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University Physics Volume 3
SENIOR CONTRIBUTING AUTHORS
SAMUEL J. LING, TRUMAN STATE UNIVERSITY JEFF SANNY, LOYOLA MARYMOUNT UNIVERSITY WILLIAM MOEBS, FORMERLY OF LOYOLA MARYMOUNT UNIVERSITY

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Contents
Preface 1
UNIT 1 OPTICS
CHAPTER 1
The Nature of Light 5
Introduction 5 1.1 The Propagation of Light 6 1.2 The Law of Reflection 10 1.3 Refraction 13 1.4 Total Internal Reflection 17 1.5 Dispersion 22 1.6 Huygens’s Principle 26 1.7 Polarization 30 Chapter Review 41
CHAPTER 2
Geometric Optics and Image Formation 51
Introduction 51 2.1 Images Formed by Plane Mirrors 52 2.2 Spherical Mirrors 54 2.3 Images Formed by Refraction 65 2.4 Thin Lenses 68 2.5 The Eye 79 2.6 The Camera 86 2.7 The Simple Magnifier 88 2.8 Microscopes and Telescopes 91 Chapter Review 100
CHAPTER 3
Interference 113
Introduction 113 3.1 Young's Double-Slit Interference 113 3.2 Mathematics of Interference 117 3.3 Multiple-Slit Interference 120 3.4 Interference in Thin Films 122 3.5 The Michelson Interferometer 127 Chapter Review 132
CHAPTER 4
Diffraction 141
Introduction 141 4.1 Single-Slit Diffraction 142 4.2 Intensity in Single-Slit Diffraction 146

4.3 Double-Slit Diffraction 151 4.4 Diffraction Gratings 153 4.5 Circular Apertures and Resolution 158 4.6 X-Ray Diffraction 164 4.7 Holography 166 Chapter Review 169
UNIT 2 MODERN PHYSICS

CHAPTER 5
Relativity 179
Introduction 179 5.1 Invariance of Physical Laws 180 5.2 Relativity of Simultaneity 182 5.3 Time Dilation 185 5.4 Length Contraction 194 5.5 The Lorentz Transformation 199 5.6 Relativistic Velocity Transformation 209 5.7 Doppler Effect for Light 213 5.8 Relativistic Momentum 216 5.9 Relativistic Energy 218 Chapter Review 227

CHAPTER 6
Photons and Matter Waves
Introduction 239 6.1 Blackbody Radiation 240 6.2 Photoelectric Effect 248 6.3 The Compton Effect 254 6.4 Bohr’s Model of the Hydrogen Atom 6.5 De Broglie’s Matter Waves 268 6.6 Wave-Particle Duality 275 Chapter Review 282

239 259

CHAPTER 7
Quantum Mechanics 295
Introduction 295 7.1 Wave Functions 296 7.2 The Heisenberg Uncertainty Principle 307 7.3 The Schrӧdinger Equation 311 7.4 The Quantum Particle in a Box 314 7.5 The Quantum Harmonic Oscillator 320 7.6 The Quantum Tunneling of Particles through Potential Barriers 326 Chapter Review 338

CHAPTER 8
Atomic Structure 347
Introduction 347
8.1 The Hydrogen Atom 348

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8.2 Orbital Magnetic Dipole Moment of the Electron 8.3 Electron Spin 361 8.4 The Exclusion Principle and the Periodic Table 8.5 Atomic Spectra and X-rays 371 8.6 Lasers 381 Chapter Review 385

357 366

CHAPTER 9
Condensed Matter Physics 393
Introduction 393 9.1 Types of Molecular Bonds 394 9.2 Molecular Spectra 399 9.3 Bonding in Crystalline Solids 402 9.4 Free Electron Model of Metals 409 9.5 Band Theory of Solids 414 9.6 Semiconductors and Doping 417 9.7 Semiconductor Devices 420 9.8 Superconductivity 426 Chapter Review 433

CHAPTER 10
Nuclear Physics 441
Introduction 441 10.1 Properties of Nuclei 442 10.2 Nuclear Binding Energy 447 10.3 Radioactive Decay 450 10.4 Nuclear Reactions 456 10.5 Fission 464 10.6 Nuclear Fusion 470 10.7 Medical Applications and Biological Effects of Nuclear Radiation 475 Chapter Review 484

CHAPTER 11
Particle Physics and Cosmology 493
Introduction 493 11.1 Introduction to Particle Physics 494 11.2 Particle Conservation Laws 499 11.3 Quarks 504 11.4 Particle Accelerators and Detectors 508 11.5 The Standard Model 516 11.6 The Big Bang 521 11.7 Evolution of the Early Universe 525 Chapter Review 533

Appendix A Units 541 Appendix B Conversion Factors 545 Appendix C Fundamental Constants 549

Appendix D Astronomical Data 551 Appendix E Mathematical Formulas 553 Appendix F Chemistry 557 Appendix G The Greek Alphabet 559 Answer Key 560 Index 593
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Preface

1

PREFACE

Welcome to University Physics, an OpenStax resource. This textbook was written to increase student access to high-quality learning materials, maintaining highest standards of academic rigor at little to no cost.
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About University Physics
University Physics is designed for the two- or threesemester calculus-based physics course. The text has been developed to meet the scope and sequence of most university physics courses and provides a foundation for a career in mathematics, science, or engineering. The book provides an important opportunity for students to learn the core concepts of physics and understand how those concepts apply to their lives and to the world around them.
Due to the comprehensive nature of the material, we are offering the book in three volumes for flexibility and efficiency.
Coverage and scope
Our University Physics textbook adheres to the scope and sequence of most two- and threesemester physics courses nationwide. We have worked to make physics interesting and accessible to students while maintaining the mathematical rigor inherent in the subject. With this objective in mind, the content of this textbook has been developed and arranged to provide a logical progression from fundamental to more advanced concepts, building upon what students have already learned and emphasizing connections between topics and between theory and applications. The goal of each section is to enable students not just to recognize concepts, but to work with them in ways that will be useful in later courses and future careers. The organization and pedagogical features were developed and vetted with feedback from science educators dedicated to the project.
VOLUME I
Unit 1: Mechanics

2

Preface

• Chapter 1: Units and Measurement • Chapter 2: Vectors • Chapter 3: Motion Along a Straight Line • Chapter 4: Motion in Two and Three Dimensions • Chapter 5: Newton’s Laws of Motion • Chapter 6: Applications of Newton’s Laws • Chapter 7: Work and Kinetic Energy • Chapter 8: Potential Energy and Conservation of
Energy • Chapter 9: Linear Momentum and Collisions • Chapter 10: Fixed-Axis Rotation • Chapter 11: Angular Momentum • Chapter 12: Static Equilibrium and Elasticity • Chapter 13: Gravitation • Chapter 14: Fluid Mechanics
Unit 2: Waves and Acoustics
• Chapter 15: Oscillations • Chapter 16: Waves • Chapter 17: Sound
VOLUME II
Unit 1: Thermodynamics
• Chapter 1: Temperature and Heat • Chapter 2: The Kinetic Theory of Gases • Chapter 3: The First Law of Thermodynamics • Chapter 4: The Second Law of Thermodynamics
Unit 2: Electricity and Magnetism
• Chapter 5: Electric Charges and Fields • Chapter 6: Gauss’s Law • Chapter 7: Electric Potential • Chapter 8: Capacitance • Chapter 9: Current and Resistance • Chapter 10: Direct-Current Circuits • Chapter 11: Magnetic Forces and Fields • Chapter 12: Sources of Magnetic Fields • Chapter 13: Electromagnetic Induction • Chapter 14: Inductance • Chapter 15: Alternating-Current Circuits • Chapter 16: Electromagnetic Waves
VOLUME III
Unit 1: Optics
• Chapter 1: The Nature of Light • Chapter 2: Geometric Optics and Image
Formation • Chapter 3: Interference • Chapter 4: Diffraction
Unit 2: Modern Physics
• Chapter 5: Relativity

• Chapter 6: Photons and Matter Waves • Chapter 7: Quantum Mechanics • Chapter 8: Atomic Structure • Chapter 9: Condensed Matter Physics • Chapter 10: Nuclear Physics • Chapter 11: Particle Physics and Cosmology
Pedagogical foundation
Throughout University Physics you will find derivations of concepts that present classical ideas and techniques, as well as modern applications and methods. Most chapters start with observations or experiments that place the material in a context of physical experience. Presentations and explanations rely on years of classroom experience on the part of long-time physics professors, striving for a balance of clarity and rigor that has proven successful with their students. Throughout the text, links enable students to review earlier material and then return to the present discussion, reinforcing connections between topics. Key historical figures and experiments are discussed in the main text (rather than in boxes or sidebars), maintaining a focus on the development of physical intuition. Key ideas, definitions, and equations are highlighted in the text and listed in summary form at the end of each chapter. Examples and chapter-opening images often include contemporary applications from daily life or modern science and engineering that students can relate to, from smart phones to the internet to GPS devices.
Assessments that reinforce key concepts
In-chapter Examples generally follow a three-part format of Strategy, Solution, and Significance to emphasize how to approach a problem, how to work with the equations, and how to check and generalize the result. Examples are often followed by Check Your Understanding questions and answers to help reinforce for students the important ideas of the examples. Problem-Solving Strategies in each chapter break down methods of approaching various types of problems into steps students can follow for guidance. The book also includes exercises at the end of each chapter so students can practice what they’ve learned.
• Conceptual questions do not require calculation but test student learning of the key concepts.
• Problems categorized by section test student problem-solving skills and the ability to apply ideas to practical situations.
• Additional Problems apply knowledge across the chapter, forcing students to identify what

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Preface

3

concepts and equations are appropriate for solving given problems. Randomly located throughout the problems are Unreasonable Results exercises that ask students to evaluate the answer to a problem and explain why it is not reasonable and what assumptions made might not be correct. • Challenge Problems extend text ideas to interesting but difficult situations.
Answers for selected exercises are available in an Answer Key at the end of the book.
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About the authors
Senior contributing authors
Samuel J. Ling, Truman State University

Dr. Samuel Ling has taught introductory and advanced physics for over 25 years at Truman State University, where he is currently Professor of Physics and the Department Chair. Dr. Ling has two PhDs from Boston University, one in Chemistry and the other in Physics, and he was a Research Fellow at the Indian Institute of Science, Bangalore, before joining Truman. Dr. Ling is also an author of A First Course in Vibrations and Waves, published by Oxford University Press. Dr. Ling has considerable experience with research in Physics Education and has published research on collaborative learning methods in physics teaching. He was awarded a Truman Fellow and a Jepson fellow in recognition of his innovative teaching methods. Dr. Ling’s research publications have spanned Cosmology, Solid State Physics, and Nonlinear Optics.
Jeff Sanny, Loyola Marymount University Dr. Jeff Sanny earned a BS in Physics from Harvey Mudd College in 1974 and a PhD in Solid State Physics from the University of California–Los Angeles in 1980. He joined the faculty at Loyola Marymount University in the fall of 1980. During his tenure, he has served as department Chair as well as Associate Dean. Dr. Sanny enjoys teaching introductory physics in particular. He is also passionate about providing students with research experience and has directed an active undergraduate student research group in space physics for many years.
William Moebs, Formerly of Loyola Marymount University Dr. William Moebs earned a BS and PhD (1959 and 1965) from the University of Michigan. He then joined their staff as a Research Associate for one year, where he continued his doctoral research in particle physics. In 1966, he accepted an appointment to the Physics Department of Indiana Purdue Fort Wayne (IPFW), where he served as Department Chair from 1971 to 1979. In 1979, he moved to Loyola Marymount University (LMU), where he served as Chair of the Physics Department from 1979 to 1986. He retired from LMU in 2000. He has published research in particle physics, chemical kinetics, cell division, atomic physics, and physics teaching.
Contributing authors
Stephen D. Druger, Northwestern University Alice Kolakowska, University of Memphis David Anderson, Albion College Daniel Bowman, Ferrum College Dedra Demaree, Georgetown University

4

Preface

Edw. S. Ginsberg, University of Massachusetts Joseph Trout, Richard Stockton College Kevin Wheelock, Bellevue College David Smith, University of the Virgin Islands Takashi Sato, Kwantlen Polytechnic University Gerald Friedman, Santa Fe Community College Lev Gasparov, University of North Florida Lee LaRue, Paris Junior College Mark Lattery, University of Wisconsin Richard Ludlow, Daniel Webster College Patrick Motl, Indiana University Kokomo Tao Pang, University of Nevada, Las Vegas Kenneth Podolak, Plattsburgh State University
Reviewers
Salameh Ahmad, Rochester Institute of Technology–Dubai John Aiken, University of Colorado–Boulder Raymond Benge, Terrant County College Gavin Buxton, Robert Morris University Erik Christensen, South Florida State College Clifton Clark, Fort Hays State University Nelson Coates, California Maritime Academy Herve Collin, Kapi’olani Community College Carl Covatto, Arizona State University Alejandro Cozzani, Imperial Valley College Danielle Dalafave, The College of New Jersey Nicholas Darnton, Georgia Institute of Technology Ethan Deneault, University of Tampa Kenneth DeNisco, Harrisburg Area Community College Robert Edmonds, Tarrant County College William Falls, Erie Community College Stanley Forrester, Broward College Umesh Garg, University of Notre Dame Maurizio Giannotti, Barry University

Bryan Gibbs, Dallas County Community College Lynn Gillette, Pima Community College–West Campus Mark Giroux, East Tennessee State University Matthew Griffiths, University of New Haven Alfonso Hinojosa, University of Texas–Arlington Steuard Jensen, Alma College David Kagan, University of Massachusetts Sergei Katsev, University of Minnesota–Duluth Gregory Lapicki, East Carolina University Jill Leggett, Florida State College–Jacksonville Alfredo Louro, University of Calgary James Maclaren, Tulane University Ponn Maheswaranathan, Winthrop University Seth Major, Hamilton College Oleg Maksimov, Excelsior College Aristides Marcano, Delaware State University James McDonald, University of Hartford Ralph McGrew, SUNY–Broome Community College Paul Miller, West Virginia University Tamar More, University of Portland Farzaneh Najmabadi, University of Phoenix Richard Olenick, The University of Dallas Christopher Porter, Ohio State University Liza Pujji, Manakau Institute of Technology Baishali Ray, Young Harris University Andrew Robinson, Carleton University Aruvana Roy, Young Harris University Gajendra Tulsian, Daytona State College Adria Updike, Roger Williams University Clark Vangilder, Central Arizona University Steven Wolf, Texas State University Alexander Wurm, Western New England University Lei Zhang, Winston Salem State University Ulrich Zurcher, Cleveland State University

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CHAPTER 1
The Nature of Light
Figure 1.1 Due to total internal reflection, an underwater swimmer’s image is reflected back into the water where the camera is located. The circular ripple in the image center is actually on the water surface. Due to the viewing angle, total internal reflection is not occurring at the top edge of this image, and we can see a view of activities on the pool deck. (credit: modification of work by “jayhem”/Flickr)
Chapter Outline
1.1 The Propagation of Light 1.2 The Law of Reflection 1.3 Refraction 1.4 Total Internal Reflection 1.5 Dispersion 1.6 Huygens’s Principle 1.7 Polarization INTRODUCTION Our investigation of light revolves around two questions of fundamental importance: (1) What is the nature of light, and (2) how does light behave under various circumstances? Answers to these questions can be found in Maxwell’s equations (in Electromagnetic Waves), which predict the existence of electromagnetic waves and their behavior. Examples of light include radio and infrared waves, visible light, ultraviolet radiation, and X-rays. Interestingly, not all light phenomena can be explained by Maxwell’s theory.

6

1 • The Nature of Light

Experiments performed early in the twentieth century showed that light has corpuscular, or particle-like, properties. The idea that light can display both wave and particle characteristics is called wave-particle duality, which is examined in Photons and Matter Waves.

In this chapter, we study the basic properties of light. In the next few chapters, we investigate the behavior of light when it interacts with optical devices such as mirrors, lenses, and apertures.
1.1 The Propagation of Light
Learning Objectives By the end of this section, you will be able to:
• Determine the index of refraction, given the speed of light in a medium • List the ways in which light travels from a source to another location

The speed of light in a vacuum c is one of the fundamental constants of physics. As you will see when you reach Relativity, it is a central concept in Einstein’s theory of relativity. As the accuracy of the measurements of the speed of light improved, it was found that different observers, even those moving at large velocities with respect to each other, measure the same value for the speed of light. However, the speed of light does vary in a precise manner with the material it traverses. These facts have far-reaching implications, as we will see in later chapters.

The Speed of Light: Early Measurements

The first measurement of the speed of light was made by the Danish astronomer Ole Roemer (1644–1710) in 1675. He studied the orbit of Io, one of the four large moons of Jupiter, and found that it had a period of revolution of 42.5 h around Jupiter. He also discovered that this value fluctuated by a few seconds, depending on the position of Earth in its orbit around the Sun. Roemer realized that this fluctuation was due to the finite speed of light and could be used to determine c.

Roemer found the period of revolution of Io by measuring the time interval between successive eclipses by

Jupiter. Figure 1.2(a) shows the planetary configurations when such a measurement is made from Earth in the

part of its orbit where it is receding from Jupiter. When Earth is at point A, Earth, Jupiter, and Io are aligned.

The next time this alignment occurs, Earth is at point B, and the light carrying that information to Earth must

travel to that point. Since B is farther from Jupiter than A, light takes more time to reach Earth when Earth is at

B. Now imagine it is about 6 months later, and the planets are arranged as in part (b) of the figure. The

measurement of Io’s period begins with Earth at point and Io eclipsed by Jupiter. The next eclipse then

occurs when Earth is at point , to which the light carrying the information of this eclipse must travel. Since

is closer to Jupiter than , light takes less time to reach Earth when it is at . This time interval between

the successive eclipses of Io seen at and is therefore less than the time interval between the eclipses seen

at A and B. By measuring the difference in these time intervals and with appropriate knowledge of the distance

between Jupiter and Earth, Roemer calculated that the speed of light was

which is 33% below

the value accepted today.

Figure 1.2 Roemer’s astronomical method for determining the speed of light. Measurements of Io’s period done with the configurations of parts (a) and (b) differ, because the light path length and associated travel time increase from A to B (a) but decrease from to (b).
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1.1 • The Propagation of Light

7

The first successful terrestrial measurement of the speed of light was made by Armand Fizeau (1819–1896) in

1849. He placed a toothed wheel that could be rotated very rapidly on one hilltop and a mirror on a second

hilltop 8 km away (Figure 1.3). An intense light source was placed behind the wheel, so that when the wheel

rotated, it chopped the light beam into a succession of pulses. The speed of the wheel was then adjusted until

no light returned to the observer located behind the wheel. This could only happen if the wheel rotated

through an angle corresponding to a displacement of

teeth, while the pulses traveled down to the

mirror and back. Knowing the rotational speed of the wheel, the number of teeth on the wheel, and the

distance to the mirror, Fizeau determined the speed of light to be

which is only 5% too high.

Figure 1.3 Fizeau’s method for measuring the speed of light. The teeth of the wheel block the reflected light upon return when the wheel is rotated at a rate that matches the light travel time to and from the mirror.

The French physicist Jean Bernard Léon Foucault (1819–1868) modified Fizeau’s apparatus by replacing the

toothed wheel with a rotating mirror. In 1862, he measured the speed of light to be

which is

within 0.6% of the presently accepted value. Albert Michelson (1852–1931) also used Foucault’s method on

several occasions to measure the speed of light. His first experiments were performed in 1878; by 1926, he

had refined the technique so well that he found c to be

Today, the speed of light is 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 value

1.1

where the approximate value of

is used whenever three-digit accuracy is sufficient.

Speed of Light in Matter

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 varies with different atoms, crystal lattices, and other substructures. We can define a constant of a material that describes the speed of light in it, called the index of refraction n:

1.2

where v is the observed speed of light in the material.

Since the speed of light is always less than c in matter and equals c only in a vacuum, the index of refraction is

always greater than or equal to one; that is,

. Table 1.1 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 separated by a prism, as we will see in

8

1 • The Nature of Light

Dispersion.) Note that for gases, n is close to 1.0. This seems reasonable, since atoms in gases are widely

separated, and light travels at c in the vacuum between atoms. It is common to take

for gases unless

great precision is needed. Although the speed of light v in a medium varies considerably from its value c in a

vacuum, it is still a large speed.

Medium

n

Gases at , 1 atm

Air

1.000293

Carbon dioxide

1.00045

Hydrogen

1.000139

Oxygen

1.000271

Liquids at

Benzene

1.501

Carbon disulfide

1.628

Carbon tetrachloride 1.461

Ethanol

1.361

Glycerine

1.473

Water, fresh

1.333

Solids at

Diamond

2.419

Fluorite

1.434

Glass, crown

1.52

Glass, flint

1.66

Ice (at

1.309

Polystyrene

1.49

Plexiglas

1.51

Quartz, crystalline 1.544

Quartz, fused

1.458

Sodium chloride

1.544

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1.1 • The Propagation of Light

9

Medium

n

Zircon

1.923

Table 1.1 Index of Refraction in Various Media For light with a wavelength of 589 nm in a vacuum

EXAMPLE 1.1

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

Strategy We can calculate the speed of light in a material v from the index of refraction n of the material, using the equation

Solution Rearranging the equation

for v gives us

The index of refraction for zircon is given as 1.923 in Table 1.1, and c is given in Equation 1.1. Entering these values in the equation gives

Significance 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 1.1 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.
CHECK YOUR UNDERSTANDING 1.1
Table 1.1 shows that ethanol and fresh water have very similar indices of refraction. By what percentage do the speeds of light in these liquids differ?
The Ray Model of Light
You have already studied some of the wave characteristics of light in the previous chapter on Electromagnetic Waves. In this chapter, we start mainly with the ray characteristics. There are three ways in which light can travel from a source to another location (Figure 1.4). 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 observer. Light can also arrive after being reflected, such as by a mirror. In all of these cases, we can model the path of light as a straight line called a ray.

10

1 • The Nature of Light

Figure 1.4 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 by traveling through media like air and glass. (c) Light 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 show that when light interacts with an object several times larger than 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 visible light is less than a micron (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 visible light encounters anything large enough that we can observe it with unaided eyes, such as a coin, it acts like a ray, with generally negligible wave characteristics.
In all of these cases, we can model the path of light as straight lines. 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. The ray model of light describes the path of light as straight lines.
Since light moves in straight lines, changing directions when it interacts with materials, its path is described by geometry and simple trigonometry. This part of optics, where the ray aspect of light dominates, is therefore called geometric optics. Two laws 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. We will examine more about each of these laws in upcoming sections of this chapter.
1.2 The Law of Reflection
Learning Objectives By the end of this section, you will be able to:
• Explain the reflection of light from polished and rough surfaces • Describe the principle and applications of corner reflectors
Whenever we look into a mirror, or squint at sunlight glinting from a lake, we are seeing a reflection. When you look at a piece of white paper, you are seeing light scattered from it. Large telescopes use reflection to form an image of stars and other astronomical objects.
The law of reflection states that the angle of reflection equals the angle of incidence, or
1.3
The law of reflection is illustrated in Figure 1.5, which also shows how the angle of incidence and angle of reflection are measured relative to the perpendicular to the surface at the point where the light ray strikes.
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1.2 • The Law of Reflection

11

Figure 1.5 The law of reflection states that the angle of reflection equals the angle of incidence— relative to the perpendicular to the surface at the point where the ray strikes the surface.

The angles are measured

We expect to see reflections from smooth surfaces, but Figure 1.6 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 shown in Figure 1.7(a). People, clothing, leaves, and walls all 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 1.7(b). When the Moon reflects from a lake, as shown in Figure 1.7(c), a combination of these effects takes place.

Figure 1.6 Light is diffused when it reflects from a rough surface. Here, many parallel rays are incident, but they are reflected at many different angles, because the surface is rough.

Figure 1.7 (a) 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. (b) A mirror illuminated by many parallel rays reflects them in only one direction, because its surface is very smooth. Only the observer at a particular angle sees the reflected light. (c) Moonlight is spread out when it is reflected by the lake, because the surface is shiny but uneven. (credit c: modification of work by Diego Torres Silvestre)

12

1 • The Nature of Light

When you see yourself in a mirror, it appears that the image is actually behind the mirror (Figure 1.8). We see the light coming from a direction determined by the law of reflection. The angles are such that the image is exactly the same distance behind the mirror as you stand in front of 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 your imagination. Mirror images can be photographed and videotaped by instruments and look just as they do with our eyes (which are optical instruments themselves). The precise manner in which images are formed by mirrors and lenses is discussed in an upcoming chapter on Geometric Optics and Image Formation.

Figure 1.8 (a) Your 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 behind the mirror at the same distance away as (b) if you were looking at your twin directly, with no mirror.
Corner Reflectors (Retroreflectors)
A light ray that strikes an object consisting of two mutually perpendicular reflecting surfaces is reflected back exactly parallel to the direction from which it came (Figure 1.9). This is true whenever the reflecting surfaces are perpendicular, and it is independent of the angle of incidence. (For proof, see Exercise 1.34 at the end of this section.) Such an object is called a corner reflector, since the light bounces from its inside corner. Corner reflectors are a subclass of retroreflectors, which all reflect rays back in the directions from which they came. Although the geometry of the proof is much more complex, corner reflectors can also be built with three mutually perpendicular reflecting surfaces and are useful in three-dimensional applications.
Figure 1.9 A light ray that strikes two mutually perpendicular reflecting surfaces is reflected back exactly parallel to the direction from which it came.
Many inexpensive reflector buttons on bicycles, cars, and warning signs have corner reflectors designed to
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1.3 • Refraction

13

return light in the direction from which it originated. Rather than simply reflecting light over a wide angle, retroreflection ensures high visibility if the observer and the light source are located together, such as a car’s driver and headlights. The Apollo astronauts placed a true corner reflector on the Moon (Figure 1.10). Laser signals from Earth can be bounced from that corner reflector to measure the gradually increasing distance to the Moon of a few centimeters per year.

Figure 1.10 (a) Astronauts placed a corner reflector on the Moon to measure its gradually increasing orbital distance. (b) The bright spots on these bicycle safety reflectors are reflections of the flash of the camera that took this picture on a dark night. (credit a: modification of work by NASA; credit b: modification of work by “Julo”/Wikimedia Commons)
Working on the same principle as these optical reflectors, corner reflectors are routinely used as radar reflectors (Figure 1.11) for radio-frequency applications. Under most circumstances, small boats made of fiberglass or wood do not strongly reflect radio waves emitted by radar systems. To make these boats visible to radar (to avoid collisions, for example), radar reflectors are attached to boats, usually in high places.
Figure 1.11 A radar reflector hoisted on a sailboat is a type of corner reflector. (credit: Tim Sheerman-Chase)
As a counterexample, if you are interested in building a stealth airplane, radar reflections should be minimized to evade detection. One of the design considerations would then be to avoid building corners into the airframe.
1.3 Refraction
Learning Objectives By the end of this section, you will be able to:
• Describe how rays change direction upon entering a medium • Apply the law of refraction in problem solving You may often notice some odd things when looking into a fish tank. For example, you may see the same fish

14

1 • The Nature of Light

appearing to be in two different places (Figure 1.12). This happens because light coming from the fish to you

changes direction when it leaves the tank, and in this case, it can travel two different paths to get to your eyes.

The changing of a light ray’s direction (loosely called bending) when it passes through substances of different

refractive indices is called refraction and is related to changes in the speed of light,

. Refraction is

responsible for a tremendous range of optical phenomena, from the action of lenses to data transmission

through optical fibers.

Figure 1.12 (a) 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, so the fish seems to be in two different places. This bending of light is called refraction and is responsible for many optical phenomena. (b) This image shows refraction of light from a fish near the top of a fish tank.
Figure 1.13 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 is reflected from the surface, but for now we concentrate on the light that is transmitted.) The change in direction of the light ray depends on the relative values of the indices of refraction (The Propagation of Light) of the two media involved. In the situations shown, medium 2 has a greater index of refraction than medium 1. Note that as shown in Figure 1.13(a), the direction of the ray moves closer to the perpendicular when it progresses from a medium with a lower index of refraction to one with a higher index of refraction. Conversely, as shown in Figure 1.13(b), the direction of the ray moves away from the perpendicular when it progresses from a medium with a higher index of refraction to one with a lower index of refraction. The path is exactly reversible.
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1.3 • Refraction

15

Figure 1.13 The change in direction of a light ray depends on how the index of refraction changes when it crosses from one medium to another. In the situations shown here, the index of refraction is greater in medium 2 than in medium 1. (a) A ray of light moves closer to the perpendicular when entering a medium with a higher index of refraction. (b) A ray of light moves away from the perpendicular when entering a medium with a lower index of refraction.
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 refraction, or Snell’s law, after the Dutch mathematician Willebrord Snell (1591–1626), who discovered it in 1621. While the law has been named after Snell, the Arabian physicist Ibn Sahl found the law of refraction in 984 and used it in his work On Burning Mirrors and Lenses. The law of refraction is stated in equation form as
1.4
Here and are the indices of refraction for media 1 and 2, and and are the angles between the rays and the perpendicular in media 1 and 2. The incoming ray is called the incident ray, the outgoing ray is called the refracted ray, and the associated angles are the incident angle and the refracted angle, respectively.
Snell’s experiments showed that the law of refraction is obeyed and that a characteristic index of refraction n could be assigned to a given medium and its value measured. Snell was not aware that the speed of light varied in different media, a key fact used when we derive the law of refraction theoretically using Huygens’s principle in Huygens’s Principle.

EXAMPLE 1.2

Determining the Index of Refraction

Find the index of refraction for medium 2 in Figure 1.13(a), assuming medium 1 is air and given that the

incident angle is

and the angle of refraction is

.

Strategy

The index of refraction for air is taken to be 1 in most cases (and up to four significant figures, it is 1.000).

Thus,

here. From the given information,

and

With this information, the only

unknown in Snell’s law is so we can use Snell’s law to find it.

Solution From Snell’s law we have

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1 • The Nature of Light

Entering known values,

Significance 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.
INTERACTIVE
Explore bending of light (https://openstax.org/l/21bendoflight) between two media with different indices of refraction. Use the “Intro” simulation and see how changing from air to water to glass changes the bending angle. Use the protractor tool to measure the angles and see if you can recreate the configuration in Example 1.2. Also by measurement, confirm that the angle of reflection equals the angle of incidence.

EXAMPLE 1.3

A Larger Change in Direction

Suppose that in a situation like that in Example 1.2, light goes from air to diamond and that the incident angle

is

. Calculate the angle of refraction in the diamond.

Strategy
Again, the index of refraction for air is taken to be index of refraction for diamond in Table 1.1, finding we wish to determine.

, and we are given

. We can look up the

. The only unknown in Snell’s law is , which

Solution Solving Snell’s law for

yields

Entering known values,

The angle is thus

Significance

For the same

angle of incidence, the angle of refraction in diamond is significantly smaller than in water

rather than

—see Example 1.2). 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.

CHECK YOUR UNDERSTANDING 1.2
In Table 1.1, the solid with the next highest index of refraction after diamond is zircon. If the diamond in Example 1.3 were replaced with a piece of zircon, what would be the new angle of refraction?

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1.4 • Total Internal Reflection

17

1.4 Total Internal Reflection
Learning Objectives By the end of this section, you will be able to:
• Explain the phenomenon of total internal reflection • Describe the workings and uses of optical fibers • Analyze the reason for the sparkle of diamonds

A good-quality mirror may reflect more than of the light that falls on it, absorbing the rest. But it would be useful to have a mirror that reflects all of the light that falls on it. Interestingly, we can produce total reflection using an aspect of refraction.

Consider what happens when a ray of light strikes the surface between two materials, as shown in Figure

1.14(a). Part of the light crosses the boundary and is refracted; the rest is reflected. If, as shown in the figure,

the index of refraction for the second medium is less than for the first, the ray bends away from the

perpendicular. (Since

the angle of refraction is greater than the angle of incidence—that is,

Now imagine what happens as the incident angle increases. This causes to increase also. The largest the

angle of refraction can be is , as shown in part (b). The critical angle for a combination of materials is

defined to be the incident angle that produces an angle of refraction of . That is, is the incident angle

for which

. If the incident angle is greater than the critical angle, as shown in Figure 1.14(c), then

all of the light is reflected back into medium 1, a condition called total internal reflection. (As the figure

shows, the reflected rays obey the law of reflection so that the angle of reflection is equal to the angle of

incidence in all three cases.)

Figure 1.14 (a) A ray of light crosses a boundary where the index of refraction decreases. That is, perpendicular. (b) The critical angle is the angle of incidence for which the angle of refraction is when the incident angle is greater than the critical angle.

The ray bends away from the (c) Total internal reflection occurs

Snell’s law states the relationship between angles and indices of refraction. It is given by

When the incident angle equals the critical angle

that

Snell’s law in this case becomes

, the angle of refraction is

. Noting

The critical angle for a given combination of materials is thus
1.5
Total internal reflection occurs for any incident angle greater than the critical angle , and it can only occur when the second medium has an index of refraction less than the first. Note that this equation is written for a light ray that travels in medium 1 and reflects from medium 2, as shown in Figure 1.14.

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1 • The Nature of Light

EXAMPLE 1.4
Determining a Critical Angle
What is the critical angle for light traveling in a polystyrene (a type of plastic) pipe surrounded by air? The index of refraction for polystyrene is 1.49.
Strategy The index of refraction of air can be taken to be 1.00, as before. Thus, the condition that the second medium (air) has an index of refraction less than the first (plastic) is satisfied, and we can use the equation

to find the critical angle where

and

Solution Substituting the identified values gives

Significance

This result means that any ray of light inside the plastic that strikes the surface at an angle greater than

is totally reflected. This makes the inside surface of the clear plastic a perfect mirror for such rays, without any

need for the silvering used on common mirrors. Different combinations of materials have different critical

angles, but any combination with

can produce total internal reflection. The same calculation as made

here shows that the critical angle for a ray going from water to air is

, whereas that from diamond to air is

, and that from flint glass to crown glass is

.

CHECK YOUR UNDERSTANDING 1.3
At the surface between air and water, light rays can go from air to water and from water to air. For which ray is there no possibility of total internal reflection?

In the photo that opens this chapter, the image of a swimmer underwater is captured by a camera that is also underwater. The swimmer in the upper half of the photograph, apparently facing upward, is, in fact, a reflected image of the swimmer below. The circular ripple near the photograph’s center is actually on the water surface. The undisturbed water surrounding it makes a good reflecting surface when viewed from below, thanks to total internal reflection. However, at the very top edge of this photograph, rays from below strike the surface with incident angles less than the critical angle, allowing the camera to capture a view of activities on the pool deck above water.
Fiber Optics: Endoscopes to Telephones
Fiber optics is one application of total internal reflection that is in wide use. In communications, it is used to transmit telephone, internet, and cable TV signals. Fiber optics employs the transmission of light down fibers of plastic or glass. Because the fibers are thin, light entering one is likely to strike the inside surface at an angle greater than the critical angle and, thus, be totally reflected (Figure 1.15). The index of refraction outside the fiber must be smaller than inside. In fact, most fibers have a varying refractive index to allow more light to be guided along the fiber through total internal refraction. Rays are reflected around corners as shown, making the fibers into tiny light pipes.

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1.4 • Total Internal Reflection

19

Figure 1.15 Light entering a thin optic fiber may strike the inside surface at large or grazing angles and is completely reflected if these angles exceed the critical angle. Such rays continue down the fiber, even following it around corners, since the angles of reflection and incidence remain large.
Bundles of fibers can be used to transmit an image without a lens, as illustrated in Figure 1.16. The output of a device called an endoscope is shown in Figure 1.16(b). Endoscopes are used to explore the interior of the body through its natural orifices or minor incisions. Light is transmitted down one fiber bundle to illuminate internal parts, and the reflected light is transmitted back out through another bundle to be observed.

Figure 1.16 (a) An image “A” is transmitted by a bundle of optical fibers. (b) An endoscope is used to probe the body, both transmitting light to the interior and returning an image such as the one shown of a human epiglottis (a structure at the base of the tongue). (credit b: modification of work by “Med_Chaos”/Wikimedia Commons)
Fiber optics has revolutionized surgical techniques and observations within the body, with a host of medical diagnostic and therapeutic uses. Surgery can be performed, such as arthroscopic surgery on a knee or shoulder joint, employing cutting tools attached to and observed with the endoscope. Samples can also be obtained, such as by lassoing an intestinal polyp for external examination. The flexibility of the fiber optic bundle allows doctors to navigate it around small and difficult-to-reach regions in the body, such as the intestines, the heart, blood vessels, and joints. Transmission of an intense laser beam to burn away obstructing plaques in major arteries, as well as delivering light to activate chemotherapy drugs, are becoming

20

1 • The Nature of Light

commonplace. Optical fibers have in fact enabled microsurgery and remote surgery where the incisions are small and the surgeon’s fingers do not need to touch the diseased tissue.
Optical fibers in bundles are surrounded by a cladding material that has a lower index of refraction than the core (Figure 1.17). The cladding prevents light from being transmitted between fibers in a bundle. Without cladding, light could pass between fibers in contact, since their indices of refraction are identical. Since no light gets into the cladding (there is total internal reflection back into the core), none can be transmitted between clad fibers that are in contact with one another. Instead, the light is propagated along the length of the fiber, minimizing the loss of signal and ensuring that a quality image is formed at the other end. The cladding and an additional protective layer make optical fibers durable as well as flexible.

Figure 1.17 Fibers in bundles are clad by a material that has a lower index of refraction than the core to ensure total internal reflection, even when fibers are in contact with one another.
Special tiny lenses that can be attached to the ends of bundles of fibers have been designed and fabricated. Light emerging from a fiber bundle can be focused through such a lens, imaging a tiny spot. In some cases, the spot can be scanned, allowing quality imaging of a region inside the body. Special minute optical filters inserted at the end of the fiber bundle have the capacity to image the interior of organs located tens of microns below the surface without cutting the surface—an area known as nonintrusive diagnostics. This is particularly useful for determining the extent of cancers in the stomach and bowel. In another type of application, optical fibers are commonly used to carry signals for telephone conversations and internet communications. Extensive optical fiber cables have been placed on the ocean floor and underground to enable optical communications. Optical fiber communication systems offer several advantages over electrical (copper)-based systems, particularly for long distances. The fibers can be made so transparent that light can travel many kilometers before it becomes dim enough to require amplification—much superior to copper conductors. This property of optical fibers is called low loss. Lasers emit light with characteristics that allow far more conversations in one fiber than are possible with electric signals on a single conductor. This property of optical fibers is called high bandwidth. Optical signals in one fiber do not produce undesirable effects in other adjacent fibers. This property of optical fibers is called reduced crosstalk. We shall explore the unique characteristics of laser radiation in a later chapter.
Corner Reflectors and Diamonds
Corner reflectors (The Law of Reflection) are perfectly efficient when the conditions for total internal reflection are satisfied. With common materials, it is easy to obtain a critical angle that is less than One use of these perfect mirrors is in binoculars, as shown in Figure 1.18. Another use is in periscopes found in submarines.
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1.4 • Total Internal Reflection

21

Figure 1.18 These binoculars employ corner reflectors (prisms) with total internal reflection to get light to the observer’s eyes.

Total internal reflection, coupled with a large index of refraction, explains why diamonds sparkle more than

other materials. The critical angle for a diamond-to-air surface is only

, so when light enters a diamond, it

has trouble getting back out (Figure 1.19). Although light freely enters the diamond, it can exit only if it makes

an angle less than

. Facets on diamonds are specifically intended to make this unlikely. Good diamonds

are very clear, so that the light makes many internal reflections and is concentrated before exiting—hence the

bright sparkle. (Zircon is a natural gemstone that has an exceptionally large index of refraction, but it is not as

large as diamond, so it is not as highly prized. Cubic zirconia is manufactured and has an even higher index of

refraction

, but it is still less than that of diamond.) The colors you see emerging from a clear diamond

are not due to the diamond’s color, which is usually nearly colorless. The colors result from dispersion, which

we discuss in Dispersion. Colored diamonds get their color from structural defects of the crystal lattice and the

inclusion of minute quantities of graphite and other materials. The Argyle Mine in Western Australia produces

around 90% of the world’s pink, red, champagne, and cognac diamonds, whereas around 50% of the world’s

clear diamonds come from central and southern Africa.

Figure 1.19 Light cannot easily escape a diamond, because its critical angle with air is so small. Most reflections are total, and the facets are placed so that light can exit only in particular ways—thus concentrating the light and making the diamond sparkle brightly.
INTERACTIVE
Explore refraction and reflection of light (https://openstax.org/l/21bendoflight) between two media with

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1 • The Nature of Light

different indices of refraction. Try to make the refracted ray disappear with total internal reflection. Use the protractor tool to measure the critical angle and compare with the prediction from Equation 1.5.
1.5 Dispersion
Learning Objectives By the end of this section, you will be able to:
• Explain the cause of dispersion in a prism • Describe the effects of dispersion in producing rainbows • Summarize the advantages and disadvantages of dispersion
Everyone enjoys the spectacle of a rainbow glimmering against a dark stormy sky. How does sunlight falling on clear drops of rain get broken into the rainbow of colors we see? The same process causes white light to be broken into colors by a clear glass prism or a diamond (Figure 1.20).

Figure 1.20 The colors of the rainbow (a) and those produced by a prism (b) are identical. (credit a: modification of work by “Alfredo55”/Wikimedia Commons; credit b: modification of work by NASA)
We see about six colors in a rainbow—red, orange, yellow, green, blue, and violet; sometimes indigo is listed, too. These colors are associated with different wavelengths of light, as shown in Figure 1.21. When our eye receives pure-wavelength light, we tend to see only one of the six colors, depending on wavelength. The thousands of other hues we can sense in other situations are our eye’s response to various mixtures of wavelengths. White light, in particular, is a fairly uniform mixture of all visible wavelengths. Sunlight, considered to be white, actually appears to be a bit yellow, because of its mixture of wavelengths, but it does contain all visible wavelengths. The sequence of colors in rainbows is the same sequence as the colors shown in the figure. This implies that white light is spread out in a rainbow according to wavelength. Dispersion is defined as the spreading of white light into its full spectrum of wavelengths. More technically, dispersion occurs whenever the propagation of light depends on wavelength.
Figure 1.21 Even though rainbows are associated with six colors, the rainbow is a continuous distribution of colors according to wavelengths.
Any type of wave can exhibit dispersion. For example, sound waves, all types of electromagnetic waves, and water waves can be dispersed according to wavelength. Dispersion may require special circumstances and can result in spectacular displays such as in the production of a rainbow. This is also true for sound, since all frequencies ordinarily travel at the same speed. If you listen to sound through a long tube, such as a vacuum cleaner hose, you can easily hear it dispersed by interaction with the tube. Dispersion, in fact, can reveal a great deal about what the wave has encountered that disperses its wavelengths. The dispersion of electromagnetic radiation from outer space, for example, has revealed much about what exists between the
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1.5 • Dispersion

23

stars—the so-called interstellar medium.
INTERACTIVE
Nick Moore’s video (https://openstax.org/l/21nickmoorevid) discusses dispersion of a pulse as he taps a long spring. Follow his explanation as Moore replays the high-speed footage showing high frequency waves outrunning the lower frequency waves.

Refraction is responsible for dispersion in rainbows and many other situations. The angle of refraction depends on the index of refraction, as we know from Snell’s law. We know that the index of refraction n depends on the medium. But for a given medium, n also depends on wavelength (Table 1.2). Note that for a given medium, n increases as wavelength decreases and is greatest for violet light. Thus, violet light is bent more than red light, as shown for a prism in Figure 1.22(b). White light is dispersed into the same sequence of wavelengths as seen in Figure 1.20 and Figure 1.21.

Medium

Red

Orange Yellow

Green

Blue

Violet

(660 nm) (610 nm) (580 nm) (550 nm) (470 nm) (410 nm)

Water

1.331

1.332

1.333

1.335

1.338

1.342

Diamond

2.410

2.415

2.417

2.426

2.444

2.458

Glass, crown 1.512

1.514

1.518

1.519

1.524

1.530

Glass, flint

1.662

1.665

1.667

1.674

1.684

1.698

Polystyrene 1.488

1.490

1.492

1.493

1.499

1.506

Quartz, fused 1.455

1.456

1.458

1.459

1.462

1.468

Table 1.2 Index of Refraction n in Selected Media at Various Wavelengths

Figure 1.22 (a) A pure wavelength of light falls onto a prism and is refracted at both surfaces. (b) White light is dispersed by the prism (shown exaggerated). Since the index of refraction varies with wavelength, the angles of refraction vary with wavelength. A sequence of red to violet is produced, because the index of refraction increases steadily with decreasing wavelength.

EXAMPLE 1.5
Dispersion of White Light by Crown Glass
A beam of white light goes from air into crown glass at an incidence angle of the red (660 nm) and violet (410 nm) parts of the refracted light?

. What is the angle between

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1 • The Nature of Light

Strategy Values for the indices of refraction for crown glass at various wavelengths are listed in Table 1.2. Use these values for calculate the angle of refraction for each color and then take the difference to find the dispersion angle. Solution Applying the law of refraction for the red part of the beam
we can solve for the angle of refraction as
Similarly, the angle of incidence for the violet part of the beam is
The difference between these two angles is
Significance Although may seem like a negligibly small angle, if this beam is allowed to propagate a long enough distance, the dispersion of colors becomes quite noticeable.
CHECK YOUR UNDERSTANDING 1.4
In the preceding example, how much distance inside the block of crown glass would the red and the violet rays have to progress before they are separated by 1.0 mm?
Rainbows are produced by a combination of refraction and reflection. You may have noticed that you see a rainbow only when you look away from the Sun. Light enters a drop of water and is reflected from the back of the drop (Figure 1.23). The light is refracted both as it enters and as it leaves the drop. Since the index of refraction of water varies with wavelength, the light is dispersed, and a rainbow is observed (Figure 1.24(a)).
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1.5 • Dispersion

25

(No dispersion occurs at the back surface, because the law of reflection does not depend on wavelength.) The actual rainbow of colors seen by an observer depends on the myriad rays being refracted and reflected toward the observer’s eyes from numerous drops of water. The effect is most spectacular when the background is dark, as in stormy weather, but can also be observed in waterfalls and lawn sprinklers. The arc of a rainbow comes from the need to be looking at a specific angle relative to the direction of the Sun, as illustrated in part (b). If two reflections of light occur within the water drop, another “secondary” rainbow is produced. This rare event produces an arc that lies above the primary rainbow arc, as in part (c), and produces colors in the reverse order of the primary rainbow, with red at the lowest angle and violet at the largest angle.

Figure 1.23 A ray of light falling on this water drop enters and is reflected from the back of the drop. This light is refracted and dispersed both as it enters and as it leaves the drop.
Figure 1.24 (a) Different colors emerge in different directions, and so you must look at different locations to see the various colors of a rainbow. (b) The arc of a rainbow results from the fact that a line between the observer and any point on the arc must make the correct angle with the parallel rays of sunlight for the observer to receive the refracted rays. (c) Double rainbow. (credit c: modification of work by “Nicholas”/Wikimedia Commons)
Dispersion may produce beautiful rainbows, but it can cause problems in optical systems. White light used to transmit messages in a fiber is dispersed, spreading out in time and eventually overlapping with other messages. Since a laser produces a nearly pure wavelength, its light experiences little dispersion, an advantage over white light for transmission of information. In contrast, dispersion of electromagnetic waves coming to us from outer space can be used to determine the amount of matter they pass through.

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1 • The Nature of Light

1.6 Huygens’s Principle
Learning Objectives By the end of this section, you will be able to:
• Describe Huygens’s principle • Use Huygens’s principle to explain the law of reflection • Use Huygens’s principle to explain the law of refraction • Use Huygens’s principle to explain diffraction
So far in this chapter, we have been discussing optical phenomena using the ray model of light. However, some phenomena require analysis and explanations based on the wave characteristics of light. This is particularly true when the wavelength is not negligible compared to the dimensions of an optical device, such as a slit in the case of diffraction. Huygens’s principle is an indispensable tool for this analysis.
Figure 1.25 shows how a transverse wave looks as viewed from above and from the side. A light wave can be imagined to propagate like this, although we do not actually see it wiggling through space. From above, we view the wave fronts (or wave crests) as if we were looking down on ocean waves. The side view would be a graph of the electric or magnetic field. The view from above is perhaps more useful in developing concepts about wave optics.

Figure 1.25 A transverse wave, such as an electromagnetic light wave, as viewed from above and from the side. The direction of propagation is perpendicular to the wave fronts (or wave crests) and is represented by a ray.
The Dutch scientist Christiaan Huygens (1629–1695) developed a useful technique for determining in detail how and where waves propagate. Starting from some known position, Huygens’s principle states that every point on a wave front is a source of wavelets that spread out in the forward direction at the same speed as the wave itself. The new wave front is tangent to all of the wavelets.
Figure 1.26 shows how Huygens’s principle is applied. A wave front is the long edge that moves, for example, with the crest or the trough. Each point on the wave front emits a semicircular wave that moves at the propagation speed v. We can draw these wavelets at a time t later, so that they have moved a distance The new wave front is a plane tangent to the wavelets and is where we would expect the wave to be a time t later. Huygens’s principle works for all types of waves, including water waves, sound waves, and light waves. It is useful not only in describing how light waves propagate but also in explaining the laws of reflection and refraction. In addition, we will see that Huygens’s principle tells us how and where light rays interfere.

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1.6 • Huygens’s Principle

27

Figure 1.26 Huygens’s principle applied to a straight wave front. Each point on the wave front emits a semicircular wavelet that moves a

distance

The new wave front is a line tangent to the wavelets.

Reflection

Figure 1.27 shows how a mirror reflects an incoming wave at an angle equal to the incident angle, verifying the law of reflection. As the wave front strikes the mirror, wavelets are first emitted from the left part of the mirror and then from the right. The wavelets closer to the left have had time to travel farther, producing a wave front traveling in the direction shown.

Figure 1.27 Huygens’s principle applied to a plane wave front striking a mirror. The wavelets shown were emitted as each point on the wave front struck the mirror. The tangent to these wavelets shows that the new wave front has been reflected at an angle equal to the incident angle. The direction of propagation is perpendicular to the wave front, as shown by the downward-pointing arrows.
Refraction
The law of refraction can be explained by applying Huygens’s principle to a wave front passing from one medium to another (Figure 1.28). Each wavelet in the figure was emitted when the wave front crossed the interface between the media. Since the speed of light is smaller in the second medium, the waves do not travel as far in a given time, and the new wave front changes direction as shown. This explains why a ray changes direction to become closer to the perpendicular when light slows down. Snell’s law can be derived from the geometry in Figure 1.28 (Example 1.6).

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1 • The Nature of Light

Figure 1.28 Huygens’s principle applied to a plane wave front traveling from one medium to another, where its speed is less. The ray bends toward the perpendicular, since the wavelets have a lower speed in the second medium.

EXAMPLE 1.6

Deriving the Law of Refraction
By examining the geometry of the wave fronts, derive the law of refraction.

Strategy

Consider Figure 1.29, which expands upon Figure 1.28. It shows the incident wave front just reaching the

surface at point A, while point B is still well within medium 1. In the time it takes for a wavelet from B to

reach on the surface at speed

a wavelet from A travels into medium 2 a distance of

where

Note that in this example, is slower than because

Figure 1.29 Geometry of the law of refraction from medium 1 to medium 2.

Solution The segment on the surface is shared by both the triangle inside medium 2. Note that from the geometry, the angle

inside medium 1 and the triangle is equal to the angle of incidence, .

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Similarly, The length of

is . is given in two ways as

1.6 • Huygens’s Principle

29

Inverting the equation and substituting

from above and similarly

, we obtain

Cancellation of allows us to simplify this equation into the familiar form

Significance Although the law of refraction was established experimentally by Snell and stated in Refraction, its derivation here requires Huygens’s principle and the understanding that the speed of light is different in different media.

CHECK YOUR UNDERSTANDING 1.5

In Example 1.6, we had

. If were decreased such that

and the speed of light in medium 2 is

faster than in medium 1, what would happen to the length of ? What would happen to the wave front

and the direction of the refracted ray?

INTERACTIVE
This applet (https://openstax.org/l/21walfedaniref) by Walter Fendt shows an animation of reflection and refraction using Huygens’s wavelets while you control the parameters. Be sure to click on “Next step” to display the wavelets. You can see the reflected and refracted wave fronts forming.

Diffraction
What happens when a wave passes through an opening, such as light shining through an open door into a dark room? For light, we observe a sharp shadow of the doorway on the floor of the room, and no visible light bends around corners into other parts of the room. When sound passes through a door, we hear it everywhere in the room and thus observe that sound spreads out when passing through such an opening (Figure 1.30). What is the difference between the behavior of sound waves and light waves in this case? The answer is that light has very short wavelengths and acts like a ray. Sound has wavelengths on the order of the size of the door and bends around corners (for frequency of 1000 Hz,

about three times smaller than the width of the doorway).

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1 • The Nature of Light

Figure 1.30 (a) Light passing through a doorway makes a sharp outline on the floor. Since light’s wavelength is very small compared with the size of the door, it acts like a ray. (b) Sound waves bend into all parts of the room, a wave effect, because their wavelength is similar to the size of the door.
If we pass light through smaller openings such as slits, we can use Huygens’s principle to see that light bends as sound does (Figure 1.31). The bending of a wave around the edges of an opening or an obstacle is called diffraction. Diffraction is a wave characteristic and occurs for all types of waves. If diffraction is observed for some phenomenon, it is evidence that the phenomenon is a wave. Thus, the horizontal diffraction of the laser beam after it passes through the slits in Figure 1.31 is evidence that light is a wave. You will learn about diffraction in much more detail in the chapter on Diffraction.
Figure 1.31 Huygens’s principle applied to a plane wave front striking an opening. The edges of the wave front bend after passing through the opening, a process called diffraction. The amount of bending is more extreme for a small opening, consistent with the fact that wave characteristics are most noticeable for interactions with objects about the same size as the wavelength.
1.7 Polarization
Learning Objectives By the end of this section, you will be able to:
• Explain the change in intensity as polarized light passes through a polarizing filter • Calculate the effect of polarization by reflection and Brewster’s angle • Describe the effect of polarization by scattering • Explain the use of polarizing materials in devices such as LCDs Polarizing sunglasses are familiar to most of us. They have a special ability to cut the glare of light reflected
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1.7 • Polarization

31

from water or glass (Figure 1.32). They have this ability because of a wave characteristic of light called polarization. What is polarization? How is it produced? What are some of its uses? The answers to these questions are related to the wave character of light.

Figure 1.32 These two photographs of a river show the effect of a polarizing filter in reducing glare in light reflected from the surface of water. Part (b) of this figure was taken with a polarizing filter and part (a) was not. As a result, the reflection of clouds and sky observed in part (a) is not observed in part (b). Polarizing sunglasses are particularly useful on snow and water. (credit a and credit b: modifications of work by “Amithshs”/Wikimedia Commons)
Malus’s Law
Light is one type of electromagnetic (EM) wave. As noted in the previous chapter on Electromagnetic Waves, EM waves are transverse waves consisting of varying electric and magnetic fields that oscillate perpendicular to the direction of propagation (Figure 1.33). However, in general, there are no specific directions for the oscillations of the electric and magnetic fields; they vibrate in any randomly oriented plane perpendicular to the direction of propagation. Polarization is the attribute that a wave’s oscillations do have a definite direction relative to the direction of propagation of the wave. (This is not the same type of polarization as that discussed for the separation of charges.) Waves having such a direction are said to be polarized. For an EM wave, we define the direction of polarization to be the direction parallel to the electric field. Thus, we can think of the electric field arrows as showing the direction of polarization, as in Figure 1.33.

Figure 1.33 An EM wave, such as light, is a transverse wave. The electric and magnetic propagation. The direction of polarization of the wave is the direction of the electric field.

fields are perpendicular to the direction of

To examine this further, consider the transverse waves in the ropes shown in Figure 1.34. The oscillations in one rope are in a vertical plane and are said to be vertically polarized. Those in the other rope are in a horizontal plane and are horizontally polarized. If a vertical slit is placed on the first rope, the waves pass through. However, a vertical slit blocks the horizontally polarized waves. For EM waves, the direction of the electric field is analogous to the disturbances on the ropes.

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1 • The Nature of Light

Figure 1.34 The transverse oscillations in one rope (a) are in a vertical plane, and those in the other rope (b) are in a horizontal plane. The first is said to be vertically polarized, and the other is said to be horizontally polarized. Vertical slits pass vertically polarized waves and block horizontally polarized waves.
The Sun and many other light sources produce waves that have the electric fields in random directions (Figure 1.35(a)). Such light is said to be unpolarized, because it is composed of many waves with all possible directions of polarization. Polaroid materials—which were invented by the founder of the Polaroid Corporation, Edwin Land—act as a polarizing slit for light, allowing only polarization in one direction to pass through. Polarizing filters are composed of long molecules aligned in one direction. If we think of the molecules as many slits, analogous to those for the oscillating ropes, we can understand why only light with a specific polarization can get through. The axis of a polarizing filter is the direction along which the filter passes the electric field of an EM wave.
Figure 1.35 The slender arrow represents a ray of unpolarized light. The bold arrows represent the direction of polarization of the individual waves composing the ray. (a) If the light is unpolarized, the arrows point in all directions. (b) A polarizing filter has a polarization axis that acts as a slit passing through electric fields parallel to its direction. The direction of polarization of an EM wave is defined to be the direction of its electric field.
Figure 1.36 shows the effect of two polarizing filters on originally unpolarized light. The first filter polarizes the light along its axis. When the axes of the first and second filters are aligned (parallel), then all of the polarized light passed by the first filter is also passed by the second filter. If the second polarizing filter is rotated, only the component of the light parallel to the second filter’s axis is passed. When the axes are perpendicular, no light is passed by the second filter.
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1.7 • Polarization

33

Figure 1.36 The effect of rotating two polarizing filters, where the first polarizes the light. (a) All of the polarized light is passed by the second polarizing filter, because its axis is parallel to the first. (b) As the second filter is rotated, only part of the light is passed. (c) When the second filter is perpendicular to the first, no light is passed. (d) In this photograph, a polarizing filter is placed above two others. Its axis is perpendicular to the filter on the right (dark area) and parallel to the filter on the left (lighter area). (credit d: modification of work by P.P. Urone)

Only the component of the EM wave parallel to the axis of a filter is passed. Let us call the angle between the

direction of polarization and the axis of a filter . If the electric field has an amplitude E, then the transmitted

part of the wave has an amplitude

(Figure 1.37). Since the intensity of a wave is proportional to its

amplitude squared, the intensity I of the transmitted wave is related to the incident wave by

1.6

where is the intensity of the polarized wave before passing through the filter. This equation is known as Malus’s law.

Figure 1.37 A polarizing filter transmits only the component of the wave parallel to its axis, reducing the intensity of any light not

34

1 • The Nature of Light

polarized parallel to its axis.
INTERACTIVE
This Open Source Physics animation (https://openstax.org/l/21phyanielefie) helps you visualize the electric field vectors as light encounters a polarizing filter. You can rotate the filter—note that the angle displayed is in radians. You can also rotate the animation for 3D visualization.

EXAMPLE 1.7

Calculating Intensity Reduction by a Polarizing Filter

What angle is needed between the direction of polarized light and the axis of a polarizing filter to reduce its

intensity by

?

Strategy When the intensity is reduced by Using this information, the equation

, it is

or 0.100 times its original value. That is, can be used to solve for the needed angle.

Solution Solving the equation

for

and substituting with the relationship between I and gives

Solving for yields

Significance

A fairly large angle between the direction of polarization and the filter axis is needed to reduce the intensity to

of its original value. This seems reasonable based on experimenting with polarizing films. It is

interesting that at an angle of , the intensity is reduced to of its original value. Note that

is

from reducing the intensity to zero, and that at an angle of

, the intensity is reduced to

of its

original value, giving evidence of symmetry.

CHECK YOUR UNDERSTANDING 1.6

Although we did not specify the direction in Example 1.7, let’s say the polarizing filter was rotated clockwise by

to reduce the light intensity by

. What would be the intensity reduction if the polarizing filter were

rotated counterclockwise by

?

Polarization by Reflection
By now, you can probably guess that polarizing sunglasses cut the glare in reflected light, because that light is polarized. You can check this for yourself by holding polarizing sunglasses in front of you and rotating them while looking at light reflected from water or glass. As you rotate the sunglasses, you will notice the light gets bright and dim, but not completely black. This implies the reflected light is partially polarized and cannot be completely blocked by a polarizing filter.
Figure 1.38 illustrates what happens when unpolarized light is reflected from a surface. Vertically polarized light is preferentially refracted at the surface, so the reflected light is left more horizontally polarized. The reasons for this phenomenon are beyond the scope of this text, but a convenient mnemonic for remembering this is to imagine the polarization direction to be like an arrow. Vertical polarization is like an arrow perpendicular to the surface and is more likely to stick and not be reflected. Horizontal polarization is like an

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1.7 • Polarization

35

arrow bouncing on its side and is more likely to be reflected. Sunglasses with vertical axes thus block more reflected light than unpolarized light from other sources.

Figure 1.38 Polarization by reflection. Unpolarized light has equal amounts of vertical and horizontal polarization. After interaction with a surface, the vertical components are preferentially absorbed or refracted, leaving the reflected light more horizontally polarized. This is akin to arrows striking on their sides and bouncing off, whereas arrows striking on their tips go into the surface.
Since the part of the light that is not reflected is refracted, the amount of polarization depends on the indices of refraction of the media involved. It can be shown that reflected light is completely polarized at an angle of reflection given by
1.7
where is the medium in which the incident and reflected light travel and is the index of refraction of the medium that forms the interface that reflects the light. This equation is known as Brewster’s law and is known as Brewster’s angle, named after the nineteenth-century Scottish physicist who discovered them.
INTERACTIVE
This Open Source Physics animation (https://openstax.org/l/21phyaniincref) shows incident, reflected, and refracted light as rays and EM waves. Try rotating the animation for 3D visualization and also change the angle of incidence. Near Brewster’s angle, the reflected light becomes highly polarized.

EXAMPLE 1.8

Calculating Polarization by Reflection
(a) At what angle will light traveling in air be completely polarized horizontally when reflected from water? (b) From glass?

Strategy

All we need to solve these problems are the indices of refraction. Air has

water has

and

crown glass has

The equation

can be directly applied to find in each case.

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1 • The Nature of Light

Solution a. Putting the known quantities into the equation

gives

Solving for the angle yields b. Similarly, for crown glass and air,

Thus,

Significance
Light reflected at these angles could be completely blocked by a good polarizing filter held with its axis vertical. Brewster’s angle for water and air are similar to those for glass and air, so that sunglasses are equally effective for light reflected from either water or glass under similar circumstances. Light that is not reflected is refracted into these media. Therefore, at an incident angle equal to Brewster’s angle, the refracted light is slightly polarized vertically. It is not completely polarized vertically, because only a small fraction of the incident light is reflected, so a significant amount of horizontally polarized light is refracted.

CHECK YOUR UNDERSTANDING 1.7
What happens at Brewster’s angle if the original incident light is already

vertically polarized?

Atomic Explanation of Polarizing Filters
Polarizing filters have a polarization axis that acts as a slit. This slit passes EM waves (often visible light) that have an electric field parallel to the axis. This is accomplished with long molecules aligned perpendicular to the axis, as shown in Figure 1.39.

Figure 1.39 Long molecules are aligned perpendicular to the axis of a polarizing filter. In an EM wave, the component of the electric field perpendicular to these molecules passes through the filter, whereas the component parallel to the molecules is absorbed.
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1.7 • Polarization

37

Figure 1.40 illustrates how the component of the electric field parallel to the long molecules is absorbed. An EM wave is composed of oscillating electric and magnetic fields. The electric field is strong compared with the magnetic field and is more effective in exerting force on charges in the molecules. The most affected charged particles are the electrons, since electron masses are small. If an electron is forced to oscillate, it can absorb energy from the EM wave. This reduces the field in the wave and, hence, reduces its intensity. In long molecules, electrons can more easily oscillate parallel to the molecule than in the perpendicular direction. The electrons are bound to the molecule and are more restricted in their movement perpendicular to the molecule. Thus, the electrons can absorb EM waves that have a component of their electric field parallel to the molecule. The electrons are much less responsive to electric fields perpendicular to the molecule and allow these fields to pass. Thus, the axis of the polarizing filter is perpendicular to the length of the molecule.

Figure 1.40 Diagram of an electron in a long molecule oscillating parallel to the molecule. The oscillation of the electron absorbs energy
and reduces the intensity of the component of the EM wave that is parallel to the molecule.
Polarization by Scattering
If you hold your polarizing sunglasses in front of you and rotate them while looking at blue sky, you will see the sky get bright and dim. This is a clear indication that light scattered by air is partially polarized. Figure 1.41 helps illustrate how this happens. Since light is a transverse EM wave, it vibrates the electrons of air molecules perpendicular to the direction that it is traveling. The electrons then radiate like small antennae. Since they are oscillating perpendicular to the direction of the light ray, they produce EM radiation that is polarized perpendicular to the direction of the ray. When viewing the light along a line perpendicular to the original ray, as in the figure, there can be no polarization in the scattered light parallel to the original ray, because that would require the original ray to be a longitudinal wave. Along other directions, a component of the other polarization can be projected along the line of sight, and the scattered light is only partially polarized. Furthermore, multiple scattering can bring light to your eyes from other directions and can contain different polarizations.

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1 • The Nature of Light

Figure 1.41 Polarization by scattering. Unpolarized light scattering from air molecules shakes their electrons perpendicular to the direction of the original ray. The scattered light therefore has a polarization perpendicular to the original direction and none parallel to the original direction.
Photographs of the sky can be darkened by polarizing filters, a trick used by many photographers to make clouds brighter by contrast. Scattering from other particles, such as smoke or dust, can also polarize light. Detecting polarization in scattered EM waves can be a useful analytical tool in determining the scattering source. A range of optical effects are used in sunglasses. Besides being polarizing, sunglasses may have colored pigments embedded in them, whereas others use either a nonreflective or reflective coating. A recent development is photochromic lenses, which darken in the sunlight and become clear indoors. Photochromic lenses are embedded with organic microcrystalline molecules that change their properties when exposed to UV in sunlight, but become clear in artificial lighting with no UV.
Liquid Crystals and Other Polarization Effects in Materials
Although you are undoubtedly aware of liquid crystal displays (LCDs) found in watches, calculators, computer screens, cellphones, flat screen televisions, and many other places, you may not be aware that they are based on polarization. Liquid crystals are so named because their molecules can be aligned even though they are in a liquid. Liquid crystals have the property that they can rotate the polarization of light passing through them by
. Furthermore, this property can be turned off by the application of a voltage, as illustrated in Figure 1.42. It is possible to manipulate this characteristic quickly and in small, well-defined regions to create the contrast patterns we see in so many LCD devices. In flat screen LCD televisions, a large light is generated at the back of the TV. The light travels to the front screen through millions of tiny units called pixels (picture elements). One of these is shown in Figure 1.42(a) and (b). Each unit has three cells, with red, blue, or green filters, each controlled independently. When the voltage across a liquid crystal is switched off, the liquid crystal passes the light through the particular filter. We can vary the picture contrast by varying the strength of the voltage applied to the liquid crystal.
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1.7 • Polarization

39

Figure 1.42 (a) Polarized light is rotated by a liquid crystal and then passed by a polarizing filter that has its axis perpendicular to the direction of the original polarization. (b) When a voltage is applied to the liquid crystal, the polarized light is not rotated and is blocked by the filter, making the region dark in comparison with its surroundings. (c) LCDs can be made color specific, small, and fast enough to use in laptop computers and TVs. (credit c: modification of work by Jane Whitney)
Many crystals and solutions rotate the plane of polarization of light passing through them. Such substances are said to be optically active. Examples include sugar water, insulin, and collagen (Figure 1.43). In addition to depending on the type of substance, the amount and direction of rotation depend on several other factors. Among these is the concentration of the substance, the distance the light travels through it, and the wavelength of light. Optical activity is due to the asymmetrical shape of molecules in the substance, such as being helical. Measurements of the rotation of polarized light passing through substances can thus be used to measure concentrations, a standard technique for sugars. It can also give information on the shapes of molecules, such as proteins, and factors that affect their shapes, such as temperature and pH.
Figure 1.43 Optical activity is the ability of some substances to rotate the plane of polarization of light passing through them. The rotation is detected with a polarizing filter or analyzer.
Glass and plastic become optically active when stressed: the greater the stress, the greater the effect. Optical stress analysis on complicated shapes can be performed by making plastic models of them and observing them through crossed filters, as seen in Figure 1.44. It is apparent that the effect depends on wavelength as

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1 • The Nature of Light

well as stress. The wavelength dependence is sometimes also used for artistic purposes.

Figure 1.44 Optical stress analysis of a plastic lens placed between crossed polarizers. (credit: “Infopro”/Wikimedia Commons)
Another interesting phenomenon associated with polarized light is the ability of some crystals to split an unpolarized beam of light into two polarized beams. This occurs because the crystal has one value for the index of refraction of polarized light but a different value for the index of refraction of light polarized in the perpendicular direction, so that each component has its own angle of refraction. Such crystals are said to be birefringent, and, when aligned properly, two perpendicularly polarized beams will emerge from the crystal (Figure 1.45). Birefringent crystals can be used to produce polarized beams from unpolarized light. Some birefringent materials preferentially absorb one of the polarizations. These materials are called dichroic and can produce polarization by this preferential absorption. This is fundamentally how polarizing filters and other polarizers work.
Figure 1.45 Birefringent materials, such as the common mineral calcite, split unpolarized beams of light into two with two different values of index of refraction.
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1 • Chapter Review 41

CHAPTER REVIEW

Key Terms

birefringent refers to crystals that split an

unpolarized beam of light into two beams

Brewster’s angle angle of incidence at which the

reflected light is completely polarized

Brewster’s law

, where is the

medium in which the incident and reflected light travel and is the index of refraction of the medium that forms the interface that reflects the light corner reflector object consisting of two (or three) mutually perpendicular reflecting surfaces, so that the light that enters is reflected back exactly parallel to the direction from which it came critical angle incident angle that produces an angle of refraction of direction of polarization direction parallel to the electric field for EM waves dispersion spreading of light into its spectrum of wavelengths fiber optics field of study of the transmission of light down fibers of plastic or glass, applying the principle of total internal reflection geometric optics part of optics dealing with the ray aspect of light horizontally polarized oscillations are in a horizontal plane Huygens’s principle every point on a wave front is a source of wavelets that spread out in the forward direction at the same speed as the wave itself; the new wave front is a plane tangent to all of the wavelets

Key Equations

Speed of light

Index of refraction

Law of reflection

Law of refraction (Snell’s law)

index of refraction for a material, the ratio of the speed of light in a vacuum to that in a material
law of reflection angle of reflection equals the angle of incidence
law of refraction when a light ray crosses from one medium to another, it changes direction by an amount that depends on the index of refraction of each medium and the sines of the angle of incidence and angle of refraction
Malus’s law where is the intensity of the polarized wave before passing through the filter
optically active substances that rotate the plane of polarization of light passing through them
polarization attribute that wave oscillations have a definite direction relative to the direction of propagation of the wave
polarized refers to waves having the electric and magnetic field oscillations in a definite direction
ray straight line that originates at some point refraction changing of a light ray’s direction when
it passes through variations in matter total internal reflection phenomenon at the
boundary between two media such that all the light is reflected and no refraction occurs unpolarized refers to waves that are randomly polarized vertically polarized oscillations are in a vertical plane wave optics part of optics dealing with the wave aspect of light

Critical angle

Malus’s law

Brewster’s law

42 1 • Chapter Review

Summary
1.1 The Propagation of Light
• The speed of light in a vacuum is .
• The index of refraction of a material is where v is the speed of light in a material and c is the speed of light in a vacuum.
• The ray model of light describes the path of light as straight lines. 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; and (3) after being reflected from a mirror.
1.2 The Law of Reflection
• When a light ray strikes a smooth surface, the angle of reflection equals the angle of incidence.
• A mirror has a smooth surface and reflects light at specific angles.
• Light is diffused when it reflects from a rough surface.
1.3 Refraction
• The change of a light ray’s direction when it passes through variations in matter is called refraction.
• The law of refraction, also called Snell’s law, relates the indices of refraction for two media at an interface to the change in angle of a light ray passing through that interface.
1.4 Total Internal Reflection
• The incident angle that produces an angle of refraction of is called the critical angle.
• Total internal reflection is a phenomenon that occurs at the boundary between two media, such that if the incident angle in the first medium is greater than the critical angle, then all the light is reflected back into that medium.
• Fiber optics involves the transmission of light down fibers of plastic or glass, applying the principle of total internal reflection.
• Cladding prevents light from being transmitted between fibers in a bundle.
• Diamonds sparkle due to total internal reflection coupled with a large index of refraction.
1.5 Dispersion
• The spreading of white light into its full

spectrum of wavelengths is called dispersion. • Rainbows are produced by a combination of
refraction and reflection, and involve the dispersion of sunlight into a continuous distribution of colors. • Dispersion produces beautiful rainbows but also causes problems in certain optical systems.

1.6 Huygens’s Principle

• According to Huygens’s principle, every point on a wave front is a source of wavelets that spread out in the forward direction at the same speed as the wave itself. The new wave front is tangent to all of the wavelets.
• A mirror reflects an incoming wave at an angle equal to the incident angle, verifying the law of reflection.
• The law of refraction can be explained by applying Huygens’s principle to a wave front passing from one medium to another.
• The bending of a wave around the edges of an opening or an obstacle is called diffraction.

1.7 Polarization

• Polarization is the attribute that wave

oscillations have a definite direction relative to

the direction of propagation of the wave. The

direction of polarization is defined to be the

direction parallel to the electric field of the EM

wave.

• Unpolarized light is composed of many rays

having random polarization directions.

• Unpolarized light can be polarized by passing it

through a polarizing filter or other polarizing

material. The process of polarizing light

decreases its intensity by a factor of 2.

• The intensity, I, of polarized light after passing

through a polarizing filter is

,

where is the incident intensity and is the

angle between the direction of polarization and

the axis of the filter.

• Polarization is also produced by reflection.

• Brewster’s law states that reflected light is

completely polarized at the angle of reflection

, known as Brewster’s angle.

• Polarization can also be produced by scattering.

• Several types of optically active substances

rotate the direction of polarization of light

passing through them.

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1 • Chapter Review 43

Conceptual Questions
1.1 The Propagation of Light
1. Under what conditions can light be modeled like a ray? Like a wave?
2. Why is the index of refraction always greater than or equal to 1?
3. Does the fact that the light flash from lightning reaches you before its sound prove that the speed of light is extremely large or simply that it is greater than the speed of sound? Discuss how you could use this effect to get an estimate of the speed of light.
4. Speculate as to what physical process might be responsible for light traveling more slowly in a medium than in a vacuum.
1.2 The Law of Reflection
5. Using the law of reflection, explain how powder takes the shine off of a person’s nose. What is the name of the optical effect?
1.3 Refraction
6. Diffusion by reflection from a rough surface is described in this chapter. Light can also be diffused by refraction. Describe how this occurs in a specific situation, such as light interacting with crushed ice.
7. Will light change direction toward or away from the perpendicular when it goes from air to water? Water to glass? Glass to air?
8. Explain why an object in water always appears to be at a depth shallower than it actually is?
9. Explain why a person’s legs appear very short when wading in a pool. Justify your explanation with a ray diagram showing the path of rays from the feet to the eye of an observer who is out of the water.
10. Explain why an oar that is partially submerged in water appears bent.
1.4 Total Internal Reflection
11. A ring with a colorless gemstone is dropped into water. The gemstone becomes invisible when submerged. Can it be a diamond? Explain.
12. The most common type of mirage is an illusion that light from faraway objects is reflected by a pool of water that is not really there. Mirages are generally observed in deserts, when there is a hot layer of air near the ground. Given that the refractive index of air is lower for air at higher

temperatures, explain how mirages can be formed. 13. How can you use total internal reflection to estimate the index of refraction of a medium?
1.5 Dispersion
14. Is it possible that total internal reflection plays a role in rainbows? Explain in terms of indices of refraction and angles, perhaps referring to that shown below. Some of us have seen the formation of a double rainbow; is it physically possible to observe a triple rainbow? (credit: "Chad"/Flickr)
15. A high-quality diamond may be quite clear and colorless, transmitting all visible wavelengths with little absorption. Explain how it can sparkle with flashes of brilliant color when illuminated by white light.
1.6 Huygens’s Principle
16. How do wave effects depend on the size of the object with which the wave interacts? For example, why does sound bend around the corner of a building while light does not?
17. Does Huygens’s principle apply to all types of waves?
18. If diffraction is observed for some phenomenon, it is evidence that the phenomenon is a wave. Does the reverse hold true? That is, if diffraction is not observed, does that mean the phenomenon is not a wave?
1.7 Polarization
19. Can a sound wave in air be polarized? Explain. 20. No light passes through two perfect polarizing
filters with perpendicular axes. However, if a third polarizing filter is placed between the original two, some light can pass. Why is this?

44 1 • Chapter Review

Under what circumstances does most of the light pass? 21. Explain what happens to the energy carried by light that it is dimmed by passing it through two crossed polarizing filters. 22. When particles scattering light are much smaller than its wavelength, the amount of scattering is proportional to . Does this mean
there is more scattering for small than large ? How does this relate to the fact that the sky is
blue? 23. Using the information given in the preceding
question, explain why sunsets are red.

Problems

1.1 The Propagation of Light

26. What is the speed of light in water? In glycerine?

27. What is the speed of light in air? In crown glass?

28. Calculate the index of refraction for a medium

in which the speed of light is

and identify the most likely substance based on

Table 1.1.

29. In what substance in Table 1.1 is the speed of

light

30. There was a major collision of an asteroid with

the Moon in medieval times. It was described by

monks at Canterbury Cathedral in England as a

red glow on and around the Moon. How long

after the asteroid hit the Moon, which is

away, would the light first arrive

on Earth?

31. Components of some computers communicate

with each other through optical fibers having an

index of refraction

What time in

nanoseconds is required for a signal to travel

0.200 m through such a fiber?

32. Compare the time it takes for light to travel

1000 m on the surface of Earth and in outer

space.

33. How far does light travel underwater during a

time interval of

?

1.2 The Law of Reflection
34. Suppose a man stands in front of a mirror as shown below. His eyes are 1.65 m above the floor and the top of his head is 0.13 m higher. Find the height above the floor of the top and bottom of the smallest mirror in which he can see both the top of his head and his feet. How is this distance related to the man’s height?

24. When light is reflected at Brewster’s angle from

a smooth surface, it is

polarized parallel

to the surface. Part of the light will be refracted

into the surface. Describe how you would do an

experiment to determine the polarization of the

refracted light. What direction would you expect

the polarization to have and would you expect it

to be

?

25. If you lie on a beach looking at the water with

your head tipped slightly sideways, your

polarized sunglasses do not work very well. Why

not?

35. Show that when light reflects from two mirrors that meet each other at a right angle, the outgoing ray is parallel to the incoming ray, as illustrated below.
36. On the Moon’s surface, lunar astronauts placed a corner reflector, off which a laser beam is

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1 • Chapter Review 45

periodically reflected. The distance to the Moon

angle of incidence of to an angle of

is calculated from the round-trip time. What

refraction of when it enters a rectangular

percent correction is needed to account for the

block of plastic. What is the index of refraction

delay in time due to the slowing of light in

of the plastic?

Earth’s atmosphere? Assume the distance to the

44. A scuba diver training in a pool looks at his

Moon is precisely

and Earth’s

instructor as shown below. What angle does the ray

atmosphere (which varies in density with

from the instructor’s face make with the

altitude) is equivalent to a layer 30.0 km thick

perpendicular to the water at the point where the ray

with a constant index of refraction

enters? The angle between the ray in the water and

the perpendicular to the water is

.

37. A flat mirror is neither converging nor diverging. To

prove this, consider two rays originating from the

same point and diverging at an angle (see below).

Show that after striking a plane mirror, the angle

between their directions remains

1.3 Refraction
Unless otherwise specified, for problems 1 through 10, the indices of refraction of glass and water should be taken to be 1.50 and 1.333, respectively.
38. A light beam in air has an angle of incidence of at the surface of a glass plate. What are the
angles of reflection and refraction? 39. A light beam in air is incident on the surface of
a pond, making an angle of with respect to the surface. What are the angles of reflection and refraction? 40. When a light ray crosses from water into glass, it emerges at an angle of with respect to the normal of the interface. What is its angle of incidence? 41. A pencil flashlight submerged in water sends a light beam toward the surface at an angle of incidence of . What is the angle of refraction in air? 42. Light rays from the Sun make a angle to the vertical when seen from below the surface of a body of water. At what angle above the horizon is the Sun? 43. The path of a light beam in air goes from an

45. (a) Using information in the preceding problem, find the height of the instructor’s head above the water, noting that you will first have to calculate the angle of incidence. (b) Find the apparent depth of the diver’s head below water as seen by the instructor.

1.4 Total Internal Reflection

46. Verify that the critical angle for light going from

water to air is

, as discussed at the end of

Example 1.4, regarding the critical angle for

light traveling in a polystyrene (a type of plastic)

pipe surrounded by air.

47. (a) At the end of Example 1.4, it was stated that

the critical angle for light going from diamond

to air is

Verify this. (b) What is the critical

angle for light going from zircon to air?

48. An optical fiber uses flint glass clad with crown

glass. What is the critical angle?

49. At what minimum angle will you get total

internal reflection of light traveling in water and

reflected from ice?

50. Suppose you are using total internal reflection

to make an efficient corner reflector. If there is

air outside and the incident angle is

, what

46 1 • Chapter Review

must be the minimum index of refraction of the

material from which the reflector is made?

51. You can determine the index of refraction of a

substance by determining its critical angle. (a)

What is the index of refraction of a substance

that has a critical angle of

when

submerged in water? What is the substance,

based on Table 1.1? (b) What would the critical

angle be for this substance in air?

52. A ray of light, emitted beneath the surface of an

unknown liquid with air above it, undergoes total

internal reflection as shown below. What is the

index of refraction for the liquid and its likely

identification?

53. Light rays fall normally on the vertical surface

of the glass prism

shown below. (a)

What is the largest value for such that the ray

is totally reflected at the slanted face? (b) Repeat

the calculation of part (a) if the prism is

immersed in water.

more dispersive?

55. A beam of white light goes from air into water at

an incident angle of

. At what angles are

the red (660 nm) and violet (410 nm) parts of

the light refracted?

56. By how much do the critical angles for red (660

nm) and violet (410 nm) light differ in a

diamond surrounded by air?

57. (a) A narrow beam of light containing yellow

(580 nm) and green (550 nm) wavelengths goes

from polystyrene to air, striking the surface at a

incident angle. What is the angle between

the colors when they emerge? (b) How far would

they have to travel to be separated by 1.00 mm?

58. A parallel beam of light containing orange (610

nm) and violet (410 nm) wavelengths goes from

fused quartz to water, striking the surface

between them at a

incident angle. What is

the angle between the two colors in water?

59. A ray of 610-nm light goes from air into fused

quartz at an incident angle of

. At what

incident angle must 470 nm light enter flint

glass to have the same angle of refraction?

60. A narrow beam of light containing red (660 nm)

and blue (470 nm) wavelengths travels from air

through a 1.00-cm-thick flat piece of crown

glass and back to air again. The beam strikes at

a

incident angle. (a) At what angles do the

two colors emerge? (b) By what distance are the

red and blue separated when they emerge?

61. A narrow beam of white light enters a prism

made of crown glass at a

incident angle, as

shown below. At what angles, and , do the

red (660 nm) and violet (410 nm) components

of the light emerge from the prism?

1.5 Dispersion
54. (a) What is the ratio of the speed of red light to violet light in diamond, based on Table 1.2? (b) What is this ratio in polystyrene? (c) Which is

1.7 Polarization

62. What angle is needed between the direction of

polarized light and the axis of a polarizing filter

to cut its intensity in half?

63. The angle between the axes of two polarizing

filters is

. By how much does the second

filter reduce the intensity of the light coming

through the first?

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1 • Chapter Review 47

64. Two polarizing sheets and are placed

together with their transmission axes oriented

at an angle to each other. What is when only

of the maximum transmitted light intensity

passes through them?

65. Suppose that in the preceding problem the light

incident on is unpolarized. At the

determined value of , what fraction of the

incident light passes through the combination?

66. If you have completely polarized light of

intensity

, what will its intensity be

after passing through a polarizing filter with its

axis at an

angle to the light’s polarization

direction?

67. What angle would the axis of a polarizing filter

need to make with the direction of polarized

light of intensity

to reduce the

intensity to

?

68. At the end of Example 1.7, it was stated that the

intensity of polarized light is reduced to

of its original value by passing through a

polarizing filter with its axis at an angle of

to the direction of polarization. Verify this

statement.

69. Show that if you have three polarizing filters,

with the second at an angle of

to the first

and the third at an angle of

to the first, the

intensity of light passed by the first will be

reduced to

of its value. (This is in

contrast to having only the first and third, which

reduces the intensity to zero, so that placing the

Additional Problems
76. From his measurements, Roemer estimated that it took 22 min for light to travel a distance equal to the diameter of Earth’s orbit around the Sun. (a) Use this estimate along with the known diameter of Earth’s orbit to obtain a rough value of the speed of light. (b) Light actually takes 16.5 min to travel this distance. Use this time to calculate the speed of light.
77. Cornu performed Fizeau’s measurement of the speed of light using a wheel of diameter 4.00 cm that contained 180 teeth. The distance from the wheel to the mirror was 22.9 km. Assuming he measured the speed of light accurately, what was the angular velocity of the wheel?

second between them increases the intensity of

the transmitted light.)

70. Three polarizing sheets are placed together

such that the transmission axis of the second

sheet is oriented at

to the axis of the first,

whereas the transmission axis of the third sheet

is oriented at

(in the same sense) to the

axis of the first. What fraction of the intensity of

an incident unpolarized beam is transmitted by

the combination?

71. In order to rotate the polarization axis of a beam

of linearly polarized light by

, a student

places sheets and with their transmission

axes at

and

, respectively, to the

beam’s axis of polarization. (a) What fraction of

the incident light passes through and (b)

through the combination? (c) Repeat your

calculations for part (b) for transmission-axis

angles of

and

, respectively.

72. It is found that when light traveling in water

falls on a plastic block, Brewster’s angle is

.

What is the refractive index of the plastic?

73. At what angle will light reflected from diamond

be completely polarized?

74. What is Brewster’s angle for light traveling in

water that is reflected from crown glass?

75. A scuba diver sees light reflected from the

water’s surface. At what angle relative to the

water’s surface will this light be completely

polarized?

78. 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

,

and you observe the angle of refraction to be

. What is the index of refraction of the

substance and its likely identity?

48 1 • Chapter Review

79. Shown below is a ray of light going from air through 85. Unreasonable results Suppose light travels

crown glass into water, such as going into a fish tank. from water to another substance, with an angle

Calculate the amount the ray is displaced by the

of incidence of

and an angle of refraction

glass

given that the incident angle is

and of

. (a) What is the index of refraction of the

the glass is 1.00 cm thick.

other substance? (b) What is unreasonable

about this result? (c) Which assumptions are

unreasonable or inconsistent?

86. Unreasonable results Light traveling from

water to a gemstone strikes the surface at an

angle of

and has an angle of refraction of

. (a) What is the speed of light in the

gemstone? (b) What is unreasonable about this

result? (c) Which assumptions are unreasonable

or inconsistent?

87. If a polarizing filter reduces the intensity of

polarized light to

of its original value, by

how much are the electric and magnetic fields

reduced?

88. Suppose you put on two pairs of polarizing

sunglasses with their axes at an angle of

.

How much longer will it take the light to deposit

a given amount of energy in your eye compared

80. Considering the previous problem, show that

with a single pair of sunglasses? Assume the

is the same as it would be if the second medium

lenses are clear except for their polarizing

were not present.

characteristics.

81. At what angle is light inside crown glass

89. (a) On a day when the intensity of sunlight is

completely polarized when reflected from water,

, a circular lens 0.200 m in

as in a fish tank?

diameter focuses light onto water in a black

82. Light reflected at

from a window is

beaker. Two polarizing sheets of plastic are

completely polarized. What is the window’s

placed in front of the lens with their axes at an

index of refraction and the likely substance of

angle of

. Assuming the sunlight is

which it is made?

unpolarized and the polarizers are

83. (a) Light reflected at

from a gemstone in a

efficient, what is the initial rate of heating of the

ring is completely polarized. Can the gem be a

water in , assuming it is

absorbed?

diamond? (b) At what angle would the light be

The aluminum beaker has a mass of 30.0 grams

completely polarized if the gem was in water?

and contains 250 grams of water. (b) Do the

84. If is Brewster’s angle for light reflected from

polarizing filters get hot? Explain.

the top of an interface between two substances,

and is Brewster’s angle for light reflected

from below, prove that

.

Challenge Problems
90. Light shows staged with lasers use moving mirrors to swing beams and create colorful effects. Show that a light ray reflected from a mirror changes direction by when the mirror is rotated by an angle .

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1 • Chapter Review 49

91. Consider sunlight entering Earth’s atmosphere at sunrise and sunset—that is, at a incident angle. Taking the boundary between nearly empty space and the atmosphere to be sudden, calculate the angle of refraction for sunlight. This lengthens the time the Sun appears to be above the horizon, both at sunrise and sunset. Now construct a problem in which you determine the angle of refraction for different models of the atmosphere, such as various layers of varying density. Your instructor may wish to guide you on the level of complexity to consider and on how the index of refraction varies with air density.
92. A light ray entering an optical fiber surrounded by air is first refracted and then reflected as shown below. Show that if the fiber is made from crown glass, any incident ray will be totally internally reflected.

93. A light ray falls on the left face of a prism (see below) at the angle of incidence for which the emerging beam has an angle of refraction at the right face. Show that the index of refraction n of the glass prism is given by
where is the vertex angle of the prism and is the angle through which the beam has been deviated. If
and the base angles of the prism are each what is n?

94. If the apex angle in the previous problem is

and

, what is the value of ?

95. The light incident on polarizing sheet is

linearly polarized at an angle of

with

respect to the transmission axis of . Sheet

is placed so that its axis is parallel to the

polarization axis of the incident light, that is,

also at

with respect to . (a) What

fraction of the incident light passes through ?

(b) What fraction of the incident light is passed

by the combination? (c) By rotating , a

maximum in transmitted intensity is obtained.

What is the ratio of this maximum intensity to

the intensity of transmitted light when is at

with respect to ?

96. Prove that if I is the intensity of light

transmitted by two polarizing filters with axes at

an angle and is the intensity when the axes

are at an angle

then

the

original intensity. (Hint: Use the trigonometric

identities

and

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CHAPTER 2
Geometric Optics and Image Formation
Figure 2.1 Cloud Gate is a public sculpture by Anish Kapoor located in Millennium Park in Chicago. Its stainless steel plates reflect and distort images around it, including the Chicago skyline. Dedicated in 2006, it has become a popular tourist attraction, illustrating how art can use the principles of physical optics to startle and entertain. (credit: modification of work by Dhilung Kirat)
Chapter Outline
2.1 Images Formed by Plane Mirrors 2.2 Spherical Mirrors 2.3 Images Formed by Refraction 2.4 Thin Lenses 2.5 The Eye 2.6 The Camera 2.7 The Simple Magnifier 2.8 Microscopes and Telescopes INTRODUCTION This chapter introduces the major ideas of geometric optics, which describe the formation

52

2 • Geometric Optics and Image Formation

of images due to reflection and refraction. It is called “geometric” optics because the images can be characterized using geometric constructions, such as ray diagrams. We have seen that visible light is an electromagnetic wave; however, its wave nature becomes evident only when light interacts with objects with dimensions comparable to the wavelength (about 500 nm for visible light). Therefore, the laws of geometric optics only apply to light interacting with objects much larger than the wavelength of the light.
2.1 Images Formed by Plane Mirrors
Learning Objectives By the end of this section, you will be able to:
• Describe how an image is formed by a plane mirror. • Distinguish between real and virtual images. • Find the location and characterize the orientation of an image created by a plane mirror.
You only have to look as far as the nearest bathroom to find an example of an image formed by a mirror. Images in a plane mirror are the same size as the object, are located behind the mirror, and are oriented in the same direction as the object (i.e., “upright”).
To understand how this happens, consider Figure 2.2. Two rays emerge from point P, strike the mirror, and reflect into the observer’s eye. Note that we use the law of reflection to construct the reflected rays. If the reflected rays are extended backward behind the mirror (see dashed lines in Figure 2.2), they seem to originate from point Q. This is where the image of point P is located. If we repeat this process for point , we obtain its image at point . You should convince yourself by using basic geometry that the image height (the distance from Q to ) is the same as the object height (the distance from P to ). By forming images of all points of the object, we obtain an upright image of the object behind the mirror.

Figure 2.2 Two light rays originating from point P on an object are reflected by a flat mirror into the eye of an observer. The reflected rays are obtained by using the law of reflection. Extending these reflected rays backward, they seem to come from point Q behind the mirror, which is where the virtual image is located. Repeating this process for point gives the image point . The image height is thus the same as the object height, the image is upright, and the object distance is the same as the image distance . (credit: modification of work by Kevin Dufendach)
Notice that the reflected rays appear to the observer to come directly from the image behind the mirror. In reality, these rays come from the points on the mirror where they are reflected. The image behind the mirror is called a virtual image because it cannot be projected onto a screen—the rays only appear to originate from a common point behind the mirror. If you walk behind the mirror, you cannot see the image, because the rays do not go there. However, in front of the mirror, the rays behave exactly as if they come from behind the mirror, so that is where the virtual image is located.
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2.1 • Images Formed by Plane Mirrors

53

Later in this chapter, we discuss real images; a real image can be projected onto a screen because the rays physically go through the image. You can certainly see both real and virtual images. The difference is that a virtual image cannot be projected onto a screen, whereas a real image can.
Locating an Image in a Plane Mirror
The law of reflection tells us that the angle of incidence is the same as the angle of reflection. Applying this to triangles PAB and QAB in Figure 2.2 and using basic geometry shows that they are congruent triangles. This means that the distance PB from the object to the mirror is the same as the distance BQ from the mirror to the image. The object distance (denoted ) is the distance from the mirror to the object (or, more generally, from the center of the optical element that creates its image). Similarly, the image distance (denoted ) is the distance from the mirror to the image (or, more generally, from the center of the optical element that creates it). If we measure distances from the mirror, then the object and image are in opposite directions, so for a plane mirror, the object and image distances should have the opposite signs:
2.1
An extended object such as the container in Figure 2.2 can be treated as a collection of points, and we can apply the method above to locate the image of each point on the extended object, thus forming the extended image.
Multiple Images
If an object is situated in front of two mirrors, you may see images in both mirrors. In addition, the image in the first mirror may act as an object for the second mirror, so the second mirror may form an image of the image. If the mirrors are placed parallel to each other and the object is placed at a point other than the midpoint between them, then this process of image-of-an-image continues without end, as you may have noticed when standing in a hallway with mirrors on each side. This is shown in Figure 2.3, which shows three images produced by the blue object. Notice that each reflection reverses front and back, just like pulling a right-hand glove inside out produces a left-hand glove (this is why a reflection of your right hand is a left hand). Thus, the fronts and backs of images 1 and 2 are both inverted with respect to the object, and the front and back of image 3 is inverted with respect to image 2, which is the object for image 3.

Figure 2.3 Two parallel mirrors can produce, in theory, an infinite number of images of an object placed off center between the mirrors. Three of these images are shown here. The front and back of each image is inverted with respect to its object. Note that the colors are only to identify the images. For normal mirrors, the color of an image is essentially the same as that of its object.
You may have noticed that image 3 is smaller than the object, whereas images 1 and 2 are the same size as the object. The ratio of the image height with respect to the object height is called magnification. More will be said about magnification in the next section.
Infinite reflections may terminate. For instance, two mirrors at right angles form three images, as shown in part (a) of Figure 2.4. Images 1 and 2 result from rays that reflect from only a single mirror, but image 1,2 is formed by rays that reflect from both mirrors. This is shown in the ray-tracing diagram in part (b) of Figure 2.4. To find image 1,2, you have to look behind the corner of the two mirrors.

54

2 • Geometric Optics and Image Formation

Figure 2.4 Two mirrors can produce multiple images. (a) Three images of a plastic head are visible in the two mirrors at a right angle. (b) A single object reflecting from two mirrors at a right angle can produce three images, as shown by the green, purple, and red images.
2.2 Spherical Mirrors
Learning Objectives By the end of this section, you will be able to:
• Describe image formation by spherical mirrors. • Use ray diagrams and the mirror equation to calculate the properties of an image in a spherical mirror.
The image in a plane mirror has the same size as the object, is upright, and is the same distance behind the mirror as the object is in front of the mirror. A curved mirror, on the other hand, can form images that may be larger or smaller than the object and may form either in front of the mirror or behind it. In general, any curved surface will form an image, although some images make be so distorted as to be unrecognizable (think of fun house mirrors).
Because curved mirrors can create such a rich variety of images, they are used in many optical devices that find many uses. We will concentrate on spherical mirrors for the most part, because they are easier to manufacture than mirrors such as parabolic mirrors and so are more common.
Curved Mirrors
We can define two general types of spherical mirrors. If the reflecting surface is the outer side of the sphere, the mirror is called a convex mirror. If the inside surface is the reflecting surface, it is called a concave mirror.
Symmetry is one of the major hallmarks of many optical devices, including mirrors and lenses. The symmetry axis of such optical elements is often called the principal axis or optical axis. For a spherical mirror, the optical axis passes through the mirror’s center of curvature and the mirror’s vertex, as shown in Figure 2.5.

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2.2 • Spherical Mirrors

55

Figure 2.5 A spherical mirror is formed by cutting out a piece of a sphere and silvering either the inside or outside surface. A concave mirror has silvering on the interior surface (think “cave”), and a convex mirror has silvering on the exterior surface.
Consider rays that are parallel to the optical axis of a parabolic mirror, as shown in part (a) of Figure 2.6. Following the law of reflection, these rays are reflected so that they converge at a point, called the focal point. Part (b) of this figure shows a spherical mirror that is large compared with its radius of curvature. For this mirror, the reflected rays do not cross at the same point, so the mirror does not have a well-defined focal point. This is called spherical aberration and results in a blurred image of an extended object. Part (c) shows a spherical mirror that is small compared to its radius of curvature. This mirror is a good approximation of a parabolic mirror, so rays that arrive parallel to the optical axis are reflected to a well-defined focal point. The distance along the optical axis from the mirror to the focal point is called the focal length of the mirror.
Figure 2.6 (a) Parallel rays reflected from a parabolic mirror cross at a single point called the focal point F. (b) Parallel rays reflected from a large spherical mirror do not cross at a common point. (c) If a spherical mirror is small compared with its radius of curvature, it better approximates the central part of a parabolic mirror, so parallel rays essentially cross at a common point. The distance along the optical axis from the mirror to the focal point is the focal length f of the mirror.
A convex spherical mirror also has a focal point, as shown in Figure 2.7. Incident rays parallel to the optical axis are reflected from the mirror and seem to originate from point F at focal length f behind the mirror. Thus, the focal point is virtual because no real rays actually pass through it; they only appear to originate from it.

56

2 • Geometric Optics and Image Formation

Figure 2.7 (a) Rays reflected by a convex spherical mirror: Incident rays of light parallel to the optical axis are reflected from a convex spherical mirror and seem to originate from a well-defined focal point at focal distance f on the opposite side of the mirror. The focal point is virtual because no real rays pass through it. (b) Photograph of a virtual image formed by a convex mirror. (credit b: modification of work by Jenny Downing)

How does the focal length of a mirror relate to the mirror’s radius of curvature? Figure 2.8 shows a single ray

that is reflected by a spherical concave mirror. The incident ray is parallel to the optical axis. The point at

which the reflected ray crosses the optical axis is the focal point. Note that all incident rays that are parallel to

the optical axis are reflected through the focal point—we only show one ray for simplicity. We want to find how

the focal length FP (denoted by f) relates to the radius of curvature of the mirror, R, whose length is

. The law of reflection tells us that angles OXC and CXF are the same, and because the incident

ray is parallel to the optical axis, angles OXC and XCP are also the same. Thus, triangle CXF is an isosceles

triangle with

. If the angle is small (so that

; this is called the “small-angle

approximation”), then

or

. Inserting this into the equation for the radius R, we get

Figure 2.8 Reflection in a concave mirror. In the small-angle approximation, a ray that is parallel to the optical axis CP is reflected through the focal point F of the mirror.
In other words, in the small-angle approximation, the focal length f of a concave spherical mirror is half of its radius of curvature, R:
2.2
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2.2 • Spherical Mirrors

57

In this chapter, we assume that the small-angle approximation (also called the paraxial approximation) is

always valid. In this approximation, all rays are paraxial rays, which means that they make a small angle with

the optical axis and are at a distance much less than the radius of curvature from the optical axis. In this case,

their angles of reflection are small angles, so

.

Using Ray Tracing to Locate Images

To find the location of an image formed by a spherical mirror, we first use ray tracing, which is the technique of drawing rays and using the law of reflection to determine the reflected rays (later, for lenses, we use the law of refraction to determine refracted rays). Combined with some basic geometry, we can use ray tracing to find the focal point, the image location, and other information about how a mirror manipulates light. In fact, we already used ray tracing above to locate the focal point of spherical mirrors, or the image distance of flat mirrors. To locate the image of an object, you must locate at least two points of the image. Locating each point requires drawing at least two rays from a point on the object and constructing their reflected rays. The point at which the reflected rays intersect, either in real space or in virtual space, is where the corresponding point of the image is located. To make ray tracing easier, we concentrate on four “principal” rays whose reflections are easy to construct.

Figure 2.9 shows a concave mirror and a convex mirror, each with an arrow-shaped object in front of it. These are the objects whose images we want to locate by ray tracing. To do so, we draw rays from point Q that is on the object but not on the optical axis. We choose to draw our ray from the tip of the object. Principal ray 1 goes from point Q and travels parallel to the optical axis. The reflection of this ray must pass through the focal point, as discussed above. Thus, for the concave mirror, the reflection of principal ray 1 goes through focal point F, as shown in part (b) of the figure. For the convex mirror, the backward extension of the reflection of principal ray 1 goes through the focal point (i.e., a virtual focus). Principal ray 2 travels first on the line going through the focal point and then is reflected back along a line parallel to the optical axis. Principal ray 3 travels toward the center of curvature of the mirror, so it strikes the mirror at normal incidence and is reflected back along the line from which it came. Finally, principal ray 4 strikes the vertex of the mirror and is reflected symmetrically about the optical axis.

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2 • Geometric Optics and Image Formation

Figure 2.9 The four principal rays shown for both (a) a concave mirror and (b) a convex mirror. The image forms where the rays intersect (for real images) or where their backward extensions intersect (for virtual images).
The four principal rays intersect at point , which is where the image of point Q is located. To locate point , drawing any two of these principle rays would suffice. We are thus free to choose whichever of the principal rays we desire to locate the image. Drawing more than two principal rays is sometimes useful to verify that the ray tracing is correct.
To completely locate the extended image, we need to locate a second point in the image, so that we know how the image is oriented. To do this, we trace the principal rays from the base of the object. In this case, all four principal rays run along the optical axis, reflect from the mirror, and then run back along the optical axis. The difficulty is that, because these rays are collinear, we cannot determine a unique point where they intersect. All we know is that the base of the image is on the optical axis. However, because the mirror is symmetrical from top to bottom, it does not change the vertical orientation of the object. Thus, because the object is vertical, the image must be vertical. Therefore, the image of the base of the object is on the optical axis directly above the image of the tip, as drawn in the figure.
For the concave mirror, the extended image in this case forms between the focal point and the center of curvature of the mirror. It is inverted with respect to the object, is a real image, and is smaller than the object. Were we to move the object closer to or farther from the mirror, the characteristics of the image would change. For example, we show, as a later exercise, that an object placed between a concave mirror and its focal point leads to a virtual image that is upright and larger than the object. For the convex mirror, the extended image forms between the focal point and the mirror. It is upright with respect to the object, is a virtual image, and is smaller than the object.
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Summary of Ray-Tracing Rules

Ray tracing is very useful for mirrors. The rules for ray tracing are summarized here for reference:

• A ray travelling parallel to the optical axis of a spherical mirror is reflected along a line that goes through the focal point of the mirror (ray 1 in Figure 2.9).
• A ray travelling along a line that goes through the focal point of a spherical mirror is reflected along a line parallel to the optical axis of the mirror (ray 2 in Figure 2.9).
• A ray travelling along a line that goes through the center of curvature of a spherical mirror is reflected back along the same line (ray 3 in Figure 2.9).
• A ray that strikes the vertex of a spherical mirror is reflected symmetrically about the optical axis of the mirror (ray 4 in Figure 2.9).

We use ray tracing to illustrate how images are formed by mirrors and to obtain numerical information about optical properties of the mirror. If we assume that a mirror is small compared with its radius of curvature, we can also use algebra and geometry to derive a mirror equation, which we do in the next section. Combining ray tracing with the mirror equation is a good way to analyze mirror systems.

Image Formation by Reflection—The Mirror Equation

For a plane mirror, we showed that the image formed has the same height and orientation as the object, and it is located at the same distance behind the mirror as the object is in front of the mirror. Although the situation is a bit more complicated for curved mirrors, using geometry leads to simple formulas relating the object and image distances to the focal lengths of concave and convex mirrors.

Consider the object OP shown in Figure 2.10. The center of curvature of the mirror is labeled C and is a

distance R from the vertex of the mirror, as marked in the figure. The object and image distances are labeled

and , and the object and image heights are labeled and , respectively. Because the angles and are

alternate interior angles, we know that they have the same magnitude. However, they must differ in sign if we

measure angles from the optical axis, so

. An analogous scenario holds for the angles and . The law

of reflection tells us that they have the same magnitude, but their signs must differ if we measure angles from

the optical axis. Thus,

. Taking the tangent of the angles and , and using the property that

, gives us

2.3

Figure 2.10 Image formed by a concave mirror.
Similarly, taking the tangent of and gives

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2 • Geometric Optics and Image Formation

Combining these two results gives

After a little algebra, this becomes
2.4
No approximation is required for this result, so it is exact. However, as discussed above, in the small-angle approximation, the focal length of a spherical mirror is one-half the radius of curvature of the mirror, or
. Inserting this into Equation 2.3 gives the mirror equation:
2.5
The mirror equation relates the image and object distances to the focal distance and is valid only in the smallangle approximation. Although it was derived for a concave mirror, it also holds for convex mirrors (proving this is left as an exercise). We can extend the mirror equation to the case of a plane mirror by noting that a plane mirror has an infinite radius of curvature. This means the focal point is at infinity, so the mirror equation simplifies to
2.6 which is the same as Equation 2.1 obtained earlier.
Notice that we have been very careful with the signs in deriving the mirror equation. For a plane mirror, the image distance has the opposite sign of the object distance. Also, the real image formed by the concave mirror in Figure 2.10 is on the opposite side of the optical axis with respect to the object. In this case, the image height should have the opposite sign of the object height. To keep track of the signs of the various quantities in the mirror equation, we now introduce a sign convention.
Sign convention for spherical mirrors
Using a consistent sign convention is very important in geometric optics. It assigns positive or negative values for the quantities that characterize an optical system. Understanding the sign convention allows you to describe an image without constructing a ray diagram. This text uses the following sign convention:
1. The focal length f is positive for concave mirrors and negative for convex mirrors. 2. The image distance is positive for real images and negative for virtual images.
Notice that rule 1 means that the radius of curvature of a spherical mirror can be positive or negative. What does it mean to have a negative radius of curvature? This means simply that the radius of curvature for a convex mirror is defined to be negative.
Image magnification
Let’s use the sign convention to further interpret the derivation of the mirror equation. In deriving this equation, we found that the object and image heights are related by
2.7
See Equation 2.3. Both the object and the image formed by the mirror in Figure 2.10 are real, so the object and image distances are both positive. The highest point of the object is above the optical axis, so the object height is positive. The image, however, is below the optical axis, so the image height is negative. Thus, this sign convention is consistent with our derivation of the mirror equation.
Equation 2.7 in fact describes the linear magnification (often simply called “magnification”) of the image in terms of the object and image distances. We thus define the dimensionless magnification m as follows:

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2.8

If m is positive, the image is upright, and if m is negative, the image is inverted. If

, the image is larger

than the object, and if

, the image is smaller than the object. With this definition of magnification, we

get the following relation between the vertical and horizontal object and image distances:

2.9

This is a very useful relation because it lets you obtain the magnification of the image from the object and image distances, which you can obtain from the mirror equation.
EXAMPLE 2.1
Solar Electric Generating System
One of the solar technologies used today for generating electricity involves a device (called a parabolic trough or concentrating collector) that concentrates sunlight onto a blackened pipe that contains a fluid. This heated fluid is pumped to a heat exchanger, where the thermal energy is transferred to another system that is used to generate steam and eventually generates electricity through a conventional steam cycle. Figure 2.11 shows such a working system in southern California. The real mirror is a parabolic cylinder with its focus located at the pipe; however, we can approximate the mirror as exactly one-quarter of a circular cylinder.

Figure 2.11 Parabolic trough collectors are used to generate electricity in southern California. (credit: “kjkolb”/Wikimedia Commons)

a. If we want the rays from the sun to focus at 40.0 cm from the mirror, what is the radius of the mirror?

b. What is the amount of sunlight concentrated onto the pipe, per meter of pipe length, assuming the

insolation (incident solar radiation) is 900

?

c. If the fluid-carrying pipe has a 2.00-cm diameter, what is the temperature increase of the fluid per meter

of pipe over a period of 1 minute? Assume that all solar radiation incident on the reflector is absorbed by

the pipe, and that the fluid is mineral oil.

Strategy First identify the physical principles involved. Part (a) is related to the optics of spherical mirrors. Part (b) involves a little math, primarily geometry. Part (c) requires an understanding of heat and density.

Solution

a. The sun is the object, so the object distance is essentially infinity:

. The desired image distance is

. We use the mirror equation to find the focal length of the mirror:

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2 • Geometric Optics and Image Formation

Thus, the radius of the mirror is

.

b. The insolation is 900

. You must find the cross-sectional area A of the concave mirror, since the

power delivered is 900

. The mirror in this case is estimated as a quarter-section of a cylinder,

so the area for a length L of the mirror is

. The area for a length of 1.00 m is then

The insolation on the 1.00-m length of pipe is then

c. The increase in temperature is given by section of pipe is

. The mass m of the mineral oil in the one-meter

Therefore, the increase in temperature in one minute is

Significance

An array of such pipes in the California desert can provide a thermal output of 250 MW on a sunny day, with

fluids reaching temperatures as high as

. We are considering only one meter of pipe here and ignoring

heat losses along the pipe.

EXAMPLE 2.2

Image in a Convex Mirror
A keratometer is a device used to measure the curvature of the cornea of the eye, particularly for fitting contact lenses. Light is reflected from the cornea, which acts like a convex mirror, and the keratometer measures the magnification of the image. The smaller the magnification, the smaller the radius of curvature of the cornea. If the light source is 12 cm from the cornea and the image magnification is 0.032, what is the radius of curvature of the cornea?

Strategy

If you find the focal length of the convex mirror formed by the cornea, then you know its radius of curvature

(it’s twice the focal length). The object distance is

and the magnification is

. First find the

image distance and then solve for the focal length f.

Solution Start with the equation for magnification,

. Solving for and inserting the given values yields

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63

where we retained an extra significant figure because this is an intermediate step in the calculation. Solve the mirror equation for the focal length f and insert the known values for the object and image distances. The result is

The radius of curvature is twice the focal length, so
Significance The focal length is negative, so the focus is virtual, as expected for a concave mirror and a real object. The radius of curvature found here is reasonable for a cornea. The distance from cornea to retina in an adult eye is about 2.0 cm. In practice, corneas may not be spherical, which complicates the job of fitting contact lenses. Note that the image distance here is negative, consistent with the fact that the image is behind the mirror. Thus, the image is virtual because no rays actually pass through it. In the problems and exercises, you will show that, for a fixed object distance, a smaller radius of curvature corresponds to a smaller the magnification.
PROBLEM-SOLVING STRATEGY
Spherical Mirrors
Step 1. First make sure that image formation by a spherical mirror is involved. Step 2. Determine whether ray tracing, the mirror equation, or both are required. A sketch is very useful even if ray tracing is not specifically required by the problem. Write symbols and known values on the sketch. Step 3. Identify exactly what needs to be determined in the problem (identify the unknowns). Step 4. Make a list of what is given or can be inferred from the problem as stated (identify the knowns). Step 5. If ray tracing is required, use the ray-tracing rules listed near the beginning of this section. Step 6. Most quantitative problems require using the mirror equation. Use the examples as guides for using the mirror equation. Step 7. Check to see whether the answer makes sense. Do the signs of object distance, image distance, and focal length correspond with what is expected from ray tracing? Is the sign of the magnification correct? Are the object and image distances reasonable?
Departure from the Small-Angle Approximation
The small-angle approximation is a cornerstone of the above discussion of image formation by a spherical mirror. When this approximation is violated, then the image created by a spherical mirror becomes distorted. Such distortion is called aberration. Here we briefly discuss two specific types of aberrations: spherical aberration and coma.
Spherical aberration
Consider a broad beam of parallel rays impinging on a spherical mirror, as shown in Figure 2.12.

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2 • Geometric Optics and Image Formation

Figure 2.12 (a) With spherical aberration, the rays that are farther from the optical axis and the rays that are closer to the optical axis are focused at different points. Notice that the aberration gets worse for rays farther from the optical axis. (b) For comatic aberration, parallel rays that are not parallel to the optical axis are focused at different heights and at different focal lengths, so the image contains a “tail” like a comet (which is “coma” in Latin). Note that the colored rays are only to facilitate viewing; the colors do not indicate the color of the light.
The farther from the optical axis the rays strike, the worse the spherical mirror approximates a parabolic mirror. Thus, these rays are not focused at the same point as rays that are near the optical axis, as shown in the figure. Because of spherical aberration, the image of an extended object in a spherical mirror will be blurred. Spherical aberrations are characteristic of the mirrors and lenses that we consider in the following section of this chapter (more sophisticated mirrors and lenses are needed to eliminate spherical aberrations).
Coma or comatic aberration
Coma is similar to spherical aberration, but arises when the incoming rays are not parallel to the optical axis, as shown in part (b) of Figure 2.12. Recall that the small-angle approximation holds for spherical mirrors that are small compared to their radius. In this case, spherical mirrors are good approximations of parabolic mirrors. Parabolic mirrors focus all rays that are parallel to the optical axis at the focal point. However, parallel rays that are not parallel to the optical axis are focused at different heights and at different focal lengths, as show in part (b) of Figure 2.12. Because a spherical mirror is symmetric about the optical axis, the various colored rays in this figure create circles of the corresponding color on the focal plane. Although a spherical mirror is shown in part (b) of Figure 2.12, comatic aberration occurs also for parabolic mirrors—it does not result from a breakdown in the small-angle approximation. Spherical aberration, however, occurs only for spherical mirrors and is a result of a breakdown in the small-angle approximation. We will discuss both coma and spherical aberration later in this chapter, in connection with telescopes.
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2.3 • Images Formed by Refraction

65

2.3 Images Formed by Refraction
Learning Objectives By the end of this section, you will be able to:
• Describe image formation by a single refracting surface • Determine the location of an image and calculate its properties by using a ray diagram • Determine the location of an image and calculate its properties by using the equation for a single refracting
surface
When rays of light propagate from one medium to another, these rays undergo refraction, which is when light waves are bent at the interface between two media. The refracting surface can form an image in a similar fashion to a reflecting surface, except that the law of refraction (Snell’s law) is at the heart of the process instead of the law of reflection.
Refraction at a Plane Interface—Apparent Depth
If you look at a straight rod partially submerged in water, it appears to bend at the surface (Figure 2.13). The reason behind this curious effect is that the image of the rod inside the water forms a little closer to the surface than the actual position of the rod, so it does not line up with the part of the rod that is above the water. The same phenomenon explains why a fish in water appears to be closer to the surface than it actually is.

Figure 2.13 Bending of a rod at a water-air interface. Point P on the rod appears to be at point Q, which is where the image of point P forms due to refraction at the air-water interface.
To study image formation as a result of refraction, consider the following questions:
1. What happens to the rays of light when they enter or pass through a different medium? 2. Do the refracted rays originating from a single point meet at some point or diverge away from each other?
To be concrete, we consider a simple system consisting of two media separated by a plane interface (Figure 2.14). The object is in one medium and the observer is in the other. For instance, when you look at a fish from above the water surface, the fish is in medium 1 (the water) with refractive index 1.33, and your eye is in medium 2 (the air) with refractive index 1.00, and the surface of the water is the interface. The depth that you “see” is the image height and is called the apparent depth. The actual depth of the fish is the object height
.

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2 • Geometric Optics and Image Formation

Figure 2.14 Apparent depth due to refraction. The real object at point P creates an image at point Q. The image is not at the same depth as the object, so the observer sees the image at an “apparent depth.”

The apparent depth depends on the angle at which you view the image. For a view from above (the so-called

“normal” view), we can approximate the refraction angle to be small, and replace sin in Snell’s law by tan .

With this approximation, you can use the triangles

and

to show that the apparent depth is given

by

2.10

The derivation of this result is left as an exercise. Thus, a fish appears at 3/4 of the real depth when viewed from above.
Refraction at a Spherical Interface
Spherical shapes play an important role in optics primarily because high-quality spherical shapes are far easier to manufacture than other curved surfaces. To study refraction at a single spherical surface, we assume that the medium with the spherical surface at one end continues indefinitely (a “semi-infinite” medium).
Refraction at a convex surface
Consider a point source of light at point P in front of a convex surface made of glass (see Figure 2.15). Let R be the radius of curvature, be the refractive index of the medium in which object point P is located, and be the refractive index of the medium with the spherical surface. We want to know what happens as a result of refraction at this interface.

Figure 2.15 Refraction at a convex surface

.

Because of the symmetry involved, it is sufficient to examine rays in only one plane. The figure shows a ray of light that starts at the object point P, refracts at the interface, and goes through the image point . We derive a formula relating the object distance , the image distance , and the radius of curvature R.

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Applying Snell’s law to the ray emanating from point P gives

approximation, so

and Snell’s law then takes the form

From the geometry of the figure, we see that

Inserting these expressions into Snell’s law gives

Using the diagram, we calculate the tangent of the angles

:

2.3 • Images Formed by Refraction

67

. We work in the small-angle

Again using the small-angle approximation, we find that

, so the above relationships become

Putting these angles into Snell’s law gives

We can write this more conveniently as
2.11
If the object is placed at a special point called the first focus, or the object focus , then the image is formed at infinity, as shown in part (a) of Figure 2.16.

Figure 2.16 (a) First focus (called the “object focus”) for refraction at a convex surface. (b) Second focus (called “image focus”) for refraction at a convex surface.

We can find the location of the first focus by setting

in the preceding equation.

2.12

2.13

Similarly, we can define a second focus or image focus where the image is formed for an object that is far

away [part (b)]. The location of the second focus is obtained from Equation 2.11 by setting

:

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2 • Geometric Optics and Image Formation

Note that the object focus is at a different distance from the vertex than the image focus because

.

Sign convention for single refracting surfaces
Although we derived this equation for refraction at a convex surface, the same expression holds for a concave surface, provided we use the following sign convention:

1.

if surface is convex toward object; otherwise,

2.

if image is real and on opposite side from the object; otherwise,

2.4 Thin Lenses

Learning Objectives By the end of this section, you will be able to:
• Use ray diagrams to locate and describe the image formed by a lens
• Employ the thin-lens equation to describe and locate the image formed by a lens

Lenses are found in a huge array of optical instruments, ranging from a simple magnifying glass to a camera’s zoom lens to the eye itself. In this section, we use the Snell’s law to explore the properties of lenses and how they form images.

The word “lens” derives from the Latin word for a lentil bean, the shape of which is similar to a convex lens. However, not all lenses have the same shape. Figure 2.17 shows a variety of different lens shapes. The vocabulary used to describe lenses is the same as that used for spherical mirrors: The axis of symmetry of a lens is called the optical axis, where this axis intersects the lens surface is called the vertex of the lens, and so forth.

Figure 2.17 Various types of lenses: Note that a converging lens has a thicker “waist,” whereas a diverging lens has a thinner waist.
A convex or converging lens is shaped so that all light rays that enter it parallel to its optical axis intersect (or focus) at a single point on the optical axis on the opposite side of the lens, as shown in part (a) of Figure 2.18. Likewise, a concave or diverging lens is shaped so that all rays that enter it parallel to its optical axis diverge, as shown in part (b). To understand more precisely how a lens manipulates light, look closely at the top ray that goes through the converging lens in part (a). Because the index of refraction of the lens is greater than that of air, Snell’s law tells us that the ray is bent toward the perpendicular to the interface as it enters the lens. Likewise, when the ray exits the lens, it is bent away from the perpendicular. The same reasoning applies to the diverging lenses, as shown in part (b). The overall effect is that light rays are bent toward the optical axis for a converging lens and away from the optical axis for diverging lenses. For a converging lens, the point at which
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69

the rays cross is the focal point F of the lens. For a diverging lens, the point from which the rays appear to originate is the (virtual) focal point. The distance from the center of the lens to its focal point is the focal length f of the lens.

Figure 2.18 Rays of light entering (a) a converging lens and (b) a diverging lens, parallel to its axis, converge at its focal point F. The distance from the center of the lens to the focal point is the lens’s focal length f. Note that the light rays are bent upon entering and exiting the lens, with the overall effect being to bend the rays toward the optical axis.
A lens is considered to be thin if its thickness t is much less than the radii of curvature of both surfaces, as shown in Figure 2.19. In this case, the rays may be considered to bend once at the center of the lens. For the case drawn in the figure, light ray 1 is parallel to the optical axis, so the outgoing ray is bent once at the center of the lens and goes through the focal point. Another important characteristic of thin lenses is that light rays that pass through the center of the lens are undeviated, as shown by light ray 2.

Figure 2.19 In the thin-lens approximation, the thickness t of the lens is much, much less than the radii and of curvature of the surfaces of the lens. Light rays are considered to bend at the center of the lens, such as light ray 1. Light ray 2 passes through the center of the lens and is undeviated in the thin-lens approximation.
As noted in the initial discussion of Snell’s law, the paths of light rays are exactly reversible. This means that the direction of the arrows could be reversed for all of the rays in Figure 2.18. For example, if a point-light source is placed at the focal point of a convex lens, as shown in Figure 2.20, parallel light rays emerge from the other side.

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2 • Geometric Optics and Image Formation

Figure 2.20 A small light source, like a light bulb filament, placed at the focal point of a convex lens results in parallel rays of light emerging from the other side. The paths are exactly the reverse of those shown in Figure 2.18 in converging and diverging lenses. This technique is used in lighthouses and sometimes in traffic lights to produce a directional beam of light from a source that emits light in all directions.
Ray Tracing and Thin Lenses
Ray tracing is the technique of determining or following (tracing) the paths taken by light rays.
Ray tracing for thin lenses is very similar to the technique we used with spherical mirrors. As for mirrors, ray tracing can accurately describe the operation of a lens. The rules for ray tracing for thin lenses are similar to those of spherical mirrors:
1. A ray entering a converging lens parallel to the optical axis passes through the focal point on the other side of the lens (ray 1 in part (a) of Figure 2.21). A ray entering a diverging lens parallel to the optical axis exits along the line that passes through the focal point on the same side of the lens (ray 1 in part (b) of the figure).
2. A ray passing through the center of either a converging or a diverging lens is not deviated (ray 2 in parts (a) and (b)).
3. For a converging lens, a ray that passes through the focal point exits the lens parallel to the optical axis (ray 3 in part (a)). For a diverging lens, a ray that approaches along the line that passes through the focal point on the opposite side exits the lens parallel to the axis (ray 3 in part (b)).

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71

Figure 2.21 Thin lenses have the same focal lengths on either side. (a) Parallel light rays from the object toward a converging lens cross at its focal point on the right. (b) Parallel light rays from the object entering a diverging lens from the left seem to come from the focal point on the left.
Thin lenses work quite well for monochromatic light (i.e., light of a single wavelength). However, for light that contains several wavelengths (e.g., white light), the lenses work less well. The problem is that, as we learned in the previous chapter, the index of refraction of a material depends on the wavelength of light. This phenomenon is responsible for many colorful effects, such as rainbows. Unfortunately, this phenomenon also leads to aberrations in images formed by lenses. In particular, because the focal distance of the lens depends on the index of refraction, it also depends on the wavelength of the incident light. This means that light of different wavelengths will focus at different points, resulting is so-called “chromatic aberrations.” In particular, the edges of an image of a white object will become colored and blurred. Special lenses called doublets are capable of correcting chromatic aberrations. A doublet is formed by gluing together a converging lens and a diverging lens. The combined doublet lens produces significantly reduced chromatic aberrations.
Image Formation by Thin Lenses
We use ray tracing to investigate different types of images that can be created by a lens. In some circumstances, a lens forms a real image, such as when a movie projector casts an image onto a screen. In other cases, the image is a virtual image, which cannot be projected onto a screen. Where, for example, is the image formed by eyeglasses? We use ray tracing for thin lenses to illustrate how they form images, and then we develop equations to analyze quantitatively the properties of thin lenses.
Consider an object some distance away from a converging lens, as shown in Figure 2.22. To find the location and size of the image, we trace the paths of selected light rays originating from one point on the object, in this case, the tip of the arrow. The figure shows three rays from many rays that emanate from the tip of the arrow. These three rays can be traced by using the ray-tracing rules given above.
• Ray 1 enters the lens parallel to the optical axis and passes through the focal point on the opposite side (rule 1).
• Ray 2 passes through the center of the lens and is not deviated (rule 2).

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2 • Geometric Optics and Image Formation

• Ray 3 passes through the focal point on its way to the lens and exits the lens parallel to the optical axis (rule 3).
The three rays cross at a single point on the opposite side of the lens. Thus, the image of the tip of the arrow is located at this point. All rays that come from the tip of the arrow and enter the lens are refracted and cross at the point shown.
After locating the image of the tip of the arrow, we need another point of the image to orient the entire image of the arrow. We chose to locate the image base of the arrow, which is on the optical axis. As explained in the section on spherical mirrors, the base will be on the optical axis just above the image of the tip of the arrow (due to the top-bottom symmetry of the lens). Thus, the image spans the optical axis to the (negative) height shown. Rays from another point on the arrow, such as the middle of the arrow, cross at another common point, thus filling in the rest of the image.
Although three rays are traced in this figure, only two are necessary to locate a point of the image. It is best to trace rays for which there are simple ray-tracing rules.

Figure 2.22 Ray tracing is used to locate the image formed by a lens. Rays originating from the same point on the object are traced—the three chosen rays each follow one of the rules for ray tracing, so that their paths are easy to determine. The image is located at the point where the rays cross. In this case, a real image—one that can be projected on a screen—is formed.
Several important distances appear in the figure. As for a mirror, we define to be the object distance, or the distance of an object from the center of a lens. The image distance is defined to be the distance of the image from the center of a lens. The height of the object and the height of the image are indicated by and , respectively. Images that appear upright relative to the object have positive heights, and those that are inverted have negative heights. By using the rules of ray tracing and making a scale drawing with paper and pencil, like that in Figure 2.22, we can accurately describe the location and size of an image. But the real benefit of ray tracing is in visualizing how images are formed in a variety of situations.
Oblique Parallel Rays and Focal Plane
We have seen that rays parallel to the optical axis are directed to the focal point of a converging lens. In the case of a diverging lens, they come out in a direction such that they appear to be coming from the focal point on the opposite side of the lens (i.e., the side from which parallel rays enter the lens). What happens to parallel rays that are not parallel to the optical axis (Figure 2.23)? In the case of a converging lens, these rays do not converge at the focal point. Instead, they come together on another point in the plane called the focal plane. The focal plane contains the focal point and is perpendicular to the optical axis. As shown in the figure, parallel rays focus where the ray through the center of the lens crosses the focal plane.
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2.4 • Thin Lenses

73

Figure 2.23 Parallel oblique rays focus on a point in a focal plane.

Thin-Lens Equation

Ray tracing allows us to get a qualitative picture of image formation. To obtain numeric information, we derive a pair of equations from a geometric analysis of ray tracing for thin lenses. These equations, called the thinlens equation and the lens maker’s equation, allow us to quantitatively analyze thin lenses.

Consider the thick bi-convex lens shown in Figure 2.24. The index of refraction of the surrounding medium is

(if the lens is in air, then

) and that of the lens is . The radii of curvatures of the two sides are

. We wish to find a relation between the object distance , the image distance , and the

parameters of the lens.

Figure 2.24 Figure for deriving the lens maker’s equation. Here, t is the thickness of lens, is the index of refraction of the exterior

medium, and is the index of refraction of the lens. We take the limit of

to obtain the formula for a thin lens.

To derive the thin-lens equation, we consider the image formed by the first refracting surface (i.e., left surface) and then use this image as the object for the second refracting surface. In the figure, the image from the first refracting surface is , which is formed by extending backwards the rays from inside the lens (these rays result from refraction at the first surface). This is shown by the dashed lines in the figure. Notice that this image is virtual because no rays actually pass through the point . To find the image distance
corresponding to the image , we use Equation 2.11. In this case, the object distance is , the image distance is , and the radius of curvature is . Inserting these into Equation 2.3 gives

2.14

The image is virtual and on the same side as the object, so

and

. The first surface is convex

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2 • Geometric Optics and Image Formation

toward the object, so

.

To find the object distance for the object Q formed by refraction from the second interface, note that the role of the indices of refraction and are interchanged in Equation 2.11. In Figure 2.24, the rays originate in the medium with index , whereas in Figure 2.15, the rays originate in the medium with index . Thus, we must interchange and in Equation 2.11. In addition, by consulting again Figure 2.24, we see that the object distance is and the image distance is . The radius of curvature is Inserting these quantities into Equation 2.11 gives

2.15

The image is real and on the opposite side from the object, so

and

. The second surface is convex

away from the object, so

. Equation 2.15 can be simplified by noting that

, where we have

taken the absolute value because is a negative number, whereas both and t are positive. We can dispense

with the absolute value if we negate , which gives

. Inserting this into Equation 2.15 gives

2.16

Summing Equation 2.14 and Equation 2.16 gives

2.17

In the thin-lens approximation, we assume that the lens is very thin compared to the first image distance, or

(or, equivalently,

). In this case, the third and fourth terms on the left-hand side of

Equation 2.17 cancel, leaving us with

Dividing by gives us finally
2.18
The left-hand side looks suspiciously like the mirror equation that we derived above for spherical mirrors. As done for spherical mirrors, we can use ray tracing and geometry to show that, for a thin lens,

2.19

where f is the focal length of the thin lens (this derivation is left as an exercise). This is the thin-lens equation. The focal length of a thin lens is the same to the left and to the right of the lens. Combining Equation 2.18 and Equation 2.19 gives

2.20

which is called the lens maker’s equation. It shows that the focal length of a thin lens depends only of the radii

of curvature and the index of refraction of the lens and that of the surrounding medium. For a lens in air,

and

, so the lens maker’s equation reduces to

2.21

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Sign conventions for lenses
To properly use the thin-lens equation, the following sign conventions must be obeyed:
1. is positive if the image is on the side opposite the object (i.e., real image); otherwise, is negative (i.e., virtual image).
2. f is positive for a converging lens and negative for a diverging lens. 3. R is positive for a surface convex toward the object, and negative for a surface concave toward object.
Magnification
By using a finite-size object on the optical axis and ray tracing, you can show that the magnification m of an image is

2.22

(where the three lines mean “is defined as”). This is exactly the same equation as we obtained for mirrors (see

Equation 2.8). If

, then the image has the same vertical orientation as the object (called an “upright”

image). If

, then the image has the opposite vertical orientation as the object (called an “inverted” image).

Using the Thin-Lens Equation

The thin-lens equation and the lens maker’s equation are broadly applicable to situations involving thin lenses. We explore many features of image formation in the following examples.

Consider a thin converging lens. Where does the image form and what type of image is formed as the object approaches the lens from infinity? This may be seen by using the thin-lens equation for a given focal length to plot the image distance as a function of object distance. In other words, we plot

for a given value of f. For

, the result is shown in part (a) of Figure 2.25.

Figure 2.25 (a) Image distance for a thin converging lens with

diverging lens with

.

as a function of object distance. (b) Same thing but for a

An object much farther than the focal length f from the lens should produce an image near the focal plane,

because the second term on the right-hand side of the equation above becomes negligible compared to the first

term, so we have

This can be seen in the plot of part (a) of the figure, which shows that the image

distance approaches asymptotically the focal length of 1 cm for larger object distances. As the object

approaches the focal plane, the image distance diverges to positive infinity. This is expected because an object

at the focal plane produces parallel rays that form an image at infinity (i.e., very far from the lens). When the

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2 • Geometric Optics and Image Formation

object is farther than the focal length from the lens, the image distance is positive, so the image is real, on the

opposite side of the lens from the object, and inverted (because

). When the object is closer than

the focal length from the lens, the image distance becomes negative, which means that the image is virtual, on

the same side of the lens as the object, and upright.

For a thin diverging lens of focal length

, a similar plot of image distance vs. object distance is

shown in part (b). In this case, the image distance is negative for all positive object distances, which means that

the image is virtual, on the same side of the lens as the object, and upright. These characteristics may also be

seen by ray-tracing diagrams (see Figure 2.26).

Figure 2.26 The red dots show the focal points of the lenses. (a) A real, inverted image formed from an object that is farther than the focal length from a converging lens. (b) A virtual, upright image formed from an object that is closer than a focal length from the lens. (c) A virtual, upright image formed from an object that is farther than a focal length from a diverging lens.
To see a concrete example of upright and inverted images, look at Figure 2.27, which shows images formed by converging lenses when the object (the person’s face in this case) is place at different distances from the lens. In part (a) of the figure, the person’s face is farther than one focal length from the lens, so the image is inverted. In part (b), the person’s face is closer than one focal length from the lens, so the image is upright.

Figure 2.27 (a) When a converging lens is held farther than one focal length from the man’s face, an inverted image is formed. Note that the image is in focus but the face is not, because the image is much closer to the camera taking this photograph than the face. (b) An upright image of the man’s face is produced when a converging lens is held at less than one focal length from his face. (credit a: modification of work by “DaMongMan”/Flickr; credit b: modification of work by Casey Fleser)
Work through the following examples to better understand how thin lenses work.
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77

PROBLEM-SOLVING STRATEGY
Lenses
Step 1. Determine whether ray tracing, the thin-lens equation, or both would be useful. Even if ray tracing is not used, a careful sketch is always very useful. Write symbols and values on the sketch. Step 2. Identify what needs to be determined in the problem (identify the unknowns). Step 3. Make a list of what is given or can be inferred from the problem (identify the knowns). Step 4. If ray tracing is required, use the ray-tracing rules listed near the beginning of this section. Step 5. Most quantitative problems require the use of the thin-lens equation and/or the lens maker’s equation. Solve these for the unknowns and insert the given quantities or use both together to find two unknowns. Step 7. Check to see if the answer is reasonable. Are the signs correct? Is the sketch or ray tracing consistent with the calculation?
EXAMPLE 2.3
Using the Lens Maker’s Equation
Find the radius of curvature of a biconcave lens symmetrically ground from a glass with index of refractive 1.55 so that its focal length in air is 20 cm (for a biconcave lens, both surfaces have the same radius of curvature). Strategy Use the thin-lens form of the lens maker’s equation:

where

and

. Since we are making a symmetric biconcave lens, we have

.

Solution We can determine the radius R of curvature from

Solving for R and inserting

gives

EXAMPLE 2.4

Converging Lens and Different Object Distances

Find the location, orientation, and magnification of the image for an 3.0 cm high object at each of the following

positions in front of a convex lens of focal length 10.0 cm.

,

, and

.

Strategy We start with the thin-lens equation

. Solve this for the image distance and insert the given

object distance and focal length.

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2 • Geometric Optics and Image Formation

Solution a. For

, this gives

The image is positive, so the image, is real, is on the opposite side of the lens from the object, and is 12.6 cm from the lens. To find the magnification and orientation of the image, use

The negative magnification means that the image is inverted. Since object. The size of the image is given by
b. For

, the image is smaller than the

The image distance is negative, so the image is virtual, is on the same side of the lens as the object, and is 10 cm from the lens. The magnification and orientation of the image are found from

The positive magnification means that the image is upright (i.e., it has the same orientation as the object).

Since

, the image is larger than the object. The size of the image is

c. For

The image distance is positive, so the image is real, is on the opposite side of the lens from the object, and is 20.0 cm from the lens. The magnification is

The negative magnification means that the image is inverted. Since the object.

, the image is the same size as

When solving problems in geometric optics, we often need to combine ray tracing and the lens equations. The following example demonstrates this approach.

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79

EXAMPLE 2.5

Choosing the Focal Length and Type of Lens
To project an image of a light bulb on a screen 1.50 m away, you need to choose what type of lens to use (converging or diverging) and its focal length (Figure 2.28). The distance between the lens and the lightbulb is fixed at 0.75 m. Also, what is the magnification and orientation of the image?

Strategy

The image must be real, so you choose to use a converging lens. The focal length can be found by using the

thin-lens equation and solving for the focal length. The object distance is

and the image distance

is

.

Solution Solve the thin lens for the focal length and insert the desired object and image distances:

The magnification is
Significance The minus sign for the magnification means that the image is inverted. The focal length is positive, as expected for a converging lens. Ray tracing can be used to check the calculation (see Figure 2.28). As expected, the image is inverted, is real, and is larger than the object.

Figure 2.28 A light bulb placed 0.75 m from a lens having a 0.50-m focal length produces a real image on a screen, as discussed in the example. Ray tracing predicts the image location and size.
2.5 The Eye
Learning Objectives By the end of this section, you will be able to:
• Understand the basic physics of how images are formed by the human eye • Recognize several conditions of impaired vision as well as the optics principles for treating these conditions
The human eye is perhaps the most interesting and important of all optical instruments. Our eyes perform a vast number of functions: They allow us to sense direction, movement, colors, and distance. In this section, we

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explore the geometric optics of the eye.

Physics of the Eye

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 2.29 shows the basic anatomy of the eye. The cornea and 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

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 2.29 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 2.1 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 2.30 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.

Material

Index of Refraction

Water

1.33

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2.5 • The Eye

81

Material

Index of Refraction

Air

1.0

Cornea

1.38

Aqueous humor 1.34

Lens

1.41*

Vitreous humor 1.34

Table 2.1 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.

Figure 2.30 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 must equal the lens-to-retina distance. Because the lens-to-retina distance does not change, the image distance 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 can use the thin-lens equations to quantitatively examine image formation by the eye. First, we define the optical power of a lens as
2.23
with the focal length f given in meters. The units of optical power are called “diopters” (D). That is, . Optometrists prescribe common eyeglasses and contact lenses in units of diopters. With
this definition of optical power, we can rewrite the thin-lens equations as

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2 • Geometric Optics and Image Formation

2.24
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:
2.25

EXAMPLE 2.6

Effective Focal Length of the Eye
The cornea and eye lens have focal lengths of 2.3 and 6.4 cm, respectively. Find the net focal length and optical power of the eye.

Strategy

The optical powers of the closely spaced lenses add, so

.

Solution Writing the equation for power in terms of the focal lengths gives

Hence, the focal length of the eye (cornea and lens together) is The optical power of the eye is

For clear vision, the image distance must equal the lens-to-retina distance. Normal vision is possible for

objects at distances

to infinity. The following example shows how to calculate the image distance

for an object placed at the near point of the eye.

EXAMPLE 2.7

Image of an object placed at the near point
The net focal length of a particular human eye is 1.7 cm. An object is placed at the near point of the eye. How far behind the lens is a focused image formed?

Strategy The near point is 25 cm from the eye, so the object distance is from the lens equation:

. We determine the image distance

Solution

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2.5 • The Eye

83

Therefore, the image is formed 1.8 cm behind the lens.

Significance

From the magnification formula, we find

. Since

, the image is inverted in

orientation with respect to the object. From the absolute value of m we see that the image is much smaller than the object; in fact, it is only 7% of the size of the object.

Vision Correction
The need for some type of vision correction is very common. Typical vision defects are easy to understand with geometric optics, and some are simple to correct. Figure 2.31 illustrates two common vision defects. Nearsightedness, or myopia, is the ability to see near objects, whereas distant objects are blurry. The eye overconverges the nearly parallel rays from a distant object, and the rays cross in front of the retina. More divergent rays from a close object are converged on the retina for a clear image. The distance to the farthest object that can be seen clearly is called the far point of the eye (normally the far point is at infinity). Farsightedness, or hyperopia, is the ability to see far objects clearly, whereas near objects are blurry. A farsighted eye does not sufficiently converge the rays from a near object to make the rays meet on the retina.

Figure 2.31 (a) The nearsighted (myopic) eye converges rays from a distant object in front of the retina, so they have diverged when they strike the retina, producing a blurry image. An eye lens that is too powerful can cause nearsightedness, or the eye may be too long. (b) The farsighted (hyperopic) eye is unable to converge the rays from a close object on the retina, producing blurry near-field vision. An eye lens with insufficient optical power or an eye that is too short can cause farsightedness.
Since the nearsighted eye overconverges light rays, the correction for nearsightedness consists of placing a diverging eyeglass lens in front of the eye, as shown in Figure 2.32. This reduces the optical power of an eye that is too powerful (recall that the focal length of a diverging lens is negative, so its optical power is negative). Another way to understand this correction is that a diverging lens will cause the incoming rays to diverge more to compensate for the excessive convergence caused by the lens system of the eye. The image produced by the

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2 • Geometric Optics and Image Formation

diverging eyeglass lens serves as the (optical) object for the eye, and because the eye cannot focus on objects beyond its far point, the diverging lens must form an image of distant (physical) objects at a point that is closer than the far point.

Figure 2.32 Correction of nearsightedness requires a diverging lens that compensates for overconvergence by the eye. The diverging lens produces an image closer to the eye than the physical object. This image serves as the optical object for the eye, and the nearsighted person can see it clearly because it is closer than their far point.

EXAMPLE 2.8

Correcting Nearsightedness
What optical power of eyeglass lens is needed to correct the vision of a nearsighted person whose far point is 30.0 cm? Assume the corrective lens is fixed 1.50 cm away from the eye.

Strategy

You want this nearsighted person to be able to see distant objects clearly, which means that the eyeglass lens

must produce an image 30.0 cm from the eye for an object at infinity. An image 30.0 cm from the eye will be

from the eyeglass lens. Therefore, we must have

when

.

The image distance is negative because it is on the same side of the eyeglass lens as the object.

Solution Since and are known, we can find the optical power of the eyeglass lens by using Equation 2.24:

Significance The negative optical power indicates a diverging (or concave) lens, as expected. If you examine eyeglasses for nearsighted people, you will find the lenses are thinnest in the center. Additionally, if you examine a prescription for eyeglasses for nearsighted people, you will find that the prescribed optical power is negative
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2.5 • The Eye

85

and given in units of diopters.
Correcting farsightedness consists simply of using the opposite type of lens as for nearsightedness (i.e., a converging lens), as shown in Figure 2.33. Such a lens will produce an image of physical objects that are closer than the near point at a distance that is between the near point and the far point, so that the person can see the image clearly. To determine the optical power needed for correction, you must therefore know the person’s near point, as explained in Example 2.9.

Figure 2.33 Correction of farsightedness uses a converging lens that compensates for the underconvergence by the eye. The converging lens produces an image farther from the eye than the object, so that the farsighted person can see it clearly.

EXAMPLE 2.9

Correcting Farsightedness
What optical power of eyeglass lens is needed to allow a farsighted person, whose near point is 1.00 m, to see an object clearly that is 25.0 cm from the eye? Assume the corrective lens is fixed 1.5 cm from the eye.

Strategy

When an object is 25.0 cm from the person’s eyes, the eyeglass lens must produce an image 1.00 m away (the

near point), so that the person can see it clearly. An image 1.00 m from the eye will be

from the eyeglass lens because the eyeglass lens is 1.5 cm from the eye. Therefore,

, where the minus sign indicates that the image is on the same side of the lens as the object. The

object is

from the eyeglass lens, so

.

Solution Since and are known, we can find the optical power of the eyeglass lens by using Equation 2.24:

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Significance The positive optical power indicates a converging (convex) lens, as expected. If you examine eyeglasses of farsighted people, you will find the lenses to be thickest in the center. In addition, prescription eyeglasses for farsighted people have a prescribed optical power that is positive.
2.6 The Camera
Learning Objectives By the end of this section, you will be able to:
• Describe the optics of a camera • Characterize the image created by a camera
Cameras are very common in our everyday life. Between 1825 and 1827, French inventor Nicéphore Niépce successfully photographed an image created by a primitive camera. Since then, enormous progress has been achieved in the design of cameras and camera-based detectors.
Initially, photographs were recorded by using the light-sensitive reaction of silver-based compounds such as silver chloride or silver bromide. Silver-based photographic paper was in common use until the advent of digital photography in the 1980s, which is intimately connected to charge-coupled device (CCD) detectors. In a nutshell, a CCD is a semiconductor chip that records images as a matrix of tiny pixels, each pixel located in a “bin” in the surface. Each pixel is capable of detecting the intensity of light impinging on it. Color is brought into play by putting red-, blue-, and green-colored filters over the pixels, resulting in colored digital images (Figure 2.34). At its best resolution, one CCD pixel corresponds to one pixel of the image. To reduce the resolution and decrease the size of the file, we can “bin” several CCD pixels into one, resulting in a smaller but “pixelated” image.

Figure 2.34 A charge-coupled device (CCD) converts light signals into electronic signals, enabling electronic processing and storage of visual images. This is the basis for electronic imaging in all digital cameras, from cell phones to movie cameras. (credit left: modification of work by Bruce Turner)
Clearly, electronics is a big part of a digital camera; however, the underlying physics is basic optics. As a matter of fact, the optics of a camera are pretty much the same as those of a single lens with an object distance that is significantly larger than the lens’s focal distance (Figure 2.35).
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2.6 • The Camera

87

Figure 2.35 Modern digital cameras have several lenses to produce a clear image with minimal aberration and use red, blue, and green filters to produce a color image.
For instance, let us consider the camera in a smartphone. An average smartphone camera is equipped with a stationary wide-angle lens with a focal length of about 4–5 mm. (This focal length is about equal to the thickness of the phone.) The image created by the lens is focused on the CCD detector mounted at the opposite side of the phone. In a cell phone, the lens and the CCD cannot move relative to each other. So how do we make sure that both the images of a distant and a close object are in focus?
Recall that a human eye can accommodate for distant and close images by changing its focal distance. A cell phone camera cannot do that because the distance from the lens to the detector is fixed. Here is where the small focal distance becomes important. Let us assume we have a camera with a 5-mm focal distance. What is the image distance for a selfie? The object distance for a selfie (the length of the hand holding the phone) is about 50 cm. Using the thin-lens equation, we can write

We then obtain the image distance:

Note that the object distance is 100 times larger than the focal distance. We can clearly see that the 1/(500 mm)

term is significantly smaller than 1/(5 mm), which means that the image distance is pretty much equal to the

lens’s focal length. An actual calculation gives us the image distance

. This value is extremely

close to the lens’s focal distance.

Now let us consider the case of a distant object. Let us say that we would like to take a picture of a person standing about 5 m from us. Using the thin-lens equation again, we obtain the image distance of 5.005 mm. The farther the object is from the lens, the closer the image distance is to the focal distance. At the limiting case of an infinitely distant object, we obtain the image distance exactly equal to the focal distance of the lens.

As you can see, the difference between the image distance for a selfie and the image distance for a distant object is just about 0.05 mm or 50 microns. Even a short object distance such as the length of your hand is two orders of magnitude larger than the lens’s focal length, resulting in minute variations of the image distance.

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2 • Geometric Optics and Image Formation

(The 50-micron difference is smaller than the thickness of an average sheet of paper.) Such a small difference can be easily accommodated by the same detector, positioned at the focal distance of the lens. Image analysis software can help improve image quality.
Conventional point-and-shoot cameras often use a movable lens to change the lens-to-image distance. Complex lenses of the more expensive mirror reflex cameras allow for superb quality photographic images. The optics of these camera lenses is beyond the scope of this textbook.
2.7 The Simple Magnifier
Learning Objectives By the end of this section, you will be able to:
• Understand the optics of a simple magnifier • Characterize the image created by a simple magnifier
The apparent size of an object perceived by the eye depends on the angle the object subtends from the eye. As shown in Figure 2.36, the object at A subtends a larger angle from the eye than when it is position at point B. Thus, the object at A forms a larger image on the retina (see ) than when it is positioned at B (see ). Thus, objects that subtend large angles from the eye appear larger because they form larger images on the retina.

Figure 2.36 Size perceived by an eye is determined by the angle subtended by the object. An image formed on the retina by an object at A

is larger than an image formed on the retina by the same object positioned at B (compared image heights

to ).

We have seen that, when an object is placed within a focal length of a convex lens, its image is virtual, upright, and larger than the object (see part (b) of Figure 2.26). Thus, when such an image produced by a convex lens serves as the object for the eye, as shown in Figure 2.37, the image on the retina is enlarged, because the image produced by the lens subtends a larger angle in the eye than does the object. A convex lens used for this purpose is called a magnifying glass or a simple magnifier.

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2.7 • The Simple Magnifier

89

Figure 2.37 The simple magnifier is a convex lens used to produce an enlarged image of an object on the retina. (a) With no convex lens,

the object subtends an angle

from the eye. (b) With the convex lens in place, the image produced by the convex lens subtends an

angle

from the eye, with

. Thus, the image on the retina is larger with the convex lens in place.

To account for the magnification of a magnifying lens, we compare the angle subtended by the image (created

by the lens) with the angle subtended by the object (viewed with no lens), as shown in Figure 2.37. We assume

that the object is situated at the near point of the eye, because this is the object distance at which the unaided

eye can form the largest image on the retina. We will compare the magnified images created by a lens with this

maximum image size for the unaided eye. The magnification of an image when observed by the eye is the

angular magnification M, which is defined by the ratio of the angle

subtended by the image to the angle

subtended by the object:

2.26

Consider the situation shown in Figure 2.37. The magnifying lens is held a distance l from the eye, and the

image produced by the magnifier forms a distance L from the eye. We want to calculate the angular

magnification for any arbitrary L and l . In the small-angle approximation, the angular size

of the image

is . The angular size

of the object at the near point is

. The angular magnification

is then

2.27

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2 • Geometric Optics and Image Formation

Using Equation 2.8 for linear magnification

and the thin-lens equation

in Equation 2.27, we arrive at the following expression for the angular magnification of a magnifying lens:

2.28

From part (b) of the figure, we see that the absolute value of the image distance is

l . Note that

because the image is virtual, so we can dispense with the absolute value by explicitly inserting the

minus sign:

l . Inserting this into Equation 2.28 gives us the final equation for the angular

magnification of a magnifying lens:

l
2.29

Note that all the quantities in this equation have to be expressed in centimeters. Often, we want the image to be

at the near-point distance (

) to get maximum magnification, and we hold the magnifying lens close

to the eye (l ). In this case, Equation 2.29 gives

2.30

which shows that the greatest magnification occurs for the lens with the shortest focal length. In addition,

when the image is at the near-point distance and the lens is held close to the eye l

, then

and Equation 2.27 becomes

2.31

where m is the linear magnification (Equation 2.32) derived for spherical mirrors and thin lenses. Another

useful situation is when the image is at infinity

. Equation 2.29 then takes the form

2.32

The resulting magnification is simply the ratio of the near-point distance to the focal length of the magnifying lens, so a lens with a shorter focal length gives a stronger magnification. Although this magnification is smaller by 1 than the magnification obtained with the image at the near point, it provides for the most comfortable viewing conditions, because the eye is relaxed when viewing a distant object.
By comparing Equation 2.29 with Equation 2.32, we see that the range of angular magnification of a given converging lens is

2.33

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