must be unitless

Problem 18.

Problem 19.

β must be unitless.

16 The following data give the results of three experiments inwhich charged particles were released from the same point in space,and the forces on them were measured:

q1 = 1 C , v1 = 0 , F1 = (1 N)yq2 = 1 C , v2 = (1 m/s)x , F2 = (1 N)yq3 = 1 C , v3 = (1 m/s)z , F3 = 0

Determine the electric and magnetic fields.√

17 If you put four times more current through a solenoid, howmany times more energy is stored in its magnetic field?

√

18 A Helmholtz coil is defined as a pair of identical circularcoils lying in parallel planes and separated by a distance, h, equalto their radius, b. (Each coil may have more than one turn of wire.)Current circulates in the same direction in each coil, so the fieldstend to reinforce each other in the interior region. This configurationhas the advantage of being fairly open, so that other apparatus canbe easily placed inside and subjected to the field while remainingvisible from the outside. The choice of h = b results in the mostuniform possible field near the center. A photograph of a Helmholtzcoil is shown in example 4 on page 684.(a) Find the percentage drop in the field at the center of one coil,compared to the full strength at the center of the whole apparatus.√

(b) What value of h (not equal to b) would make this differenceequal to zero?

√

19 The figure shows a nested pair of circular wire loops usedto create magnetic fields. (The twisting of the leads is a practicaltrick for reducing the magnetic fields they contribute, so the fieldsare very nearly what we would expect for an ideal circular currentloop.) The coordinate system below is to make it easier to discussdirections in space. One loop is in the y − z plane, the other in thex − y plane. Each of the loops has a radius of 1.0 cm, and carries1.0 A in the direction indicated by the arrow.(a) Calculate the magnetic field that would be produced by one suchloop, at its center.

√

(b) Describe the direction of the magnetic field that would be pro-duced, at its center, by the loop in the x− y plane alone.(c) Do the same for the other loop.(d) Calculate the magnitude of the magnetic field produced by thetwo loops in combination, at their common center. Describe itsdirection.

√

20 Four long wires are arranged, as shown, so that their cross-section forms a square, with connections at the ends so that current

Problems 751

Problem 25.

Problem 20.

flows through all four before exiting. Note that the current is to theright in the two back wires, but to the left in the front wires. If thedimensions of the cross-sectional square (height and front-to-back)are b, find the magnetic field (magnitude and direction) along thelong central axis.

√

21 In problem 16, the three experiments gave enough informationto determine both fields. Is it possible to design a procedure so that,using only two such experiments, we can always find E and B? Ifso, design it. If not, why not?

22 Use the Biot-Savart law to derive the magnetic field of a long,straight wire, and show that this reproduces the result of example6 on page 686.

23 (a) Modify the calculation on page 691 to determine thecomponent of the magnetic field of a sheet of charge that is perpen-dicular to the sheet.

√

(b) Show that your answer has the right units.(c) Show that your answer approaches zero as z approaches infinity.(d) What happens to your answer in the case of a = b? Explainwhy this makes sense.

24 Consider two solenoids, one of which is smaller so that itcan be put inside the other. Assume they are long enough so thateach one only contributes significantly to the field inside itself, andthe interior fields are nearly uniform. Consider the configurationwhere the small one is inside the big one with their currents circu-lating in the same direction, and a second configuration in whichthe currents circulate in opposite directions. Compare the energiesof these configurations with the energy when the solenoids are farapart. Based on this reasoning, which configuration is stable, andin which configuration will the little solenoid tend to get twistedaround or spit out? . Hint, p. 1037

25 (a) A solenoid can be imagined as a series of circular currentloops that are spaced along their common axis. Integrate the resultof example 12 on page 700 to show that the field on the axis of asolenoid can be written as B = (2πkη/c2)(cosβ + cos γ), where theangles β and γ are defined in the figure.(b) Show that in the limit where the solenoid is very long, thisexact result agrees with the approximate one derived in example 13on page 703 using Ampere’s law.

752 Chapter 11 Electromagnetism

A nautilus shell is approxi-mately a logarithmic spiral, of thetype in problem 28.

(c) Note that, unlike the calculation using Ampere’s law, this oneis valid at points that are near the mouths of the solenoid, or evenoutside it entirely. If the solenoid is long, at what point on the axisis the field equal to one half of its value at the center of the solenoid?(d) What happens to your result when you apply it to points thatare very far away from the solenoid? Does this make sense?

26 The first step in the proof of Ampere’s law on page 704 is toshow that Ampere’s law holds in the case shown in figure f/1, wherea circular Amperian loop is centered on a long, straight wire that isperpendicular to the plane of the loop. Carry out this calculation,using the result for the field of a wire that was established withoutusing Ampere’s law.

27 A certain region of space has a magnetic field given by B =bxy. Find the electric current flowing through the square definedby z = 0, 0 ≤ x ≤ a, and 0 ≤ y ≤ a.

√

28 Perform a calculation similar to the one in problem 54, butfor a logarithmic spiral, defined by r = weuθ, and show that thefield is B = (kI/c2u)(1/a− 1/b). Note that the solution to problem54 is given in the back of the book.

29 (a) For the geometry described in example 8 on page 689,find the field at a point the lies in the plane of the wires, but notbetween the wires, at a distance b from the center line. Use thesame technique as in that example.(b) Now redo the calculation using the technique demonstrated onpage 694. The integrals are nearly the same, but now the reasoningis reversed: you already know β = 1, and you want to find anunknown field. The only difference in the integrals is that you aretiling a different region of the plane in order to mock up the currentsin the two wires. Note that you can’t tile a region that containsa point of interest, since the technique uses the field of a distantdipole.

√

30 (a) A long, skinny solenoid consists of N turns of wirewrapped uniformly around a hollow cylinder of length ` and cross-sectional area A. Find its inductance.

√

(b) Show that your answer has the right units to be an inductance.

31 Consider two solenoids, one of which is smaller so that itcan be put inside the other. Assume they are long enough to actlike ideal solenoids, so that each one only contributes significantlyto the field inside itself, and the interior fields are nearly uniform.Consider the configuration where the small one is partly inside andpartly hanging out of the big one, with their currents circulating inthe same direction. Their axes are constrained to coincide.(a) Find the difference in the magnetic energy between the configu-ration where the solenoids are separate and the configuration where

Problems 753

Problem 32.

Problem 35.

the small one is inserted into the big one. Your equation will in-clude the length x of the part of the small solenoid that is insidethe big one, as well as other relevant variables describing the twosolenoids.

√

(b) Based on your answer to part a, find the force acting

32 Verify Ampere’s law in the case shown in the figure, as-suming the known equation for the field of a wire. A wire carryingcurrent I passes perpendicularly through the center of the rectan-gular Amperian surface. The length of the rectangle is infinite, soit’s not necessary to compute the contributions of the ends.

33 The purpose of this problem is to find how the gain of atransformer depends on its construction.(a) The number of loops of wire, N , in a solenoid is changed, whilekeeping the length constant. How does the impedance depend on N?State your answer as a proportionality, e.g., Z ∝ N3 or Z ∝ N−5.(b) For a given AC voltage applied across the inductor, how doesthe magnetic field depend on N? You need to take into accountboth the dependence of a solenoid’s field on N for a given currentand your answer to part a, which affects the current.(c) Now consider a transformer consisting of two solenoids. Theinput side has N1 loops, and the output N2. We wish to find how theoutput voltage V2 depends on N1, N2, and the input voltage V1. Thetext has already established V2 ∝ V1N2, so it only remains to findthe dependence on N1. Use your result from part b to accomplishthis. The ratio V2/V1 is called the voltage gain.

34 Problem 33 dealt with the dependence of a transformer’s gainon the number of loops of wire in the input solenoid. Carry out asimilar analysis of how the gain depends on the frequency at whichthe circuit is operated.

35 A U-shaped wire makes electrical contact with a second,straight wire, which rolls along it to the right, as shown in thefigure. The whole thing is immersed in a uniform magnetic field,which is perpendicular to the plane of the circuit. The resistance ofthe rolling wire is much greater than that of the U.(a) Find the direction of the force on the wire based on conservationof energy.(b) Verify the direction of the force using right-hand rules.(c) Find the magnitude of the force acting on the wire. There ismore than one way to do this, but please do it using Faraday’s law(which works even though it’s the Amperian surface itself that ischanging, rather than the field).

√

(d) Consider how the answer to part a would have changed if thedirection of the field had been reversed, and also do the case wherethe direction of the rolling wire’s motion is reversed. Verify thatthis is in agreement with your answer to part c.


Problem 37.

36 A charged particle is in motion at speed v, in a region ofvacuum through which an electromagnetic wave is passing. In whatdirection should the particle be moving in order to minimize thetotal force acting on it? Consider both possibilities for the sign ofthe charge. (Based on a problem by David J. Raymond.)

37 A wire loop of resistance R and area A, lying in the y − zplane, falls through a nonuniform magnetic field B = kzx, where kis a constant. The z axis is vertical.(a) Find the direction of the force on the wire based on conservationof energy.(b) Verify the direction of the force using right-hand rules.(c) Find the magnetic force on the wire.

√

38 A capacitor has parallel plates of area A, separated by adistance h. If there is a vacuum between the plates, then Gauss’slaw gives E = 4πkσ = 4πkq/A for the field between the plates, andcombining this with E = V/h, we find C = q/V = (1/4πk)A/h.(a) Generalize this derivation to the case where there is a dielectricbetween the plates. (b) Suppose we have a list of possible materialswe could choose as dielectrics, and we wish to construct a capacitorthat will have the highest possible energy density, Ue/v, where v isthe volume. For each dielectric, we know its permittivity ε, and alsothe maximum electric field E it can sustain without breaking downand allowing sparks to cross between the plates. Write the maximumenergy density in terms of these two variables, and determine a figureof merit that could be used to decide which material would be thebest choice.

39 (a) For each term appearing on the right side of Maxwell’sequations, give an example of an everyday situation it describes.(b) Most people doing calculations in the SI system of units don’tuse k and k/c2. Instead, they express everything in terms of theconstants

εo =1

4πkand

µo =4πk

c2.

Rewrite Maxwell’s equations in terms of these constants, eliminatingk and c everywhere.

Problems 755

Problem 42.

40 (a) Prove that in an electromagnetic plane wave, half theenergy is in the electric field and half in the magnetic field.(b) Based on your result from part a, find the proportionality con-stant in the relation dp ∝ E×B dv, where dp is the momentum ofthe part of a plane light wave contained in the volume dv. The vec-tor E×B, multiplied by the appropriate constant, is known as thePoynting vector, and even outside the context of an electromagneticplane wave, it can be interpreted as a momentum density or rate ofenergy flow. (To do this problem, you need to know the relativisticrelationship between the energy and momentum of a beam of lightfrom problem 11 on p. 460.)

√

41 (a) A beam of light has cross-sectional area A and powerP , i.e., P is the number of joules per second that enter a windowthrough which the beam passes. Find the energy density U/v interms of P , A, and universal constants.(b) Find E and B, the amplitudes of the electric and magnetic fields,in terms of P , A, and universal constants (i.e., your answer shouldnot include U or v). You will need the result of problem 40a. A realbeam of light usually consists of many short wavetrains, not one bigsine wave, but don’t worry about that.

√. Hint, p. 1037

(c) A beam of sunlight has an intensity of P/A = 1.35× 103 W/m2,assuming no clouds or atmospheric absorption. This is known as thesolar constant. Compute E and B, and compare with the strengthsof static fields you experience in everyday life: E ∼ 106 V/m in athunderstorm, and B ∼ 10−3 T for the Earth’s magnetic field.√

42 The circular parallel-plate capacitor shown in the figure isbeing charged up over time, with the voltage difference across theplates varying as V = st, where s is a constant. The plates haveradius b, and the distance between them is d. We assume d � b,so that the electric field between the plates is uniform, and parallelto the axis. Find the induced magnetic field at a point between theplates, at a distance R from the axis. . Hint, p. 1037

√

43 A positively charged particle is released from rest at the originat t = 0, in a region of vacuum through which an electromagneticwave is passing. The particle accelerates in response to the wave.In this region of space, the wave varies as E = xE sinωt, B =yB sinωt, and we assume that the particle has a relatively largevalue of m/q, so that its response to the wave is sluggish, and itnever ends up moving at any speed comparable to the speed of light.Therefore we don’t have to worry about the spatial variation of thewave; we can just imagine that these are uniform fields imposed bysome external mechanism on this region of space.(a) Find the particle’s coordinates as functions of time.

√

(b) Show that the motion is confined to −zmax ≤ z ≤ zmax, where

zmax = 1.101(q2EB/m2ω3

).


44 Electromagnetic waves are supposed to have their electricand magnetic fields perpendicular to each other. (Throughout thisproblem, assume we’re talking about waves traveling through a vac-uum, and that there is only a single sine wave traveling in a singledirection, not a superposition of sine waves passing through eachother.) Suppose someone claims they can make an electromagneticwave in which the electric and magnetic fields lie in the same plane.Prove that this is impossible based on Maxwell’s equations.

45 Repeat the self-check on page 739, but with one change inthe procedure: after we charge the capacitor, we open the circuit,and then continue with the observations.

46 On page 742, I proved that H‖,1 = H‖,2 at the boundarybetween two substances if there is no free current and the fields arestatic. In fact, each of Maxwell’s four equations implies a constraintwith a similar structure. Some are constraints on the field compo-nents parallel to the boundary, while others are constraints on theperpendicular parts. Since some of the fields referred to in Maxwell’sequations are the electric and magnetic fields E and B, while othersare the auxiliary fields D and H, some of the constraints deal withE and B, others with D and H. Find the other three constraints.

47 (a) Figure j on page 743 shows a hollow sphere with µ/µo =x, inner radius a, and outer radius b, which has been subjectedto an external field Bo. Finding the fields on the exterior, in theshell, and on the interior requires finding a set of fields that satisfiesfive boundary conditions: (1) far from the sphere, the field mustapproach the constant Bo; (2) at the outer surface of the sphere,the field must have H‖,1 = H‖,2, as discussed on page 742; (3) thesame constraint applies at the inner surface of the sphere; (4) and(5) there is an additional constraint on the fields at the inner andouter surfaces, as found in problem 46. The goal of this problemis to find the solution for the fields, and from it, to prove that theinterior field is uniform, and given by

B =

[9x

(2x+ 1)(x+ 2)− 2a3

b3(x− 1)2

]Bo.

This is a very difficult problem to solve from first principles, becauseit’s not obvious what form the fields should have, and if you hadn’tbeen told, you probably wouldn’t have guessed that the interior fieldwould be uniform. We could, however, guess that once the spherebecomes polarized by the external field, it would become a dipole,and at r � b, the field would be a uniform field superimposed onthe field of a dipole. It turns out that even close to the sphere, thesolution has exactly this form. In order to complete the solution,we need to find the field in the shell (a < r < b), but the only waythis field could match up with the detailed angular variation of the

Problems 757

interior and exterior fields would be if it was also a superposition ofa uniform field with a dipole field. The final result is that we havefour unknowns: the strength of the dipole component of the externalfield, the strength of the uniform and dipole components of the fieldwithin the shell, and the strength of the uniform interior field. Thesefour unknowns are to be determined by imposing constraints (2)through (5) above.(b) Show that the expression from part a has physically reasonablebehavior in its dependence on x and a/b.

48 Two long, parallel strips of thin metal foil form a configurationlike a long, narrow sandwich. The air gap between them has heighth, the width of each strip is w, and their length is `. Each stripcarries current I, and we assume for concreteness that the currentsare in opposite directions, so that the magnetic force, F , betweenthe strips is repulsive.(a) Find the force in the limit of w � h.

√

(b) Find the force in the limit of w � h, which is like two ordinarywires.(c) Discuss the relationship between the two results.

49 Suppose we are given a permanent magnet with a compli-cated, asymmetric shape. Describe how a series of measurementswith a magnetic compass could be used to determine the strengthand direction of its magnetic field at some point of interest. Assumethat you are only able to see the direction to which the compassneedle settles; you cannot measure the torque acting on it.

50 On page 709, the curl of xy was computed. Now considerthe fields xx and yy.(a) Sketch these fields.(b) Using the same technique of explicitly constructing a smallsquare, prove that their curls are both zero. Do not use the compo-nent form of the curl; this was one step in deriving the componentform of the curl.

51 If you watch a movie played backwards, some vectors reversetheir direction. For instance, people walk backwards, with theirvelocity vectors flipped around. Other vectors, such as forces, keepthe same direction, e.g., gravity still pulls down. An electric fieldis another example of a vector that doesn’t turn around: positivecharges are still positive in the time-reversed universe, so they stillmake diverging electric fields, and likewise for the converging fieldsaround negative charges.(a) How does the momentum of a material object behave undertime-reversal? . Solution, p. 1049(b) The laws of physics are still valid in the time-reversed universe.For example, show that if two material objects are interacting, andmomentum is conserved, then momentum is still conserved in thetime-reversed universe. . Solution, p. 1049


(c) Discuss how currents and magnetic fields would behave undertime reversal. . Hint, p. 1037(d) Similarly, show that the equation dp ∝ E×B is still valid undertime reversal.

52 This problem is a more advanced exploration of the time-reversal ideas introduced in problem 51.(a) In that problem, we assumed that charge did not flip its sign un-der time reversal. Suppose we make the opposite assumption, thatcharge does change its sign. This is an idea introduced by RichardFeynman: that antimatter is really matter traveling backward intime! Determine the time-reversal properties of E and B under thisnew assumption, and show that dp ∝ E × B is still valid undertime-reversal.(b) Show that Maxwell’s equations are time-reversal symmetric, i.e.,that if the fields E(x, y, z, t) and B(x, y, z, t) satisfy Maxwell’s equa-tions, then so do E(x, y, z,−t) and B(x, y, z,−t). Demonstrate thisunder both possible assumptions about charge, q → q and q → −q.

53 The purpose of this problem is to prove that the constantof proportionality a in the equation dUm = aB2 dv, for the energydensity of the magnetic field, is given by a = c2/8πk as asserted onpage 693. The geometry we’ll use consists of two sheets of current,like a sandwich with nothing in between but some vacuum in whichthere is a magnetic field. The currents are in opposite directions,and we can imagine them as being joined together at the ends toform a complete circuit, like a tube made of paper that has beensquashed almost flat. The sheets have lengths L in the directionparallel to the current, and widths w. They are separated by a dis-tance d, which, for convenience, we assume is small compared to Land w. Thus each sheet’s contribution to the field is uniform, andcan be approximated by the expression 2πkη/c2.(a) Make a drawing similar to the one in figure 11.2.1 on page 692,and show that in this opposite-current configuration, the magneticfields of the two sheets reinforce in the region between them, pro-ducing double the field, but cancel on the outside.(b) By analogy with the case of a single strand of wire, one sheet’sforce on the other is ILB1, were I = ηw is the total current in onesheet, and B1 = B/2 is the field contributed by only one of thesheets, since the sheet can’t make any net force on itself. Based onyour drawing and the right-hand rule, show that this force is repul-sive.For the rest of the problem, consider a process in which the sheetsstart out touching, and are then separated to a distance d. Sincethe force between the sheets is repulsive, they do mechanical workon the outside world as they are separated, in much the same waythat the piston in an engine does work as the gases inside the cylin-der expand. At the same time, however, there is an induced emf

Problems 759

Problem 54.

which would tend to extinguish the current, so in order to maintaina constant current, energy will have to be drained from a battery.There are three types of energy involved: the increase in the mag-netic field energy, the increase in the energy of the outside world,and the decrease in energy as the battery is drained. (We assumethe sheets have very little resistance, so there is no ohmic heatinginvolved.)

√

(c) Find the mechanical work done by the sheets, which equals theincrease in the energy of the outside world. Show that your resultcan be stated in terms of η, the final volume v = wLd, and nothingelse but numerical and physical constants.

√

(d) The power supplied by the battery is P = IΓE (like P = I∆V ,but with an emf instead of a voltage difference), and the circulationis given by Γ = −dΦB/dt. The negative sign indicates that thebattery is being drained. Calculate the energy supplied by the bat-tery, and, as in part c, show that the result can be stated in termsof η, v, and universal constants.

√

(e) Find the increase in the magnetic-field energy, in terms of η, v,and the unknown constant a.

√

(f) Use conservation of energy to relate your answers from parts c,d, and e, and solve for a.

√

54 Magnet coils are often wrapped in multiple layers. Thefigure shows the special case where the layers are all confined to asingle plane, forming a spiral. Since the thickness of the wires (plustheir insulation) is fixed, the spiral that results is a mathematicaltype known as an Archimedean spiral, in which the turns are evenlyspaced. The equation of the spiral is r = wθ, where w is a constant.For a spiral that starts from r = a and ends at r = b, show that thefield at the center is given by (kI/c2w) ln b/a.

. Solution, p. 1049

55 Resolve the following paradox. A capacitance C is initiallycharged, and is then connected to another capacitance C, forming aloop. With the charge now shared equally, the energy is halved. Ifthe connection is made using wires that have finite resistance, thenthis energy loss could be explained through resistive heating. Buthow is conservation of energy satisfied if the resistance of the wiresis zero?


Problem 56.

56 The figure shows a simplified example of a device called asector mass spectrometer. In an oven near the bottom, positivelyionized atoms are produced. For simplicity, we assume that theatoms are all singly ionized. They may have different masses, how-ever, and the goal is to separate them according to these masses.In the example shown in the figure, there are two different massespresent. The reason this is called a “sector” mass spectrometer isthat it contains two regions of uniform fields.

In the first sector, between the two long capacitor plates, thereis an electric field E in the x direction. Superimposed on this is auniform magnetic field B in the negative z direction (into the page).As analyzed in problem 7, these fields are chosen so that ions at acertain velocity v are not deflected. You will need the result ofthat problem in order to do this problem. Only the ions with thecorrect velocity make it out through the slits at the upper end ofthe capacitor.

In the second sector, at the top, there is no electric field, only amagnetic field, which we assume for simplicity to have the same mag-nitude and direction as in the first sector. This causes the beam tobend into a semicircular arc and hit a detector. In the first such spec-trometers, this detector was simply some photographic film, whereasin modern ones it would probably be a silicon chip similar to thesensor of a camera.

The diameter h of the semicircle depends on the mass m of theion. The quantity ∆h/∆m tells us how good the spectrometer is atseparating similar masses.

(a) Express ∆h/∆m in terms of E, B, and e, eliminating v (whichwe can neither control nor measure directly).

√

(b) Show that the units of your answer make sense.(c) You will have found that increasing E makes the spectrometermore sensitive, while increasing B makes it less so. Explain physi-cally why this is so. What stops us from getting an arbitrarily largesensitivity simply by making B small enough?

Remark: This design makes inefficient use of the ion source’s intensity, becauseany ions with the wrong velocity are wasted. For this reason, real-world spec-trometers of this type include complicated focusing elements.

Key to symbols:easy typical challenging difficult very difficult√

An answer check is available at www.lightandmatter.com.

Problems 761

ExercisesExercise 11B: Polarization

Apparatus:

calcite (Iceland spar) crystal

polaroid film

1. Lay the crystal on a piece of paper that has print on it. You will observe a double image.See what happens if you rotate the crystal.

Evidently the crystal does something to the light that passes through it on the way from thepage to your eye. One beam of light enters the crystal from underneath, but two emerge fromthe top; by conservation of energy the energy of the original beam must be shared betweenthem. Consider the following three possible interpretations of what you have observed:

(a) The two new beams differ from each other, and from the original beam, only in energy.Their other properties are the same.

(b) The crystal adds to the light some mysterious new property (not energy), which comes intwo flavors, X and Y. Ordinary light doesn’t have any of either. One beam that emerges fromthe crystal has some X added to it, and the other beam has Y.

(c) There is some mysterious new property that is possessed by all light. It comes in two flavors,X and Y, and most ordinary light sources make an equal mixture of type X and type Y light.The original beam is an even mixture of both types, and this mixture is then split up by thecrystal into the two purified forms.

In parts 2 and 3 you’ll make observations that will allow you to figure out which of these iscorrect.

2. Now place a polaroid film over the crystal and see what you observe. What happens whenyou rotate the film in the horizontal plane? Does this observation allow you to rule out any ofthe three interpretations?

3. Now put the polaroid film under the crystal and try the same thing. Putting together allyour observations, which interpretation do you think is correct?

4. Look at an overhead light fixture through the polaroid, and try rotating it. What do youobserve? What does this tell you about the light emitted by the lightbulb?


5. Now position yourself with your head under a light fixture and directly over a shiny surface,such as a glossy tabletop. You’ll see the lamp’s reflection, and the light coming from the lampto your eye will have undergone a reflection through roughly a 180-degree angle (i.e., it verynearly reversed its direction). Observe this reflection through the polaroid, and try rotating it.Finally, position yourself so that you are seeing glancing reflections, and try the same thing.Summarize what happens to light with properties X and Y when it is reflected. (This is theprinciple behind polarizing sunglasses.)

Exercises 763

Chapter 12

Optics

12.1 The ray model of lightAds for one Macintosh computer bragged that it could do an arith-metic calculation in less time than it took for the light to get from thescreen to your eye. We find this impressive because of the contrastbetween the speed of light and the speeds at which we interact withphysical objects in our environment. Perhaps it shouldn’t surpriseus, then, that Newton succeeded so well in explaining the motion ofobjects, but was far less successful with the study of light.

The climax of our study of electricity and magnetism was discov-ery that light is an electromagnetic wave. Knowing this, however, isnot the same as knowing everything about eyes and telescopes. Infact, the full description of light as a wave can be rather cumber-some. We will instead spend most of our treatment of optics makinguse of a simpler model of light, the ray model, which does a fine jobin most practical situations. Not only that, but we will even back-track a little and start with a discussion of basic ideas about lightand vision that predated the discovery of electromagnetic waves.

765

12.1.1 The nature of light

The cause and effect relationship in vision

Despite its title, this chapter is far from your first look at light.That familiarity might seem like an advantage, but most people havenever thought carefully about light and vision. Even smart peoplewho have thought hard about vision have come up with incorrectideas. The ancient Greeks, Arabs and Chinese had theories of lightand vision, all of which were mostly wrong, and all of which wereaccepted for thousands of years.

One thing the ancients did get right is that there is a distinctionbetween objects that emit light and objects that don’t. When yousee a leaf in the forest, it’s because three different objects are doingtheir jobs: the leaf, the eye, and the sun. But luminous objectslike the sun, a flame, or the filament of a light bulb can be seen bythe eye without the presence of a third object. Emission of lightis often, but not always, associated with heat. In modern times,we are familiar with a variety of objects that glow without beingheated, including fluorescent lights and glow-in-the-dark toys.

How do we see luminous objects? The Greek philosophers Pythago-ras (b. ca. 560 BC) and Empedocles of Acragas (b. ca. 492BC), who unfortunately were very influential, claimed that whenyou looked at a candle flame, the flame and your eye were bothsending out some kind of mysterious stuff, and when your eye’s stuffcollided with the candle’s stuff, the candle would become evident toyour sense of sight.

Bizarre as the Greek “collision of stuff theory” might seem, ithad a couple of good features. It explained why both the candleand your eye had to be present for your sense of sight to function.The theory could also easily be expanded to explain how we seenonluminous objects. If a leaf, for instance, happened to be presentat the site of the collision between your eye’s stuff and the candle’sstuff, then the leaf would be stimulated to express its green nature,allowing you to perceive it as green.

Modern people might feel uneasy about this theory, since it sug-gests that greenness exists only for our seeing convenience, implyinga human precedence over natural phenomena. Nowadays, peoplewould expect the cause and effect relationship in vision to be theother way around, with the leaf doing something to our eye ratherthan our eye doing something to the leaf. But how can you tell?The most common way of distinguishing cause from effect is to de-termine which happened first, but the process of seeing seems tooccur too quickly to determine the order in which things happened.Certainly there is no obvious time lag between the moment whenyou move your head and the moment when your reflection in themirror moves.

766 Chapter 12 Optics

a / Light from a candle is bumpedoff course by a piece of glass.Inserting the glass causes theapparent location of the candleto shift. The same effect canbe produced by taking off youreyeglasses and looking at whichyou see near the edge of thelens, but a flat piece of glassworks just as well as a lens forthis purpose.

Today, photography provides the simplest experimental evidencethat nothing has to be emitted from your eye and hit the leaf in orderto make it “greenify.” A camera can take a picture of a leaf evenif there are no eyes anywhere nearby. Since the leaf appears greenregardless of whether it is being sensed by a camera, your eye, oran insect’s eye, it seems to make more sense to say that the leaf’sgreenness is the cause, and something happening in the camera oreye is the effect.

Light is a thing, and it travels from one point to another.

Another issue that few people have considered is whether a can-dle’s flame simply affects your eye directly, or whether it sends outlight which then gets into your eye. Again, the rapidity of the effectmakes it difficult to tell what’s happening. If someone throws a rockat you, you can see the rock on its way to your body, and you cantell that the person affected you by sending a material substanceyour way, rather than just harming you directly with an arm mo-tion, which would be known as “action at a distance.” It is not easyto do a similar observation to see whether there is some “stuff” thattravels from the candle to your eye, or whether it is a case of actionat a distance.

Newtonian physics includes both action at a distance (e.g., theearth’s gravitational force on a falling object) and contact forcessuch as the normal force, which only allow distant objects to exertforces on each other by shooting some substance across the spacebetween them (e.g., a garden hose spraying out water that exerts aforce on a bush).

One piece of evidence that the candle sends out stuff that travelsto your eye is that as in figure a, intervening transparent substancescan make the candle appear to be in the wrong location, suggestingthat light is a thing that can be bumped off course. Many peo-ple would dismiss this kind of observation as an optical illusion,however. (Some optical illusions are purely neurological or psycho-logical effects, although some others, including this one, turn out tobe caused by the behavior of light itself.)

A more convincing way to decide in which category light belongsis to find out if it takes time to get from the candle to your eye; inNewtonian physics, action at a distance is supposed to be instan-taneous. The fact that we speak casually today of “the speed oflight” implies that at some point in history, somebody succeeded inshowing that light did not travel infinitely fast. Galileo tried, andfailed, to detect a finite speed for light, by arranging with a personin a distant tower to signal back and forth with lanterns. Galileouncovered his lantern, and when the other person saw the light, heuncovered his lantern. Galileo was unable to measure any time lagthat was significant compared to the limitations of human reflexes.

Section 12.1 The ray model of light 767

b / An image of Jupiter andits moon Io (left) from the Cassiniprobe.

c / The earth is moving to-ward Jupiter and Io. Since thedistance is shrinking, it is takingless and less time for the light toget to us from Io, and Io appearsto circle Jupiter more quickly thannormal. Six months later, theearth will be on the opposite sideof the sun, and receding fromJupiter and Io, so Io will appearto revolve around Jupiter moreslowly.

The first person to prove that light’s speed was finite, and todetermine it numerically, was Ole Roemer, in a series of measure-ments around the year 1675. Roemer observed Io, one of Jupiter’smoons, over a period of several years. Since Io presumably took thesame amount of time to complete each orbit of Jupiter, it could bethought of as a very distant, very accurate clock. A practical and ac-curate pendulum clock had recently been invented, so Roemer couldcheck whether the ratio of the two clocks’ cycles, about 42.5 hoursto 1 orbit, stayed exactly constant or changed a little. If the processof seeing the distant moon was instantaneous, there would be noreason for the two to get out of step. Even if the speed of light wasfinite, you might expect that the result would be only to offset onecycle relative to the other. The earth does not, however, stay at aconstant distance from Jupiter and its moons. Since the distance ischanging gradually due to the two planets’ orbital motions, a finitespeed of light would make the “Io clock” appear to run faster as theplanets drew near each other, and more slowly as their separationincreased. Roemer did find a variation in the apparent speed of Io’sorbits, which caused Io’s eclipses by Jupiter (the moments when Iopassed in front of or behind Jupiter) to occur about 7 minutes earlywhen the earth was closest to Jupiter, and 7 minutes late when itwas farthest. Based on these measurements, Roemer estimated thespeed of light to be approximately 2×108 m/s, which is in the rightballpark compared to modern measurements of 3×108 m/s. (I’m notsure whether the fairly large experimental error was mainly due toimprecise knowledge of the radius of the earth’s orbit or limitationsin the reliability of pendulum clocks.)

Light can travel through a vacuum.

Many people are confused by the relationship between soundand light. Although we use different organs to sense them, there aresome similarities. For instance, both light and sound are typicallyemitted in all directions by their sources. Musicians even use visualmetaphors like “tone color,” or “a bright timbre” to describe sound.One way to see that they are clearly different phenomena is to notetheir very different velocities. Sure, both are pretty fast compared toa flying arrow or a galloping horse, but as we have seen, the speed oflight is so great as to appear instantaneous in most situations. Thespeed of sound, however, can easily be observed just by watching agroup of schoolchildren a hundred feet away as they clap their handsto a song. There is an obvious delay between when you see theirpalms come together and when you hear the clap.

The fundamental distinction between sound and light is thatsound is an oscillation in air pressure, so it requires air (or someother medium such as water) in which to travel. Today, we knowthat outer space is a vacuum, so the fact that we get light from thesun, moon and stars clearly shows that air is not necessary for the


propagation of light.

Discussion Questions

A If you observe thunder and lightning, you can tell how far away thestorm is. Do you need to know the speed of sound, of light, or of both?

B When phenomena like X-rays and cosmic rays were first discovered,suggest a way one could have tested whether they were forms of light.

C Why did Roemer only need to know the radius of the earth’s orbit,not Jupiter’s, in order to find the speed of light?

12.1.2 Interaction of light with matter

Absorption of light

The reason why the sun feels warm on your skin is that thesunlight is being absorbed, and the light energy is being transformedinto heat energy. The same happens with artificial light, so the netresult of leaving a light turned on is to heat the room. It doesn’tmatter whether the source of the light is hot, like the sun, a flame,or an incandescent light bulb, or cool, like a fluorescent bulb. (Ifyour house has electric heat, then there is absolutely no point infastidiously turning off lights in the winter; the lights will help toheat the house at the same dollar rate as the electric heater.)

This process of heating by absorption is entirely different fromheating by thermal conduction, as when an electric stove heatsspaghetti sauce through a pan. Heat can only be conducted throughmatter, but there is vacuum between us and the sun, or between usand the filament of an incandescent bulb. Also, heat conduction canonly transfer heat energy from a hotter object to a colder one, but acool fluorescent bulb is perfectly capable of heating something thathad already started out being warmer than the bulb itself.

How we see nonluminous objects

Not all the light energy that hits an object is transformed intoheat. Some is reflected, and this leads us to the question of howwe see nonluminous objects. If you ask the average person how wesee a light bulb, the most likely answer is “The light bulb makeslight, which hits our eyes.” But if you ask how we see a book, theyare likely to say “The bulb lights up the room, and that lets mesee the book.” All mention of light actually entering our eyes hasmysteriously disappeared.

Most people would disagree if you told them that light was re-flected from the book to the eye, because they think of reflection assomething that mirrors do, not something that a book does. Theyassociate reflection with the formation of a reflected image, whichdoes not seem to appear in a piece of paper.

Imagine that you are looking at your reflection in a nice smoothpiece of aluminum foil, fresh off the roll. You perceive a face, not a


d / Two self-portraits of theauthor, one taken in a mirror andone with a piece of aluminum foil.

e / Specular and diffuse re-flection.

piece of metal. Perhaps you also see the bright reflection of a lampover your shoulder behind you. Now imagine that the foil is justa little bit less smooth. The different parts of the image are nowa little bit out of alignment with each other. Your brain can stillrecognize a face and a lamp, but it’s a little scrambled, like a Picassopainting. Now suppose you use a piece of aluminum foil that hasbeen crumpled up and then flattened out again. The parts of theimage are so scrambled that you cannot recognize an image. Instead,your brain tells you you’re looking at a rough, silvery surface.

Mirror-like reflection at a specific angle is known as specularreflection, and random reflection in many directions is called diffusereflection. Diffuse reflection is how we see nonluminous objects.Specular reflection only allows us to see images of objects otherthan the one doing the reflecting. In top part of figure d, imaginethat the rays of light are coming from the sun. If you are lookingdown at the reflecting surface, there is no way for your eye-brainsystem to tell that the rays are not really coming from a sun downbelow you.

Figure f shows another example of how we can’t avoid the con-clusion that light bounces off of things other than mirrors. Thelamp is one I have in my house. It has a bright bulb, housed in acompletely opaque bowl-shaped metal shade. The only way lightcan get out of the lamp is by going up out of the top of the bowl.The fact that I can read a book in the position shown in the figuremeans that light must be bouncing off of the ceiling, then bouncingoff of the book, then finally getting to my eye.

This is where the shortcomings of the Greek theory of visionbecome glaringly obvious. In the Greek theory, the light from thebulb and my mysterious “eye rays” are both supposed to go to thebook, where they collide, allowing me to see the book. But we nowhave a total of four objects: lamp, eye, book, and ceiling. Wheredoes the ceiling come in? Does it also send out its own mysterious“ceiling rays,” contributing to a three-way collision at the book?That would just be too bizarre to believe!

The differences among white, black, and the various shades ofgray in between is a matter of what percentage of the light theyabsorb and what percentage they reflect. That’s why light-coloredclothing is more comfortable in the summer, and light-colored up-holstery in a car stays cooler that dark upholstery.


f / Light bounces off of theceiling, then off of the book.

g / Discussion question C.

Numerical measurement of the brightness of light

We have already seen that the physiological sensation of loudnessrelates to the sound’s intensity (power per unit area), but is notdirectly proportional to it. If sound A has an intensity of 1 nW/m2,sound B is 10 nW/m2, and sound C is 100 nW/m2, then the increasein loudness from B to C is perceived to be the same as the increasefrom A to B, not ten times greater. That is, the sensation of loudnessis logarithmic.

The same is true for the brightness of light. Brightness is re-lated to power per unit area, but the psychological relationship isa logarithmic one rather than a proportionality. For doing physics,it’s the power per unit area that we’re interested in. The relevantunit is W/m2. One way to determine the brightness of light is tomeasure the increase in temperature of a black object exposed tothe light. The light energy is being converted to heat energy, andthe amount of heat energy absorbed in a given amount of time canbe related to the power absorbed, using the known heat capacityof the object. More practical devices for measuring light intensity,such as the light meters built into some cameras, are based on theconversion of light into electrical energy, but these meters have tobe calibrated somehow against heat measurements.


A The curtains in a room are drawn, but a small gap lets light through,illuminating a spot on the floor. It may or may not also be possible to seethe beam of sunshine crossing the room, depending on the conditions.What’s going on?

B Laser beams are made of light. In science fiction movies, laserbeams are often shown as bright lines shooting out of a laser gun on aspaceship. Why is this scientifically incorrect?

C A documentary film-maker went to Harvard’s 1987 graduation cer-emony and asked the graduates, on camera, to explain the cause of theseasons. Only two out of 23 were able to give a correct explanation, butyou now have all the information needed to figure it out for yourself, as-suming you didn’t already know. The figure shows the earth in its winterand summer positions relative to the sun. Hint: Consider the units usedto measure the brightness of light, and recall that the sun is lower in thesky in winter, so its rays are coming in at a shallower angle.

12.1.3 The ray model of light

Models of light

Note how I’ve been casually diagramming the motion of lightwith pictures showing light rays as lines on the page. More formally,this is known as the ray model of light. The ray model of lightseems natural once we convince ourselves that light travels throughspace, and observe phenomena like sunbeams coming through holesin clouds. Having already been introduced to the concept of lightas an electromagnetic wave, you know that the ray model is not the


ultimate truth about light, but the ray model is simpler, and in anycase science always deals with models of reality, not the ultimatenature of reality. The following table summarizes three models oflight.

h / Three models of light.

The ray model is a generic one. By using it we can discuss thepath taken by the light, without committing ourselves to any specificdescription of what it is that is moving along that path. We willuse the nice simple ray model for most of our treatment of optics,and with it we can analyze a great many devices and phenomena.Not until section 12.5 will we concern ourselves specifically withwave optics, although in the intervening chapters I will sometimesanalyze the same phenomenon using both the ray model and thewave model.

Note that the statements about the applicability of the variousmodels are only rough guides. For instance, wave interference effectsare often detectable, if small, when light passes around an obstaclethat is quite a bit bigger than a wavelength. Also, the criterion forwhen we need the particle model really has more to do with energyscales than distance scales, although the two turn out to be related.

The alert reader may have noticed that the wave model is re-quired at scales smaller than a wavelength of light (on the order of amicrometer for visible light), and the particle model is demanded onthe atomic scale or lower (a typical atom being a nanometer or so insize). This implies that at the smallest scales we need both the wavemodel and the particle model. They appear incompatible, so howcan we simultaneously use both? The answer is that they are notas incompatible as they seem. Light is both a wave and a particle,


but a full understanding of this apparently nonsensical statement isa topic for section 13.2.

i / Examples of ray diagrams.

Ray diagrams

Without even knowing how to use the ray model to calculateanything numerically, we can learn a great deal by drawing raydiagrams. For instance, if you want to understand how eyeglasseshelp you to see in focus, a ray diagram is the right place to start.Many students under-utilize ray diagrams in optics and instead relyon rote memorization or plugging into formulas. The trouble withmemorization and plug-ins is that they can obscure what’s reallygoing on, and it is easy to get them wrong. Often the best plan is todo a ray diagram first, then do a numerical calculation, then checkthat your numerical results are in reasonable agreement with whatyou expected from the ray diagram.

j / 1. Correct. 2. Incorrect: im-plies that diffuse reflection onlygives one ray from each reflectingpoint. 3. Correct, but unneces-sarily complicated

Figure j shows some guidelines for using ray diagrams effectively.The light rays bend when they pass out through the surface of thewater (a phenomenon that we’ll discuss in more detail later). Therays appear to have come from a point above the goldfish’s actuallocation, an effect that is familiar to people who have tried spear-fishing.

• A stream of light is not really confined to a finite number ofnarrow lines. We just draw it that way. In j/1, it has beennecessary to choose a finite number of rays to draw (five),rather than the theoretically infinite number of rays that willdiverge from that point.


• There is a tendency to conceptualize rays incorrectly as ob-jects. In his Optics, Newton goes out of his way to cautionthe reader against this, saying that some people “consider ...the refraction of ... rays to be the bending or breaking of themin their passing out of one medium into another.” But a rayis a record of the path traveled by light, not a physical thingthat can be bent or broken.

• In theory, rays may continue infinitely far into the past andfuture, but we need to draw lines of finite length. In j/1, ajudicious choice has been made as to where to begin and endthe rays. There is no point in continuing the rays any fartherthan shown, because nothing new and exciting is going tohappen to them. There is also no good reason to start themearlier, before being reflected by the fish, because the directionof the diffusely reflected rays is random anyway, and unrelatedto the direction of the original, incoming ray.

• When representing diffuse reflection in a ray diagram, manystudents have a mental block against drawing many rays fan-ning out from the same point. Often, as in example j/2, theproblem is the misconception that light can only be reflectedin one direction from one point.

• Another difficulty associated with diffuse reflection, examplej/3, is the tendency to think that in addition to drawing manyrays coming out of one point, we should also be drawing manyrays coming from many points. In j/1, drawing many rayscoming out of one point gives useful information, telling us,for instance, that the fish can be seen from any angle. Drawingmany sets of rays, as in j/3, does not give us any more usefulinformation, and just clutters up the picture in this example.The only reason to draw sets of rays fanning out from morethan one point would be if different things were happening tothe different sets.

Discussion Question

A Suppose an intelligent tool-using fish is spear-hunting for humans.Draw a ray diagram to show how the fish has to correct its aim. Notethat although the rays are now passing from the air to the water, the samerules apply: the rays are closer to being perpendicular to the surface whenthey are in the water, and rays that hit the air-water interface at a shallowangle are bent the most.

12.1.4 Geometry of specular reflection

To change the motion of a material object, we use a force. Isthere any way to exert a force on a beam of light? Experimentsshow that electric and magnetic fields do not deflect light beams, soapparently light has no electric charge. Light also has no mass, so


k / The geometry of specularreflection.

until the twentieth century it was believed to be immune to gravityas well. Einstein predicted that light beams would be very slightlydeflected by strong gravitational fields, and he was proved correctby observations of rays of starlight that came close to the sun, butobviously that’s not what makes mirrors and lenses work!

If we investigate how light is reflected by a mirror, we will findthat the process is horrifically complex, but the final result is sur-prisingly simple. What actually happens is that the light is madeof electric and magnetic fields, and these fields accelerate the elec-trons in the mirror. Energy from the light beam is momentarilytransformed into extra kinetic energy of the electrons, but becausethe electrons are accelerating they re-radiate more light, convert-ing their kinetic energy back into light energy. We might expectthis to result in a very chaotic situation, but amazingly enough, theelectrons move together to produce a new, reflected beam of light,which obeys two simple rules:

• The angle of the reflected ray is the same as that of the incidentray.

• The reflected ray lies in the plane containing the incident rayand the normal (perpendicular) line. This plane is known asthe plane of incidence.

The two angles can be defined either with respect to the normal,like angles B and C in the figure, or with respect to the reflectingsurface, like angles A and D. There is a convention of several hundredyears’ standing that one measures the angles with respect to thenormal, but the rule about equal angles can logically be stated eitheras B=C or as A=D.

The phenomenon of reflection occurs only at the boundary be-tween two media, just like the change in the speed of light thatpasses from one medium to another. As we have seen in section 6.2,this is the way all waves behave.

Most people are surprised by the fact that light can be reflectedback from a less dense medium. For instance, if you are diving andyou look up at the surface of the water, you will see a reflection ofyourself.


self-check AEach of these diagrams is supposed to show two different rays beingreflected from the same point on the same mirror. Which are correct,and which are incorrect?

. Answer, p. 1065

Reversibility of light rays

The fact that specular reflection displays equal angles of inci-dence and reflection means that there is a symmetry: if the ray hadcome in from the right instead of the left in the figure above, the an-gles would have looked exactly the same. This is not just a pointlessdetail about specular reflection. It’s a manifestation of a very deepand important fact about nature, which is that the laws of physicsdo not distinguish between past and future. Cannonballs and plan-ets have trajectories that are equally natural in reverse, and so dolight rays. This type of symmetry is called time-reversal symmetry.

Typically, time-reversal symmetry is a characteristic of any pro-cess that does not involve heat. For instance, the planets do notexperience any friction as they travel through empty space, so thereis no frictional heating. We should thus expect the time-reversedversions of their orbits to obey the laws of physics, which they do.In contrast, a book sliding across a table does generate heat fromfriction as it slows down, and it is therefore not surprising that thistype of motion does not appear to obey time-reversal symmetry. Abook lying still on a flat table is never observed to spontaneouslystart sliding, sucking up heat energy and transforming it into kineticenergy.

Similarly, the only situation we’ve observed so far where lightdoes not obey time-reversal symmetry is absorption, which involvesheat. Your skin absorbs visible light from the sun and heats up,but we never observe people’s skin to glow, converting heat energyinto visible light. People’s skin does glow in infrared light, butthat doesn’t mean the situation is symmetric. Even if you absorbinfrared, you don’t emit visible light, because your skin isn’t hotenough to glow in the visible spectrum.

These apparent heat-related asymmetries are not actual asym-metries in the laws of physics. The interested reader may wish tolearn more about this from optional chapter 5 on thermodynamics.

Ray tracing on a computer example 1A number of techniques can be used for creating artificial visualscenes in computer graphics. Figure l shows such a scene, whichwas created by the brute-force technique of simply constructinga very detailed ray diagram on a computer. This technique re-quires a great deal of computation, and is therefore too slow to


be used for video games and computer-animated movies. Onetrick for speeding up the computation is to exploit the reversibilityof light rays. If one was to trace every ray emitted by every illu-minated surface, only a tiny fraction of those would actually endup passing into the virtual “camera,” and therefore almost all ofthe computational effort would be wasted. One can instead starta ray at the camera, trace it backward in time, and see where itwould have come from. With this technique, there is no wastedeffort.

l / This photorealistic image of a nonexistent countertop was pro-duced completely on a computer, by computing a complicated raydiagram.


m / Discussion question B.

n / Discussion question C.

o / The solid lines are physi-cally possible paths for light raystraveling from A to B and fromA to C. They obey the principleof least time. The dashed linesdo not obey the principle ofleast time, and are not physicallypossible.


A If a light ray has a velocity vector with components cx and cy , whatwill happen when it is reflected from a surface that lies along the y axis?Make sure your answer does not imply a change in the ray’s speed.

B Generalizing your reasoning from discussion question A, what willhappen to the velocity components of a light ray that hits a corner, asshown in the figure, and undergoes two reflections?

C Three pieces of sheet metal arranged perpendicularly as shown inthe figure form what is known as a radar corner. Let’s assume that theradar corner is large compared to the wavelength of the radar waves, sothat the ray model makes sense. If the radar corner is bathed in radarrays, at least some of them will undergo three reflections. Making a fur-ther generalization of your reasoning from the two preceding discussionquestions, what will happen to the three velocity components of such aray? What would the radar corner be useful for?

12.1.5 ? The principle of least time for reflection

We had to choose between an unwieldy explanation of reflectionat the atomic level and a simpler geometric description that wasnot as fundamental. There is a third approach to describing theinteraction of light and matter which is very deep and beautiful.Emphasized by the twentieth-century physicist Richard Feynman,it is called the principle of least time, or Fermat’s principle.

Let’s start with the motion of light that is not interacting withmatter at all. In a vacuum, a light ray moves in a straight line. Thiscan be rephrased as follows: of all the conceivable paths light couldfollow from P to Q, the only one that is physically possible is thepath that takes the least time.

What about reflection? If light is going to go from one point toanother, being reflected on the way, the quickest path is indeed theone with equal angles of incidence and reflection. If the starting andending points are equally far from the reflecting surface, o, it’s nothard to convince yourself that this is true, just based on symmetry.There is also a tricky and simple proof, shown in figure p, for themore general case where the points are at different distances fromthe surface.

Not only does the principle of least time work for light in a


p / Paths AQB and APB aretwo conceivable paths that a raycould follow to get from A to Bwith one reflection, but only AQBis physically possible. We wishto prove that the path AQB, withequal angles of incidence andreflection, is shorter than anyother path, such as APB. Thetrick is to construct a third point,C, lying as far below the surfaceas B lies above it. Then pathAQC is a straight line whoselength is the same as AQB’s, andpath APC has the same length aspath APB. Since AQC is straight,it must be shorter than any otherpath such as APC that connectsA and C, and therefore AQB mustbe shorter than any path such asAPB.

q / Light is emitted at the centerof an elliptical mirror. There arefour physically possible paths bywhich a ray can be reflected andreturn to the center.

vacuum and light undergoing reflection, we will also see in a laterchapter that it works for the bending of light when it passes fromone medium into another.

Although it is beautiful that the entire ray model of light canbe reduced to one simple rule, the principle of least time, it mayseem a little spooky to speak as if the ray of light is intelligent,and has carefully planned ahead to find the shortest route to itsdestination. How does it know in advance where it’s going? Whatif we moved the mirror while the light was en route, so conditionsalong its planned path were not what it “expected?” The answeris that the principle of least time is really a shortcut for findingcertain results of the wave model of light, which is the topic of thelast chapter of this book.

There are a couple of subtle points about the principle of leasttime. First, the path does not have to be the quickest of all pos-sible paths; it only needs to be quicker than any path that differsinfinitesimally from it. In figure p, for instance, light could get fromA to B either by the reflected path AQB or simply by going straightfrom A to B. Although AQB is not the shortest possible path, itcannot be shortened by changing it infinitesimally, e.g., by movingQ a little to the right or left. On the other hand, path APB is phys-ically impossible, because it is possible to improve on it by movingpoint P infinitesimally to the right.

It’s not quite right to call this the principle of least time. In fig-ure q, for example, the four physically possible paths by which a raycan return to the center consist of two shortest-time paths and twolongest-time paths. Strictly speaking, we should refer to the prin-ciple of least or greatest time, but most physicists omit the niceties,and assume that other physicists understand that both maxima andminima are possible.


a / An image formed by amirror.

12.2 Images by reflectionInfants are always fascinated by the antics of the Baby in the Mirror.Now if you want to know something about mirror images that mostpeople don’t understand, try this. First bring this page closer andcloser to your eyes, until you can no longer focus on it withoutstraining. Then go in the bathroom and see how close you canget your face to the surface of the mirror before you can no longereasily focus on the image of your own eyes. You will find thatthe shortest comfortable eye-mirror distance is much less than theshortest comfortable eye-paper distance. This demonstrates thatthe image of your face in the mirror acts as if it had depth andexisted in the space behind the mirror. If the image was like a flatpicture in a book, then you wouldn’t be able to focus on it fromsuch a short distance.

In this chapter we will study the images formed by flat andcurved mirrors on a qualitative, conceptual basis. Although thistype of image is not as commonly encountered in everyday life asimages formed by lenses, images formed by reflection are simpler tounderstand, so we discuss them first. In section 12.3 we will turnto a more mathematical treatment of images made by reflection.Surprisingly, the same equations can also be applied to lenses, whichare the topic of section 12.4.

12.2.1 A virtual image

We can understand a mirror image using a ray diagram. Figurea shows several light rays, 1, that originated by diffuse reflection atthe person’s nose. They bounce off the mirror, producing new rays,2. To anyone whose eye is in the right position to get one of theserays, they appear to have come from a behind the mirror, 3, wherethey would have originated from a single point. This point is wherethe tip of the image-person’s nose appears to be. A similar analysisapplies to every other point on the person’s face, so it looks asthough there was an entire face behind the mirror. The customaryway of describing the situation requires some explanation:

Customary description in physics: There is an image of the facebehind the mirror.

Translation: The pattern of rays coming from the mirror is exactlythe same as it would be if there were a face behind the mirror.Nothing is really behind the mirror.

This is referred to as a virtual image, because the rays do notactually cross at the point behind the mirror. They only appear tohave originated there.

self-check B


c / The praxinoscope.

Imagine that the person in figure a moves his face down quite a bit —a couple of feet in real life, or a few inches on this scale drawing. Themirror stays where it is. Draw a new ray diagram. Will there still be animage? If so, where is it visible from? . Answer, p. 1065

The geometry of specular reflection tells us that rays 1 and 2are at equal angles to the normal (the imaginary perpendicular linepiercing the mirror at the point of reflection). This means thatray 2’s imaginary continuation, 3, forms the same angle with themirror as ray 1. Since each ray of type 3 forms the same angles withthe mirror as its partner of type 1, we see that the distance of theimage from the mirror is the same as that of the actual face fromthe mirror, and it lies directly across from it. The image thereforeappears to be the same size as the actual face.

b / Example 2.

An eye exam example 2Figure b shows a typical setup in an optometrist’s examinationroom. The patient’s vision is supposed to be tested at a distanceof 6 meters (20 feet in the U.S.), but this distance is larger thanthe amount of space available in the room. Therefore a mirror isused to create an image of the eye chart behind the wall.

The Praxinoscope example 3Figure c shows an old-fashioned device called a praxinoscope,which displays an animated picture when spun. The removablestrip of paper with the pictures printed on it has twice the radiusof the inner circle made of flat mirrors, so each picture’s virtualimage is at the center. As the wheel spins, each picture’s imageis replaced by the next.

Section 12.2 Images by reflection 781

Discussion Question

A The figure shows an object that is off to one side of a mirror. Drawa ray diagram. Is an image formed? If so, where is it, and from whichdirections would it be visible?


d / An image formed by acurved mirror.

e / The image is magnifiedby the same factor in depth andin its other dimensions.

f / Increased magnificationalways comes at the expense ofdecreased field of view.

12.2.2 Curved mirrors

An image in a flat mirror is a pretechnological example: evenanimals can look at their reflections in a calm pond. We now passto our first nontrivial example of the manipulation of an image bytechnology: an image in a curved mirror. Before we dive in, let’sconsider why this is an important example. If it was just a ques-tion of memorizing a bunch of facts about curved mirrors, then youwould rightly rebel against an effort to spoil the beauty of your lib-erally educated brain by force-feeding you technological trivia. Thereason this is an important example is not that curved mirrors areso important in and of themselves, but that the results we derive forcurved bowl-shaped mirrors turn out to be true for a large class ofother optical devices, including mirrors that bulge outward ratherthan inward, and lenses as well. A microscope or a telescope is sim-ply a combination of lenses or mirrors or both. What you’re reallylearning about here is the basic building block of all optical devicesfrom movie projectors to octopus eyes.

Because the mirror in figure d is curved, it bends the rays backcloser together than a flat mirror would: we describe it as converging.Note that the term refers to what it does to the light rays, not to thephysical shape of the mirror’s surface . (The surface itself would bedescribed as concave. The term is not all that hard to remember,because the hollowed-out interior of the mirror is like a cave.) Itis surprising but true that all the rays like 3 really do converge ona point, forming a good image. We will not prove this fact, but itis true for any mirror whose curvature is gentle enough and thatis symmetric with respect to rotation about the perpendicular linepassing through its center (not asymmetric like a potato chip). Theold-fashioned method of making mirrors and lenses is by grindingthem in grit by hand, and this automatically tends to produce analmost perfect spherical surface.

Bending a ray like 2 inward implies bending its imaginary contin-uation 3 outward, in the same way that raising one end of a seesawcauses the other end to go down. The image therefore forms deeperbehind the mirror. This doesn’t just show that there is extra dis-tance between the image-nose and the mirror; it also implies thatthe image itself is bigger from front to back. It has been magnifiedin the front-to-back direction.

It is easy to prove that the same magnification also applies to theimage’s other dimensions. Consider a point like E in figure e. Thetrick is that out of all the rays diffusely reflected by E, we pick theone that happens to head for the mirror’s center, C. The equal-angleproperty of specular reflection plus a little straightforward geometryeasily leads us to the conclusion that triangles ABC and CDE arethe same shape, with ABC being simply a scaled-up version of CDE.The magnification of depth equals the ratio BC/CD, and the up-


down magnification is AB/DE. A repetition of the same proof showsthat the magnification in the third dimension (out of the page) isalso the same. This means that the image-head is simply a largerversion of the real one, without any distortion. The scaling factoris called the magnification, M . The image in the figure is magnifiedby a factor M = 1.9.

Note that we did not explicitly specify whether the mirror wasa sphere, a paraboloid, or some other shape. However, we assumedthat a focused image would be formed, which would not necessarilybe true, for instance, for a mirror that was asymmetric or very deeplycurved.

12.2.3 A real image

If we start by placing an object very close to the mirror, g/1,and then move it farther and farther away, the image at first behavesas we would expect from our everyday experience with flat mirrors,receding deeper and deeper behind the mirror. At a certain point,however, a dramatic change occurs. When the object is more thana certain distance from the mirror, g/2, the image appears upside-down and in front of the mirror.

g / 1. A virtual image. 2. A realimage. As you’ll verify in home-work problem 12, the image isupside-down

Here’s what’s happened. The mirror bends light rays inward, butwhen the object is very close to it, as in g/1, the rays coming from a


h / A Newtonian telescopebeing used with a camera.

given point on the object are too strongly diverging (spreading) forthe mirror to bring them back together. On reflection, the rays arestill diverging, just not as strongly diverging. But when the objectis sufficiently far away, g/2, the mirror is only intercepting the raysthat came out in a narrow cone, and it is able to bend these enoughso that they will reconverge.

Note that the rays shown in the figure, which both originated atthe same point on the object, reunite when they cross. The pointwhere they cross is the image of the point on the original object.This type of image is called a real image, in contradistinction to thevirtual images we’ve studied before.

Definition: A real image is one where rays actually cross. A virtualimage is a point from which rays only appear to have come.

The use of the word “real” is perhaps unfortunate. It soundsas though we are saying the image was an actual material object,which of course it is not.

The distinction between a real image and a virtual image is animportant one, because a real image can be projected onto a screenor photographic film. If a piece of paper is inserted in figure g/2at the location of the image, the image will be visible on the paper(provided the object is bright and the room is dark). Your eye usesa lens to make a real image on the retina.

self-check CSketch another copy of the face in figure g/1, even farther from themirror, and draw a ray diagram. What has happened to the location ofthe image? . Answer, p. 1066

12.2.4 Images of images

If you are wearing glasses right now, then the light rays from thepage are being manipulated first by your glasses and then by the lensof your eye. You might think that it would be extremely difficultto analyze this, but in fact it is quite easy. In any series of opticalelements (mirrors or lenses or both), each element works on the raysfurnished by the previous element in exactly the same manner as ifthe image formed by the previous element was an actual object.

Figure h shows an example involving only mirrors. The Newto-nian telescope, invented by Isaac Newton, consists of a large curvedmirror, plus a second, flat mirror that brings the light out of thetube. (In very large telescopes, there may be enough room to puta camera or even a person inside the tube, in which case the sec-ond mirror is not needed.) The tube of the telescope is not vital; itis mainly a structural element, although it can also be helpful forblocking out stray light. The lens has been removed from the frontof the camera body, and is not needed for this setup. Note that the


i / A Newtonian telescopebeing used for visual rather thanphotographic observing. In reallife, an eyepiece lens is normallyused for additional magnification,but this simpler setup will alsowork.

two sample rays have been drawn parallel, because an astronomicaltelescope is used for viewing objects that are extremely far away.These two “parallel” lines actually meet at a certain point, say acrater on the moon, so they can’t actually be perfectly parallel, butthey are parallel for all practical purposes since we would have tofollow them upward for a quarter of a million miles to get to thepoint where they intersect.

The large curved mirror by itself would form an image I, but thesmall flat mirror creates an image of the image, I′. The relationshipbetween I and I′ is exactly the same as it would be if I was an actualobject rather than an image: I and I′ are at equal distances fromthe plane of the mirror, and the line between them is perpendicularto the plane of the mirror.

One surprising wrinkle is that whereas a flat mirror used by itselfforms a virtual image of an object that is real, here the mirror isforming a real image of virtual image I. This shows how pointless itwould be to try to memorize lists of facts about what kinds of imagesare formed by various optical elements under various circumstances.You are better off simply drawing a ray diagram.

j / The angular size of the flowerdepends on its distance from theeye.

Although the main point here was to give an example of an imageof an image, figure i also shows an interesting case where we needto make the distinction between magnification and angular mag-nification. If you are looking at the moon through this telescope,then the images I and I′ are much smaller than the actual moon.Otherwise, for example, image I would not fit inside the telescope!However, these images are very close to your eye compared to theactual moon. The small size of the image has been more than com-pensated for by the shorter distance. The important thing here isthe amount of angle within your field of view that the image covers,and it is this angle that has been increased. The factor by which itis increased is called the angular magnification, Ma.


k / The person uses a mirror toget a view of both sides of theladybug. Although the flat mirrorhas M = 1, it doesn’t give an an-gular magnification of 1. The im-age is farther from the eye thanthe object, so the angular magni-fication Ma = αi/αo is less thanone.


A Locate the images of you that will be formed if you stand betweentwo parallel mirrors.


B Locate the images formed by two perpendicular mirrors, as in thefigure. What happens if the mirrors are not perfectly perpendicular?


C Locate the images formed by the periscope.


a / The relationship betweenthe object’s position and theimage’s can be expressed interms of the angles θo and θi .

12.3 Images, quantitativelyIt sounds a bit odd when a scientist refers to a theory as “beauti-ful,” but to those in the know it makes perfect sense. One markof a beautiful theory is that it surprises us by being simple. Themathematical theory of lenses and curved mirrors gives us just sucha surprise. We expect the subject to be complex because there areso many cases: a converging mirror forming a real image, a diverg-ing lens that makes a virtual image, and so on for a total of sixpossibilities. If we want to predict the location of the images in allthese situations, we might expect to need six different equations,and six more for predicting magnifications. Instead, it turns outthat we can use just one equation for the location of the image andone equation for its magnification, and these two equations workin all the different cases with no changes except for plus and minussigns. This is the kind of thing the physicist Eugene Wigner referredto as “the unreasonable effectiveness of mathematics.” Sometimeswe can find a deeper reason for this kind of unexpected simplicity,but sometimes it almost seems as if God went out of Her way tomake the secrets of universe susceptible to attack by the humanthought-tool called math.

12.3.1 A real image formed by a converging mirror

Location of the image

We will now derive the equation for the location of a real imageformed by a converging mirror. We assume for simplicity that themirror is spherical, but actually this isn’t a restrictive assumption,because any shallow, symmetric curve can be approximated by asphere. The shape of the mirror can be specified by giving thelocation of its center, C. A deeply curved mirror is a sphere with asmall radius, so C is close to it, while a weakly curved mirror hasC farther away. Given the point O where the object is, we wish tofind the point I where the image will be formed.

To locate an image, we need to track a minimum of two rayscoming from the same point. Since we have proved in the previouschapter that this type of image is not distorted, we can use an on-axispoint, O, on the object, as in figure a/1. The results we derive willalso hold for off-axis points, since otherwise the image would haveto be distorted, which we know is not true. We let one of the rays bethe one that is emitted along the axis; this ray is especially easy totrace, because it bounces straight back along the axis again. As oursecond ray, we choose one that strikes the mirror at a distance of 1from the axis. “One what?” asks the astute reader. The answer isthat it doesn’t really matter. When a mirror has shallow curvature,all the reflected rays hit the same point, so 1 could be expressedin any units you like. It could, for instance, be 1 cm, unless yourmirror is smaller than 1 cm!


b / The geometrical interpre-tation of the focal angle.

c / Example 4, an alternativetest for finding the focal angle.The mirror is the same as infigure b.

The only way to find out anything mathematical about the raysis to use the sole mathematical fact we possess concerning specularreflection: the incident and reflected rays form equal angles withrespect to the normal, which is shown as a dashed line. Thereforethe two angles shown in figure a/2 are the same, and skipping somestraightforward geometry, this leads to the visually reasonable resultthat the two angles in figure a/3 are related as follows:

θi + θo = constant

(Note that θi and θo, which are measured from the image and theobject, not from the eye like the angles we referred to in discussingangular magnification on page 786.) For example, move O fartherfrom the mirror. The top angle in figure a/2 is increased, so thebottom angle must increase by the same amount, causing the imagepoint, I, to move closer to the mirror. In terms of the angles shown infigure a/3, the more distant object has resulted in a smaller angle θo,while the closer image corresponds to a larger θi; One angle increasesby the same amount that the other decreases, so their sum remainsconstant. These changes are summarized in figure a/4.

The sum θi + θo is a constant. What does this constant repre-sent? Geometrically, we interpret it as double the angle made bythe dashed radius line. Optically, it is a measure of the strength ofthe mirror, i.e., how strongly the mirror focuses light, and so we callit the focal angle, θf ,

θi + θo = θf .

Suppose, for example, that we wish to use a quick and dirty opticaltest to determine how strong a particular mirror is. We can layit on the floor as shown in figure c, and use it to make an imageof a lamp mounted on the ceiling overhead, which we assume isvery far away compared to the radius of curvature of the mirror,so that the mirror intercepts only a very narrow cone of rays fromthe lamp. This cone is so narrow that its rays are nearly parallel,and θo is nearly zero. The real image can be observed on a piece ofpaper. By moving the paper nearer and farther, we can bring theimage into focus, at which point we know the paper is located atthe image point. Since θo ≈ 0, we have θi ≈ θf , and we can thendetermine this mirror’s focal angle either by measuring θi directlywith a protractor, or indirectly via trigonometry. A strong mirrorwill bring the rays together to form an image close to the mirror,and these rays will form a blunt-angled cone with a large θi and θf .

An alternative optical test example 4. Figure c shows an alternative optical test. Rather than placingthe object at infinity as in figure b, we adjust it so that the imageis right on top of the object. Points O and I coincide, and the raysare reflected right back on top of themselves. If we measure theangle θ shown in figure c, how can we find the focal angle?

Section 12.3 Images, quantitatively 791

d / The object and image dis-tances

e / Mirror 1 is weaker thanmirror 2. It has a shallowercurvature, a longer focal length,and a smaller focal angle. Itreflects rays at angles not muchdifferent than those that would beproduced with a flat mirror.

. The object and image angles are the same; the angle labeledθ in the figure equals both of them. We therefore have θi + θo =θ = θf . Comparing figures b and c, it is indeed plausible that theangles are related by a factor of two.

At this point, we could consider our work to be done. Typically,we know the strength of the mirror, and we want to find the imagelocation for a given object location. Given the mirror’s focal angleand the object location, we can determine θo by trigonometry, sub-tract to find θi = θf − θo, and then do more trig to find the imagelocation.

There is, however, a shortcut that can save us from doing somuch work. Figure a/3 shows two right triangles whose legs oflength 1 coincide and whose acute angles are θo and θi. These canbe related by trigonometry to the object and image distances shownin figure d:

tan θo = 1/do tan θi = 1/di

Ever since section 12.2, we’ve been assuming small angles. For smallangles, we can use the small-angle approximation tanx ≈ x (for xin radians), giving simply

θo = 1/do θi = 1/di.

We likewise define a distance called the focal length, f according toθf = 1/f . In figure b, f is the distance from the mirror to the placewhere the rays cross. We can now reexpress the equation relatingthe object and image positions as

1

f=

1

di+

1

do.

Figure e summarizes the interpretation of the focal length and focalangle.1

Which form is better, θf = θi + θo or 1/f = 1/di + 1/do? Theangular form has in its favor its simplicity and its straightforwardvisual interpretation, but there are two reasons why we might preferthe second version. First, the numerical values of the angles dependon what we mean by “one unit” for the distance shown as 1 in

1There is a standard piece of terminology which is that the “focal point” isthe point lying on the optical axis at a distance from the mirror equal to the focallength. This term isn’t particularly helpful, because it names a location wherenothing normally happens. In particular, it is not normally the place where therays come to a focus! — that would be the image point. In other words, wedon’t normally have di = f , unless perhaps do =∞. A recent online discussionamong some physics teachers (https://carnot.physics.buffalo.edu/archives, Feb.2006) showed that many disliked the terminology, felt it was misleading, or didn’tknow it and would have misinterpreted it if they had come across it. That is, itappears to be what grammarians call a “skunked term” — a word that bothershalf the population when it’s used incorrectly, and the other half when it’s usedcorrectly.


figure a/1. Second, it is usually easier to measure distances ratherthan angles, so the distance form is more convenient for numbercrunching. Neither form is superior overall, and we will often needto use both to solve any given problem.2

A searchlight example 5Suppose we need to create a parallel beam of light, as in a search-light. Where should we place the lightbulb? A parallel beam haszero angle between its rays, so θi = 0. To place the lightbulbcorrectly, however, we need to know a distance, not an angle:the distance do between the bulb and the mirror. The probleminvolves a mixture of distances and angles, so we need to geteverything in terms of one or the other in order to solve it. Sincethe goal is to find a distance, let’s figure out the image distancecorresponding to the given angle θi = 0. These are related bydi = 1/θi , so we have di = ∞. (Yes, dividing by zero gives infin-ity. Don’t be afraid of infinity. Infinity is a useful problem-solvingdevice.) Solving the distance equation for do, we have

do = (1/f − 1/di )−1

= (1/f − 0)−1

= f

The bulb has to be placed at a distance from the mirror equal toits focal point.

Diopters example 6An equation like di = 1/θi really doesn’t make sense in terms ofunits. Angles are unitless, since radians aren’t really units, sothe right-hand side is unitless. We can’t have a left-hand sidewith units of distance if the right-hand side of the same equationis unitless. This is an artifact of my cavalier statement that theconical bundles of rays spread out to a distance of 1 from the axiswhere they strike the mirror, without specifying the units used tomeasure this 1. In real life, optometrists define the thing we’recalling θi = 1/di as the “dioptric strength” of a lens or mirror,and measure it in units of inverse meters (m−1), also known asdiopters (1 D=1 m−1).

Magnification

We have already discussed in the previous chapter how to findthe magnification of a virtual image made by a curved mirror. Theresult is the same for a real image, and we omit the proof, whichis very similar. In our new notation, the result is M = di/do. Anumerical example is given in subsection 12.3.2.

2I would like to thank Fouad Ajami for pointing out the pedagogical advan-tages of using both equations side by side.


12.3.2 Other cases with curved mirrors

The equation di = (1/f − 1/do)−1 can easily produce a negative

result, but we have been thinking of di as a distance, and distancescan’t be negative. A similar problem occurs with θi = θf − θo forθo > θf . What’s going on here?

The interpretation of the angular equation is straightforward.As we bring the object closer and closer to the image, θo gets biggerand bigger, and eventually we reach a point where θo = θf andθi = 0. This large object angle represents a bundle of rays forminga cone that is very broad, so broad that the mirror can no longerbend them back so that they reconverge on the axis. The imageangle θi = 0 represents an outgoing bundle of rays that are parallel.The outgoing rays never cross, so this is not a real image, unless wewant to be charitable and say that the rays cross at infinity. If wego on bringing the object even closer, we get a virtual image.

f / A graph of the image distancedi as a function of the object dis-tance do.

To analyze the distance equation, let’s look at a graph of di asa function of do. The branch on the upper right corresponds to thecase of a real image. Strictly speaking, this is the only part of thegraph that we’ve proven corresponds to reality, since we never didany geometry for other cases, such as virtual images. As discussed inthe previous section, making do bigger causes di to become smaller,and vice-versa.


Letting do be less than f is equivalent to θo > θf : a virtual imageis produced on the far side of the mirror. This is the first exampleof Wigner’s “unreasonable effectiveness of mathematics” that wehave encountered in optics. Even though our proof depended onthe assumption that the image was real, the equation we derivedturns out to be applicable to virtual images, provided that we eitherinterpret the positive and negative signs in a certain way, or elsemodify the equation to have different positive and negative signs.

self-check DInterpret the three places where, in physically realistic parts of the graph,the graph approaches one of the dashed lines. [This will come morenaturally if you have learned the concept of limits in a math class.] .

Answer, p. 1066

A flat mirror example 7We can even apply the equation to a flat mirror. As a sphere getsbigger and bigger, its surface is more and more gently curved.The planet Earth is so large, for example, that we cannot evenperceive the curvature of its surface. To represent a flat mirror, welet the mirror’s radius of curvature, and its focal length, becomeinfinite. Dividing by infinity gives zero, so we have

1/do = −1/di ,

or

do = −di .

If we interpret the minus sign as indicating a virtual image on thefar side of the mirror from the object, this makes sense.

It turns out that for any of the six possible combinations ofreal or virtual images formed by converging or diverging lenses ormirrors, we can apply equations of the form

θf = θi + θo

and

1

f=

1

di+

1

do,

with only a modification of plus or minus signs. There are two pos-sible approaches here. The approach we have been using so far isthe more popular approach in American textbooks: leave the equa-tion the same, but attach interpretations to the resulting negativeor positive values of the variables. The trouble with this approachis that one is then forced to memorize tables of sign conventions,e.g., that the value of di should be negative when the image is avirtual image formed by a converging mirror. Positive and negative


signs also have to be memorized for focal lengths. Ugh! It’s highlyunlikely that any student has ever retained these lengthy tables inhis or her mind for more than five minutes after handing in the finalexam in a physics course. Of course one can always look such thingsup when they are needed, but the effect is to turn the whole thinginto an exercise in blindly plugging numbers into formulas.

As you have gathered by now, there is another method which Ithink is better, and which I’ll use throughout the rest of this book.In this method, all distances and angles are positive by definition,and we put in positive and negative signs in the equations dependingon the situation. (I thought I was the first to invent this method, butI’ve been told that this is known as the European sign convention,and that it’s fairly common in Europe.) Rather than memorizingthese signs, we start with the generic equations

θf = ±θi ± θo1

f= ± 1

di± 1

do,

and then determine the signs by a two-step method that depends onray diagrams. There are really only two signs to determine, not four;the signs in the two equations match up in the way you’d expect.The method is as follows:

1. Use ray diagrams to decide whether θo and θi vary in the sameway or in opposite ways. (In other words, decide whether making θogreater results in a greater value of θi or a smaller one.) Based onthis, decide whether the two signs in the angle equation are the sameor opposite. If the signs are opposite, go on to step 2 to determinewhich is positive and which is negative.

2. If the signs are opposite, we need to decide which is thepositive one and which is the negative. Since the focal angle is nevernegative, the smaller angle must be the one with a minus sign.

In step 1, many students have trouble drawing the ray diagramcorrectly. For simplicity, you should always do your diagram for apoint on the object that is on the axis of the mirror, and let oneof your rays be the one that is emitted along the axis and reflectedstraight back on itself, as in the figures in subsection 12.3.1. Asshown in figure a/4 in subsection 12.3.1, there are four angles in-volved: two at the mirror, one at the object (θo), and one at theimage (θi). Make sure to draw in the normal to the mirror so thatyou can see the two angles at the mirror. These two angles areequal, so as you change the object position, they fan out or fan in,like opening or closing a book. Once you’ve drawn this effect, youshould easily be able to tell whether θo and θi change in the sameway or in opposite ways.

Although focal lengths are always positive in the method usedin this book, you should be aware that diverging mirrors and lenses


g / Example 8.

are assigned negative focal lengths in the other method, so if yousee a lens labeled f = −30 cm, you’ll know what it means.

An anti-shoplifting mirror example 8. Convenience stores often install a diverging mirror so that theclerk has a view of the whole store and can catch shoplifters. Usea ray diagram to show that the image is reduced, bringing moreinto the clerk’s field of view. If the focal length of the mirror is 3.0m, and the mirror is 7.0 m from the farthest wall, how deep is theimage of the store?

. As shown in ray diagram g/1, di is less than do. The magnifica-tion, M = di/do, will be less than one, i.e., the image is actuallyreduced rather than magnified.

Apply the method outlined above for determining the plus andminus signs. Step 1: The object is the point on the oppositewall. As an experiment, g/2, move the object closer. I did thesedrawings using illustration software, but if you were doing themby hand, you’d want to make the scale much larger for greateraccuracy. Also, although I split figure g into two separate drawingsin order to make them easier to understand, you’re less likely tomake a mistake if you do them on top of each other.

The two angles at the mirror fan out from the normal. Increasingθo has clearly made θi larger as well. (All four angles got big-ger.) There must be a cancellation of the effects of changing thetwo terms on the right in the same way, and the only way to getsuch a cancellation is if the two terms in the angle equation haveopposite signs:

θf = +θi − θo

orθf = −θi + θo.

Step 2: Now which is the positive term and which is negative?Since the image angle is bigger than the object angle, the angleequation must be

θf = θi − θo,

in order to give a positive result for the focal angle. The signs ofthe distance equation behave the same way:

1f

=1di− 1

do.

Solving for di , we find

di =(

1f

+1do

)−1

= 2.1 m.

The image of the store is reduced by a factor of 2.1/7.0 = 0.3,i.e., it is smaller by 70%.


h / A diverging mirror in the shapeof a sphere. The image is re-duced (M < 1). This is similarto example 8, but here the imageis distorted because the mirror’scurve is not shallow.

A shortcut for real images example 9In the case of a real image, there is a shortcut for step 1, thedetermination of the signs. In a real image, the rays cross atboth the object and the image. We can therefore time-reverse theray diagram, so that all the rays are coming from the image andreconverging at the object. Object and image swap roles. Dueto this time-reversal symmetry, the object and image cannot betreated differently in any of the equations, and they must thereforehave the same signs. They are both positive, since they must addup to a positive result.

12.3.3 ? Aberrations

An imperfection or distortion in an image is called an aberra-tion. An aberration can be produced by a flaw in a lens or mirror,but even with a perfect optical surface some degree of aberration isunavoidable. To see why, consider the mathematical approximationwe’ve been making, which is that the depth of the mirror’s curveis small compared to do and di. Since only a flat mirror can sat-isfy this shallow-mirror condition perfectly, any curved mirror willdeviate somewhat from the mathematical behavior we derived byassuming that condition. There are two main types of aberration incurved mirrors, and these also occur with lenses.

(1) An object on the axis of the lens or mirror may be imagedcorrectly, but off-axis objects may be out of focus or distorted. Ina camera, this type of aberration would show up as a fuzziness orwarping near the sides of the picture when the center was perfectlyfocused. An example of this is shown in figure i, and in that partic-ular example, the aberration is not a sign that the equipment wasof low quality or wasn’t right for the job but rather an inevitableresult of trying to flatten a panoramic view; in the limit of a 360-


degree panorama, the problem would be similar to the problem ofrepresenting the Earth’s surface on a flat map, which can’t be ac-complished without distortion.

i / This photo was taken using a“fish-eye lens,” which gives an ex-tremely large field of view.

(2) The image may be sharp when the object is at certain dis-tances and blurry when it is at other distances. The blurrinessoccurs because the rays do not all cross at exactly the same point.If we know in advance the distance of the objects with which themirror or lens will be used, then we can optimize the shape of theoptical surface to make in-focus images in that situation. For in-stance, a spherical mirror will produce a perfect image of an objectthat is at the center of the sphere, because each ray is reflected di-rectly onto the radius along which it was emitted. For objects atgreater distances, however, the focus will be somewhat blurry. Inastronomy the objects being used are always at infinity, so a spher-ical mirror is a poor choice for a telescope. A different shape (aparabola) is better specialized for astronomy.

One way of decreasing aberration is to use a small-diameter mir-ror or lens, or block most of the light with an opaque screen with ahole in it, so that only light that comes in close to the axis can get


j / Spherical mirrors are thecheapest to make, but parabolicmirrors are better for makingimages of objects at infinity.A sphere has equal curvatureeverywhere, but a parabola hastighter curvature at its center andgentler curvature at the sides.

through. Either way, we are using a smaller portion of the lens ormirror whose curvature will be more shallow, thereby making theshallow-mirror (or thin-lens) approximation more accurate. Youreye does this by narrowing down the pupil to a smaller hole. Ina camera, there is either an automatic or manual adjustment, andnarrowing the opening is called “stopping down.” The disadvantageof stopping down is that light is wasted, so the image will be dimmeror a longer exposure must be used.

k / Even though the spherical mir-ror (solid line) is not well adaptedfor viewing an object at infinity,we can improve its performancegreatly by stopping it down. Nowthe only part of the mirror be-ing used is the central portion,where its shape is virtually in-distinguishable from a parabola(dashed line).

What I would suggest you take away from this discussion for thesake of your general scientific education is simply an understandingof what an aberration is, why it occurs, and how it can be reduced,not detailed facts about specific types of aberrations.


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