the original object and the same distance behind the mirror as the object is in front of it.
Figure 3: Dr. Toggerson’s reflections in a spoon.
Dr. Toggerson in the back of a spoon. His image is behind the spoon and definitely smaller than he is!
Figure 3a: Dr. Toggerson in the back of a spoon. His image is behind the spoon and definitely smaller than he is!
Dr. Toggerson's reflection on the inside of a spoon. Again, the image is smaller than the object. This time, the image is also inverted. Finally, if you look closely, the image appears to hover in front of the spoon!
Figure 3b: Dr. Toggerson’s reflection on the inside of a spoon. Again, the image is smaller than the object. This time, the image is also inverted. Finally, if you look closely, the image appears to hover in front of the spoon!

Magnification in a Flat Mirror

As we have seen, in a flat mirror, your image is the same height as you. What is the magnification of a flat mirror?

Solution:

Well, we know the definition of magnification:

m = \frac{h_i}{h_o}

Let’s use Dr. Toggerson as our example. Dr. Toggerson is about 1.7m tall, which means his reflection in a flat mirror is also 1.7m tall. Using these values we see:

m = \frac{1.7 \, \mathrm{m}}{1.7 \, \mathrm{m}} = 1

Discussion:

Note, the answer is just 1, no units!

Magnification in a Spoon

Dr. Toggerson is 1.7m tall. His image in the back of a spoon is about 0.5cm. The image in the front of the spoon is about 1cm. What is the magnification of each?

Solution for the back of the spoon:

Again, we know the definition of magnification:

m = \frac{h_i}{h_o}

All we need to do is substitute the known values. However, for the result to be unit-less we need to express both the image height and the object height in the same units; doesn’t matter what units we use, they just have to be the same. Let’s use meters:

m = \frac{0.005 \, m}{1.7 \, \mathrm{m}} = 0.0029

The image is 0.3% of Dr. Toggerson’s height!

Solution for the front of the spoon:

The procedure is essentially the same, the only difference this time is that the image height should be considered to be negative h_i = 0.5 \, \mathrm{cm} as the image is inverted. Also, for fun, lets work in cm this time:

m = \frac{h_i}{h_0}

m = \frac{-1.0 \, \mathrm{cm}}{170 \, \mathrm{cm}} = -0.0059

The magnification is negative, again representing the fact that the image is inverted.

Introduction to Lenses

Instructor’s Note

 

By the end of this section you should:

  • Describe what a lens is and describe the two kinds of lenses, converging and diverging.
  • Describe what each type does and how these two types of lenses are different.
  • Define focal length and focal point for any lens.
  • Describe the idea behind the thin lens approximation.
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What is a lens?

A lens is a piece of transparent material designed to take incoming parallel photons and can bend them to a point which we call a converging lens, pictured below, where you have incoming parallel photons from the right passing into the lens and converging to a point on the left.

Light Rays
Figure 1.

A common application that you might have experienced with this is taking a regular magnifying glass and focusing the sun’s light to a point. The incoming light from the Sun is effectively parallel and the magnifying glass focuses it to a point.

The other thing a lens can do is spread incoming parallel photons as if they came from a point. This is known as a diverging lens. When the light passes through the lens, the light spreads out and from the left side it appears that the light originated at this point.

 

Figure 2.

A common application of diverging lenses is on your face if you happen to be nearsighted, the glasses you’re wearing if you’re nearsighted are a diverging lens.

Let’s talk a little bit about the properties of converging versus diverging lenses. Converging lenses are always thicker at the center than at the edges. On the flip side, diverging lenses are thicker at the edges than they are in the middle.

Types of lenses
Figure 3.

Instructor’s Note

 

You don’t need to know the names of these particular shapes you just need to know that converging lenses are thicker at the middle and diverging lenses are thicker at the edges.

Let’s move on to probably the most important property of lens, the focal length and focal points. Here are our two lenses, the converging lens on the left and the diverging lens on the right, in both cases you can see the light coming from the right passing through the lens on its way to the left.

Converging and Diverging lenses
Figure 4.

The focal point is defined as the point at which the parallel photons either converge or appear to. Meanwhile the focal length is the distance from the center of the lens to that point, since the focal length is a distance the unit of focal length is meters.

For the converging lens, the point where the photons appear to intersect is the focal point. Meanwhile for the diverging lens we trace the rays back through the lens and they appear to come from this point behind the lens, this is the focal point of a diverging lens.

Converging and Diverging lens
Figure 5.

We always measure the focal length from the center of the lens to the focal point, so you can see it here on the left for the converging lens and here on the right for the diverging lens. Every lens has only one focal length, but two focal points with one on each side.

Below we can see the full set up for converging lens. In this case we’ve got the light coming in from the left and moving towards the right. We can see the two different focal points, one on each side of the lens, each a focal length from the center of the lens and these two focal lengths are the same.

Convex lens
Figure 6.

Similarly, for a diverging lens we have two focal points one on each side of the lens, each the focal length from the center. The convention is that focal lengths for converging lenses are positive while diverging lenses have negative focal lengths.

Concave lens
Figure 7.

Instructor’s Note

 

In addition to a lot of vocabulary optics also has a lot of sign conventions that I am expecting you to learn from this prep, there is a set on flashcards on Quizlet to help you.

Finally let’s move on to the thin lens approximation. In a real lens, light will bend at each surface, it will bend as it goes from the air to glass, or whatever material the lens is made of, and then it will bend again by Snell’s law when the light moves from the glass back into the air.

Figure 8.

However, we will assume that the lens is very thin. What do we mean by thin? We mean that the thickness of the lens is very small relative to the focal length. Under this approximation it’s essentially as if all the light bending happens at the center. Now this is not actually what happens, remember the light does bend at each interface, air to glass and glass to air, however, if the lens is very thin these two interfaces are so close together and so close to the center of the lens that we can ignore it.

Figure 9.

Section Summary

Lenses Specifically as Applied to the Human Eye

Instructor’s Note

 

A Venn-diagram showing the different sciences

In this section, you will return to the overview of the human eye that was introduced in the motivation for this unit. Here, you will begin to see how to connect what you may know from biology, to what you have just read about lenses. What we expect you to know from this reading is:

 

  • Most of the focusing of the eye does NOT happen at the lens, but instead happens at the air-to-water interface at the front of the cornea.
  • The lens of the eye provides the fine-tuning of the focus to get the image distance to land on the retina.
  • The cornea and lens together act as a single lens system.
  • In medicine, people often talk of the power of a lens: this is 1/f where f is the focal length. Your glasses prescription, if you have one, is listed in diopters.
    • Note, that while this power shares the same name as the power discussed in Unit I, it is a different quantity (Homonyms – they are not just for English!)
    • The unit of optical power is 1/m or a diopter.
    • This is consistent as a lens with a shorter focal length, bends the light more strongly – has a higher power.

The eye is remarkable in how it forms images and in the richness of detail and color it can detect. However, our eyes often need some correction to reach what is called “normal” vision. Actually, normal vision should be called “ideal” vision because nearly one-half of the human population requires some sort of eyesight correction, so requiring glasses is by no means “abnormal.” Image formation by our eyes and common vision correction can be analyzed with the optics discussed earlier in this chapter.

Figure 1 shows the basic anatomy of the eye. The corneaand lens form a system that, to a good approximation, acts as a single thin lens. For clear vision, a real image must be projected onto the light-sensitive retina, which lies a fixed distance from the lens. The flexible lens of the eye allows it to adjust the radius of curvature of the lens to produce an image on the retina for objects at different distances. The center of the image falls on the fovea, which has the greatest density of light receptors and the greatest acuity (sharpness) in the visual field. The variable opening (i.e., the pupil) of the eye, along with chemical adaptation, allows the eye to detect light intensities from the lowest observable to 10101010">1010 times greater (without damage). This is an incredible range of detection. Processing of visual nerve impulses begins with interconnections in the retina and continues in the brain. The optic nerve conveys the signals received by the eye to the brain.

Figure shows the cross section of a human eye. At the very front is the cornea, followed by a bulging part called aqueous humor. At the top and bottom of the aqueous humor, towards the back is the iris. Between this and the vitreous humor are Ciliary fibers. The vitreous humor forms the bulk of the eye, which is roughly round in shape. At the back, the outermost layer is labeled sclera followed by retina. There is a small pit in the retina labeled fovea. The eye is connected to the optic nerve at the back and at the junction is a small circle labeled optic disc.
Figure 1: The cornea and lens of the eye act together to form a real image on the light-sensing retina, which has its densest concentration of receptors in the fovea and a blind spot over the optic nerve. The radius of curvature of the lens of an eye is adjustable to form an image on the retina for different object distances. Layers of tissues with varying indices of refraction in the lens are shown here. However, they have been omitted from other pictures for clarity.
 

The indices of refraction in the eye are crucial to its ability to form images. Table 1 lists the indices of refraction relevant to the eye. The biggest change in the index of refraction, which is where the light rays are most bent, occurs at the air-cornea interface rather than at the aqueous humor-lens interface. The ray diagram in Figure 2 shows image formation by the cornea and lens of the eye. The cornea, which is itself a converging lens with a focal length of approximately 2.3 cm, provides most of the focusing power of the eye. The lens, which is a converging lens with a focal length of about 6.4 cm, provides the finer focus needed to produce a clear image on the retina. The cornea and lens can be treated as a single thin lens, even though the light rays pass through several layers of material (such as cornea, aqueous humor, several layers in the lens, and vitreous humor), changing direction at each interface. The image formed is much like the one produced by a single convex lens (i.e., a real, inverted image). Although images formed in the eye are inverted, the brain inverts them once more to make them seem upright.

>Table 2 Refractive Indices Relevant to the Eye *This is an average value. The actual index of refraction varies throughout the lens and is greatest in center of the lens.
Material Index of Refraction
Water 1.33
Air 1.0
Cornea 1.38
Aqueous humor 1.34
Lens 1.41*
Vitreous humor 1.34
 
Figure shows a tree in front of an eye. Rays from the top and bottom of the tree strike the cornea of the eye. They are refracted, intersect in the middle of the vitreous humor and reach the retina. The image formed on the retina is tiny and inverted.
Figure 2: In the human eye, an image forms on the retina. Rays from the top and bottom of the object are traced to show how a real, inverted image is produced on the retina. The distance to the object is not to scale.

As noted, the image must fall precisely on the retina to produce clear vision—that is, the image distance i must equal the lens-to-retina distance. Because the lens-to-retina distance does not change, the image distance i must be the same for objects at all distances. The ciliary muscles adjust the shape of the eye lens for focusing on nearby or far objects. By changing the shape of the eye lens, the eye changes the focal length of the lens. This mechanism of the eye is called accommodation.

The nearest point an object can be placed so that the eye can form a clear image on the retina is called the near point of the eye. Similarly, the far point is the farthest distance at which an object is clearly visible. A person with normal vision can see objects clearly at distances ranging from 25 cm to essentially infinity. The near point increases with age, becoming several meters for some older people. In this text, we consider the near point to be 25 cm.

We define the optical power of a lens as

P = \frac{1}{f}

with the focal length f given in meters. The units of optical power are called “diopters” (D). That is \mathrm{D} = 1/\mathrm{m} = \mathrm{m}^{-1}. Optometrists prescribe common eyeglasses and contact lenses in units of diopters.

Working with optical power is convenient because, for two or more lenses close together, the effective optical power of the lens system is approximately the sum of the optical power of the individual lenses:

P_{\mathrm{total}} = P_{\mathrm{lens_1}} + P_{\mathrm{lens_2}} + P_{\mathrm{lens_3}} + ...

Introduction to Mirrors

Instructor’s Note

 

By the end of this section you should:

  • Be able to look at a mirror and say if it’s either converging or diverging.
  • Know the connection between a mirrors radius and its focal length.
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Just like lenses, curved mirrors can be used to focus light to a point or spread it out. Mirrors come in two basic shapes, we have concave mirrors, such as this mirror below on the left, which has a concave shape bending towards the light source, or we can have a convex mirror, pictured below on the right, that curves away from the light source.

Converging and Diverging Mirrors
Figure 1.

Concave mirrors are known also as converging mirrors and convex mirrors are also known as diverging mirrors. We can see for the concave mirror that the two light rays have been converged to a point. In this sense, the concave converging mirror is functioning similar to a converging lens, it’s taking incoming parallel light rays and converging them to a point. A common example of this particular type of mirror in use is the shaving mirror or makeup mirror that you might have in your bathroom that lets you see your face a little bit larger.

On the flip side, a convex mirror takes the incoming light rays and causes them to spread out. Just as we did with the lens, if we sort of imagine what someone to the left of the light source sees, our eyes kind of assume that light travels in straight lines so we trace the rays back and the light rays appear to originate from a point behind the mirror.

In this case, a convex mirror is providing a very similar function to a diverging lens, taking incoming parallel light and producing outgoing light that’s diverging as if it came from some point. You can see convex mirrors quite frequently as security mirrors in buildings.

Instructor’s Note

 

All the mirrors we will talk about in this particular class will be called spherical mirrors, meaning that the mirror is part of a sphere, while most mirrors are not actually spherical, it turns out that studying spherical mirrors is a very good approximation for most real curved mirrors.

Let’s talk about focal lengths and focal points for mirrors. Just like lenses, the focal point is the point where the photons either converge to or appear to emanate from. For the concave mirror, which is converging on the left, we’ve already seen that light comes in and converges to some point, this is known as the focal point. For the diverging mirror, the light comes in and bounces off, light comes in and bounces off, and the light appears to emanate from some point behind the mirror, so this would be the focal point for this diverging mirror.

Unlike lenses, mirrors only have a single focal point, and this stems from the fact that light can go through a lens from either direction, we can put the light on either side of the lens and the light will go through. For a mirror on the other hand, a mirror only has one reflective surface, there’s only one side that will act as a mirror, and therefore mirrors only have a single focal point.

As with lenses, the focal length, which we designate f is the distance in meters from the surface of the mirror to the focal point. For the converging lens, below on the left, this blue arrow represents the focal length f and for the diverging mirror we have this focal length below on the right.fff">

 

 

Concave Mirror
Figure 2. Concave Mirror (Credit: Cronholm144)
Convex Mirror
Figure 3. Convex Mirror (Credit: Cronholm144)

And just as with lenses, converging mirrors have a positive focal length and diverging mirrors have a negative focal length. You can start to see some similarity between mirrors and lenses. When things are converging you have positive focal lengths, and when things are diverging you have negative focal lengths.

Now let’s talk a little bit about focal lengths and the radius of curvature.

Instructor’s Note

 

I want you to remember that all the mirrors we will be talking about here are parts of a sphere.

We need two new terms, the center of curvature, which represents the center of the sphere of which the mirror is a part, and the radius of curvature, which is the radius of which the mirror is a part.

The focal length of any spherical mirror is half the radius, so above we have a concave mirror on the top with positive focal length and a convex mirror on the bottom with negative focal length, the distance R is the radius of the sphere of which this mirror is a part and the focal length is half that radius

f = \frac{R}{2}.rrr">

Section Summary

 

 

Ray Tracing

14

Ray Tracing

We now know what images are: apparent reproductions of objects. In the last chapter, we characterized images as erect and inverted, and discussed the idea of magnification. Our next goal is to determine where these images will be for a given optical system. To solve this, we will use a new type of problem solving called ray tracing. Ray tracing is a type of problem solving quite different than what you are used to for physics classes; this method of problem solving uses a lot more diagrams and, while equations will still be used, they will play a comparatively smaller role. These multiple problem solving approaches goes back to Physics Goal #2: Representing physics Ideas in different ways.

So what is ray tracing? If an object emits light, it emits light in all directions. If the object is visible by reflecting light from another source, your face is visible because it reflects light from the surroundings, that reflected light is diffuse and goes in all directions. Also keep in mind, that typical objects emit huge (1030) numbers of photons. Thus, there are effectively photons going in every conceivable direction as shown in Figure 1.

While rays come off in all directions, we follow the rays which are easy - like one that goes through the lens parallel to the optical axis and then through the focal point.
Figure 1: While rays come off in all directions, we follow the rays which are easy – like one that goes through the lens parallel to the optical axis and then through the focal point.

Ray tracing allows us to follow very specific photons: photons which will easy to follow because of the paths they take. For example, we know from the last chapter, that a photon that enters a converging lens parallel to the optical axis will go through the focal point as shown in Figure 2 below. Since there are so many photons, leaving your face, one will go parallel to the optical axis and then through the lens and towards the focal point as shown in Figure 1.

Incoming parallel rays converge to the focal point of a converging lens.
Figure 2: Incoming parallel rays converge to the focal point of a converging lens.

Throughout the next few sections, we will go through the particular rays to follow for the different optical elements:

For each optical element, there will be three rays to follow for any object a finite distance away (if the object is infinitely far away, then the rays come in parallel and we saw what happens with parallel rays in the previous chapter). The three rays to follow are:

Our three rays that we will follow

  1. A ray that comes in parallel to the optical axis leaves using a focal point. If the optical element is converging (like a concave mirror or convex lens) then use a focal point that brings the ray towards the optical axis.
  2. A ray that aims for the center of a lens or mirror will go straight.
    • The center of a lens is the middle: this ray will travel un-deflected as though the lens was not there.
    • The center of a mirror is the geometric center: this ray will hit the mirror at a 90o angle and bounce straight back.
  3. A ray that comes in using a focal point will exit parallel to the optical axis.

 

Instructor’s Note

 

Your quiz will cover:

  • For a given ray, you need to be able to determine where it will go for each of these basic optical elements

We will NOT expect you to be able to interpret the results, do calculations, or consider multiple elements. We will do that in class.

Ray Tracing for Converging Lenses

Instructor’s Note

 

All of the following sections are presented both as video and with a text-based transcript. Normally, I feel that the two presentations are equivalent. In this particular case, however, I feel that most students will learn more by watching the videos particularly if you follow along on your own sheet of paper.

 

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In this section,  we are going to explore how to draw ray diagrams for the first of our four types of optical elements: the converging lens. Now,  these videos use pictures of paper drawings to really show you the mechanics of how to draw these things out by hand. You will be expected to draw ray diagrams on an exam. To draw ray diagrams, it really helps to have a few things: one it helps to have pens in a couple of different colors: black, red, and blue. I also like to have a pencil which gives me four colors. You also
really need a protractor, which was labeled on the syllabus as one of the things you need for this course.
An optical axis, a protractor, a pencil, and pens of different colors: red, blue, and black.
Get started drawing a ray diagram for converging lens, by drawing an optical axis. Now, put your lens on this axis somewhere, kind of in the middle to give yourself some room to work. Draw the center of the lens first, using your protractor to make sure it is perpendicular to your optical axis and that all your lines are straight. Trying to ray diagrams freehand will NOT work! You need your lines to be straight you need your angles to be precise there’s no way you’ll ever get these diagrams to work properly if you’re trying to draw them freehand.
Since I’m not an artist, and drawing the lens itself takes time, you’ll often see me identify a converging lens just with a symbol that looks like a line with arrows
on both ends. These arrows are meant to indicate that the lens gets thicker towards the middle than it does towards the ends. You’ll see when we do the diverging lens, we’ll use a different symbol. Now put an object about six centimeters away from my lens. Your object might be a little face or something like that. Again I’m no artist, often you’ll see people put a little arrow just to help identify which way is up. We also need to know the focal lengths for our lens: let’s put our focal length for this example at two centimeters. It’s going to be a positive focal length because our lens is converging. Measure out your two focal points and label them with f. Now, we have a set-up, and at this point we can go through and actually start drawing our rays.
Setup for a converging lens ray diagram showing lens, symbol for lens, object, and focal points.
Ray number one is in parallel out through focal point. Recall, light rays are going in all directions off the top of this object. One of these is a ray going off towards the lens parallel to the optical axis. When this ray hits the lens, it’s going to go out through the focal point because that’s what a focal point is for a converging lens: converging lenses convert incoming light that’s parallel and bend them to their focal points. You’ll notice we’re pretending that all the bending happens here at the center of the lens and that’s due to the thin lens approximation that was discussed in the previous chapter.
Ray 1 for a converging lens: in parallel, out through the focal point on the far side.
The second ray is the rule that if a ray through the center goes straight. In the case of a lens, the center is where the lens meets the optical axis: by definition the optical axis goes through the center of the lens (It may not look like it in my drawings because I’m not much of an artist!). So ray number two goes straight through the middle.
Ray 2 for a converging lens: straight through the center.
Ray number three, remember, is in using focal point out parallel. Now in this case, we’ve already used the focal point on the far side of the lens from the object. Therefore, this time we’re going to come in using the focal point we haven’t used yet, the one on the same side as the object. Out of all the infinite numbers of rays, we are going to follow the one which just happens to come in towards the lens going for this near focal point. This ray  is going to go out parallel to our optical axis.
Final ray for a converging lens: in through near focal point, out parallel. All rays intersect on far side, this will be the location of the image.
You can see that in this particular case, all of our ray’s happen to converge on the far side of the lens. As we’ll discuss in class, that is going to be the location of our image. We’ll talk more about that in class. What you need to know right now is how to draw these three rays:
  1. In parallel, out using the focal point.
  2. Straight through the center.
  3. In through the focal point, and out parallel

Simulation

Below is a flash simulation (you may need to click and allow flash) that shows the paths of rays through a converging lens. A few things to explore:

  • Click “Principal Rays” to see the rays used in the ray diagrams discussed in the video.
  • Click “Many Rays” to see the fact that there are a bunch of rays coming from the object that all converge to the point, not just the ones we saw in the ray diagram. You will see that some even miss the lens entirely and just go straight!

An interactive or media element has been excluded from this version of the text. You can view it online here: http://openbooks.library.umass.edu/toggerson-132/?p=501

Ray tracing for Diverging Lenses

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In this section, we will be drawing ray diagrams with diverging lenses. To get setup draw your optical axis and put your lens in the middle: make it pretty big. Always give yourself room to work when you’re drawing these things. This time we’re doing not a converging lens but a diverging lens. A diverging lens is, as you know, much thinner in the middle than it is on the outside. Because clearly my ability to draw diverging lens is even worse than my ability to draw converging lenses, I’ll often use arrow heads pointing towards the center of the lens to indicate that it’s thinner in the middle than it is on the outside. For this example, we’re going to use a focal length of 4 centimeters: one focal point on each side labelled f. We will put our object 7 centimeters away. Give it a pretty good height: make it like three centimeters tall.
Setup of diverging lens: focal length -4cm and a 3cm object 7cm away. Arrows pointing in symbolize a diverging lens.
Now we once again go with our same process: ray number one is “in parallel out using a focal point.” You will notice notice I’m using the word using not the word through because the ray will not actually go through a focal point. The ray comes in parallel, use your protractor to make sure that the ray is parallel to the optical axis. Since this is a
diverging lens, it’s not going to bring the ray towards the optical axis; it’s going to cause the ray to diverge away from the optical axis as if the light had come from this focal point on the same side as the object. That is why I say not through but using. The light is going to diverge as if it came from this focal point on the left. I usually include a dashed line to help me get my line straight.’
Ray 1 for a diverging lens: in parallel out as if it had come from the left focal point.
Ray number two, “straight through the middle,” which is pretty straightforward. The middle is where our optical axis meets our optical element.
Ray 2 for a diverging lens: straight through the center
Then we’ve got ray number three, “in using a focal point, out parallel.” Again, you’ll notice I’m using the word using. In this particular case, I’ve already used the left focal point, and I can’t use it again. Therefore, I have to use the other one on the far side from the object. So I’m gonna come in as if I were going for that focal point, but then I hit the lens and, instead of continuing on, I go out parallel to the optical axis. That is ray number 3.
Ray 3 for a diverging lens: in going for far focal point, out parallel.
You’ll notice these rays are spraying out, which they should. This is a diverging lens after all, the rays should diverge, and they do. Thus, they don’t converge to a point anywhere like we did saw in the last example. However, what if your eye were over here on the right looking through the lens at the object, what would you see? Well your brain assumes that light travels in straight lines, because in most of your experience it does. So your brain is going to assume that all these light rays emanated from this point and traveled in a straight lines. Thus, we will have our image over here on the left, the same side as the object! We’ll discuss this part more in class, but I thought I would just expose you to it while we’re here. What you need to know right now is how to draw these three rays.
Diverging lens final image on the same side as object.

Ray Tracing for Concave Mirrors

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In this section, I am going to show you how to draw the ray diagram for a concave mirror now. In this particular class, we’re only interested in mirrors that are parts of circles or flat. In class, we will provide you a nice little drafting tool to help you draw mirrors that are circles but for the purposes of this video, I’m going to use a compass. We will begin by drawing a circular mirror using my compass. Note that for mirrors, the center is the geometric center of the circle. For lenses, in contrast, the center is the middle part so it’s a slight difference in terminology. Now I will add my optical axis which connects the center of the mirror and goes away from it (you’re going to really need a protractor to do all of this stuff!). Next, I need to measure how big my circle actually is. I can see from measuring it that my center is 9.2 cm away from the vertex, which by the sign conventions discussed in the last chapter, since this is a concave mirror, we’re going to think of our radius as being positive.
Now, a fundamental property of all spherical mirrors is that the focal length is half of the radius f = R/2. That is generally true for all spherical mirrors. Therefore, in this case, the focal length is going to be 4.6cm. I can, therefore, measure 4.6 cm away from the vertex and mark a little point which will my focal point. Now, all we need is an object. In this example, we’re going to place our object nice and and far outside our center. It will be outside the center of curvature. Let’s make an object that’s gonna have some some height to it. I’m actually gonna make it as tall as my protractor ruler for a reason that’ll become hopefully apparent in a moment. As usual, we like to put a little arrow on our object so we know which way is up but you can think of it as being a face, or a tree, or a candle, or whatever you want. Arrows are just easy to draw. Recall, there are photons coming off of this object in all directions and we’re just going to choose the three photons that are easiest for us to follow.
setup for the concave mirror example showing the focal point and object outside the center of curvature.
The nice thing about phrasing the ray diagram rules the way I have at the beginning of this chapter is the same three rays that we’ve been doing for lenses still work: we just have to think about them a little bit differently. The first ray, just as for lenses, is “in parallel, out using the focal point.” We are therefore, going to draw our first ray coming in parallel (now you can see why I’ve made my object the exact same thickness as my protractor – it makes drawing a parallel ray pretty easy) and it’s going to then go out through my focal point. You can see why by zooming in over here, our law of reflection: the incoming angle and the outgoing angle are the same.
Ray 1 for the concave mirror: in parallel to the optical axis, out through the focal point.
Ray number two: “rays that go through the center travel straight.” Notice again that the phrasing of how we describe our rays is exactly the same for mirrors and lenses. In this case, however, the center is the center of the circle, not physically on the mirror. That point, where the mirror meets the optical axis, is called the vertex. A ray that goes through the center will travel in a straight line: that ray will come in hit the mirror and end up bouncing to travel straight back the way it came. You can see that such a ray meets the mirror at 90 degrees so it bounces straight in-and-out.
Concave mirror ray 2 comes in through the center and bounces back the way it came.
Finally, ray number three again follow the same rules as the lens: “in using the focal point out parallel.” Now, lenses have two focal points, one on each side, but mirrors only only have the one. Thus, we don’t have to worry about which focal point to use: mirror only has one, so that makes it maybe a little bit easier. Our ray is going to come in using our focal point and then going to go out parallel to our optical axis.
Ray 3 for a concave mirror comes in through the focal point and goes out parallel.
You will see that all of these rays actually converge at this point right in front of the mirror. That point is where our image is going to be. Thus, we are
going to have an image distance that’s positive as it is on the side of the outgoing light. The thing with mirrors is that the incoming side and the outgoing side are the same side of the mirror. We are therefore going to have a positive image distance and a positive object distance. Moreover, our image is going to be in the previous chapter when we discussed looking in the inside of a spoon at some length. We mentioned that the image appears to hover in front of the spoon a little bit: here we see exactly where that image is. We will talk more in class about why this is the image, how to characterize that image, and some other aspects of interpreting this diagram.  What I really need you to know for right now is how to draw these three rays.
The three rays for a concave mirror converge to a point creating an image.

Ray Tracing for Convex Mirrors

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Now for the last optical element, where we draw the ray diagram for a convex mirror. Once again I’m going to use my compass to draw the spherical mirror to begin. In class, I’ll give you a tool to help with this but here I’m going to use a compass to help me draw a nice circle. Again, I use the center of the circle to define my optical axis. My circle has a radius of 9.2cm, but since it’s a convex mirror our sign conventions say that the radius should be negative: R = - 9.2 \, \mathrm{cm}. Just as with a
concave mirror, the focal length is half the radius f = R/2. Therefore, the focal length is going to be negative: f = -4.6 \, \mathrm{cm} measured from the vertex. Now go and put an object on the other side from the center, because, remember we’re looking at a convex mirror – we’re looking at the back of a spoon. In this example, let’s put it 5 cm outside. Just as in the last section, I am going to make my object the thickness of my protractor ruler. That will make drawing a nice incoming
parallel right easy. As always, we draw it as an arrow just so it’s easy to tell which way is up but you can think of it as a little face, a tree, a candle, it could be anything.
Setup for the convex mirror example.
Once again, our rules are the same: the first ray is, as always, “in parallel out using the focal point.” Thus, I draw my incoming ray parallel. now I can’t go through because it’s a mirror, the ray has to bounce instead. You will also notice that again I use the verb using instead of through to help me remember that I can’t go throughthe focal point that’s behind the mirror. Thus, the ray is going to go out using the focal point: this ray is gonna bounce as if it had come from the focal point. That is what the focal point for a convex mirror means. Thus, we have ray number one and, as always, we can see that \theta_i = \theta_f.
Ray 1 for the convex mirror: in parallel exiting as if it had come from the focal point of the mirror.
Ray number two: “a ray using the center travels straight.” For this ray, we are going to aim for the center of the mirror’s circle, but it can’t go through (because that’s a mirror) so instead it will bounce and head back off directly the way it came.
Ray 2 of a convex mirror aims for the center and bounces back the way it came.
Ray number three: “in using the focal point out parallel.” This time we’re going to consider a ray that comes in as if it were going for the focal point behind the mirror and is going to go out parallel to the optical axis.
Convex mirror's ray 3: in aiming for focal point bounces parallel to the optical axis.
Now, as with the diverging lens describe above, these three rays don’t converge anywhere; they’re spraying out as they leave the mirror. This is a diverging optical element so they should spread out. However as with the diverging lens, if your eye is behind the object looking at this whole thing what your eye going to see is these three photons and it’s going to assume these photons traveled in straight lines. All the photons appear to have originated from a point behind the mirror. That point is where our image is going to be. In this particular case, we have an object distance to the right that is positive, because the object is on the same side is the incoming light but we have a negative image distance because the light while it comes in on the right, also goes out on the right because of the mirror. The image is on the reverse side from this outgoing light, so the image distance is negative. Ultimately, we see a little upright erect image behind the mirror. If you look in the back of a spoon, as was shown in the last chapter, that’s what you see you see: a littlemini version of yourself erect behind the spoon.
All the rays from a convex mirror appear to originate from a point behind the mirror - the location of the image.

 

Homework Problems

15

  1. Identify the portion of the eye responsible for most of the focusing of light.
  2. What is the purpose of the iris?
  3. In geometric optics, we do analyses using similar triangles. This problem is here to help you practice working on these again.
  4. Look at this map and determine the angle.
  5. For this set of intersecting lines, use the following information to find the missing values.
  6. Indicate where the outgoing ray from a mirror intersects the dotted line.
  7. What is the speed of light in water? In glycerine? The indices of refraction for water is 1.333 and for glycerine is 1.473.
  8. Calculate the index of refraction for a medium in which the speed of light is 1.416×108 m/s.
  9. Consider two materials. When light passes through the space between the two materials at 0°<θ<90°, there is no change in the direction of the propagation of the light. What can you infer about the two materials?
  10. Which of the properties of a light ray change as it goes from glass to vacuum?
  11. What are the wavelengths of visible light in crown glass?
  12. Suppose you have an unknown clear substance immersed in water, and you wish to identify it by finding its index of refraction. You arrange to have a beam of light enter it at an angle of 48.6∘, and you observe the angle of refraction to be 32.4∘. What is the index of refraction of the substance? Water has an index of refraction equal to 1.333.
  13. A beam of white light goes from air into water at an incident angle of 83.0∘. At what angles are the red 660 nm and violet 410 nm parts of the light refracted? Red light in water has an index of refraction equal to 1.331 and that of violet light is 1.342.
  14. Given that the angle between the ray in the water and the perpendicular to the water is 28.3, and using information in the figure above, find the height of the instructor’s head above the water. Water has an index of refraction equal to 1.333.
  15. Sign conventions for object and image distances: objects and images on opposite sides of the optical element.
  16. Sign conventions for object and image distances: objects and images on same side of the optical element.
  17. What characterizes an object with a negative magnification?
  18. Calculating magnification of a gemstone.
  19. Characterizing lenses.
  20. Diopters to focal length.
  21. Focal length to diopters.
  22. Focal lengths of mirrors.
  23. Ray diagrams: converging lenses.
  24. Ray diagrams: diverging lenses.
  25. Ray diagrams: concave mirrors.
  26. Ray diagrams: convex mirrors.

Unit III

III

Unit III On-a-Page

16

Instructor’s Note

This chapter very strongly follows the ideas of a few distinct principles discussed in Unit I On-a-Page.

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Electric fields, forces, potentials, and potential energies from a positive point charge.
This image effectively summarizes the connections between the two new ideas in this chapter, electric fields and electric potentials, as well as showing how they connect to ideas with which you are already familiar: forces and potential energy.
  • All charges generate electric fields \vec{E} (units N/C orV/m)
  • For a point charge \vert \vec{E} \vert = \frac{1}{4 \pi \varepsilon_0} \frac{q}{r^2}
  • Point away from positive, towards negative
  • These are real and exist regardless if something is there to feel or not!
  • Put a charge in an electric field ⇒ it will feel a force \vec {F } =q \vec {E}
  • Can also think in terms of energy: All charges generate electric potentials V (units Volts)
  • For a point charge V = \frac{1}{4 \pi \varepsilon_0} \frac{q}{r}
  • Exists if something is there to feel or not!
  • Put a charge in a potential it has a potential energy U = qV
  • Electric potentials and electric fields are two sides of the same coin
  • Just like in P131: Can think of forces or energy
  • \vec{E} points ‘down the potential hill’
  • \vec{E} = - \frac{\Delta V} {\Delta x}

 

Introduction

17

We know that electrons have electric charge. Having charge is one of the key ways in which electrons differ from photons. Up to this point, however, we have been completely neglecting the fact that electrons have charge. How does having charge impact how an electron behaves? Exploring this charged aspect of an electron’s identity will be the focus of this chapter. Along the way, we will need to introduce two other concepts: the electric field and the electric potential.

As you may recall from Physics 131, there are four fundamental forces in the Universe: gravity, electricity/magnetism, the weak nuclear force, and the strong nuclear force. With the exception of gravity, all of the other forces in your everyday life are either electrical or magnetic in nature. The normal force that keeps you from falling through your chair is, in reality, electrical in origin: the electrons in the chair are electrically repelling the electrons in your body. The tension forces in ropes are also electrical: they arise from the chemical bonds in the rope which ultimately arise from the electrical attractions between protons and electrons. Since the electrical force is the ultimate source of all electrical bonds, that means that it is also the most relevant force for biology and chemistry!

The readings for this unit are ultimately divided up into four main parts. First we motivate the study of electricity with some applications to biology and chemistry. The second part is review. Our study of electrical forces will be, as all forces are, heavily dependent on being able to use vectors. Since it may have been a while since you studied vectors, we have included the chapter on vectors from the 131 textbook for your reference. If you feel comfortable with vectors, feel free to skip this section: it is just there for your review, but there are some homework problems to make sure you are fresh. Afterwards there are some problems reviewing charge conservation. Following these refreshers, we get to the meat of this unit: we will introduce the idea of electric field  \vec{E} which is the ultimate source of electrical forces. Finally, just as you can think about a falling ball in terms of the gravitational force or in terms of the gravitational potential energy, the same is true for electrical forces. We will, therefore, then consider the interactions between charges from an energy perspective and introduce the idea of electric potential  V Note, electric potential and electric potential energy are two different, but related, ideas – be careful with your vocabulary here!

You may have seen a Coulomb’s Law for electrical forces in other courses, but I really want you to try to think of electrical forces as arising from fields and potentials. Try to visualize them. Fields and potentials are just as real as the electrons and photons we have been studying. Being able to think of the interactions between charged particles in terms of electric fields and electric potentials will be key to being successful in the rest of this course.

Motivating Context for Unit III

18

A Venn-diagram showing the different sciences

This unit will focus on the electric field \vec{E} and the electric potential V. As described in the introduction, the electric force is the underpinning of chemistry, so I would like you to refresh some ideas about chemical bonds that you probably saw in your chemistry classes that we will use in this unit.

Another case that we will explore in some detail in this unit is gel electrophoresis: a process you have probably discussed in your biology class. This laboratory technique is fundamentally based upon the ideas of electric field and potential we will discuss in this unit. Thus, a review of the procedure based upon OpenStax Microbiology chapter 12.2 is included below for your review.

Molecular Bond Basics

Valence bond theory describes a covalent bond as the overlap of half-filled atomic orbitals (each containing a single electron) that yield a pair of electrons shared between the two bonded atoms. We say that orbitals on two different atoms overlap when a portion of one orbital and a portion of a second orbital occupy the same region of space. According to valence bond theory, a covalent bond results when two conditions are met: (1) an orbital on one atom overlaps an orbital on a second atom and (2) the single electrons in each orbital combine to form an electron pair. The mutual attraction between this negatively charged electron pair and the two atoms’ positively charged nuclei serves to physically link the two atoms through a force we define as a covalent bond. The strength of a covalent bond depends on the extent of overlap of the orbitals involved. Orbitals that overlap extensively form bonds that are stronger than those that have less overlap. We calculated the shapes of these orbitals for simple molecules such as 1,3-butadine in-class during Unit I by modeling the electrons as a particle in a box!

The energy of the system depends on how much the orbitals overlap. Figure 1 illustrates how the sum of the energies of two hydrogen atoms (the colored curve) changes as they approach each other. When the atoms are far apart there is no overlap, and by the convention described in Unit I – Chapter 5 Some Energy-Related Ideas that Might be New or Are Particularly Important: The Potential Energy of Atoms and Molecules the potential energy of each atom is zero. As the atoms move together, the electron waves (orbitals) begin to overlap. Each electron begins to feel the attraction of the nucleus in the other atom. In addition, the electrons begin to repel each other, as do the nuclei. The result is the electron waves change shape in response.

While the atoms are still widely separated, the attraction is slightly stronger than the repulsion, and the energy of the system decreases. (A bond begins to form.) As the atoms move closer together, the overlap increases, so the attraction of the nuclei for the electrons continues to increase (as do the repulsions among electrons and between the nuclei). At some specific distance between the atoms, which varies depending on the atoms involved, the energy reaches its lowest (most stable) value. This optimum distance between the two bonded nuclei is the bond distance between the two atoms. The bond is stable because at this point, the attractive and repulsive forces combine to create the lowest possible energy configuration. If the distance between the nuclei were to decrease further, the repulsions between nuclei and the repulsions as electrons are confined in closer proximity to each other would become stronger than the attractive forces. The energy of the system would then rise (making the system destabilized), as shown at the far left of Figure 1.

A pair of diagrams are shown and labeled “a” and “b”. Diagram a shows three consecutive images. The first image depicts two separated blurry circles, each labeled with a positive sign and the term “H atom.” The phrase written under them reads, “Sufficiently far apart to have no interaction.” The second image shows the same two circles, but this time they are much closer together and are labeled, “Atoms begin to interact as they move closer together.” The third image shows the two circles overlapping, labeled, “H subscript 2,” and, “Optimum distance to achieve lowest overall energy of system.” Diagram b shows a graph on which the y-axis is labeled “Energy ( J ),” and the x-axis is labeled, “Internuclear distance ( p m ).” The midpoint of the y-axis is labeled as zero. The curve on the graph begins at zero p m and high on the y-axis. The graph slopes downward steeply to a point far below the zero joule line on the y-axis and the lowest point reads “0.74 p m” and “H bonded to H bond length.” It is also labeled “ negative 7.24 times 10 superscript negative 19 J.” The graph then rises again to zero J. The graph is accompanied by the same images from diagram a; the first image correlates to the point in the graph where it crosses the zero point on the y-axis, the third image where the graph is lowest.
Figure 1 (a) The interaction of two hydrogen atoms changes as a function of distance. (b) The energy of the system changes as the atoms interact. The lowest (most stable) energy occurs at a distance of 74 pm, which is the bond length observed for the H2 molecule.

Play with the energies of atoms and bonds

Below is a simulation where you can see the energies of atoms and their bonds. Play around with it. In this simulation, one atom is “pinned down” and the other is free to move.

  • Drag the free atom wherever you wish and let it go. It will move according to its potential energy.
  • You can see the overlap in the electron waves (electron clouds) at the bottom.
  • You can change the types of atoms using the menu in the upper right.
  • You can turn on the forces as well to see how the forces are responding to the potential energy.

You will notice that the atoms do not stay at a fixed distance; they bounce as skateboarders on a hill! You can even get neon atoms to “bond” although there is not a lot of room to do so! This is true; at small enough temperatures you can get neon to bond. At 24.56K it will actually become a solid. You can see O=O, on the other hand, has a much deeper well and is therefore a much stronger bond.

An interactive or media element has been excluded from this version of the text. You can view it online here:
http://openbooks.library.umass.edu/toggerson-132/?p=1470

Basic Description of Gel Electrophoresis

We will explore this device in class, so it is probably beneficial if you have already familiar with how it works. This material is from OpenStax Microbiology – Chapter 12.2 Visualizing and Characterizing DNA, RNA, and Protein.

There are a number of situations in which a researcher might want to physically separate a collection of DNA fragments of different sizes. A researcher may also digest a DNA sample with a restriction enzyme to form fragments. The resulting size and fragment distribution pattern can often yield useful information about the sequence of DNA bases that can be used, much like a bar-code scan, to identify the individual or species to which the DNA belongs.

Gel electrophoresis is a technique commonly used to separate biological molecules based on size and biochemical characteristics, such as charge and polarity. Agarose gel electrophoresis is widely used to separate DNA (or RNA) of varying sizes that may be generated by restriction enzyme digestion or by other means, such as the PCR (Figure 2).

Due to its negatively charged backbone, DNA is strongly attracted to a positive electrode. In agarose gel electrophoresis, the gel is oriented horizontally in a buffer solution. Samples are loaded into sample wells on the side of the gel closest to the negative electrode, then drawn through the molecular sieve of the agarose matrix toward the positive electrode. The agarose matrix impedes the movement of larger molecules through the gel, whereas smaller molecules pass through more readily. Thus, the distance of migration is inversely correlated to the size of the DNA fragment, with smaller fragments traveling a longer distance through the gel. Sizes of DNA fragments within a sample can be estimated by comparison to fragments of known size in a DNA ladder also run on the same gel. To separate very large DNA fragments, such as chromosomes or viral genomes, agarose gel electrophoresis can be modified by periodically alternating the orientation of the electric field during pulsed-field gel electrophoresis (PFGE). In PFGE, smaller fragments can reorient themselves and migrate slightly faster than larger fragments and this technique can thus serve to separate very large fragments that would otherwise travel together during standard agarose gel electrophoresis. In any of these electrophoresis techniques, the locations of the DNA or RNA fragments in the gel can be detected by various methods. One common method is adding ethidium bromide, a stain that inserts into the nucleic acids at non-specific locations and can be visualized when exposed to ultraviolet light. Other stains that are safer than ethidium bromide, a potential carcinogen, are now available.

a) A diagram of the process of agarose gel electrophoresis. 1 – An agarose and buffer solution is heated and poured into a form. This result in a rectangular block with indents along one end labeled “S=sample wells”. 2 – When cooled, the agarose gel block contains small wells (S) where the sample will be place. 3 – Each sample is added to a separate well. Then the agarose gel is placed in a chamber that generates a charge across the gel. The samples are added using micropipettes. 4 – The solution within the chamber conducts the electric current generated by the chamber. The side nearest the sample well will have a negative charge; the other side will have a positive charge. 5 – DNA has a negative charge and will be drawn to the positive pole of the gel. Smaller DNA molecules will be able to travel faster through the matrix of the gel. 6 – One well will contain a DNA ladder, which has fragments of known size. This ladder is used to identify the sizes of the bands in the sample. The ladder looks like many bands in the gel; from top to bottom the sizes of the bands are – 2000 bp, 15000 bp, 1000 bp, 750 bp, 500 bp, 250 bp. The other lanes have a few bands of various sizes.
Figure 2 (a) The process of agarose gel electrophoresis. (b) A researcher loading samples into a gel. (c) This photograph shows a completed electrophoresis run on an agarose gel. The DNA ladder is located in lanes 1 and 9. Seven samples are located in lanes 2 through 8. The gel was stained with ethidium bromide and photographed under ultraviolet light. (credit a: modification of work by Magnus Manske; credit b: modification of work by U.S. Department of Agriculture; credit c: modification of work by James Jacob)
 

Vector Review

19

Instructor’s Note

 

You should already be familiar with vectors, but we will use them in this unit so this chapter from Physics 131 is a review. If this material is familiar, feel free to go to the homework problems at the end (there are none embedded in the sections as it is review)

Your Quiz would Cover

  • A vector is a quantity with a magnitude and direction
  • Converting between magnitude/direction and the component form for any vector. This ties into the Pythagorean Theorem

Kinematics in Two Dimensions: an Introduction

Figure 1. Walkers and drivers in a city like New York are rarely able to travel in straight lines to reach their destinations. Instead, they must follow roads and sidewalks, making two-dimensional, zigzagged paths. (credit: Margaret W. Carruthers)

Two-Dimensional Motion: Walking in a City

Suppose you want to walk from one point to another in a city with uniform square blocks, as pictured in Figure 2.

Figure 2. A pedestrian walks a two-dimensional path between two points in a city. In this scene, all blocks are square and are the same size.

The straight-line path that a helicopter might fly is blocked to you as a pedestrian, and so you are forced to take a two-dimensional path, such as the one shown. You walk 14 blocks in all, 9 east followed by 5 north. What is the straight-line distance?

An old adage states that the shortest distance between two points is a straight line. The two legs of the trip and the straight-line path form a right triangle, and so the Pythagorean theorem, a^2 + b^2 = c^2, can be used to find the straight-line distance.

Figure 3. The Pythagorean theorem relates the length of the legs of a right triangle, labeled a and b, with the hypotenuse, labeled c. The relationship is given by: a^2 + b^2 = c^2. This can be rewritten, solving for c : c = \sqrt {a^2 + b^2 }

Instructor’s Note

 

We will be using the Pythagorean Theorem all throughout two-dimensional kinematics, as well as throughout this entire course. If you are uncomfortable or unfamiliar with the Pythagorean Theorem, or even if it’s just been a long time since you’ve used it, please come see your instructor as soon as possible and they will get you up to speed.

The hypotenuse of the triangle is the straight-line path, and so in this case its length in units of city blocks is \sqrt {(9 \text{ blocks})^2 + (5 \text{ blocks})^2 } = 10.3 \text{ blocks} , considerably shorter than the 14 blocks you walked. (Note that we are using three significant figures in the answer. Although it appears that “9” and “5” have only one significant digit, they are discrete numbers. In this case “9 blocks” is the same as “9.0 or 9.00 blocks.” We have decided to use three significant figures in the answer in order to show the result more precisely.)

Figure 4. The straight-line path followed by a helicopter between the two points is shorter than the 14 blocks walked by the pedestrian. All blocks are square and the same size.

The fact that the straight-line distance (10.3 blocks) in Figure 4 is less than the total distance walked (14 blocks) is one example of a general characteristic of vectors. (Recall that vectors are quantities that have both magnitude and direction.)

As for one-dimensional kinematics, we use arrows to represent vectors. The length of the arrow is proportional to the vector’s magnitude. The arrow’s length is indicated by hash marks in Figure 2 and Figure 4. The arrow points in the same direction as the vector. For two-dimensional motion, the path of an object can be represented with three vectors: one vector shows the straight-line path between the initial and final points of the motion, one vector shows the horizontal component of the motion, and one vector shows the vertical component of the motion. The horizontal and vertical components of the motion add together to give the straight-line path. For example, observe the three vectors in Figure 4. The first represents a 9-block displacement east. The second represents a 5-block displacement north. These vectors are added to give the third vector, with a 10.3-block total displacement. The third vector is the straight-line path between the two points. Note that in this example, the vectors that we are adding are perpendicular to each other and thus form a right triangle. This means that we can use the Pythagorean theorem to calculate the magnitude of the total displacement. (Note that we cannot use the Pythagorean theorem to add vectors that are not perpendicular. We will develop techniques for adding vectors having any direction, not just those perpendicular to one another, in Vector Addition and Subtraction: Graphical Methods and Vector Addition and Subtraction: Analytical Methods.)

The Independence of Perpendicular Motions

Instructor’s Note

 

The idea of the independence of perpendicular motion is a fundamental one that you should take some time to think about, and there are some questions about this on the homework.

The person taking the path shown in Figure walks east and then north (two perpendicular directions). How far he or she walks east is only affected by his or her motion eastward. Similarly, how far he or she walks north is only affected by his or her motion northward.

Independence of Motion

The horizontal and vertical components of two-dimensional motion are independent of each other. Any motion in the horizontal direction does not affect motion in the vertical direction, and vice versa.

This is true in a simple scenario like that of walking in one direction first, followed by another. It is also true of more complicated motion involving movement in two directions at once. For example, let’s compare the motions of two baseballs. One baseball is dropped from rest. At the same instant, another is thrown horizontally from the same height and follows a curved path. A stroboscope has captured the positions of the balls at fixed time intervals as they fall.

 

Figure 5. This shows the motions of two identical balls—one falls from rest, the other has an initial horizontal velocity. Each subsequent position is an equal time interval. Arrows represent horizontal and vertical velocities at each position. The ball on the right has an initial horizontal velocity, while the ball on the left has no horizontal velocity. Despite the difference in horizontal velocities, the vertical velocities and positions are identical for both balls. This shows that the vertical and horizontal motions are independent.

Instructor’s Note

 

This graphic displays this concept quite nicely; notice how both balls fall downward at the same speed at each point, even though one of the balls has a horizontal velocity. Basically, the velocity of the ball in the x-direction has no effect on the velocity in the y-direction, and vice-versa. This will be an important idea, especially when working with vectors.

It is remarkable that for each flash of the strobe, the vertical positions of the two balls are the same. This similarity implies that the vertical motion is independent of whether or not the ball is moving horizontally. (Assuming no air resistance, the vertical motion of a falling object is influenced by gravity only, and not by any horizontal forces.) Careful examination of the ball thrown horizontally shows that it travels the same horizontal distance between flashes. This is due to the fact that there are no additional forces on the ball in the horizontal direction after it is thrown. This result means that the horizontal velocity is constant, and affected neither by vertical motion nor by gravity (which is vertical). Note that this case is true only for ideal conditions. In the real world, air resistance will affect the speed of the balls in both directions.

The two-dimensional curved path of the horizontally thrown ball is composed of two independent one-dimensional motions (horizontal and vertical). The key to analyzing such motion, called projectile motion, is to resolve (break) it into motions along perpendicular directions. Resolving two-dimensional motion into perpendicular components is possible because the components are independent. We shall see how to resolve vectors in Vector Addition and Subtraction: Graphical Methods and Vector Addition and Subtraction: Analytical Methods. We will find such techniques to be useful in many areas of physics.

Section Summary

  • The shortest path between any two points is a straight line. In two dimensions, this path can be represented by a vector with horizontal and vertical components.
  • The horizontal and vertical components of a vector are independent of one another. Motion in the horizontal direction does not affect motion in the vertical direction, and vice versa.

Vector Addition and Subtraction: Graphical Methods

Instructor’s Note

 

Your Quiz would Cover

  • Given two graphical representations of vectors, be able to draw the sum or difference. There are some simple procedures to follow. Solidify your understanding of these procedures and we can work on why this makes sense in class
  • Describe both visually and mathematically what happens when a scalar is multiplied by a vector. If I give you a vector and a number, you should be able to turn the crank and multiply them mathematically. I am NOT expecting you to be able to do this graphically and will not ask you what it means. Just focus on the mechanics of how to do it.
  • Convert between magnitude/direction and component form for any vector
Figure 1. Displacement can be determined graphically using a scale map, such as this one of the Hawaiian Islands. A journey from Hawai’i to Moloka’i has a number of legs, or journey segments. These segments can be added graphically with a ruler to determine the total two-dimensional displacement of the journey. (credit: US Geological Survey)

Vectors in Two Dimensions

A vector is a quantity that has magnitude and direction. Displacement, velocity, acceleration, and force, for example, are all vectors. In one-dimensional, or straight-line, motion, the direction of a vector can be given simply by a plus or minus sign. In two dimensions (2-d), however, we specify the direction of a vector relative to some reference frame (i.e., coordinate system), using an arrow having length proportional to the vector’s magnitude and pointing in the direction of the vector.

Figure 2 shows such a graphical representation of a vector, using as an example the total displacement for the person walking in a city considered in Kinematics in Two Dimensions: An Introduction. We shall use the notation that a boldface symbol, such as \text{D}, stands for a vector. Its magnitude is represented by the symbol in italics, D, and its direction by \theta

Instructor’s Note

 

There’s some notation in the following note that would be useful to pay attention to.

Vectors in this Text

In this text, we will represent a vector with a boldface variable. For example, we will represent the quantity force with the vector \mathbf{F}, which has both magnitude and direction. The magnitude of the vector will be represented by a variable in italics, such as F, and the direction of the variable will be given by an angle \theta.

Figure 2. A person walks 9 blocks east and 5 blocks north. The displacement is 10.3 blocks at an angle 29.1^{\circ} north of east.

 

Figure 3. To describe the resultant vector for the person walking in a city considered in Figure graphically, draw an arrow to represent the total displacement vector \mathbf{D}. Using a protractor, draw a line at an angle \theta relative to the east-west axis. The length D of the arrow is proportional to the vector’s magnitude and is measured along the line with a ruler. In this example, the magnitude D of the vector is 10.3 units, and the direction \theta is 29.1^{\circ} north of east.

Instructor’s Note

 

Taking some time to understand and practice the head-to-tail method is recommended, you’ll notice that there’s a series of algorithmic steps, so you just need to learn the process, and it will work for any two vectors.

Vector Addition: Head-to-Tail Method

The head-to-tail method is a graphical way to add vectors, described in Figure 4 below and in the steps following. The tail[/pb_glossary] of the vector is the starting point of the vector, and the head (or tip) of a vector is the final, pointed end of the arrow.

Figure 4. Head-to-Tail Method: The head-to-tail method of graphically adding vectors is illustrated for the two displacements of the person walking in a city considered in Figure. (a) Draw a vector representing the displacement to the east. (b) Draw a vector representing the displacement to the north. The tail of this vector should originate from the head of the first, east-pointing vector. (c) Draw a line from the tail of the east-pointing vector to the head of the north-pointing vector to form the sum or resultant vector \mathbf{D}. The length of the arrow D is proportional to the vector’s magnitude and is measured to be 10.3 units . Its direction, described as the angle with respect to the east (or horizontal axis) \theta is measured with a protractor to be 29.1^{\circ}.

Step 1. Draw an arrow to represent the first vector (9 blocks to the east) using a ruler and protractor.

Figure 5.

Step 2. Now draw an arrow to represent the second vector (5 blocks to the north). Place the tail of the second vector at the head of the first vector.

Figure 6.

Step 3. If there are more than two vectors, continue this process for each vector to be added. Note that in our example, we have only two vectors, so we have finished placing arrows tip to tail.

Step 4. Draw an arrow from the tail of the first vector to the head of the last vector. This is the resultant, or the sum, of the other vectors.

Figure 7.

Step 5. To get the magnitude of the resultant, measure its length with a ruler. (Note that in most calculations, we will use the Pythagorean theorem to determine this length.)

Step 6. To get the direction of the resultant, measure the angle it makes with the reference frame using a protractor. (Note that in most calculations, we will use trigonometric relationships to determine this angle.)

The graphical addition of vectors is limited in accuracy only by the precision with which the drawings can be made and the precision of the measuring tools. It is valid for any number of vectors.

Adding Vectors Graphically using the Head-to-Tail Method: A Woman Takes a Walk

Use the graphical technique for adding vectors to find the total displacement of a person who walks the following three paths (displacements) on a flat field. First, she walks 25.0 m in a direction 49.0^{\circ} north of east. Then, she walks 23.0 m heading15.0^{\circ} north of east. Finally, she turns and walks 32.0 m in a direction 68.0^{\circ} south of east.

Strategy

Represent each displacement vector graphically with an arrow, labeling the first \mathbf{A}, the second \mathbf{B}, and the third \mathbf{C}, making the lengths proportional to the distance and the directions as specified relative to an east-west line. The head-to-tail method outlined above will give a way to determine the magnitude and direction of the resultant displacement, denoted \mathbf{R}

Solution

(1) Draw the three displacement vectors.

Figure 8.

(2) Place the vectors head to tail retaining both their initial magnitude and direction.

Figure 9.

(3) Draw the resultant vector, \mathbf{R}

Figure 10.

(4) Use a ruler to measure the magnitude of \mathbf{R}, and a protractor to measure the direction of \mathbf{R}. While the direction of the vector can be specified in many ways, the easiest way is to measure the angle between the vector and the nearest horizontal or vertical axis. Since the resultant vector is south of the eastward pointing axis, we flip the protractor upside down and measure the angle between the eastward axis and the vector.

Figure 11.

In this case, the total displacement \mathbf{R} is seen to have a magnitude of 50.0 m and to lie in a direction 7.0^{\circ} south of east. By using its magnitude and direction, this vector can be expressed as R = 50.0meters and \theta = 7.0^{\circ} south of east.

Discussion

The head-to-tail graphical method of vector addition works for any number of vectors. It is also important to note that the resultant is independent of the order in which the vectors are added. Therefore, we could add the vectors in any order as illustrated in Figure 12 and we will still get the same solution.

Figure 12.

Here, we see that when the same vectors are added in a different order, the result is the same. This characteristic is true in every case and is an important characteristic of vectors. Vector addition is commutative. Vectors can be added in any order.

\mathbf{A} + \mathbf{B} = \mathbf{B} + \mathbf{A}

(This is true for the addition of ordinary numbers as well—you get the same result whether you add 2+3 or 3+2 for example).

Play with Vectors

In the simulation below, choose “Explore 2-D” you can then

  • Move vectors \vec{a}, \vec{b}, and \vec{c} on to the grid
  • Change their magnitude and direction by clicking on the tip and dragging it around.
  • Manipulate the actual size using the values at the top.
  • Turn on using the menus on the right:
    • The components
    • Angles
    • Values
  • Have the simulation draw the sum for you checking the tip-to-tail method

An interactive or media element has been excluded from this version of the text. You can view it online here:
http://openbooks.library.umass.edu/toggerson-132/?p=241

 

Vector Subtraction

Instructor’s Note

 

Understanding vector subtraction is necessary to understand other physics ideas. For example, acceleration is \frac {\Delta v}{\Delta t}, and \Delta v is v_f - v_i. Velocity is a vector, so you’re looking at a vector subtraction whenever you’re working with acceleration.

Vector subtraction is a straightforward extension of vector addition. To define subtraction (say we want to subtract \mathbf{B} from\mathbf{A}, written \mathbf{A} - \mathbf{B} , we must first define what we mean by subtraction. The negative of a vector \mathbf{B} is defined to be - \mathbf{B}; that is, graphically the negative of any vector has the same magnitude but the opposite direction, as shown in Figure 13. In other words, \mathbf{B} has the same length as - \mathbf{B}, but points in the opposite direction. Essentially, we just flip the vector so it points in the opposite direction.

Figure 13. The negative of a vector is just another vector of the same magnitude but pointing in the opposite direction. So \mathbf{B} is the negative of \mathbf{B}; it has the same length but opposite direction.

The subtraction of vector \mathbf{B} from vector \mathbf{A} is then simply defined to be the addition of - \mathbf{B} to \mathbf{A}. Note that vector subtraction is the addition of a negative vector. The order of subtraction does not affect the results.

\mathbf{A} - \mathbf{B} = \mathbf{A} + ( - \mathbf{B} ).

This is analogous to the subtraction of scalars (where, for example, 5 - 2 = 5 + (- 2)Again, the result is independent of the order in which the subtraction is made. When vectors are subtracted graphically, the techniques outlined above are used, as the following example illustrates.

Subtracting Vectors Graphically: A Woman Sailing a Boat

A woman sailing a boat at night is following directions to a dock. The instructions read to first sail 27.5 m in a direction 66.0^{\circ} north of east from her current location, and then travel 30.0 m in a direction 112^{\circ} north of east (or 22.0^{\circ} west of north). If the woman makes a mistake and travels in the opposite direction for the second leg of the trip, where will she end up? Compare this location with the location of the dock.

Figure 14.

Strategy

We can represent the first leg of the trip with a vector \mathbf{A}, and the second leg of the trip with a vector \mathbf{A}. The dock is located at a location \mathbf{A} + \mathbf{B}. If the woman mistakenly travels in the opposite direction for the second leg of the journey, she will travel a distance \mathbf{A} (30.0 m) in the direction 180^{\circ} - 112^{\circ} = 68^{\circ} south of east. We represent this as - \mathbf{B}, as shown below. The vector - \mathbf{B} has the same magnitude as \mathbf{B} but is in the opposite direction. Thus, she will end up at a location \mathbf{A} + ( - \mathbf{B} ) or \mathbf{A} - \mathbf{B}.

Figure 15.

We will perform vector addition to compare the location of the dock, \mathbf{A} + \mathbf{B}, with the location at which the woman mistakenly arrives, \mathbf{A} + ( - \mathbf{B} ).

Solution

(1) To determine the location at which the woman arrives by accident, draw vectors \mathbf{A} and - \mathbf{B}.

(2) Place the vectors head to tail.

(3) Draw the resultant vector \mathbf{R}.

(4) Use a ruler and protractor to measure the magnitude and direction of \mathbf{R}.

Figure 16.

In this case, R = 23.0 \text{m} and image = 7.5^{\circ} " title="Rendered by QuickLaTeX.com" height="16" width="488" style="vertical-align: -3px;"> south of east.

(5) To determine the location of the dock, we repeat this method to add vectors\mathbf{A}and \mathbf{B}. We obtain the resultant vector \mathbf{R}':

Figure 17.

In this case R = 52.9 \text{m} and image = 90.1^{\circ} " title="Rendered by QuickLaTeX.com" height="15" width="496" style="vertical-align: -3px;"> north of east.

We can see that the woman will end up a significant distance from the dock if she travels in the opposite direction for the second leg of the trip.

Discussion

Because subtraction of a vector is the same as addition of a vector with the opposite direction, the graphical method of subtracting vectors works the same as for addition.

Instructor’s Note

 

If you’ve taken another physics course, you’ve probably seen \mathbf{F} = m \mathbf{a}. This equation will play a significant role in this class, and you’ll notice that mass is a scalar, and acceleration is a vector, so understanding how scalars and vectors multiply will be important.

Multiplication of Vectors and Scalars

If we decided to walk three times as far on the first leg of the trip considered in the preceding example, then we would walk 3 \times 27.5 \text{m} or 82.5 m, in a direction 66^\circ north of east. This is an example of multiplying a vector by a positive scalar. Notice that the magnitude changes, but the direction stays the same.

If the scalar is negative, then multiplying a vector by it changes the vector’s magnitude and gives the new vector the opposite direction. For example, if you multiply by –2, the magnitude doubles but the direction changes. We can summarize these rules in the following way: When vector \mathbf{A} is multiplied by a scalar c,

In our case, c = 3 and A = 27.5 \text{m}. Vectors are multiplied by scalars in many situations. Note that division is the inverse of multiplication. For example, dividing by 2 is the same as multiplying by the value \frac{1}{2}. The rules for multiplication of vectors by scalars are the same for division; simply treat the divisor as a scalar between 0 and 1.

Instructor’s Note

 

When dealing with vectors using analytic methods (which is covered in the next section), you need to break down vectors into essentially x-components and y-components. This next part covers this idea, so try to familiarize yourself with breaking down vectors as you read.

Resolving a Vector into Components

In the examples above, we have been adding vectors to determine the resultant vector. In many cases, however, we will need to do the opposite. We will need to take a single vector and find what other vectors added together produce it. In most cases, this involves determining the perpendicular components of a single vector, for example the x and y-components, or the north-south and east-west components.

For example, we may know that the total displacement of a person walking in a city is 10.3 blocks in a direction 29.0^{\circ} north of east and want to find out how many blocks east and north had to be walked. This method is called finding the components (or parts) of the displacement in the east and north directions, and it is the inverse of the process followed to find the total displacement. It is one example of finding the components of a vector. There are many applications in physics where this is a useful thing to do. We will see this soon in Projectile Motion, and much more when we cover forces in Dynamics: Newton’s Laws of Motion. Most of these involve finding components along perpendicular axes (such as north and east), so that right triangles are involved. The analytical techniques presented in Vector Addition and Subtraction: Analytical Methods are ideal for finding vector components.

Section Summary

  • The graphical method of adding vectors \mathbf{A} and \mathbf{B} involves drawing vectors on a graph and adding them using the head-to-tail method. The resultant vector \mathbf{R}is defined such that \mathbf{A} + \mathbf{B} = \mathbf{R}. The magnitude and direction of \mathbf{R} are then determined with a ruler and protractor, respectively.
  • The graphical method of subtracting vector \mathbf{B} from \mathbf{A} involves adding the opposite of vector \mathbf{B}, which is defined as - \mathbf{B}. In this case, A–B=A+(–B)=RA–B=A+(–B)=R">A–B=A+(–B)=RA–B=A+(–B)=R">AB=A+(–B)=R. Then, the head-to-tail method of addition is followed in the usual way to obtain the resultant vector \mathbf{R}
  • Addition of vectors is commutative such that \text{A} + \text{B} = \text{A} + \text{B}
  • The head-to-tail method of adding vectors involves drawing the first vector on a graph and then placing the tail of each subsequent vector at the head of the previous vector. The resultant vector is then drawn from the tail of the first vector to the head of the final vector.
  • If a vector \mathbf{A} is multiplied by a scalar quantity c, the magnitude of the product is given by \lvert c \cdot \mathbf{A} \rvert. If c is positive, the direction of the product points in the same direction as \mathbf{A} ; if c is negative, the direction of the product points in the opposite direction as \mathbf{A}

Vector Addition and Subtraction: Analytical Methods

Instructor’s Note

 

Your Quiz would Cover
  • Adding vectors by components. Don’t focus too much on what it means to add vectors. Just learn the mechanics of how to do it. We will talk about the meaning in class.

Analytical methods of vector addition and subtraction employ geometry and simple trigonometry rather than the ruler and protractor of graphical methods. Part of the graphical technique is retained, because vectors are still represented by arrows for easy visualization. However, analytical methods are more concise, accurate, and precise than graphical methods, which are limited by the accuracy with which a drawing can be made. Analytical methods are limited only by the accuracy and precision with which physical quantities are known.

Resolving a Vector into Perpendicular Components

Analytical techniques and right triangles go hand-in-hand in physics because (among other things) motions along perpendicular directions are independent. We very often need to separate a vector into perpendicular components. For example, given a vector like \mathbf{A}in Figure 1, we may wish to find which two perpendicular vectors, \mathbf{A}_x and \mathbf{A}_y, add to produce it.

Figure 1. The vector A, with its tail at the origin of an x, y-coordinate system, is shown together with its x- and y-components, Ax and Ay. These vectors form a right triangle. The analytical relationships among these vectors are summarized below.

\mathbf{A}_x and \mathbf{A}_y, are defined to be the components of \mathbf{A} along the x– and y-axes. The three vectors \mathbf{A}, \mathbf{A}_x and \mathbf{A}_y, form a right triangle:

\mathbf{A}_x + \mathbf{A}_y = \mathbf{A}

Note that this relationship between vector components and the resultant vector holds only for vector quantities (which include both magnitude and direction). The relationship does not apply for the magnitudes alone. For example, if \mathbf{A}_x = 3 \text{m}  east, \mathbf{A}_y = 4 \text{m} north, and \mathbf{A} = 5 \text{m} north-east, then it is true that the vectors \mathbf{A}_x + \mathbf{A}_y = \mathbf{A}. However, it is not true that the sum of the magnitudes of the vectors is also equal. That is,

3 \text{m} + 4 \text{m} \neq 5 \text{m}

Thus,

A_x + A_y \neq A

If the vector \mathbf{A} is known, then its magnitude A(its length) and its angle \theta (its direction) are known. To find A_x and A_y, its x– and y-components, we use the following relationships for a right triangle.

A_x = A \cos \theta

and

A_y = A \sin \theta.

Figure 2. The magnitudes of the vector components A_x and A_y can be related to the resultant vector \mathbf{A} and the angle \theta with trigonometric identities. Here we see that A_x = A \cos \theta and A_y = A \sin \theta.

Suppose, for example, that \mathbf{A} is the vector representing the total displacement of the person walking in a city considered in Kinematics in Two Dimensions: An Introduction and Vector Addition and Subtraction: Graphical Methods.

Figure 3. We can use the relationships A_x = A \cos \theta and A_y = A \sin \theta to determine the magnitude of the horizontal and vertical component vectors in this example.

Then A = 10.3blocks and \theta = 29.6^{\circ}, so that

A_x = A \cos{\theta} = (10.3 \text{ blocks})(\cos{29.1 ^{\circ}}) = 9.0 \text{ blocks}

A_y = A \sin{\theta} = (10.3 \text{ blocks})(\cos{29.1 ^{\circ}}) = 5.0 \text{ blocks}

Calculating a Resultant Vector

If the perpendicular components \mathbf{A}_x and \mathbf{A}_yAyAy size 12{A rSub { size 8{y} } } {}"> of a vector are known, then \mathbf{A}AA size 12{A} {}"> AA size 12{A} {}"> can also be found analytically. To find the magnitude AAA size 12{A} {}"> and direction \theta of a vector from its perpendicular components \mathbf{A}_x and \mathbf{A}_y, we use the following relationships:θ=tan−1(Ay/Ax).θ=tan−1(Ay/Ax). size 12{θ="tan" rSup { size 8{ - 1} } \( A rSub { size 8{y} } /A rSub { size 8{x} } \) } {}">

A = \sqrt{A_x ^2 + A_y ^2 }

\theta = \tan^{-1} \frac{A_y}{A_x}.

Figure 4. The magnitude and direction of the resultant vector can be determined once the horizontal and vertical components \mathbf{A}_x and \mathbf{A}_yAyAy size 12{A rSub { size 8{y} } } {}">  have been determined.

Note that the equation A = \sqrt{A_x ^2 + A_y ^2 } is just the Pythagorean theorem relating the legs of a right triangle to the length of the hypotenuse. For example, if AxAx size 12{A rSub { size 8{x} } } {}"> A_x and A_y are 9 and 5 blocks, respectively, then A = \sqrt{9^2 + 5^2} = 10.3 \text{blocks}, again consistent with the example of the person walking in a city. Finally, the direction is \theta = \tan^{-1} \frac{5}{9} = 29.1^{\circ}, as before.

Determining Vectors and Vector Components with Analytical Methods

Equations A_x = A \cos \theta and A_y = A \sin \theta  are used to find the perpendicular components of a vector—that is, to go from \mathbf{A} and \theta to \mathbf{A}_x and \mathbf{A}_yAyAy size 12{A rSub { size 8{y} } } {}"> . Equations A = \sqrt{A_x ^2 + A_y ^2 } and \theta = \tan^{-1} \frac{A_y}{A_x} are used to find a vector from its perpendicular components—that is, to go from \mathbf{A}_x and \mathbf{A}_y to \mathbf{A} and \theta. Both processes are crucial to analytical methods of vector addition and subtraction.

Instructor’s Note

 

Now that you know how to break down vectors into components, here’s a procedure to adding vectors analytically. There’s some trigonometry involved, so, again, if you’re not familiar or comfortable with trigonometry, come see your instructor. You should be familiar with both methods. you should be able to add two vectors given their x and y components, and you should be able to draw the resulting vector of two added vectors. Also, we will go over how to use these to solve problems, so focus primarily on the methods of adding vectors.

Adding Vectors Using Analytical Methods

To see how to add vectors using perpendicular components, consider Figure 5, in which the vectors \mathbf{A} and \mathbf{B} are added to produce the resultant \mathbf{R}.

 

Figure 5. Vectors \mathbf{A} and \mathbf{B} are two legs of a walk, and \mathbf{R} is the resultant or total displacement. You can use a