In section 12.2 we've only done examples of mirrors with

Problem 13.

9 In section 12.2 we’ve only done examples of mirrors withhollowed-out shapes (called concave mirrors). Now draw a ray dia-gram for a curved mirror that has a bulging outward shape (called aconvex mirror). (a) How does the image’s distance from the mirrorcompare with the actual object’s distance from the mirror? Fromthis comparison, determine whether the magnification is greaterthan or less than one. (b) Is the image real, or virtual? Couldthis mirror ever make the other type of image?

10 As discussed in question 9, there are two types of curvedmirrors, concave and convex. Make a list of all the possible com-binations of types of images (virtual or real) with types of mirrors(concave and convex). (Not all of the four combinations are phys-ically possible.) Now for each one, use ray diagrams to determinewhether increasing the distance of the object from the mirror leadsto an increase or a decrease in the distance of the image from themirror.

Draw BIG ray diagrams! Each diagram should use up about halfa page of paper.

Some tips: To draw a ray diagram, you need two rays. For oneof these, pick the ray that comes straight along the mirror’s axis,since its reflection is easy to draw. After you draw the two rays andlocate the image for the original object position, pick a new objectposition that results in the same type of image, and start a new raydiagram, in a different color of pen, right on top of the first one. Forthe two new rays, pick the ones that just happen to hit the mirror atthe same two places; this makes it much easier to get the result rightwithout depending on extreme accuracy in your ability to draw thereflected rays.

11 If the user of an astronomical telescope moves her headcloser to or farther away from the image she is looking at, doesthe magnification change? Does the angular magnification change?Explain. (For simplicity, assume that no eyepiece is being used.)

12 In figure g/2 in on page 756, only the image of my fore-head was located by drawing rays. Either photocopy the figure ordownload the book and print out the relevant page. On this copyof the figure, make a new set of rays coming from my chin, andlocate its image. To make it easier to judge the angles accurately,draw rays from the chin that happen to hit the mirror at the samepoints where the two rays from the forehead were shown hitting it.By comparing the locations of the chin’s image and the forehead’simage, verify that the image is actually upside-down, as shown inthe original figure.

13 The figure shows four points where rays cross. Of these,which are image points? Explain.

Problems 801

Problem 18.

14 Here’s a game my kids like to play. I sit next to a sunnywindow, and the sun reflects from the glass on my watch, making adisk of light on the wall or floor, which they pretend to chase as Imove it around. Is the spot a disk because that’s the shape of thesun, or because it’s the shape of my watch? In other words, woulda square watch make a square spot, or do we just have a circularimage of the circular sun, which will be circular no matter what?

15 Apply the equation M = di/do to the case of a flat mirror.

16 Use the method described in the text to derive the equationrelating object distance to image distance for the case of a virtualimage produced by a converging mirror. . Solution, p. 943

17 Find the focal length of the mirror in problem 6 .√

18 Rank the focal lengths of the mirrors in the figure, fromshortest to longest. Explain.

19 (a) A converging mirror with a focal length of 20 cm is usedto create an image, using an object at a distance of 10 cm. Is theimage real, or is it virtual? (b) How about f = 20 cm and do = 30cm? (c) What if it was a diverging mirror with f = 20 cm anddo = 10 cm? (d) A diverging mirror with f = 20 cm and do = 30cm? . Solution, p. 944

20 (a) Make up a numerical example of a virtual image formed bya converging mirror with a certain focal length, and determine themagnification. (You will need the result of problem 16.) Make sureto choose values of do and f that would actually produce a virtualimage, not a real one. Now change the location of the object alittle bit and redetermine the magnification, showing that it changes.At my local department store, the cosmetics department sells handmirrors advertised as giving a magnification of 5 times. How wouldyou interpret this?

(b) Suppose a Newtonian telescope is being used for astronom-ical observing. Assume for simplicity that no eyepiece is used, andassume a value for the focal length of the mirror that would bereasonable for an amateur instrument that is to fit in a closet. Isthe angular magnification different for objects at different distances?For example, you could consider two planets, one of which is twiceas far as the other.

21 (a) Find a case where the magnification of a curved mirroris infinite. Is the angular magnification infinite from any realisticviewing position? (b) Explain why an arbitrarily large magnificationcan’t be achieved by having a sufficiently small value of do.

802 Chapter 12 Optics

Problem 23.

22 A concave surface that reflects sound waves can act just likea converging mirror. Suppose that, standing near such a surface,you are able to find a point where you can place your head so thatyour own whispers are focused back on your head, so that theysound loud to you. Given your distance to the surface, what is thesurface’s focal length?

23 The figure shows a device for constructing a realistic opticalillusion. Two mirrors of equal focal length are put against eachother with their silvered surfaces facing inward. A small objectplaced in the bottom of the cavity will have its image projected inthe air above. The way it works is that the top mirror produces avirtual image, and the bottom mirror then creates a real image ofthe virtual image. (a) Show that if the image is to be positionedas shown, at the mouth of the cavity, then the focal length of themirrors is related to the dimension h via the equation

1

f=

1

h+

1

h+(

1h −

1f

)−1 .

(b) Restate the equation in terms of a single variable x = h/f , andshow that there are two solutions for x. Which solution is physicallyconsistent with the assumptions of the calculation?

24 (a) A converging mirror is being used to create a virtualimage. What is the range of possible magnifications? (b) Do thesame for the other types of images that can be formed by curvedmirrors (both converging and diverging).

25 A diverging mirror of focal length f is fixed, and faces down.An object is dropped from the surface of the mirror, and falls awayfrom it with acceleration g. The goal of the problem is to find themaximum velocity of the image.(a) Describe the motion of the image verbally, and explain why weshould expect there to be a maximum velocity.(b) Use arguments based on units to determine the form of thesolution, up to an unknown unitless multiplicative constant.(c) Complete the solution by determining the unitless constant.

26 Diamond has an index of refraction of 2.42, and part ofthe reason diamonds sparkle is that this encourages a light ray toundergo many total internal reflections before it emerges. (a) Cal-culate the critical angle at which total internal reflection occurs indiamond. (b) Explain the interpretation of your result: Is it mea-sured from the normal, or from the surface? Is it a minimum anglefor total internal reflection, or is it a maximum? How would thecritical angle have been different for a substance such as glass orplastic, with a lower index of refraction?

√

Problems 803

27 Suppose a converging lens is constructed of a type of plasticwhose index of refraction is less than that of water. How will thelens’s behavior be different if it is placed underwater?

28 There are two main types of telescopes, refracting (using alens) and reflecting (using a mirror, as in figure i on p. 758). (Sometelescopes use a mixture of the two types of elements: the light firstencounters a large curved mirror, and then goes through an eyepiecethat is a lens. To keep things simple, assume no eyepiece is used.)What implications would the color-dependence of focal length havefor the relative merits of the two types of telescopes? Describe thecase where an image is formed of a white star. You may find ithelpful to draw a ray diagram.

29 Based on Snell’s law, explain why rays of light passing throughthe edges of a converging lens are bent more than rays passingthrough parts closer to the center. It might seem like it shouldbe the other way around, since the rays at the edge pass throughless glass — shouldn’t they be affected less? In your answer:

• Include a ray diagram showing a huge, full-page, close-up viewof the relevant part of the lens.

• Make use of the fact that the front and back surfaces aren’talways parallel; a lens in which the front and back surfaces arealways parallel doesn’t focus light at all, so if your explanationdoesn’t make use of this fact, your argument must be incorrect.

• Make sure your argument still works even if the rays don’tcome in parallel to the axis.

30 When you take pictures with a camera, the distance betweenthe lens and the film has to be adjusted, depending on the distanceat which you want to focus. This is done by moving the lens. Ifyou want to change your focus so that you can take a picture ofsomething farther away, which way do you have to move the lens?Explain using ray diagrams. [Based on a problem by Eric Mazur.]

31 When swimming underwater, why is your vision made muchclearer by wearing goggles with flat pieces of glass that trap airbehind them? [Hint: You can simplify your reasoning by consideringthe special case where you are looking at an object far away, andalong the optic axis of the eye.]

32 An object is more than one focal length from a converginglens. (a) Draw a ray diagram. (b) Using reasoning like that devel-oped in section 12.3, determine the positive and negative signs inthe equation 1/f = ±1/di ± 1/do. (c) The images of the rose in


Problem 33.

Problem 34.

section 4.2 were made using a lens with a focal length of 23 cm. Ifthe lens is placed 80 cm from the rose, locate the image.

√

33 The figure shows four lenses. Lens 1 has two spherical sur-faces. Lens 2 is the same as lens 1 but turned around. Lens 3 ismade by cutting through lens 1 and turning the bottom around.Lens 4 is made by cutting a central circle out of lens 1 and recessingit.

(a) A parallel beam of light enters lens 1 from the left, parallelto its axis. Reasoning based on Snell’s law, will the beam emergingfrom the lens be bent inward, or outward, or will it remain parallelto the axis? Explain your reasoning. As part of your answer, makea huge drawing of one small part of the lens, and apply Snell’s lawat both interfaces. Recall that rays are bent more if they come tothe interface at a larger angle with respect to the normal.

(b) What will happen with lenses 2, 3, and 4? Explain. Drawingsare not necessary.

34 The drawing shows the anatomy of the human eye, at twicelife size. Find the radius of curvature of the outer surface of thecornea by measurements on the figure, and then derive the focallength of the air-cornea interface, where almost all the focusing oflight occurs. You will need to use physical reasoning to modifythe lensmaker’s equation for the case where there is only a singlerefracting surface. Assume that the index of refraction of the corneais essentially that of water.

35 An object is less than one focal length from a converging lens.(a) Draw a ray diagram. (b) Using reasoning like that developedin section 12.3, determine the positive and negative signs in theequation 1/f = ±1/di ± 1/do. (c) The images of the rose in section4.2 were made using a lens with a focal length of 23 cm. If the lensis placed 10 cm from the rose, locate the image.

√

36 Nearsighted people wear glasses whose lenses are diverging.(a) Draw a ray diagram. For simplicity pretend that there is noeye behind the glasses. (b) Using reasoning like that developedin section 12.3, determine the positive and negative signs in theequation 1/f = ±1/di ± 1/do. (c) If the focal length of the lens is50.0 cm, and the person is looking at an object at a distance of 80.0cm, locate the image.

√

37 (a) Light is being reflected diffusely from an object 1.000 munderwater. The light that comes up to the surface is refracted atthe water-air interface. If the refracted rays all appear to come fromthe same point, then there will be a virtual image of the object inthe water, above the object’s actual position, which will be visibleto an observer above the water. Consider three rays, A, B and C,whose angles in the water with respect to the normal are θi = 0.000◦,1.000◦ and 20.000◦ respectively. Find the depth of the point at which

Problems 805

Problem 39.

the refracted parts of A and B appear to have intersected, and dothe same for A and C. Show that the intersections are at nearly thesame depth, but not quite. [Check: The difference in depth shouldbe about 4 cm.]

(b) Since all the refracted rays do not quite appear to have comefrom the same point, this is technically not a virtual image. Inpractical terms, what effect would this have on what you see?

(c) In the case where the angles are all small, use algebra andtrig to show that the refracted rays do appear to come from thesame point, and find an equation for the depth of the virtual im-age. Do not put in any numerical values for the angles or for theindices of refraction — just keep them as symbols. You will needthe approximation sin θ ≈ tan θ ≈ θ, which is valid for small anglesmeasured in radians.

38 Prove that the principle of least time leads to Snell’s law.

39 Two standard focal lengths for camera lenses are 50 mm(standard) and 28 mm (wide-angle). To see how the focal lengthsrelate to the angular size of the field of view, it is helpful to visualizethings as represented in the figure. Instead of showing many rayscoming from the same point on the same object, as we normally do,the figure shows two rays from two different objects. Although thelens will intercept infinitely many rays from each of these points, wehave shown only the ones that pass through the center of the lens,so that they suffer no angular deflection. (Any angular deflection atthe front surface of the lens is canceled by an opposite deflection atthe back, since the front and back surfaces are parallel at the lens’scenter.) What is special about these two rays is that they are aimedat the edges of one 35-mm-wide frame of film; that is, they showthe limits of the field of view. Throughout this problem, we assumethat do is much greater than di. (a) Compute the angular widthof the camera’s field of view when these two lenses are used. (b)Use small-angle approximations to find a simplified equation for theangular width of the field of view, θ, in terms of the focal length,f , and the width of the film, w. Your equation should not haveany trig functions in it. Compare the results of this approximationwith your answers from part a. (c) Suppose that we are holdingconstant the aperture (amount of surface area of the lens beingused to collect light). When switching from a 50-mm lens to a 28-mm lens, how many times longer or shorter must the exposure bein order to make a properly developed picture, i.e., one that is notunder- or overexposed? [Based on a problem by Arnold Arons.]

. Solution, p. 944


40 A nearsighted person is one whose eyes focus light toostrongly, and who is therefore unable to relax the lens inside hereye sufficiently to form an image on her retina of an object that istoo far away.

(a) Draw a ray diagram showing what happens when the persontries, with uncorrected vision, to focus at infinity.

(b) What type of lenses do her glasses have? Explain.

(c) Draw a ray diagram showing what happens when she wearsglasses. Locate both the image formed by the glasses and the finalimage.

(d) Suppose she sometimes uses contact lenses instead of herglasses. Does the focal length of her contacts have to be less than,equal to, or greater than that of her glasses? Explain.

41 Fred’s eyes are able to focus on things as close as 5.0 cm.Fred holds a magnifying glass with a focal length of 3.0 cm at aheight of 2.0 cm above a flatworm. (a) Locate the image, and findthe magnification. (b) Without the magnifying glass, from whatdistance would Fred want to view the flatworm to see its detailsas well as possible? With the magnifying glass? (c) Compute theangular magnification.

Problems 807

Problem 42.

42 Panel 1 of the figure shows the optics inside a pair of binoc-ulars. They are essentially a pair of telescopes, one for each eye.But to make them more compact, and allow the eyepieces to be theright distance apart for a human face, they incorporate a set of eightprisms, which fold the light path. In addition, the prisms make theimage upright. Panel 2 shows one of these prisms, known as a Porroprism. The light enters along a normal, undergoes two total internalreflections at angles of 45 degrees with respect to the back surfaces,and exits along a normal. The image of the letter R has been flippedacross the horizontal. Panel 3 shows a pair of these prisms gluedtogether. The image will be flipped across both the horizontal andthe vertical, which makes it oriented the right way for the user ofthe binoculars.(a) Find the minimum possible index of refraction for the glass usedin the prisms.(b) For a material of this minimal index of refraction, find the frac-tion of the incoming light that will be lost to reflection in the fourPorro prisms on a each side of a pair of binoculars. (See section 6.2.)In real, high-quality binoculars, the optical surfaces of the prismshave antireflective coatings, but carry out your calculation for thecase where there is no such coating.(c) Discuss the reasons why a designer of binoculars might or mightnot want to use a material with exactly the index of refraction foundin part a.

43 It would be annoying if your eyeglasses produced a magnifiedor reduced image. Prove that when the eye is very close to a lens,and the lens produces a virtual image, the angular magnification isalways approximately equal to 1 (regardless of whether the lens isdiverging or converging).


Problems 44 and 47.

44 The figure shows a diffraction pattern made by a doubleslit, along with an image of a meter stick to show the scale. Sketchthe diffraction pattern from the figure on your paper. Now considerthe four variables in the equation λ/d = sin θ/m. Which of theseare the same for all five fringes, and which are different for eachfringe? Which variable would you naturally use in order to labelwhich fringe was which? Label the fringes on your sketch using thevalues of that variable.

45 Match gratings A-C with the diffraction patterns 1-3 thatthey produce. Explain.

Problems 809

46 The figure below shows two diffraction patterns. The top onewas made with yellow light, and the bottom one with red. Couldthe slits used to make the two patterns have been the same?

47 The figure on p. 809 shows a diffraction pattern made by adouble slit, along with an image of a meter stick to show the scale.The slits were 146 cm away from the screen on which the diffractionpattern was projected. The spacing of the slits was 0.050 mm. Whatwas the wavelength of the light?

√

48 Why would blue or violet light be the best for microscopy?

49 The figure below shows two diffraction patterns, both madewith the same wavelength of red light. (a) What type of slits madethe patterns? Is it a single slit, double slits, or something else?Explain. (b) Compare the dimensions of the slits used to make thetop and bottom pattern. Give a numerical ratio, and state whichway the ratio is, i.e., which slit pattern was the larger one. Explain.

50 When white light passes through a diffraction grating, whatis the smallest value of m for which the visible spectrum of order moverlaps the next one, of order m + 1? (The visible spectrum runsfrom about 400 nm to about 700 nm.)


Problem 51. This image ofthe Pleiades star cluster showshaloes around the stars due to thewave nature of light.

51 For star images such as the ones in figure y, estimate theangular width of the diffraction spot due to diffraction at the mouthof the telescope. Assume a telescope with a diameter of 10 meters(the largest currently in existence), and light with a wavelength inthe middle of the visible range. Compare with the actual angularsize of a star of diameter 109 m seen from a distance of 1017 m.What does this tell you?

52 The figure below shows three diffraction patterns. All weremade under identical conditions, except that a different set of doubleslits was used for each one. The slits used to make the top patternhad a center-to-center separation d = 0.50 mm, and each slit wasw = 0.04 mm wide. (a) Determine d and w for the slits used tomake the pattern in the middle. (b) Do the same for the slits usedto make the bottom pattern.

Problems 811

53 The beam of a laser passes through a diffraction grating,fans out, and illuminates a wall that is perpendicular to the originalbeam, lying at a distance of 2.0 m from the grating. The beamis produced by a helium-neon laser, and has a wavelength of 694.3nm. The grating has 2000 lines per centimeter. (a) What is thedistance on the wall between the central maximum and the maximaimmediately to its right and left? (b) How much does your answerchange when you use the small-angle approximations θ ≈ sin θ ≈tan θ?

√

54 Ultrasound, i.e., sound waves with frequencies too high to beaudible, can be used for imaging fetuses in the womb or for break-ing up kidney stones so that they can be eliminated by the body.Consider the latter application. Lenses can be built to focus soundwaves, but because the wavelength of the sound is not all that smallcompared to the diameter of the lens, the sound will not be concen-trated exactly at the geometrical focal point. Instead, a diffractionpattern will be created with an intense central spot surrounded byfainter rings. About 85% of the power is concentrated within thecentral spot. The angle of the first minimum (surrounding the cen-tral spot) is given by sin θ = λ/b, where b is the diameter of the lens.This is similar to the corresponding equation for a single slit, butwith a factor of 1.22 in front which arises from the circular shape ofthe aperture. Let the distance from the lens to the patient’s kidneystone be L = 20 cm. You will want f > 20 kHz, so that the soundis inaudible. Find values of b and f that would result in a usabledesign, where the central spot is small enough to lie within a kidneystone 1 cm in diameter.

55 Under what circumstances could one get a mathematicallyundefined result by solving the double-slit diffraction equation for θ?Give a physical interpretation of what would actually be observed.

56 When ultrasound is used for medical imaging, the frequencymay be as high as 5-20 MHz. Another medical application of ul-trasound is for therapeutic heating of tissues inside the body; here,the frequency is typically 1-3 MHz. What fundamental physical rea-sons could you suggest for the use of higher frequencies for imaging?


Problem 57.

Problem 58.

57 Suppose we have a polygonal room whose walls are mirrors,and there a pointlike light source in the room. In most such exam-ples, every point in the room ends up being illuminated by the lightsource after some finite number of reflections. A difficult mathemat-ical question, first posed in the middle of the last century, is whetherit is ever possible to have an example in which the whole room isnot illuminated. (Rays are assumed to be absorbed if they strikeexactly at a vertex of the polygon, or if they pass exactly throughthe plane of a mirror.)

The problem was finally solved in 1995 by G.W. Tokarsky, whofound an example of a room that was not illuminable from a cer-tain point. Figure 57 shows a slightly simpler example found twoyears later by D. Castro. If a light source is placed at either of thelocations shown with dots, the other dot remains unilluminated, al-though every other point is lit up. It is not straightforward to proverigorously that Castro’s solution has this property. However, theplausibility of the solution can be demonstrated as follows.

Suppose the light source is placed at the right-hand dot. Locateall the images formed by single reflections. Note that they form aregular pattern. Convince yourself that none of these images illumi-nates the left-hand dot. Because of the regular pattern, it becomesplausible that even if we form images of images, images of imagesof images, etc., none of them will ever illuminate the other dot.

There are various other versions of the problem, some of whichremain unsolved. The book by Klee and Wagon gives a good in-troduction to the topic, although it predates Tokarsky and Castro’swork.

References:G.W. Tokarsky, “Polygonal Rooms Not Illuminable from Every Point.”Amer. Math. Monthly 102, 867-879, 1995.D. Castro, “Corrections.” Quantum 7, 42, Jan. 1997.V. Klee and S. Wagon, Old and New Unsolved Problems in PlaneGeometry and Number Theory. Mathematical Association of Amer-ica, 1991.

58 A mechanical linkage is a device that changes one type ofmotion into another. The most familiar example occurs in a gasolinecar’s engine, where a connecting rod changes the linear motion of thepiston into circular motion of the crankshaft. The top panel of thefigure shows a mechanical linkage invented by Peaucellier in 1864,and independently by Lipkin around the same time. It consists ofsix rods joined by hinges, the four short ones forming a rhombus.Point O is fixed in space, but the apparatus is free to rotate aboutO. Motion at P is transformed into a different motion at P′ (or viceversa).

Geometrically, the linkage is a mechanical implementation of

Problems 813

Problem 59.

the ancient problem of inversion in a circle. Considering the case inwhich the rhombus is folded flat, let the k be the distance from Oto the point where P and P′ coincide. Form the circle of radius kwith its center at O. As P and P′ move in and out, points on theinside of the circle are always mapped to points on its outside, suchthat rr′ = k2. That is, the linkage is a type of analog computerthat exactly solves the problem of finding the inverse of a numberr. Inversion in a circle has many remarkable geometrical properties,discussed in H.S.M. Coxeter, Introduction to Geometry, Wiley, 1961.If a pen is inserted through a hole at P, and P′ is traced over ageometrical figure, the Peaucellier linkage can be used to draw akind of image of the figure.

A related problem is the construction of pictures, like the one inthe bottom panel of the figure, called anamorphs. The drawing ofthe column on the paper is highly distorted, but when the reflectingcylinder is placed in the correct spot on top of the page, an undis-torted image is produced inside the cylinder. (Wide-format movietechnologies such as Cinemascope are based on similar principles.)

Show that the Peaucellier linkage does not convert correctly be-tween an image and its anamorph, and design a modified version ofthe linkage that does. Some knowledge of analytic geometry will behelpful.

59 The figure shows a lens with surfaces that are curved, butwhose thickness is constant along any horizontal line. Use the lens-maker’s equation to prove that this “lens” is not really a lens atall. . Solution, p. 944

60 Under ordinary conditions, gases have indices of refractiononly a little greater than that of vacuum, i.e., n = 1 + ε, where ε issome small number. Suppose that a ray crosses a boundary betweena region of vacuum and a region in which the index of refraction is1 + ε. Find the maximum angle by which such a ray can ever bedeflected, in the limit of small ε. . Hint, p. 925

61 A converging mirror has focal length f . An object is locatedat a distance (1 + ε)f from the mirror, where ε is small. Find thedistance of the image from the mirror, simplifying your result asmuch as possible by using the assumption that ε is small.

. Answer, p. 934

Key to symbols:easy typical challenging difficult very difficult√

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


ExercisesExercise 12A: Exploring Images With a Curved Mirror

Equipment:

concave mirrors with deep curvature

concave mirrors with gentle curvature

convex mirrors

1. Obtain a curved mirror from your instructor. If it is silvered on both sides, make sure you’reworking with the concave side, which bends light rays inward. Look at your own face in themirror. Now change the distance between your face and the mirror, and see what happens.Explore the full range of possible distances between your face and the mirror.

In these observations you’ve been changing two variables at once: the distance between theobject (your face) and the mirror, and the distance from the mirror to your eye. In general,scientific experiments become easier to interpret if we practice isolation of variables, i.e., onlychange one variable while keeping all the others constant. In parts 2 and 3 you’ll form an imageof an object that’s not your face, so that you can have independent control of the object distanceand the point of view.

2. With the mirror held far away from you, observe the image of something behind you, overyour shoulder. Now bring your eye closer and closer to the mirror. Can you see the image withyour eye very close to the mirror? See if you can explain your observation by drawing a raydiagram.

——————–> turn page

Exercises 815

3. Now imagine the following new situation, but don’t actually do it yet. Suppose you lay themirror face-up on a piece of tissue paper, put your finger a few cm above the mirror, and lookat the image of your finger. As in part 2, you can bring your eye closer and closer to the mirror.

Will you be able to see the image with your eye very close to the mirror? Draw a ray diagramto help you predict what you will observe.

Prediction:

Now test your prediction. If your prediction was incorrect, see if you can figure out what wentwrong, or ask your instructor for help.

4. For parts 4 and 5, it’s more convenient to use concave mirrors that are more gently curved;obtain one from your instructor. Lay the mirror on the tissue paper, and use it to create animage of the overhead lights on a piece of paper above it and a little off to the side. What doyou have to do in order to make the image clear? Can you explain this observation using a raydiagram?

——————–> turn page


5. Now imagine the following experiment, but don’t do it yet. What will happen to the imageon the paper if you cover half of the mirror with your hand?

Prediction:

Test your prediction. If your prediction was incorrect, can you explain what happened?

6. Now imagine forming an image with a convex mirror (one that bulges outward), and thattherefore bends light rays away from the central axis (i.e., is diverging). Draw a typical raydiagram.

Is the image real, or virtual? Will there be more than one type of image?

Prediction:

Test your prediction.

Exercises 817

Exercise 12B: Object and Image Distances

Equipment:

optical benches

converging mirrors

illuminated objects

1. Set up the optical bench with the mirror at zero on the centimeter scale. Set up theilluminated object on the bench as well.

2. Each group will locate the image for their own value of the object distance, by finding wherea piece of paper has to be placed in order to see the image on it. (The instructor will do onepoint as well.) Note that you will have to tilt the mirror a little so that the paper on which youproject the image doesn’t block the light from the illuminated object.

Is the image real or virtual? How do you know? Is it inverted, or uninverted?

Draw a ray diagram.

3. Measure the image distance and write your result in the table on the board. Do the same forthe magnification.

4. What do you notice about the trend of the data on the board? Draw a second ray diagramwith a different object distance, and show why this makes sense. Some tips for doing thiscorrectly: (1) For simplicity, use the point on the object that is on the mirror’s axis. (2) Youneed to trace two rays to locate the image. To save work, don’t just do two rays at randomangles. You can either use the on-axis ray as one ray, or do two rays that come off at the sameangle, one above and one below the axis. (3) Where each ray hits the mirror, draw the normalline, and make sure the ray is at equal angles on both sides of the normal.

5. We will find the mirror’s focal length from the instructor’s data-point. Then, using this focallength, calculate a theoretical prediction of the image distance, and write it on the board nextto the experimentally determined image distance.


Exercise 12C: How strong are your glasses?

This exercise was created by Dan MacIsaac.

Equipment:

eyeglasses

diverging lenses for students who don’t wear glasses, or who use glasses with converginglenses

rulers and metersticks

scratch paper

marking pens

Most people who wear glasses have glasses whose lenses are outbending, which allows them tofocus on objects far away. Such a lens cannot form a real image, so its focal length cannot bemeasured as easily as that of a converging lens. In this exercise you will determine the focallength of your own glasses by taking them off, holding them at a distance from your face, andlooking through them at a set of parallel lines on a piece of paper. The lines will be reduced(the lens’s magnification is less than one), and by adjusting the distance between the lens andthe paper, you can make the magnification equal 1/2 exactly, so that two spaces between linesas seen through the lens fit into one space as seen simultaneously to the side of the lens. Thisobject distance can be used in order to find the focal length of the lens.

1. Use a marker to draw three evenly spaced parallel lines on the paper. (A spacing of a fewcm works well.)

2. Does this technique really measure magnification or does it measure angular magnification?What can you do in your experiment in order to make these two quantities nearly the same, sothe math is simpler?

3. Before taking any numerical data, use algebra to find the focal length of the lens in terms ofdo, the object distance that results in a magnification of 1/2.

4. Measure the object distance that results in a magnification of 1/2, and determine the focallength of your lens.

Exercises 819

Exercise 12D: Double-Source Interference

1. Two sources separated by a distance d = 2 cm make circular ripples with a wavelength ofλ = 1 cm. On a piece of paper, make a life-size drawing of the two sources in the default setup,and locate the following points:

A. The point that is 10 wavelengths from source #1 and 10 wavelengths from source #2.

B. The point that is 10.5 wavelengths from #1 and 10.5 from #2.

C. The point that is 11 wavelengths from #1 and 11 from #2.

D. The point that is 10 wavelengths from #1 and 10.5 from #2.

E. The point that is 11 wavelengths from #1 and 11.5 from #2.

F. The point that is 10 wavelengths from #1 and 11 from #2.

G. The point that is 11 wavelengths from #1 and 12 from #2.

You can do this either using a compass or by putting the next page under your paper andtracing. It is not necessary to trace all the arcs completely, and doing so is unnecessarily time-consuming; you can fairly easily estimate where these points would lie, and just trace arcs longenough to find the relevant intersections.

What do these points correspond to in the real wave pattern?

2. Make a fresh copy of your drawing, showing only point F and the two sources, which form along, skinny triangle. Now suppose you were to change the setup by doubling d, while leaving λthe same. It’s easiest to understand what’s happening on the drawing if you move both sourcesoutward, keeping the center fixed. Based on your drawing, what will happen to the position ofpoint F when you double d? How has the angle of point F changed?

3. What would happen if you doubled both λ and d compared to the standard setup?

4. Combining the ideas from parts 2 and 3, what do you think would happen to your angles if,starting from the standard setup, you doubled λ while leaving d the same?

5. Suppose λ was a millionth of a centimeter, while d was still as in the standard setup. Whatwould happen to the angles? What does this tell you about observing diffraction of light?


Exercises 821

Exercise 12E: Single-slit diffraction

Equipment:

rulers

computer with web browser

The following page is a diagram of a single slit and a screen onto which its diffraction patternis projected. The class will make a numerical prediction of the intensity of the pattern at thedifferent points on the screen. Each group will be responsible for calculating the intensity atone of the points. (Either 11 groups or six will work nicely – in the latter case, only points a,c, e, g, i, and k are used.) The idea is to break up the wavefront in the mouth of the slit intonine parts, each of which is assumed to radiate semicircular ripples as in Huygens’ principle.The wavelength of the wave is 1 cm, and we assume for simplicity that each set of ripples hasan amplitude of 1 unit when it reaches the screen.

1. For simplicity, let’s imagine that we were only to use two sets of ripples rather than nine.You could measure the distance from each of the two points inside the slit to your point onthe screen. Suppose the distances were both 25.0 cm. What would be the amplitude of thesuperimposed waves at this point on the screen?

Suppose one distance was 24.0 cm and the other was 25.0 cm. What would happen?

What if one was 24.0 cm and the other was 26.0 cm?

What if one was 24.5 cm and the other was 25.0 cm?

In general, what combinations of distances will lead to completely destructive and completelyconstructive interference?

Can you estimate the answer in the case where the distances are 24.7 and 25.0 cm?

2. Although it is possible to calculate mathematically the amplitude of the sine wave that resultsfrom superimposing two sine waves with an arbitrary phase difference between them, the algebrais rather laborious, and it become even more tedious when we have more than two waves to super-impose. Instead, one can simply use a computer spreadsheet or some other computer program toadd up the sine waves numerically at a series of points covering one complete cycle. This is whatwe will actually do. You just need to enter the relevant data into the computer, then examine theresults and pick off the amplitude from the resulting list of numbers. You can run the softwarethrough a web interface at http://lightandmatter.com/cgi-bin/diffraction1.cgi.

3. Measure all nine distances to your group’s point on the screen, and write them on the board- that way everyone can see everyone else’s data, and the class can try to make sense of why theresults came out the way they did. Determine the amplitude of the combined wave, and writeit on the board as well.

The class will discuss why the results came out the way they did.


Exercises 823

Exercise 12F: Diffraction of Light

Equipment:

slit patterns, lasers, straight-filament bulbs

station 1You have a mask with a bunch of different double slits cut out of it. The values of w and d areas follows:

pattern A w=0.04 mm d=.250 mmpattern B w=0.04 mm d=.500 mmpattern C w=0.08 mm d=.250 mmpattern D w=0.08 mm d=.500 mm

Predict how the patterns will look different, and test your prediction. The easiest way to getthe laser to point at different sets of slits is to stick folded up pieces of paper in one side or theother of the holders.

station 2This is just like station 1, but with single slits:

pattern A w=0.02 mmpattern B w=0.04 mmpattern C w=0.08 mmpattern D w=0.16 mm

Predict what will happen, and test your predictions. If you have time, check the actual numericalratios of the w values against the ratios of the sizes of the diffraction patterns

station 3This is like station 1, but the only difference among the sets of slits is how many slits there are:

pattern A double slitpattern B 3 slitspattern C 4 slitspattern D 5 slits

station 4Hold the diffraction grating up to your eye, and look through it at the straight-filament lightbulb. If you orient the grating correctly, you should be able to see the m = 1 and m = −1diffraction patterns off the left and right. If you have it oriented the wrong way, they’ll be aboveand below the bulb instead, which is inconvenient because the bulb’s filament is vertical. Whereis the m = 0 fringe? Can you see m = 2, etc.?

Station 5 has the same equipment as station 4. If you’re assigned to station 5 first, you shouldactually do activity 4 first, because it’s easier.

station 5Use the transformer to increase and decrease the voltage across the bulb. This allows you tocontrol the filament’s temperature. Sketch graphs of intensity as a function of wavelength forvarious temperatures. The inability of the wave model of light to explain the mathematicalshapes of these curves was historically one of the reasons for creating a new model, in whichlight is both a particle and a wave.


a / In 1980, the continentalU.S. got its first taste of activevolcanism in recent memory withthe eruption of Mount St. Helens.

Chapter 13

Quantum Physics

13.1 Rules of RandomnessGiven for one instant an intelligence which could comprehend allthe forces by which nature is animated and the respective positionsof the things which compose it...nothing would be uncertain, andthe future as the past would be laid out before its eyes.

Pierre Simon de Laplace, 1776

The energy produced by the atom is a very poor kind of thing.Anyone who expects a source of power from the transformation ofthese atoms is talking moonshine.

Ernest Rutherford, 1933

The Quantum Mechanics is very imposing. But an inner voice tellsme that it is still not the final truth. The theory yields much, butit hardly brings us nearer to the secret of the Old One. In any case,I am convinced that He does not play dice.

Albert Einstein

However radical Newton’s clockwork universe seemed to his con-temporaries, by the early twentieth century it had become a sort ofsmugly accepted dogma. Luckily for us, this deterministic picture ofthe universe breaks down at the atomic level. The clearest demon-stration that the laws of physics contain elements of randomnessis in the behavior of radioactive atoms. Pick two identical atomsof a radioactive isotope, say the naturally occurring uranium 238,and watch them carefully. They will decay at different times, eventhough there was no difference in their initial behavior.

We would be in big trouble if these atoms’ behavior was as pre-dictable as expected in the Newtonian world-view, because radioac-tivity is an important source of heat for our planet. In reality, eachatom chooses a random moment at which to release its energy, re-sulting in a nice steady heating effect. The earth would be a muchcolder planet if only sunlight heated it and not radioactivity. Prob-ably there would be no volcanoes, and the oceans would never havebeen liquid. The deep-sea geothermal vents in which life first evolvedwould never have existed. But there would be an even worse conse-quence if radioactivity was deterministic: after a few billion years ofpeace, all the uranium 238 atoms in our planet would presumablypick the same moment to decay. The huge amount of stored nuclear

825

energy, instead of being spread out over eons, would all be releasedat one instant, blowing our whole planet to Kingdom Come.1

The new version of physics, incorporating certain kinds of ran-domness, is called quantum physics (for reasons that will becomeclear later). It represented such a dramatic break with the pre-vious, deterministic tradition that everything that came before isconsidered “classical,” even the theory of relativity. This chapter isa basic introduction to quantum physics.

Discussion Question

A I said “Pick two identical atoms of a radioactive isotope.” Are twoatoms really identical? If their electrons are orbiting the nucleus, can wedistinguish each atom by the particular arrangement of its electrons atsome instant in time?

13.1.1 Randomness isn’t random.

Einstein’s distaste for randomness, and his association of deter-minism with divinity, goes back to the Enlightenment conception ofthe universe as a gigantic piece of clockwork that only had to be setin motion initially by the Builder. Many of the founders of quan-tum mechanics were interested in possible links between physics andEastern and Western religious and philosophical thought, but everyeducated person has a different concept of religion and philosophy.Bertrand Russell remarked, “Sir Arthur Eddington deduces religionfrom the fact that atoms do not obey the laws of mathematics. SirJames Jeans deduces it from the fact that they do.”

Russell’s witticism, which implies incorrectly that mathemat-ics cannot describe randomness, remind us how important it is notto oversimplify this question of randomness. You should not sim-ply surmise, “Well, it’s all random, anything can happen.” Forone thing, certain things simply cannot happen, either in classicalphysics or quantum physics. The conservation laws of mass, energy,momentum, and angular momentum are still valid, so for instanceprocesses that create energy out of nothing are not just unlikelyaccording to quantum physics, they are impossible.

A useful analogy can be made with the role of randomness inevolution. Darwin was not the first biologist to suggest that specieschanged over long periods of time. His two new fundamental ideaswere that (1) the changes arose through random genetic variation,and (2) changes that enhanced the organism’s ability to survive andreproduce would be preserved, while maladaptive changes would be

1This is under the assumption that all the uranium atoms were created at thesame time. In reality, we have only a general idea of the processes that mighthave created the heavy elements in the nebula from which our solar systemcondensed. Some portion of them may have come from nuclear reactions insupernova explosions in that particular nebula, but some may have come fromprevious supernova explosions throughout our galaxy, or from exotic events likecollisions of white dwarf stars.

826 Chapter 13 Quantum Physics

eliminated by natural selection. Doubters of evolution often consideronly the first point, about the randomness of natural variation, butnot the second point, about the systematic action of natural selec-tion. They make statements such as, “the development of a complexorganism like Homo sapiens via random chance would be like a whirl-wind blowing through a junkyard and spontaneously assembling ajumbo jet out of the scrap metal.” The flaw in this type of reason-ing is that it ignores the deterministic constraints on the results ofrandom processes. For an atom to violate conservation of energy isno more likely than the conquest of the world by chimpanzees nextyear.

Discussion Question

A Economists often behave like wannabe physicists, probably becauseit seems prestigious to make numerical calculations instead of talkingabout human relationships and organizations like other social scientists.Their striving to make economics work like Newtonian physics extendsto a parallel use of mechanical metaphors, as in the concept of a mar-ket’s supply and demand acting like a self-adjusting machine, and theidealization of people as economic automatons who consistently strive tomaximize their own wealth. What evidence is there for randomness ratherthan mechanical determinism in economics?

13.1.2 Calculating randomness

You should also realize that even if something is random, wecan still understand it, and we can still calculate probabilities nu-merically. In other words, physicists are good bookmakers. A goodbookmaker can calculate the odds that a horse will win a race muchmore accurately that an inexperienced one, but nevertheless cannotpredict what will happen in any particular race.

Statistical independence

As an illustration of a general technique for calculating odds,suppose you are playing a 25-cent slot machine. Each of the threewheels has one chance in ten of coming up with a cherry. If allthree wheels come up cherries, you win $100. Even though theresults of any particular trial are random, you can make certainquantitative predictions. First, you can calculate that your oddsof winning on any given trial are 1/10 × 1/10 × 1/10 = 1/1000 =0.001. Here, I am representing the probabilities as numbers from0 to 1, which is clearer than statements like “The odds are 999 to1,” and makes the calculations easier. A probability of 0 representssomething impossible, and a probability of 1 represents somethingthat will definitely happen.

Also, you can say that any given trial is equally likely to result ina win, and it doesn’t matter whether you have won or lost in priorgames. Mathematically, we say that each trial is statistically inde-pendent, or that separate games are uncorrelated. Most gamblersare mistakenly convinced that, to the contrary, games of chance are

Section 13.1 Rules of Randomness 827

correlated. If they have been playing a slot machine all day, theyare convinced that it is “getting ready to pay,” and they do notwant anyone else playing the machine and “using up” the jackpotthat they “have coming.” In other words, they are claiming thata series of trials at the slot machine is negatively correlated, thatlosing now makes you more likely to win later. Craps players claimthat you should go to a table where the person rolling the dice is“hot,” because she is likely to keep on rolling good numbers. Crapsplayers, then, believe that rolls of the dice are positively correlated,that winning now makes you more likely to win later.

My method of calculating the probability of winning on the slotmachine was an example of the following important rule for calcu-lations based on independent probabilities:

the law of independent probabilitiesIf the probability of one event happening is PA, and the prob-ability of a second statistically independent event happeningis PB, then the probability that they will both occur is theproduct of the probabilities, PAPB. If there are more thantwo events involved, you simply keep on multiplying.

This can be taken as the definition of statistical independence.

Note that this only applies to independent probabilities. Forinstance, if you have a nickel and a dime in your pocket, and yourandomly pull one out, there is a probability of 0.5 that it will bethe nickel. If you then replace the coin and again pull one outrandomly, there is again a probability of 0.5 of coming up with thenickel, because the probabilities are independent. Thus, there is aprobability of 0.25 that you will get the nickel both times.

Suppose instead that you do not replace the first coin beforepulling out the second one. Then you are bound to pull out theother coin the second time, and there is no way you could pull thenickel out twice. In this situation, the two trials are not indepen-dent, because the result of the first trial has an effect on the secondtrial. The law of independent probabilities does not apply, and theprobability of getting the nickel twice is zero, not 0.25.

Experiments have shown that in the case of radioactive decay,the probability that any nucleus will decay during a given time in-terval is unaffected by what is happening to the other nuclei, andis also unrelated to how long it has gone without decaying. Thefirst observation makes sense, because nuclei are isolated from eachother at the centers of their respective atoms, and therefore have nophysical way of influencing each other. The second fact is also rea-sonable, since all atoms are identical. Suppose we wanted to believethat certain atoms were “extra tough,” as demonstrated by theirhistory of going an unusually long time without decaying. Thoseatoms would have to be different in some physical way, but nobody


b / Normalization: the proba-bility of picking land plus theprobability of picking water addsup to 1.

has ever succeeded in detecting differences among atoms. There isno way for an atom to be changed by the experiences it has in itslifetime.

Addition of probabilities

The law of independent probabilities tells us to use multiplica-tion to calculate the probability that both A and B will happen,assuming the probabilities are independent. What about the prob-ability of an “or” rather than an “and”? If two events A and Bare mutually exclusive, then the probability of one or the other oc-curring is the sum PA + PB. For instance, a bowler might have a30% chance of getting a strike (knocking down all ten pins) and a20% chance of knocking down nine of them. The bowler’s chance ofknocking down either nine pins or ten pins is therefore 50%.

It does not make sense to add probabilities of things that arenot mutually exclusive, i.e., that could both happen. Say I have a90% chance of eating lunch on any given day, and a 90% chance ofeating dinner. The probability that I will eat either lunch or dinneris not 180%.

Normalization

If I spin a globe and randomly pick a point on it, I have about a70% chance of picking a point that’s in an ocean and a 30% chanceof picking a point on land. The probability of picking either wa-ter or land is 70% + 30% = 100%. Water and land are mutuallyexclusive, and there are no other possibilities, so the probabilitieshad to add up to 100%. It works the same if there are more thantwo possibilities — if you can classify all possible outcomes into alist of mutually exclusive results, then all the probabilities have toadd up to 1, or 100%. This property of probabilities is known asnormalization.

Averages

Another way of dealing with randomness is to take averages.The casino knows that in the long run, the number of times you winwill approximately equal the number of times you play multipliedby the probability of winning. In the slot-machine game describedon page 827, where the probability of winning is 0.001, if you spenda week playing, and pay $2500 to play 10,000 times, you are likelyto win about 10 times (10, 000× 0.001 = 10), and collect $1000. Onthe average, the casino will make a profit of $1500 from you. Thisis an example of the following rule.

Rule for Calculating AveragesIf you conduct N identical, statistically independent trials,and the probability of success in each trial is P , then on theaverage, the total number of successful trials will be NP . If Nis large enough, the relative error in this estimate will become


c / Why are dice random?

small.

The statement that the rule for calculating averages gets moreand more accurate for larger and larger N(known popularly as the“law of averages”) often provides a correspondence principle thatconnects classical and quantum physics. For instance, the amountof power produced by a nuclear power plant is not random at anydetectable level, because the number of atoms in the reactor is solarge. In general, random behavior at the atomic level tends toaverage out when we consider large numbers of atoms, which is whyphysics seemed deterministic before physicists learned techniques forstudying atoms individually.

We can achieve great precision with averages in quantum physicsbecause we can use identical atoms to reproduce exactly the samesituation many times. If we were betting on horses or dice, we wouldbe much more limited in our precision. After a thousand races, thehorse would be ready to retire. After a million rolls, the dice wouldbe worn out.

self-check AWhich of the following things must be independent, which could be in-dependent, and which definitely are not independent? (1) the probabil-ity of successfully making two free-throws in a row in basketball; (2) theprobability that it will rain in London tomorrow and the probability that itwill rain on the same day in a certain city in a distant galaxy; (3) yourprobability of dying today and of dying tomorrow. . Answer, p. 932

Discussion Questions

A Newtonian physics is an essentially perfect approximation for de-scribing the motion of a pair of dice. If Newtonian physics is deterministic,why do we consider the result of rolling dice to be random?

B Why isn’t it valid to define randomness by saying that randomnessis when all the outcomes are equally likely?

C The sequence of digits 121212121212121212 seems clearly nonran-dom, and 41592653589793 seems random. The latter sequence, how-ever, is the decimal form of pi, starting with the third digit. There is a storyabout the Indian mathematician Ramanujan, a self-taught prodigy, that afriend came to visit him in a cab, and remarked that the number of thecab, 1729, seemed relatively uninteresting. Ramanujan replied that onthe contrary, it was very interesting because it was the smallest numberthat could be represented in two different ways as the sum of two cubes.The Argentine author Jorge Luis Borges wrote a short story called “TheLibrary of Babel,” in which he imagined a library containing every bookthat could possibly be written using the letters of the alphabet. It would in-clude a book containing only the repeated letter “a;” all the ancient Greektragedies known today, all the lost Greek tragedies, and millions of Greektragedies that were never actually written; your own life story, and variousincorrect versions of your own life story; and countless anthologies con-taining a short story called “The Library of Babel.” Of course, if you pickeda book from the shelves of the library, it would almost certainly look like a


d / Probability distribution forthe result of rolling a single die.

e / Rolling two dice and addingthem up.

nonsensical sequence of letters and punctuation, but it’s always possiblethat the seemingly meaningless book would be a science-fiction screen-play written in the language of a Neanderthal tribe, or the lyrics to a setof incomparably beautiful love songs written in a language that never ex-isted. In view of these examples, what does it really mean to say thatsomething is random?

13.1.3 Probability distributions

So far we’ve discussed random processes having only two possibleoutcomes: yes or no, win or lose, on or off. More generally, a randomprocess could have a result that is a number. Some processes yieldintegers, as when you roll a die and get a result from one to six, butsome are not restricted to whole numbers, for example the numberof seconds that a uranium-238 atom will exist before undergoingradioactive decay.

Consider a throw of a die. If the die is “honest,” then we expectall six values to be equally likely. Since all six probabilities mustadd up to 1, then probability of any particular value coming upmust be 1/6. We can summarize this in a graph, d. Areas underthe curve can be interpreted as total probabilities. For instance,the area under the curve from 1 to 3 is 1/6 + 1/6 + 1/6 = 1/2, sothe probability of getting a result from 1 to 3 is 1/2. The functionshown on the graph is called the probability distribution.

Figure e shows the probabilities of various results obtained byrolling two dice and adding them together, as in the game of craps.The probabilities are not all the same. There is a small probabilityof getting a two, for example, because there is only one way to doit, by rolling a one and then another one. The probability of rollinga seven is high because there are six different ways to do it: 1+6,2+5, etc.

If the number of possible outcomes is large but finite, for examplethe number of hairs on a dog, the graph would start to look like asmooth curve rather than a ziggurat.

What about probability distributions for random numbers thatare not integers? We can no longer make a graph with probabilityon the y axis, because the probability of getting a given exact num-ber is typically zero. For instance, there is zero probability that aradioactive atom will last for exactly 3 seconds, since there is areinfinitely many possible results that are close to 3 but not exactlythree: 2.999999999999999996876876587658465436, for example. Itdoesn’t usually make sense, therefore, to talk about the probabilityof a single numerical result, but it does make sense to talk aboutthe probability of a certain range of results. For instance, the prob-ability that an atom will last more than 3 and less than 4 seconds isa perfectly reasonable thing to discuss. We can still summarize theprobability information on a graph, and we can still interpret areasunder the curve as probabilities.


f / A probability distributionfor height of human adults (notreal data).

g / Example 1.

h / The average of a proba-bility distribution.

But the y axis can no longer be a unitless probability scale. Inradioactive decay, for example, we want the x axis to have units oftime, and we want areas under the curve to be unitless probabilities.The area of a single square on the graph paper is then

(unitless area of a square)

= (width of square with time units)

×(height of square) .

If the units are to cancel out, then the height of the square mustevidently be a quantity with units of inverse time. In other words,the y axis of the graph is to be interpreted as probability per unittime, not probability.

Figure f shows another example, a probability distribution forpeople’s height. This kind of bell-shaped curve is quite common.

self-check B

Compare the number of people with heights in the range of 130-135 cmto the number in the range 135-140. . Answer, p. 933

Looking for tall basketball players example 1. A certain country with a large population wants to find very tallpeople to be on its Olympic basketball team and strike a blowagainst western imperialism. Out of a pool of 108 people who arethe right age and gender, how many are they likely to find who areover 225 cm (7 feet 4 inches) in height? Figure g gives a close-upof the “tail” of the distribution shown previously in figure f.

. The shaded area under the curve represents the probability thata given person is tall enough. Each rectangle represents a prob-ability of 0.2× 10−7 cm−1× 1 cm = 2× 10−8. There are about 35rectangles covered by the shaded area, so the probability of hav-ing a height greater than 225 cm is 7×10−7 , or just under one ina million. Using the rule for calculating averages, the average, orexpected number of people this tall is (108)× (7× 10−7) = 70.

Average and width of a probability distribution

If the next Martian you meet asks you, “How tall is an adult hu-man?,” you will probably reply with a statement about the averagehuman height, such as “Oh, about 5 feet 6 inches.” If you wantedto explain a little more, you could say, “But that’s only an average.Most people are somewhere between 5 feet and 6 feet tall.” Withoutbothering to draw the relevant bell curve for your new extraterres-trial acquaintance, you’ve summarized the relevant information bygiving an average and a typical range of variation.

The average of a probability distribution can be defined geomet-rically as the horizontal position at which it could be balanced ifit was constructed out of cardboard. A convenient numerical mea-


i / The full width at half maxi-mum (FWHM) of a probabilitydistribution.

sure of the amount of variation about the average, or amount ofuncertainty, is the full width at half maximum, or FWHM, shownin figure i.

A great deal more could be said about this topic, and indeedan introductory statistics course could spend months on ways ofdefining the center and width of a distribution. Rather than force-feeding you on mathematical detail or techniques for calculatingthese things, it is perhaps more relevant to point out simply thatthere are various ways of defining them, and to inoculate you againstthe misuse of certain definitions.

The average is not the only possible way to say what is a typicalvalue for a quantity that can vary randomly; another possible defi-nition is the median, defined as the value that is exceeded with 50%probability. When discussing incomes of people living in a certaintown, the average could be very misleading, since it can be affectedmassively if a single resident of the town is Bill Gates. Nor is theFWHM the only possible way of stating the amount of random vari-ation; another possible way of measuring it is the standard deviation(defined as the square root of the average squared deviation fromthe average value).

13.1.4 Exponential decay and half-life

Half-life

Most people know that radioactivity “lasts a certain amount oftime,” but that simple statement leaves out a lot. As an example,consider the following medical procedure used to diagnose thyroidfunction. A very small quantity of the isotope 131I, produced in anuclear reactor, is fed to or injected into the patient. The body’sbiochemical systems treat this artificial, radioactive isotope exactlythe same as 127I, which is the only naturally occurring type. (Nu-tritionally, iodine is a necessary trace element. Iodine taken intothe body is partly excreted, but the rest becomes concentrated inthe thyroid gland. Iodized salt has had iodine added to it to pre-vent the nutritional deficiency known as goiters, in which the iodine-starved thyroid becomes swollen.) As the 131I undergoes beta decay,it emits electrons, neutrinos, and gamma rays. The gamma rays canbe measured by a detector passed over the patient’s body. As theradioactive iodine becomes concentrated in the thyroid, the amountof gamma radiation coming from the thyroid becomes greater, andthat emitted by the rest of the body is reduced. The rate at whichthe iodine concentrates in the thyroid tells the doctor about thehealth of the thyroid.

If you ever undergo this procedure, someone will presumablyexplain a little about radioactivity to you, to allay your fears thatyou will turn into the Incredible Hulk, or that your next child willhave an unusual number of limbs. Since iodine stays in your thyroid


for a long time once it gets there, one thing you’ll want to know iswhether your thyroid is going to become radioactive forever. Theymay just tell you that the radioactivity “only lasts a certain amountof time,” but we can now carry out a quantitative derivation of howthe radioactivity really will die out.

Let Psurv(t) be the probability that an iodine atom will survivewithout decaying for a period of at least t. It has been experimen-tally measured that half all 131I atoms decay in 8 hours, so we have

Psurv(8 hr) = 0.5 .

Now using the law of independent probabilities, the probabilityof surviving for 16 hours equals the probability of surviving for thefirst 8 hours multiplied by the probability of surviving for the second8 hours,

Psurv(16 hr) = 0.50× 0.50

= 0.25 .

Similarly we have

Psurv(24 hr) = 0.50× 0.5× 0.5

= 0.125 .

Generalizing from this pattern, the probability of surviving for anytime t that is a multiple of 8 hours is

Psurv(t) = 0.5t/8 hr .

We now know how to find the probability of survival at intervalsof 8 hours, but what about the points in time in between? Whatwould be the probability of surviving for 4 hours? Well, using thelaw of independent probabilities again, we have

Psurv(8 hr) = Psurv(4 hr)× Psurv(4 hr) ,

which can be rearranged to give

Psurv(4 hr) =√Psurv(8 hr)

=√

0.5

= 0.707 .

This is exactly what we would have found simply by plugging inPsurv(t) = 0.5t/8 hr and ignoring the restriction to multiples of 8hours. Since 8 hours is the amount of time required for half of theatoms to decay, it is known as the half-life, written t1/2. The generalrule is as follows:

Exponential Decay Equation

Psurv(t) = 0.5t/t1/2


Using the rule for calculating averages, we can also find the num-ber of atoms, N(t), remaining in a sample at time t:

N(t) = N(0)× 0.5t/t1/2

Both of these equations have graphs that look like dying-out expo-nentials, as in the example below.

Radioactive contamination at Chernobyl example 2.One of the most dangerous radioactive isotopes released by theChernobyl disaster in 1986 was 90Sr, whose half-life is 28 years.(a) How long will it be before the contamination is reduced to onetenth of its original level? (b) If a total of 1027 atoms was released,about how long would it be before not a single atom was left?

. (a) We want to know the amount of time that a 90Sr nucleushas a probability of 0.1 of surviving. Starting with the exponentialdecay formula,

Psur v = 0.5t/t1/2 ,

we want to solve for t . Taking natural logarithms of both sides,

ln P =t

t1/2ln 0.5 ,

so

t =t1/2

ln 0.5ln P

Plugging in P = 0.1 and t1/2 = 28 years, we get t = 93 years.

(b) This is just like the first part, but P = 10−27 . The result isabout 2500 years.

14C Dating example 3Almost all the carbon on Earth is 12C, but not quite. The isotope14C, with a half-life of 5600 years, is produced by cosmic rays inthe atmosphere. It decays naturally, but is replenished at such arate that the fraction of 14C in the atmosphere remains constant,at 1.3 × 10−12 . Living plants and animals take in both 12C and14C from the atmosphere and incorporate both into their bodies.Once the living organism dies, it no longer takes in C atoms fromthe atmosphere, and the proportion of 14C gradually falls off as itundergoes radioactive decay. This effect can be used to find theage of dead organisms, or human artifacts made from plants oranimals. Figure j on page 836 shows the exponential decay curveof 14C in various objects. Similar methods, using longer-lived iso-topes, provided the first firm proof that the earth was billions ofyears old, not a few thousand as some had claimed on religiousgrounds.


j / Calibration of the 14C dating method using tree rings and arti-facts whose ages were known from other methods. Redrawn from EmilioSegre, Nuclei and Particles, 1965.

Rate of decay

If you want to find how many radioactive decays occur within atime interval lasting from time t to time t + ∆t, the most straight-forward approach is to calculate it like this:

(number of decays between t and t+ ∆t)

= N(t)−N(t+ ∆t)

Usually we’re interested in the case where ∆t is small compared tot1/2, and in this limiting case the calculation starts to look exactlylike the limit that goes into the definition of the derivative dN/dt. Itis therefore more convenient to talk about the rate of decay −dN/dtrather than the number of decays in some finite time interval. Doingcalculus on the function ex is also easier than with 0.5x, so we rewritethe function N(t) as

N = N(0)e−t/τ ,

where τ = t1/2/ ln 2 is shown in example 6 on p. 839 to be theaverage time of survival. The rate of decay is then

−dN

dt=N(0)

τe−t/τ .


Mathematically, differentating an exponential just gives back an-other exponential. Physically, this is telling us that as N falls offexponentially, the rate of decay falls off at the same exponentialrate, because a lower N means fewer atoms that remain available todecay.

self-check CCheck that both sides of the equation for the rate of decay have units ofs−1, i.e., decays per unit time. . Answer, p. 933

The hot potato example 4. A nuclear physicist with a demented sense of humor tossesyou a cigar box, yelling “hot potato.” The label on the box says“contains 1020 atoms of 17F, half-life of 66 s, produced today inour reactor at 1 p.m.” It takes you two seconds to read the label,after which you toss it behind some lead bricks and run away. Thetime is 1:40 p.m. Will you die?

. The time elapsed since the radioactive fluorine was producedin the reactor was 40 minutes, or 2400 s. The number of elapsedhalf-lives is therefore t/t1/2 = 36. The initial number of atomswas N(0) = 1020 . The number of decays per second is nowabout 107 s−1, so it produced about 2×107 high-energy electronswhile you held it in your hands. Although twenty million electronssounds like a lot, it is not really enough to be dangerous.

By the way, none of the equations we’ve derived so far was theactual probability distribution for the time at which a particularradioactive atom will decay. That probability distribution would befound by substituting N(0) = 1 into the equation for the rate ofdecay.


A In the medical procedure involving 131I, why is it the gamma raysthat are detected, not the electrons or neutrinos that are also emitted?

B For 1 s, Fred holds in his hands 1 kg of radioactive stuff with ahalf-life of 1000 years. Ginger holds 1 kg of a different substance, with ahalf-life of 1 min, for the same amount of time. Did they place themselvesin equal danger, or not?

C How would you interpret it if you calculated N(t), and found it wasless than one?

D Does the half-life depend on how much of the substance you have?Does the expected time until the sample decays completely depend onhow much of the substance you have?

13.1.5 Applications of calculus

The area under the probability distribution is of course an in-tegral. If we call the random number x and the probability distri-bution D(x), then the probability that x lies in a certain range is


given by

(probability of a ≤ x ≤ b) =

∫ b

aD(x) dx .

What about averages? If x had a finite number of equally probablevalues, we would simply add them up and divide by how many wehad. If they weren’t equally likely, we’d make the weighted averagex1P1 + x2P2+... But we need to generalize this to a variable x thatcan take on any of a continuum of values. The continuous versionof a sum is an integral, so the average is

(average value of x) =

∫xD(x) dx ,

where the integral is over all possible values of x.

Probability distribution for radioactive decay example 5Here is a rigorous justification for the statement in subsection13.1.4 that the probability distribution for radioactive decay is foundby substituting N(0) = 1 into the equation for the rate of decay. Weknow that the probability distribution must be of the form

D(t) = k0.5t/t1/2 ,

where k is a constant that we need to determine. The atom isguaranteed to decay eventually, so normalization gives us

(probability of 0 ≤ t <∞) = 1

=∫ ∞

0D(t) dt .

The integral is most easily evaluated by converting the functioninto an exponential with e as the base

D(t) = k exp[ln(

0.5t/t1/2

)]= k exp

[t

t1/2ln 0.5

]= k exp

(− ln 2

t1/2t)

,

which gives an integral of the familiar form∫

ecx dx = (1/c)ecx .We thus have

1 = −kt1/2

ln 2exp

(− ln 2

t1/2t)]∞

0,

which gives the desired result:

k =ln 2t1/2

.


Average lifetime example 6You might think that the half-life would also be the average life-time of an atom, since half the atoms’ lives are shorter and halflonger. But the half whose lives are longer include some that sur-vive for many half-lives, and these rare long-lived atoms skew theaverage. We can calculate the average lifetime as follows:

(average lifetime) =∫ ∞

0t D(t) dt

Using the convenient base-e form again, we have

(average lifetime) =ln 2t1/2

∫ ∞0

t exp(− ln 2

t1/2t)

dt .

This integral is of a form that can either be attacked with in-tegration by parts or by looking it up in a table. The result is∫

xecx dx = xc ecx − 1

c2 ecx , and the first term can be ignored forour purposes because it equals zero at both limits of integration.We end up with

(average lifetime) =ln 2t1/2

(t1/2

ln 2

)2

=t1/2

ln 2= 1.443 t1/2 ,

which is, as expected, longer than one half-life.


k / In recent decades, a huge hole in the ozone layer has spreadout from Antarctica. Left: November 1978. Right: November 1992

13.2 Light As a ParticleThe only thing that interferes with my learning is my education.

Albert Einstein

Radioactivity is random, but do the laws of physics exhibit ran-domness in other contexts besides radioactivity? Yes. Radioactivedecay was just a good playpen to get us started with concepts ofrandomness, because all atoms of a given isotope are identical. Bystocking the playpen with an unlimited supply of identical atom-toys, nature helped us to realize that their future behavior could bedifferent regardless of their original identicality. We are now readyto leave the playpen, and see how randomness fits into the structureof physics at the most fundamental level.

The laws of physics describe light and matter, and the quantumrevolution rewrote both descriptions. Radioactivity was a good ex-ample of matter’s behaving in a way that was inconsistent withclassical physics, but if we want to get under the hood and under-stand how nonclassical things happen, it will be easier to focus onlight rather than matter. A radioactive atom such as uranium-235is after all an extremely complex system, consisting of 92 protons,143 neutrons, and 92 electrons. Light, however, can be a simple sinewave.

However successful the classical wave theory of light had been— allowing the creation of radio and radar, for example — it stillfailed to describe many important phenomena. An example thatis currently of great interest is the way the ozone layer protects usfrom the dangerous short-wavelength ultraviolet part of the sun’sspectrum. In the classical description, light is a wave. When a wave


passes into and back out of a medium, its frequency is unchanged,and although its wavelength is altered while it is in the medium,it returns to its original value when the wave reemerges. Luckilyfor us, this is not at all what ultraviolet light does when it passesthrough the ozone layer, or the layer would offer no protection atall!

13.2.1 Evidence for light as a particle

For a long time, physicists tried to explain away the problemswith the classical theory of light as arising from an imperfect under-standing of atoms and the interaction of light with individual atomsand molecules. The ozone paradox, for example, could have beenattributed to the incorrect assumption that one could think of theozone layer as a smooth, continuous substance, when in reality itwas made of individual ozone molecules. It wasn’t until 1905 thatAlbert Einstein threw down the gauntlet, proposing that the prob-lem had nothing to do with the details of light’s interaction withatoms and everything to do with the fundamental nature of lightitself.

a / Digital camera images of dimmer and dimmer sources of light.The dots are records of individual photons.

In those days the data were sketchy, the ideas vague, and theexperiments difficult to interpret; it took a genius like Einstein to cutthrough the thicket of confusion and find a simple solution. Today,however, we can get right to the heart of the matter with a piece ofordinary consumer electronics, the digital camera. Instead of film, adigital camera has a computer chip with its surface divided up into agrid of light-sensitive squares, called “pixels.” Compared to a grainof the silver compound used to make regular photographic film, adigital camera pixel is activated by an amount of light energy ordersof magnitude smaller. We can learn something new about light byusing a digital camera to detect smaller and smaller amounts oflight, as shown in figure a. Figure a/1 is fake, but a/2 and a/3 arereal digital-camera images made by Prof. Lyman Page of PrincetonUniversity as a classroom demonstration. Figure a/1 is what wewould see if we used the digital camera to take a picture of a fairly

Section 13.2 Light As a Particle 841

b / A wave is partially absorbed.

c / A stream of particles ispartially absorbed.

d / Einstein and Seurat: twinsseparated at birth? Seine GrandeJatte by Georges Seurat (19thcentury).

dim source of light. In figures a/2 and a/3, the intensity of the lightwas drastically reduced by inserting semitransparent absorbers likethe tinted plastic used in sunglasses. Going from a/1 to a/2 to a/3,more and more light energy is being thrown away by the absorbers.

The results are drastically different from what we would expectbased on the wave theory of light. If light was a wave and nothingbut a wave, b, then the absorbers would simply cut down the wave’samplitude across the whole wavefront. The digital camera’s entirechip would be illuminated uniformly, and weakening the wave withan absorber would just mean that every pixel would take a long timeto soak up enough energy to register a signal.

But figures a/2 and a/3 show that some pixels take strong hitswhile others pick up no energy at all. Instead of the wave picture,the image that is naturally evoked by the data is something morelike a hail of bullets from a machine gun, c. Each “bullet” of lightapparently carries only a tiny amount of energy, which is why de-tecting them individually requires a sensitive digital camera ratherthan an eye or a piece of film.

Although Einstein was interpreting different observations, thisis the conclusion he reached in his 1905 paper: that the pure wavetheory of light is an oversimplification, and that the energy of a beamof light comes in finite chunks rather than being spread smoothlythroughout a region of space.

We now think of these chunks as particles of light, and call them“photons,” although Einstein avoided the word “particle,” and theword “photon” was invented later. Regardless of words, the trou-ble was that waves and particles seemed like inconsistent categories.The reaction to Einstein’s paper could be kindly described as vig-orously skeptical. Even twenty years later, Einstein wrote, “Thereare therefore now two theories of light, both indispensable, and —as one must admit today despite twenty years of tremendous efforton the part of theoretical physicists — without any logical connec-tion.” In the remainder of this chapter we will learn how the seemingparadox was eventually resolved.


A Suppose someone rebuts the digital camera data in figure a, claim-ing that the random pattern of dots occurs not because of anything fun-damental about the nature of light but simply because the camera’s pixelsare not all exactly the same — some are just more sensitive than others.How could we test this interpretation?

B Discuss how the correspondence principle applies to the observa-tions and concepts discussed in this section.


e / Apparatus for observingthe photoelectric effect. A beamof light strikes a capacitor plateinside a vacuum tube, and elec-trons are ejected (black arrows).

13.2.2 How much light is one photon?

The photoelectric effect

We have seen evidence that light energy comes in little chunks,so the next question to be asked is naturally how much energy isin one chunk. The most straightforward experimental avenue foraddressing this question is a phenomenon known as the photoelec-tric effect. The photoelectric effect occurs when a photon strikesthe surface of a solid object and knocks out an electron. It occurscontinually all around you. It is happening right now at the surfaceof your skin and on the paper or computer screen from which youare reading these words. It does not ordinarily lead to any observ-able electrical effect, however, because on the average free electronsare wandering back in just as frequently as they are being ejected.(If an object did somehow lose a significant number of electrons,its growing net positive charge would begin attracting the electronsback more and more strongly.)

Figure e shows a practical method for detecting the photoelec-tric effect. Two very clean parallel metal plates (the electrodes of acapacitor) are sealed inside a vacuum tube, and only one plate is ex-posed to light. Because there is a good vacuum between the plates,any ejected electron that happens to be headed in the right direc-tion will almost certainly reach the other capacitor plate withoutcolliding with any air molecules.

The illuminated (bottom) plate is left with a net positive charge,and the unilluminated (top) plate acquires a negative charge fromthe electrons deposited on it. There is thus an electric field betweenthe plates, and it is because of this field that the electrons’ paths arecurved, as shown in the diagram. However, since vacuum is a goodinsulator, any electrons that reach the top plate are prevented fromresponding to the electrical attraction by jumping back across thegap. Instead they are forced to make their way around the circuit,passing through an ammeter. The ammeter allows a measurementof the strength of the photoelectric effect.

An unexpected dependence on frequency

The photoelectric effect was discovered serendipitously by Hein-rich Hertz in 1887, as he was experimenting with radio waves. Hewas not particularly interested in the phenomenon, but he did noticethat the effect was produced strongly by ultraviolet light and moreweakly by lower frequencies. Light whose frequency was lower thana certain critical value did not eject any electrons at all. (In factthis was all prior to Thomson’s discovery of the electron, so Hertzwould not have described the effect in terms of electrons — we arediscussing everything with the benefit of hindsight.) This depen-dence on frequency didn’t make any sense in terms of the classicalwave theory of light. A light wave consists of electric and magnetic


f / The hamster in her hamsterball is like an electron emergingfrom the metal (tiled kitchen floor)into the surrounding vacuum(wood floor). The wood floor ishigher than the tiled floor, so asshe rolls up the step, the hamsterwill lose a certain amount ofkinetic energy, analogous to Es.If her kinetic energy is too small,she won’t even make it up thestep.

fields. The stronger the fields, i.e., the greater the wave’s ampli-tude, the greater the forces that would be exerted on electrons thatfound themselves bathed in the light. It should have been amplitude(brightness) that was relevant, not frequency. The dependence onfrequency not only proves that the wave model of light needs mod-ifying, but with the proper interpretation it allows us to determinehow much energy is in one photon, and it also leads to a connec-tion between the wave and particle models that we need in order toreconcile them.

To make any progress, we need to consider the physical processby which a photon would eject an electron from the metal electrode.A metal contains electrons that are free to move around. Ordinarily,in the interior of the metal, such an electron feels attractive forcesfrom atoms in every direction around it. The forces cancel out. Butif the electron happens to find itself at the surface of the metal,the attraction from the interior side is not balanced out by anyattraction from outside. In popping out through the surface theelectron therefore loses some amount of energy Es, which dependson the type of metal used.

Suppose a photon strikes an electron, annihilating itself and giv-ing up all its energy to the electron. (We now know that this is whatalways happens in the photoelectric effect, although it had not yetbeen established in 1905 whether or not the photon was completelyannihilated.) The electron will (1) lose kinetic energy through colli-sions with other electrons as it plows through the metal on its wayto the surface; (2) lose an amount of kinetic energy equal to Es asit emerges through the surface; and (3) lose more energy on its wayacross the gap between the plates, due to the electric field betweenthe plates. Even if the electron happens to be right at the surface ofthe metal when it absorbs the photon, and even if the electric fieldbetween the plates has not yet built up very much, Es is the bareminimum amount of energy that it must receive from the photonif it is to contribute to a measurable current. The reason for usingvery clean electrodes is to minimize Es and make it have a definitevalue characteristic of the metal surface, not a mixture of valuesdue to the various types of dirt and crud that are present in tinyamounts on all surfaces in everyday life.

We can now interpret the frequency dependence of the photo-electric effect in a simple way: apparently the amount of energypossessed by a photon is related to its frequency. A low-frequencyred or infrared photon has an energy less than Es, so a beam ofthem will not produce any current. A high-frequency blue or violetphoton, on the other hand, packs enough of a punch to allow anelectron to make it to the other plate. At frequencies higher thanthe minimum, the photoelectric current continues to increase withthe frequency of the light because of effects (1) and (3).


g / A different way of study-ing the photoelectric effect.

h / The quantity Es + e∆V in-dicates the energy of one photon.It is found to be proportional tothe frequency of the light.

Numerical relationship between energy and frequency

Prompted by Einstein’s photon paper, Robert Millikan (whomwe first encountered in chapter 8) figured out how to use the pho-toelectric effect to probe precisely the link between frequency andphoton energy. Rather than going into the historical details of Mil-likan’s actual experiments (a lengthy experimental program thatoccupied a large part of his professional career) we will describe asimple version, shown in figure g, that is used sometimes in collegelaboratory courses.2 The idea is simply to illuminate one plate ofthe vacuum tube with light of a single wavelength and monitor thevoltage difference between the two plates as they charge up. Sincethe resistance of a voltmeter is very high (much higher than theresistance of an ammeter), we can assume to a good approximationthat electrons reaching the top plate are stuck there permanently,so the voltage will keep on increasing for as long as electrons aremaking it across the vacuum tube.

At a moment when the voltage difference has a reached a value∆V, the minimum energy required by an electron to make it out ofthe bottom plate and across the gap to the other plate is Es + e∆V.As ∆V increases, we eventually reach a point at which Es + e∆Vequals the energy of one photon. No more electrons can cross thegap, and the reading on the voltmeter stops rising. The quantityEs+e∆V now tells us the energy of one photon. If we determine thisenergy for a variety of wavelengths, h, we find the following simplerelationship between the energy of a photon and the frequency ofthe light:

E = hf ,

where h is a constant with the value 6.63 × 10−34 J · s. Note howthe equation brings the wave and particle models of light under thesame roof: the left side is the energy of one particle of light, whilethe right side is the frequency of the same light, interpreted as awave. The constant h is known as Planck’s constant, for historicalreasons explained in the footnote beginning on the preceding page.

2What I’m presenting in this chapter is a simplified explanation of how thephoton could have been discovered. The actual history is more complex. MaxPlanck (1858-1947) began the photon saga with a theoretical investigation ofthe spectrum of light emitted by a hot, glowing object. He introduced quanti-zation of the energy of light waves, in multiples of hf , purely as a mathematicaltrick that happened to produce the right results. Planck did not believe thathis procedure could have any physical significance. In his 1905 paper Einsteintook Planck’s quantization as a description of reality, and applied it to varioustheoretical and experimental puzzles, including the photoelectric effect. Mil-likan then subjected Einstein’s ideas to a series of rigorous experimental tests.Although his results matched Einstein’s predictions perfectly, Millikan was skep-tical about photons, and his papers conspicuously omit any reference to them.Only in his autobiography did Millikan rewrite history and claim that he hadgiven experimental proof for photons.


self-check D

How would you extract h from the graph in figure h? What if you didn’teven know Es in advance, and could only graph e∆V versus f? .

Answer, p. 933

Since the energy of a photon is hf , a beam of light can only haveenergies of hf , 2hf , 3hf , etc. Its energy is quantized — there is nosuch thing as a fraction of a photon. Quantum physics gets its namefrom the fact that it quantizes quantities like energy, momentum,and angular momentum that had previously been thought to besmooth, continuous and infinitely divisible.

Number of photons emitted by a lightbulb per second example 7. Roughly how many photons are emitted by a 100-W lightbulb in1 second?

. People tend to remember wavelengths rather than frequenciesfor visible light. The bulb emits photons with a range of frequen-cies and wavelengths, but let’s take 600 nm as a typical wave-length for purposes of estimation. The energy of a single photonis

Ephoton = hf= hc/λ

A power of 100 W means 100 joules per second, so the numberof photons is

(100 J)/Ephoton = (100 J)/(hc/λ)

≈ 3× 1020

This hugeness of this number is consistent with the correspon-dence principle. The experiments that established the classicaltheory of optics weren’t wrong. They were right, within their do-main of applicability, in which the number of photons was so largeas to be indistinguishable from a continuous beam.

Measuring the wave example 8When surfers are out on the water waiting for their chance tocatch a wave, they’re interested in both the height of the wavesand when the waves are going to arrive. In other words, they ob-serve both the amplitude and phase of the waves, and it doesn’tmatter to them that the water is granular at the molecular level.The correspondence principle requires that we be able to do thesame thing for electromagnetic waves, since the classical theoryof electricity and magnetism was all stated and verified experi-mentally in terms of the fields E and B, which are the amplitudeof an electromagnetic wave. The phase is also necessary, sincethe induction effects predicted by Maxwell’s equation would flip


their signs depending on whether an oscillating field is on its wayup or on its way back down.

This is a more demanding application of the correspondence prin-ciple than the one in example 7, since amplitudes and phasesconstitute more detailed information than the over-all intensity ofa beam of light. Eyeball measurements can’t detect this type of in-formation, since the eye is much bigger than a wavelength, but forexample an AM radio receiver can do it with radio waves, sincethe wavelength for a station at 1000 kHz is about 300 meters,which is much larger than the antenna. The correspondence prin-ciple demands that we be able to explain this in terms of the pho-ton theory, and this requires not just that we have a large numberof photons emitted by the transmitter per second, as in example 7,but that even by the time they spread out and reach the receivingantenna, there should be many photons overlapping each otherwithin a space of one cubic wavelength. Problem 47 on p. 907verifies that the number is in fact extremely large.

Momentum of a photon example 9. According to the theory of relativity, the momentum of a beamof light is given by p = E/c. Apply this to find the momentumof a single photon in terms of its frequency, and in terms of itswavelength.

. Combining the equations p = E/c and E = hf , we find

p = E/c

=hc

f .

To reexpress this in terms of wavelength, we use c = fλ:

p =hc· cλ

=hλ

The second form turns out to be simpler.


A The photoelectric effect only ever ejects a very tiny percentage ofthe electrons available near the surface of an object. How well does thisagree with the wave model of light, and how well with the particle model?Consider the two different distance scales involved: the wavelength of thelight, and the size of an atom, which is on the order of 10−10 or 10−9 m.

B What is the significance of the fact that Planck’s constant is numeri-cally very small? How would our everyday experience of light be differentif it was not so small?

C How would the experiments described above be affected if a singleelectron was likely to get hit by more than one photon?

D Draw some representative trajectories of electrons for ∆V = 0, ∆Vless than the maximum value, and ∆V greater than the maximum value.


E Explain based on the photon theory of light why ultraviolet light wouldbe more likely than visible or infrared light to cause cancer by damagingDNA molecules. How does this relate to discussion question C?

F Does E = hf imply that a photon changes its energy when it passesfrom one transparent material into another substance with a different in-dex of refraction?

13.2.3 Wave-particle duality

How can light be both a particle and a wave? We are nowready to resolve this seeming contradiction. Often in science whensomething seems paradoxical, it’s because we (1) don’t define ourterms carefully, or (2) don’t test our ideas against any specific real-world situation. Let’s define particles and waves as follows:

• Waves exhibit superposition, and specifically interference phe-nomena.

• Particles can only exist in whole numbers, not fractions.

As a real-world check on our philosophizing, there is one partic-ular experiment that works perfectly. We set up a double-slit inter-ference experiment that we know will produce a diffraction patternif light is an honest-to-goodness wave, but we detect the light witha detector that is capable of sensing individual photons, e.g., a dig-ital camera. To make it possible to pick out individual dots due toindividual photons, we must use filters to cut down the intensity ofthe light to a very low level, just as in the photos by Prof. Pageon p. 841. The whole thing is sealed inside a light-tight box. Theresults are shown in figure i. (In fact, the similar figures in on page841 are simply cutouts from these figures.)

i / Wave interference patternsphotographed by Prof. LymanPage with a digital camera. Laserlight with a single well-definedwavelength passed through aseries of absorbers to cut downits intensity, then through a set ofslits to produce interference, andfinally into a digital camera chip.(A triple slit was actually used,but for conceptual simplicity wediscuss the results in the maintext as if it was a double slit.) Inpanel 2 the intensity has beenreduced relative to 1, and evenmore so for panel 3.


Neither the pure wave theory nor the pure particle theory canexplain the results. If light was only a particle and not a wave, therewould be no interference effect. The result of the experiment wouldbe like firing a hail of bullets through a double slit, j. Only twospots directly behind the slits would be hit.

If, on the other hand, light was only a wave and not a particle,we would get the same kind of diffraction pattern that would happenwith a water wave, k. There would be no discrete dots in the photo,only a diffraction pattern that shaded smoothly between light anddark.

Applying the definitions to this experiment, light must be botha particle and a wave. It is a wave because it exhibits interferenceeffects. At the same time, the fact that the photographs containdiscrete dots is a direct demonstration that light refuses to be splitinto units of less than a single photon. There can only be wholenumbers of photons: four photons in figure i/3, for example.

A wrong interpretation: photons interfering with each other

One possible interpretation of wave-particle duality that occurredto physicists early in the game was that perhaps the interference ef-fects came from photons interacting with each other. By analogy, awater wave consists of moving water molecules, and interference ofwater waves results ultimately from all the mutual pushes and pullsof the molecules. This interpretation was conclusively disproved byG.I. Taylor, a student at Cambridge. The demonstration by Prof.Page that we’ve just been discussing is essentially a modernizedversion of Taylor’s work. Taylor reasoned that if interference effectscame from photons interacting with each other, a bare minimum oftwo photons would have to be present at the same time to produceinterference. By making the light source extremely dim, we can bevirtually certain that there are never two photons in the box at thesame time. In figure i/3, however, the intensity of the light hasbeen cut down so much by the absorbers that if it was in the open,the average separation between photons would be on the order ofa kilometer! At any given moment, the number of photons in thebox is most likely to be zero. It is virtually certain that there werenever two photons in the box at once.

The concept of a photon’s path is undefined.

If a single photon can demonstrate double-slit interference, thenwhich slit did it pass through? The unavoidable answer must be thatit passes through both! This might not seem so strange if we thinkof the photon as a wave, but it is highly counterintuitive if we tryto visualize it as a particle. The moral is that we should not thinkin terms of the path of a photon. Like the fully human and fullydivine Jesus of Christian theology, a photon is supposed to be 100%wave and 100% particle. If a photon had a well defined path, then it


j / Bullets pass through a doubleslit.

k / A water wave passes througha double slit.

l / A single photon can gothrough both slits.

would not demonstrate wave superposition and interference effects,contradicting its wave nature. (In subsection 13.3.4 we will discussthe Heisenberg uncertainty principle, which gives a numerical wayof approaching this issue.)

Another wrong interpretation: the pilot wave hypothesis

A second possible explanation of wave-particle duality was takenseriously in the early history of quantum mechanics. What if thephoton particle is like a surfer riding on top of its accompanyingwave? As the wave travels along, the particle is pushed, or “piloted”by it. Imagining the particle and the wave as two separate entitiesallows us to avoid the seemingly paradoxical idea that a photon isboth at once. The wave happily does its wave tricks, like super-position and interference, and the particle acts like a respectableparticle, resolutely refusing to be in two different places at once. Ifthe wave, for instance, undergoes destructive interference, becomingnearly zero in a particular region of space, then the particle simplyis not guided into that region.

The problem with the pilot wave interpretation is that the onlyway it can be experimentally tested or verified is if someone managesto detach the particle from the wave, and show that there really aretwo entities involved, not just one. Part of the scientific method isthat hypotheses are supposed to be experimentally testable. Sincenobody has ever managed to separate the wavelike part of a photonfrom the particle part, the interpretation is not useful or meaningfulin a scientific sense.

The probability interpretation

The correct interpretation of wave-particle duality is suggestedby the random nature of the experiment we’ve been discussing: eventhough every photon wave/particle is prepared and released in thesame way, the location at which it is eventually detected by thedigital camera is different every time. The idea of the probabilityinterpretation of wave-particle duality is that the location of thephoton-particle is random, but the probability that it is in a certainlocation is higher where the photon-wave’s amplitude is greater.

More specifically, the probability distribution of the particle mustbe proportional to the square of the wave’s amplitude,

(probability distribution) ∝ (amplitude)2 .

This follows from the correspondence principle and from the factthat a wave’s energy density is proportional to the square of its am-plitude. If we run the double-slit experiment for a long enough time,the pattern of dots fills in and becomes very smooth as would have


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In section 12.2 we've only done examples of mirrors with