7 (a) find a case where the magni cation of a curved ... · 8 as discussed in question 7, there are...

Problem 29.

7 (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.

. Solution, p. 448

8 As discussed in question 7, 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 half apage of paper.

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

9 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.)

. Solution, p. 449

10 (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).

Problems 251

11 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?

. Solution, p. 449

12 There are two main types of telescopes, refracting (using alens) and reflecting (using a mirror as in figure p on p. 224). (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.

13 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 or from a point on the axis.

. Solution, p. 449

14 When you take pictures with a camera, the distance betweenthe lens and the film or chip has to be adjusted, depending on thedistance at which you want to focus. This is done by moving thelens. If you want to change your focus so that you can take a pictureof something farther away, which way do you have to move the lens?Explain using ray diagrams. [Based on a problem by Eric Mazur.]

15 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.] . Solution, p. 450

252 Chapter 11 Images, quantitatively

Problem 18.

Problem 19.

16 An object is more than one focal length from a converginglens. (a) Draw a ray diagram. (b) Using reasoning like that devel-oped in chapter 11, 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 80 cm from the rose, locate the image.

√

17 Zahra likes to play practical jokes on the friends she goeshiking with. One night, by a blazing camp fire, she stealthily usesa lens of focal length f to gather light from the fire and make ahot spot on Becky’s neck. (a) Using the method of section section11.2, p. 237, draw a ray diagram and set up the equation for theimage location, inferring the correct plus and minus signs from thediagram. (b) Let A be the distance from the lens to the campfire,and B the distance from the lens to Becky’s neck. Consider thefollowing nine possibilities:

B< f = f > f

A< f= f> f

By reasoning about your equation from part a, determine which ofthese are possible and which are not. . Solution, p. 451

18 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. 451

19 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, parallel toits 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. . Solution, p. 452

Problems 253

Problem 22.

20 An object is less than one focal length from a converging lens.(a) Draw a ray diagram. (b) Using reasoning like that developed inchapter 11, determine the positive and negative signs in the equation1/f = ±1/di ± 1/do. (c) The images of the rose in section 4.2 weremade using a lens with a focal length of 23 cm. If the lens is placed10 cm from the rose, locate the image.

√

21 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 developed inchapter 11, determine the positive and negative signs in the equation1/f = ±1/di ± 1/do. (c) If the focal length of the lens is 50.0 cm,and the person is looking at an object at a distance of 80.0 cm,locate the image.

√

22 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. 452


23 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 fi-nal image.

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

24 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.

25 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).

26 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. 443

√

27 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. 459

Problems 255

Problem 30.

28 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.

√

29 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? ?

30 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 30 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 a


regular 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 which re-main unsolved. The book by Klee and Wagon gives a good intro-duction 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. ?

31 The intensity of a beam of light is defined as the power perunit area incident on a perpendicular surface. Suppose that a beamof light in a medium with index of refraction n reaches the surface ofthe medium, with air on the outside. Its incident angle with respectto the normal is θ. (All angles are in radians.) Only a fractionf of the energy is transmitted, the rest being reflected. Becauseof this, we might expect that the transmitted ray would always beless intense than the incident one. But because the transmitted rayis refracted, it becomes narrower, causing an additional change inintensity by a factor g > 1. The product of these factors I = fg canbe greater than one. The purpose of this problem is to estimate themaximum amount of intensification.We will use the small-angle approximation θ � 1 freely, in order tomake the math tractable. In our previous studies of waves, we haveonly studied the factor f in the one-dimensional case where θ =0. The generalization to θ 6= 0 is rather complicated and dependson the polarization, but for unpolarized light, we can use Schlick’sapproximation,

f(θ) = f(0)(1− cos θ)5,

where the value of f at θ = 0 is found as in problem 10 on p. 143.(a) Using small-angle approximations, obtain an expression for g ofthe form g ≈ 1 + Pθ2, and find the constant P . . Answer, p. 459(b) Find an expression for I that includes the two leading-orderterms in θ. We will call this expression I2. Obtain a simple expres-sion for the angle at which I2 is maximized. As a check on yourwork, you should find that for n = 1.3, θ = 63◦. (Trial-and-errormaximization of I gives 60◦.)(c) Find an expression for the maximum value of I2. You shouldfind that for n = 1.3, the maximum intensification is 31%.

?

Problems 257

Problem 32.

32 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 theancient 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 betweenan image and its anamorph, and design a modified version of thelinkage that does. Some knowledge of analytic geometry will behelpful. ?


Exercise 11A: Object and image distancesEquipment:

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 11A: Object and image distances 259

Exercise 11B: 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.


a / In this view from overhead, astraight, sinusoidal water waveencounters a barrier with twogaps in it. Strong wave vibrationoccurs at angles X and Z, butthere is none at all at angle Y.(The figure has been retouchedfrom a real photo of water waves.In reality, the waves beyond thebarrier would be much weakerthan the ones before it, and theywould therefore be difficult tosee.)

b / This doesn’t happen.

Chapter 12

Wave optics

Electron microscopes can make images of individual atoms, but whywill a visible-light microscope never be able to? Stereo speakerscreate the illusion of music that comes from a band arranged inyour living room, but why doesn’t the stereo illusion work with bassnotes? Why are computer chip manufacturers investing billions ofdollars in equipment to etch chips with x-rays instead of visiblelight?

The answers to all of these questions have to do with the subjectof wave optics. So far this book has discussed the interaction oflight waves with matter, and its practical applications to opticaldevices like mirrors, but we have used the ray model of light almostexclusively. Hardly ever have we explicitly made use of the fact thatlight is an electromagnetic wave. We were able to get away with thesimple ray model because the chunks of matter we were discussing,such as lenses and mirrors, were thousands of times larger than awavelength of light. We now turn to phenomena and devices thatcan only be understood using the wave model of light.

12.1 Diffraction

Figure a shows a typical problem in wave optics, enacted withwater waves. It may seem surprising that we don’t get a simplepattern like figure b, but the pattern would only be that simpleif the wavelength was hundreds of times shorter than the distancebetween the gaps in the barrier and the widths of the gaps.

Wave optics is a broad subject, but this example will help usto pick out a reasonable set of restrictions to make things moremanageable:

(1) We restrict ourselves to cases in which a wave travels througha uniform medium, encounters a certain area in which the mediumhas different properties, and then emerges on the other side into asecond uniform region.

(2) We assume that the incoming wave is a nice tidy sine-wavepattern with wavefronts that are lines (or, in three dimensions,planes).

(3) In figure a we can see that the wave pattern immediatelybeyond the barrier is rather complex, but farther on it sorts itself

261

c / A practical, low-tech setup forobserving diffraction of light.

out into a set of wedges separated by gaps in which the water isstill. We will restrict ourselves to studying the simpler wave patternsthat occur farther away, so that the main question of interest is howintense the outgoing wave is at a given angle.

The kind of phenomenon described by restriction (1) is calleddiffraction. Diffraction can be defined as the behavior of a wavewhen it encounters an obstacle or a nonuniformity in its medium.In general, diffraction causes a wave to bend around obstacles andmake patterns of strong and weak waves radiating out beyond theobstacle. Understanding diffraction is the central problem of waveoptics. If you understand diffraction, even the subset of diffractionproblems that fall within restrictions (2) and (3), the rest of waveoptics is icing on the cake.

Diffraction can be used to find the structure of an unknowndiffracting object: even if the object is too small to study withordinary imaging, it may be possible to work backward from thediffraction pattern to learn about the object. The structure of acrystal, for example, can be determined from its x-ray diffractionpattern.

Diffraction can also be a bad thing. In a telescope, for example,light waves are diffracted by all the parts of the instrument. This willcause the image of a star to appear fuzzy even when the focus hasbeen adjusted correctly. By understanding diffraction, one can learnhow a telescope must be designed in order to reduce this problem— essentially, it should have the biggest possible diameter.

There are two ways in which restriction (2) might commonly beviolated. First, the light might be a mixture of wavelengths. If wesimply want to observe a diffraction pattern or to use diffraction asa technique for studying the object doing the diffracting (e.g., if theobject is too small to see with a microscope), then we can pass thelight through a colored filter before diffracting it.

A second issue is that light from sources such as the sun or alightbulb does not consist of a nice neat plane wave, except oververy small regions of space. Different parts of the wave are out ofstep with each other, and the wave is referred to as incoherent. Oneway of dealing with this is shown in figure c. After filtering to selecta certain wavelength of red light, we pass the light through a smallpinhole. The region of the light that is intercepted by the pinhole isso small that one part of it is not out of step with another. Beyondthe pinhole, light spreads out in a spherical wave; this is analogousto what happens when you speak into one end of a paper towel rolland the sound waves spread out in all directions from the other end.By the time the spherical wave gets to the double slit it has spreadout and reduced its curvature, so that we can now think of it as asimple plane wave.

262 Chapter 12 Wave optics

If this seems laborious, you may be relieved to know that moderntechnology gives us an easier way to produce a single-wavelength,coherent beam of light: the laser.

The parts of the final image on the screen in c are called diffrac-tion fringes. The center of each fringe is a point of maximum bright-ness, and halfway between two fringes is a minimum.

Because the diffraction spreads out in all directions, the fringesdo not have a spacing or locations that we can define in units ofmeters — such dimensions would get bigger if we moved the screenfarther away. But a diffraction does have definable angular dimen-sions. An example would be the angle between lines X and Z infigure a. A common problem in diffraction is to find some angulardimension or position θ, given one or more linear dimensions, suchas the distance d between a pair of double slits. Conversely, one canobserve a θ and determine an unknown d. In an example such asfigure d, there could actually be many θ’s and many d’s.

d / 1. Dorothy Hodgkin was awarded the Nobel Prize in chemistry in 1964 for her work in determiningthe structures of organic molecules using x-ray diffraction. The following figures sketch the technique. 2. Theunknown molecule, in this case a natural antibacterial enzyme called lysozyme, is crystallized. Withoutcrystallization, the random orientations of all the molecules would make the diffraction pattern a blurredaverage over orientations. 3. The x-ray diffraction pattern is observed. 4. The three-dimensional structure ofthe molecule is determined. As a much simpler example of the determination of an unknown structure from itsdiffraction pattern, see problem 9, p. 278. The more general technique is called Fourier analysis.

Section 12.1 Diffraction 263

e / The bottom figure is sim-ply a copy of the middle portionof the top one, scaled up by afactor of two. All the angles arethe same. Physically, the angularpattern of the diffraction fringescan’t be any different if we scaleboth λ and d by the same factor,leaving λ/d unchanged.

12.2 Scaling of diffractionThis chapter has “optics” in its title, so it is nominally about light,but we started out with an example involving water waves. Waterwaves are certainly easier to visualize, but is this a legitimate com-parison? In fact the analogy works quite well, despite the fact thata light wave has a wavelength about a million times shorter. Thisis because diffraction effects scale uniformly. That is, if we enlargeor reduce the whole diffraction situation by the same factor, includ-ing both the wavelengths and the sizes of the obstacles the waveencounters, the result is still a valid solution.

This is unusually simple behavior! Most physical phenomena donot scale in any simple way. For example, we can’t have spiders thesize of horses, because weight scales like the cube of the dimensions,while the strength of the animal’s legs scale like the square.

Of course water waves and light waves differ in many ways, notjust in scale, but the general facts you will learn about diffractionare applicable to all waves.

Another way of stating the simple scaling behavior of diffractionis that the diffraction angles we get depend only on the unitless ratioλ/d, where λ is the wavelength of the wave and d is some dimensionof the diffracting objects, e.g., the center-to-center spacing betweenthe slits in figure a. If, for instance, we scale up both λ and d by afactor of 37, the ratio λ/d will be unchanged.

Discussion question

A Why would x-rays rather than visible light be used to find the structureof a crystal, as in figure d? Sound waves are used to make images offetuses in the womb. What would influence the choice of wavelength?

12.3 The correspondence principleThe only reason we don’t usually notice diffraction of light in ev-eryday life is that we don’t normally deal with objects that arecomparable in size to a wavelength of visible light, which is about amillionth of a meter. Does this mean that wave optics contradictsray optics, or that wave optics sometimes gives wrong results? No.If you hold three fingers out in the sunlight and cast a shadow withthem, either wave optics or ray optics can be used to predict thestraightforward result: a shadow pattern with two bright lines wherethe light has gone through the gaps between your fingers. Wave op-tics is a more general theory than ray optics, so in any case whereray optics is valid, the two theories will agree. This is an exampleof a general idea enunciated by the physicist Niels Bohr, called thecorrespondence principle: when flaws in a physical theory lead tothe creation of a new and more general theory, the new theory muststill agree with the old theory within its more restricted area of ap-plicability. After all, a theory is only created as a way of describing


f / Double-slit diffraction.

g / A wavefront can be analyzedby the principle of superposition,breaking it down into many smallparts.

h / If it was by itself, each ofthe parts would spread out as acircular ripple.

i / Adding up the ripples pro-duces a new wavefront.

experimental observations. If the original theory had not worked inany cases at all, it would never have become accepted.

In the case of optics, the correspondence principle tells us thatwhen λ/d is small, both the ray and the wave model of light mustgive approximately the same result. Suppose you spread your fingersand cast a shadow with them using a coherent light source. Thequantity λ/d is about 10−4, so the two models will agree very closely.(To be specific, the shadows of your fingers will be outlined by aseries of light and dark fringes, but the angle subtended by a fringewill be on the order of 10−4 radians, so they will be too tiny to bevisible.

self-check AWhat kind of wavelength would an electromagnetic wave have to havein order to diffract dramatically around your body? Does this contradictthe correspondence principle? . Answer, p. 457

12.4 Huygens’ principle

Returning to the example of double-slit diffraction, f, note thestrong visual impression of two overlapping sets of concentric semi-circles. This is an example of Huygens’ principle, named after aDutch physicist and astronomer. (The first syllable rhymes with“boy.”) Huygens’ principle states that any wavefront can be brokendown into many small side-by-side wave peaks, g, which then spreadout as circular ripples, h, and by the principle of superposition, theresult of adding up these sets of ripples must give the same resultas allowing the wave to propagate forward, i. In the case of soundor light waves, which propagate in three dimensions, the “ripples”are actually spherical rather than circular, but we can often imaginethings in two dimensions for simplicity.

In double-slit diffraction the application of Huygens’ principle isvisually convincing: it is as though all the sets of ripples have beenblocked except for two. It is a rather surprising mathematical fact,however, that Huygens’ principle gives the right result in the case ofan unobstructed linear wave, h and i. A theoretically infinite numberof circular wave patterns somehow conspire to add together andproduce the simple linear wave motion with which we are familiar.

Since Huygens’ principle is equivalent to the principle of super-position, and superposition is a property of waves, what Huygenshad created was essentially the first wave theory of light. However,he imagined light as a series of pulses, like hand claps, rather thanas a sinusoidal wave.

The history is interesting. Isaac Newton loved the atomic theoryof matter so much that he searched enthusiastically for evidence thatlight was also made of tiny particles. The paths of his light particles

Section 12.4 Huygens’ principle 265

j / Thomas Young

k / Double-slit diffraction.

l / Use of Huygens’ principle.

m / Constructive interferencealong the center-line.

would correspond to rays in our description; the only significantdifference between a ray model and a particle model of light wouldoccur if one could isolate individual particles and show that lighthad a “graininess” to it. Newton never did this, so although hethought of his model as a particle model, it is more accurate to sayhe was one of the builders of the ray model.

Almost all that was known about reflection and refraction oflight could be interpreted equally well in terms of a particle modelor a wave model, but Newton had one reason for strongly opposingHuygens’ wave theory. Newton knew that waves exhibited diffrac-tion, but diffraction of light is difficult to observe, so Newton be-lieved that light did not exhibit diffraction, and therefore must notbe a wave. Although Newton’s criticisms were fair enough, the de-bate also took on the overtones of a nationalistic dispute betweenEngland and continental Europe, fueled by English resentment overLeibniz’s supposed plagiarism of Newton’s calculus. Newton wrotea book on optics, and his prestige and political prominence tendedto discourage questioning of his model.

Thomas Young (1773-1829) was the person who finally, a hun-dred years later, did a careful search for wave interference effectswith light and analyzed the results correctly. He observed double-slit diffraction of light as well as a variety of other diffraction ef-fects, all of which showed that light exhibited wave interference ef-fects, and that the wavelengths of visible light waves were extremelyshort. The crowning achievement was the demonstration by the ex-perimentalist Heinrich Hertz and the theorist James Clerk Maxwellthat light was an electromagnetic wave. Maxwell is said to have re-lated his discovery to his wife one starry evening and told her thatshe was the only other person in the world who knew what starlightwas.

12.5 Double-slit diffractionLet’s now analyze double-slit diffraction, k, using Huygens’ princi-ple. The most interesting question is how to compute the anglessuch as X and Z where the wave intensity is at a maximum, andthe in-between angles like Y where it is minimized. Let’s measureall our angles with respect to the vertical center line of the figure,which was the original direction of propagation of the wave.

If we assume that the width of the slits is small (on the orderof the wavelength of the wave or less), then we can imagine only asingle set of Huygens ripples spreading out from each one, l. Whitelines represent peaks, black ones troughs. The only dimension of thediffracting slits that has any effect on the geometric pattern of theoverlapping ripples is then the center-to-center distance, d, betweenthe slits.


n / The waves travel distances Land L′ from the two slits to getto the same point in space, at anangle θ from the center line.

o / A close-up view of figuren, showing how the path lengthdifference L − L′ is related to dand to the angle θ.

We know from our discussion of the scaling of diffraction thatthere must be some equation that relates an angle like θZ to theratio λ/d,

λ

d↔ θZ .

If the equation for θZ depended on some other expression such asλ+ d or λ2/d, then it would change when we scaled λ and d by thesame factor, which would violate what we know about the scalingof diffraction.

Along the central maximum line, X, we always have positivewaves coinciding with positive ones and negative waves coincidingwith negative ones. (I have arbitrarily chosen to take a snapshot ofthe pattern at a moment when the waves emerging from the slit areexperiencing a positive peak.) The superposition of the two sets ofripples therefore results in a doubling of the wave amplitude alongthis line. There is constructive interference. This is easy to explain,because by symmetry, each wave has had to travel an equal numberof wavelengths to get from its slit to the center line, m: Becauseboth sets of ripples have ten wavelengths to cover in order to reachthe point along direction X, they will be in step when they get there.

At the point along direction Y shown in the same figure, onewave has traveled ten wavelengths, and is therefore at a positiveextreme, but the other has traveled only nine and a half wavelengths,so it is at a negative extreme. There is perfect cancellation, so pointsalong this line experience no wave motion.

But the distance traveled does not have to be equal in order toget constructive interference. At the point along direction Z, onewave has gone nine wavelengths and the other ten. They are bothat a positive extreme.

self-check BAt a point half a wavelength below the point marked along direction X,carry out a similar analysis. . Answer, p. 457

To summarize, we will have perfect constructive interference atany point where the distance to one slit differs from the distance tothe other slit by an integer number of wavelengths. Perfect destruc-tive interference will occur when the number of wavelengths of pathlength difference equals an integer plus a half.

Now we are ready to find the equation that predicts the anglesof the maxima and minima. The waves travel different distancesto get to the same point in space, n. We need to find whether thewaves are in phase (in step) or out of phase at this point in order topredict whether there will be constructive interference, destructiveinterference, or something in between.

One of our basic assumptions in this chapter is that we will onlybe dealing with the diffracted wave in regions very far away from the

Section 12.5 Double-slit diffraction 267

p / Cutting d in half doublesthe angles of the diffractionfringes.

object that diffracts it, so the triangle is long and skinny. Most real-world examples with diffraction of light, in fact, would have triangleswith even skinner proportions than this one. The two long sides aretherefore very nearly parallel, and we are justified in drawing theright triangle shown in figure o, labeling one leg of the right triangleas the difference in path length , L−L′, and labeling the acute angleas θ. (In reality this angle is a tiny bit greater than the one labeledθ in figure n.)

The difference in path length is related to d and θ by the equation

L− L′

d= sin θ.

Constructive interference will result in a maximum at angles forwhich L− L′ is an integer number of wavelengths,

L− L′ = mλ. [condition for a maximum;

m is an integer]

Here m equals 0 for the central maximum, −1 for the first maximumto its left, +2 for the second maximum on the right, etc. Puttingall the ingredients together, we find mλ/d = sin θ, or

λ

d=

sin θ

m. [condition for a maximum;

m is an integer]

Similarly, the condition for a minimum is

λ

d=

sin θ

m. [condition for a minimum;

m is an integer plus 1/2]

That is, the minima are about halfway between the maxima.

As expected based on scaling, this equation relates angles to theunitless ratio λ/d. Alternatively, we could say that we have proventhe scaling property in the special case of double-slit diffraction. Itwas inevitable that the result would have these scaling properties,since the whole proof was geometric, and would have been equallyvalid when enlarged or reduced on a photocopying machine!

Counterintuitively, this means that a diffracting object withsmaller dimensions produces a bigger diffraction pattern, p.


q / Double-slit diffraction pat-terns of long-wavelength red light(top) and short-wavelength bluelight (bottom).

r / Interpretation of the angu-lar spacing ∆θ in example 3.It can be defined either frommaximum to maximum or fromminimum to minimum. Either way,the result is the same. It does notmake sense to try to interpret ∆θas the width of a fringe; one cansee from the graph and from theimage below that it is not obviouseither that such a thing is welldefined or that it would be thesame for all fringes.

Double-slit diffraction of blue and red light example 1Blue light has a shorter wavelength than red. For a given double-slit spacing d , the smaller value of λ/d for leads to smaller valuesof sin θ, and therefore to a more closely spaced set of diffractionfringes, shown in figure q.

The correspondence principle example 2Let’s also consider how the equations for double-slit diffractionrelate to the correspondence principle. When the ratio λ/d is verysmall, we should recover the case of simple ray optics. Now if λ/dis small, sin θ must be small as well, and the spacing betweenthe diffraction fringes will be small as well. Although we have notproven it, the central fringe is always the brightest, and the fringesget dimmer and dimmer as we go farther from it. For small valuesof λ/d , the part of the diffraction pattern that is bright enough tobe detectable covers only a small range of angles. This is exactlywhat we would expect from ray optics: the rays passing throughthe two slits would remain parallel, and would continue movingin the θ = 0 direction. (In fact there would be images of the twoseparate slits on the screen, but our analysis was all in terms ofangles, so we should not expect it to address the issue of whetherthere is structure within a set of rays that are all traveling in theθ = 0 direction.)

Spacing of the fringes at small angles example 3At small angles, we can use the approximation sin θ ≈ θ, whichis valid if θ is measured in radians. The equation for double-slitdiffraction becomes simply

λ

d=θ

m,

which can be solved for θ to give

θ =mλd

.

The difference in angle between successive fringes is the changein θ that results from changing m by plus or minus one,

∆θ =λ

d.

For example, if we write θ7 for the angle of the seventh brightfringe on one side of the central maximum and θ8 for the neigh-boring one, we have

θ8 − θ7 =8λd− 7λ

d

=λ

d,

and similarly for any other neighboring pair of fringes.

Section 12.5 Double-slit diffraction 269

s / A triple slit.

t / A double-slit diffraction pattern(top), and a pattern made by fiveslits (bottom).

Although the equation λ/d = sin θ/m is only valid for a doubleslit, it is can still be a guide to our thinking even if we are observingdiffraction of light by a virus or a flea’s leg: it is always true that

(1) large values of λ/d lead to a broad diffraction pattern, and

(2) diffraction patterns are repetitive.

In many cases the equation looks just like λ/d = sin θ/m butwith an extra numerical factor thrown in, and with d interpreted assome other dimension of the object, e.g., the diameter of a piece ofwire.

12.6 RepetitionSuppose we replace a double slit with a triple slit, s. We can thinkof this as a third repetition of the structures that were present inthe double slit. Will this device be an improvement over the doubleslit for any practical reasons?

The answer is yes, as can be shown using figure u. For easeof visualization, I have violated our usual rule of only consideringpoints very far from the diffracting object. The scale of the drawingis such that a wavelengths is one cm. In u/1, all three waves travelan integer number of wavelengths to reach the same point, so thereis a bright central spot, as we would expect from our experiencewith the double slit. In figure u/2, we show the path lengths toa new point. This point is farther from slit A by a quarter of awavelength, and correspondingly closer to slit C. The distance fromslit B has hardly changed at all. Because the paths lengths traveledfrom slits A and C differ by half a wavelength, there will be perfectdestructive interference between these two waves. There is still someuncanceled wave intensity because of slit B, but the amplitude willbe three times less than in figure u/1, resulting in a factor of 9decrease in brightness. Thus, by moving off to the right a little, wehave gone from the bright central maximum to a point that is quitedark.

Now let’s compare with what would have happened if slit C hadbeen covered, creating a plain old double slit. The waves comingfrom slits A and B would have been out of phase by 0.23 wavelengths,but this would not have caused very severe interference. The pointin figure u/2 would have been quite brightly lit up.

To summarize, we have found that adding a third slit narrowsdown the central fringe dramatically. The same is true for all the


v / Single-slit diffraction ofwater waves.

w / Single-slit diffraction ofred light. Note the double widthof the central maximum.

x / A pretty good simulationof the single-slit pattern of figurev, made by using three motors toproduce overlapping ripples fromthree neighboring points in thewater.

u / 1. There is a bright central maximum. 2. At this point just off the central maximum, the path lengths traveledby the three waves have changed.

other fringes as well, and since the same amount of energy is con-centrated in narrower diffraction fringes, each fringe is brighter andeasier to see, t.

This is an example of a more general fact about diffraction: ifsome feature of the diffracting object is repeated, the locations ofthe maxima and minima are unchanged, but they become narrower.

Taking this reasoning to its logical conclusion, a diffracting ob-ject with thousands of slits would produce extremely narrow fringes.Such an object is called a diffraction grating.

12.7 Single-slit diffraction

If we use only a single slit, is there diffraction? If the slit is notwide compared to a wavelength of light, then we can approximateits behavior by using only a single set of Huygens ripples. Thereare no other sets of ripples to add to it, so there are no constructiveor destructive interference effects, and no maxima or minima. Theresult will be a uniform spherical wave of light spreading out in alldirections, like what we would expect from a tiny lightbulb. Wecould call this a diffraction pattern, but it is a completely feature-less one, and it could not be used, for instance, to determine thewavelength of the light, as other diffraction patterns could.

All of this, however, assumes that the slit is narrow compared toa wavelength of light. If, on the other hand, the slit is broader, therewill indeed be interference among the sets of ripples spreading outfrom various points along the opening. Figure v shows an examplewith water waves, and figure w with light.

self-check CHow does the wavelength of the waves compare with the width of theslit in figure v? . Answer, p. 457

Section 12.7 Single-slit diffraction 271

y / An image of the Pleiadesstar cluster. The circular ringsaround the bright stars are due tosingle-slit diffraction at the mouthof the telescope’s tube.

z / A radio telescope.

We will not go into the details of the analysis of single-slit diffrac-tion, but let us see how its properties can be related to the generalthings we’ve learned about diffraction. We know based on scalingarguments that the angular sizes of features in the diffraction pat-tern must be related to the wavelength and the width, a, of the slitby some relationship of the form

λ

a↔ θ.

This is indeed true, and for instance the angle between the maximumof the central fringe and the maximum of the next fringe on one sideequals 1.5λ/a. Scaling arguments will never produce factors such asthe 1.5, but they tell us that the answer must involve λ/a, so all thefamiliar qualitative facts are true. For instance, shorter-wavelengthlight will produce a more closely spaced diffraction pattern.

An important scientific example of single-slit diffraction is intelescopes. Images of individual stars, as in figure y, are a good wayto examine diffraction effects, because all stars except the sun are sofar away that no telescope, even at the highest magnification, canimage their disks or surface features. Thus any features of a star’simage must be due purely to optical effects such as diffraction. Aprominent cross appears around the brightest star, and dimmer onessurround the dimmer stars. Something like this is seen in most tele-scope photos, and indicates that inside the tube of the telescopethere were two perpendicular struts or supports. Light diffractedaround these struts. You might think that diffraction could be elim-inated entirely by getting rid of all obstructions in the tube, but thecircles around the stars are diffraction effects arising from single-slit diffraction at the mouth of the telescope’s tube! (Actually wehave not even talked about diffraction through a circular opening,but the idea is the same.) Since the angular sizes of the diffractedimages depend on λ/a, the only way to improve the resolution ofthe images is to increase the diameter, a, of the tube. This is oneof the main reasons (in addition to light-gathering power) why thebest telescopes must be very large in diameter.

self-check DWhat would this imply about radio telescopes as compared with visible-light telescopes? . Answer, p.457

Double-slit diffraction is easier to understand conceptually thansingle-slit diffraction, but if you do a double-slit diffraction experi-ment in real life, you are likely to encounter a complicated patternlike figure aa/1, rather than the simpler one, 2, you were expecting.This is because the slits are fairly big compared to the wavelengthof the light being used. We really have two different distances inour pair of slits: d, the distance between the slits, and w, the width


of each slit. Remember that smaller distances on the object thelight diffracts around correspond to larger features of the diffractionpattern. The pattern 1 thus has two spacings in it: a short spac-ing corresponding to the large distance d, and a long spacing thatrelates to the small dimension w.

aa / 1. A diffraction pattern formed by a real double slit. The width of each slit is fairly big compared tothe wavelength of the light. This is a real photo. 2. This idealized pattern is not likely to occur in real life. To getit, you would need each slit to be so narrow that its width was comparable to the wavelength of the light, butthat’s not usually possible. This is not a real photo. 3. A real photo of a single-slit diffraction pattern caused bya slit whose width is the same as the widths of the slits used to make the top pattern.

Discussion question

A Why is it optically impossible for bacteria to evolve eyes that usevisible light to form images?

12.8 CoherenceUp until now, we have avoided too much detailed discussion of twofacts that sometimes make interference and diffraction effects unob-servable, and that historically made them more difficult to discover.First there is the fact that white light is a mixture of all the visiblewavelengths. This is why, for example, the thin-film interferencepattern of a soap bubble looks like a rainbow. To simplify things,we need a source of light that is monochromatic, i.e., contains onlya single wavelength or a small range of wavelengths. We could dothis either by filtering a white light source or by using a source oflight that is intrinsically monochromatic, such as a laser or some gasdischarge tubes.

But even with a monochromatic light source, we encounter a sep-arate issue, which is that most light sources do not emit light wavesthat are perfect, infinitely long sine waves. Sunlight and candlelight,for example, can be thought of as being composed of separate littlespurts of light, referred to as wave packets or wave trains. Each

Section 12.8 Coherence 273

ab / 1. Interference in an airwedge. 2. Side view. 3. If thewedge is thicker than the co-herence length of the light, theinterference pattern disappears.

wave packet is emitted by a separate atom of the gas. It containssome number of wavelengths, and it has no fixed phase relationshipto any other wave packet. The wave trains emitted by a laser aremuch longer, but still not infinitely long.

As an example of an experiment that can show these effects,figure ab/1 shows a thin-film interference pattern created by the airwedge between two pieces of very flat glass, where the top piece isplaced at a very small angle relative to the bottom one, ab/2. Thephase relationship between the two reflected waves is determinedby the extra distance traveled by the ray that is reflected by thebottom plate (as well as the fact that one of the two reflections willbe inverting).

If the angle is opened up too much, ab/3, we will no longer seefringes where the air layer is too thick. This is because the incidentwave train has only a certain length, and the extra distance traveledis now so great that the two reflected wave trains no longer overlapin space. In general, if the incident wave trains are n wavelengthslong, then we can see at most n bright and n dark fringes. The factthat about 18 fringes are visible in ab/1 shows that the light sourceused (let’s say a sodium gas discharge tube) made wave trains atleast 18 wavelengths in length.

In real-world light sources, the wave packets may not be as neatand tidy as the ones in figure ab. They may not look like sinewaves with clean cut-offs at the ends, and they may overlap oneanother. The result will look more like the examples in figure ac.Such a wave pattern has a property called its coherence length L.

ac / Waves with three different co-herence lengths, indicated by thearrows. Note that although thereis a superficial similarity betweenthese pictures and figure ab/1,they represent completely differ-ent things. Figure ab/1 is anactual photograph of interferencefringes, whose brightness is pro-portional to the square of the am-plitude. This figure is a pictureof the wave’s amplitude, not thesquared amplitude, and is anal-ogous to the little sine waves inab/2 and ab/3. These are wavesthat are traveling across the pageat the speed of light.


On scales small compared to L, the wave appears like a perfect sinewave. On scales large compared to L, we lose all phase correlations.For example, the middle wave in figure ac has L ≈ 5λ. If we picktwo points within this wave separated by a distance of λ in the left-right direction, they are likely to be very nearly in phase. But if theseparation is 20λ, approximately the width of the entire figure, thephase relationship is essentially random. If the light comes from aflame or a gas discharge tube, then this lack of a phase relationshipwould be because the parts of the wave at these large separationsfrom one another probably originated from different atoms in thesource.

Section 12.8 Coherence 275

Problems 1 and 4.

ProblemsKey√

A computerized answer check is available online.? A difficult problem.

1 The figure shows a diffraction pattern made by a double slit,along with an image of a meter stick to show the scale. Sketch thediffraction 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.

2 Match gratings A-C with the diffraction patterns 1-3 that theyproduce. Explain.


3 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?

4 The figure on p. 276 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?

√

5 Why would blue or violet light be the best for microscopy?. Solution, p. 453

6 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.

. Solution, p. 453

7 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.)

Problems 277

Problem 8. This image ofthe Pleiades star cluster showshaloes around the stars due to thewave nature of light.

8 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? . Solution, p. 453

9 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.

. Solution, p. 453


10 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?

√

11 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.

12 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.

. Solution, p. 453

13 When ultrasound is used for medical imaging, the frequencymay be as high as 5-20 MHz. Another medical application of ultra-sound is for therapeutic heating of tissues inside the body; here, thefrequency is typically 1-3 MHz. What fundamental physical reasonscould you suggest for the use of higher frequencies for imaging?

Problems 279

Exercise 12A: Two-source interference1. 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?


Exercise 12A: Two-source interference 281

Exercise 12B: Single-slit interferenceEquipment:

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.


Exercise 12B: Single-slit interference 283

The microscopicdescription of matter andquantum physics

284

Chapter 13

The atom and the nucleus

In chapters 7 and 9, we have used a microscopic description of mat-ter to find out things about thermodynamics. It’s remarkable thatone can get so far with almost no detailed description of what mat-ter is actually made of at the microscopic level. For example, wewere able to deduce the heat capacities of solids and gases (example2, p. 174), without even knowing anything at all about their con-stituent particles (atoms!). This is a good-news/bad-news situation.It’s good that we can figure out all kinds of things like the heat ca-pacity of copper without even having to know that copper is madeout of atoms that have certain specific properties. But if we we’reactually trying to find out about what copper is made out of, thenit’s unfortunate that we can’t find out much from its thermodynamicproperties.

There does come a point, though, at which we need to knowwhat’s really going on at the atomic and subatomic level. In thehistory of technology, this coincided with the creation of radio andthe exploitation of nuclear power.

13.1 The electrical nature of matter andquantization of charge

In your course on electricity and magnetism, you have seen evidencethat all matter contains electrically charged parts. For example,copper is an electrical conductor, and we interpret that as evidencethat it contains positively and negatively charged stuff, with at leastsome of that stuff being free to move. Furthermore, there is no suchthing as a form of matter that is a perfect insulator. Most aren’t asgood as copper, but all can conduct electricity to some extent. Thistells us that all matter contains charged stuff.

In 1909, Robert Millikan and coworkers published experimentalresults showing that when he prepared tiny droplets of oil and ma-nipulated them under a microscope with electric fields, their electriccharge seemed to come only in integer multiples of a certain amount.We notate this basic amount of charge as e and refer to it to as thefundamental charge. Millikan is now known to have fudged his data,and his result for e is statistically inconsistent with the currently ac-cepted value, e = 1.602× 10−19 C. Quantization of charge suggeststhat the charged stuff inside matter actually consists of charged

287

particles.

Today, the standard model of particle physics includes particlescalled quarks, which have fractional charges ±(1/3)e and ±(2/3)e.However, single quarks are never observed, only clusters of them,and the clusters always have charges that are integer multiples of e.

We summarize these facts by saying that charge is “quantized”in units of e. Similarly, money in the US is quantized in units ofcents, and discerning music listeners bewail the use of software in therecording studio that quantizes rhythm, forcing notes to land exactlyon the beat rather than allowing the kinds of creative variation thatused to be common in popular music.

Sometimes we will be casual and say, for example, that a protonhas “one unit of charge,” or even “a charge of one,” but this means1e, not one coulomb.

If you mix baking soda and vinegar to get a fizzy chemical reac-tion, you don’t really care that the number of molecules is an integer.The chemicals are, for all practical purposes, continuous fluids, be-cause the number of molecules is so large. Similarly, quantizationof charge has no consequences for many electrical circuits, and thecharge flowing through a wire acts like a continuous substance. Inthe SI, this is expressed by the fact that e is a very small numberwhen measured in practical units of coulombs.

13.2 The electron13.2.1 Cathode rays

Nineteenth-century physicists spent a lot of time trying to comeup with wild, random ways to play with electricity. The best ex-periments of this kind were the ones that made big sparks or prettycolors of light.

One such parlor trick was the cathode ray. To produce it, youfirst had to hire a good glassblower and find a good vacuum pump.The glassblower would create a hollow tube and embed two pieces ofmetal in it, called the electrodes, which were connected to the out-side via metal wires passing through the glass. Before letting himseal up the whole tube, you would hook it up to a vacuum pump,and spend several hours huffing and puffing away at the pump’shand crank to get a good vacuum inside. Then, while you were stillpumping on the tube, the glassblower would melt the glass and sealthe whole thing shut. Finally, you would put a large amount of pos-itive charge on one wire and a large amount of negative charge onthe other. Metals have the property of letting charge move throughthem easily, so the charge deposited on one of the wires wouldquickly spread out because of the repulsion of each part of it forevery other part. This spreading-out process would result in nearly

288 Chapter 13 The atom and the nucleus

a / Cathode rays observed ina vacuum tube.

all the charge ending up in the electrodes, where there is more roomto spread out than there is in the wire. For obscure historical rea-sons a negative electrode is called a cathode and a positive one isan anode.

Figure a shows the light-emitting stream that was observed. If,as shown in this figure, a hole was made in the anode, the beamwould extend on through the hole until it hit the glass. Drilling ahole in the cathode, however would not result in any beam comingout on the left side, and this indicated that the stuff, whatever itwas, was coming from the cathode. The rays were therefore chris-tened “cathode rays.” (The terminology is still used today in theterm “cathode ray tube” or “CRT” for the picture tube of a TV orcomputer monitor.)

13.2.2 Were cathode rays a form of light, or of matter?

Were cathode rays a form of light, or matter? At first no one re-ally cared what they were, but as their scientific importance becamemore apparent, the light-versus-matter issue turned into a contro-versy along nationalistic lines, with the Germans advocating lightand the English holding out for matter. The supporters of the ma-terial interpretation imagined the rays as consisting of a stream ofatoms ripped from the substance of the cathode.

One of our defining characteristics of matter is that materialobjects cannot pass through each other. Experiments showed thatcathode rays could penetrate at least some small thickness of matter,such as a metal foil a tenth of a millimeter thick, implying that theywere a form of light.

Other experiments, however, pointed to the contrary conclusion.Light is a wave phenomenon, and one distinguishing property ofwaves is demonstrated by speaking into one end of a paper towelroll. The sound waves do not emerge from the other end of thetube as a focused beam. Instead, they begin spreading out in alldirections as soon as they emerge. This shows that waves do notnecessarily travel in straight lines. If a piece of metal foil in the shapeof a star or a cross was placed in the way of the cathode ray, thena “shadow” of the same shape would appear on the glass, showingthat the rays traveled in straight lines. This straight-line motionsuggested that they were a stream of small particles of matter.

These observations were inconclusive, so what was really neededwas a determination of whether the rays had mass and weight. Thetrouble was that cathode rays could not simply be collected in a cupand put on a scale. When the cathode ray tube is in operation, onedoes not observe any loss of material from the cathode, or any crustbeing deposited on the anode.

Nobody could think of a good way to weigh cathode rays, so thenext most obvious way of settling the light/matter debate was to

Section 13.2 The electron 289

b / J.J. Thomson in the lab.

check whether the cathode rays possessed electrical charge. Lightwas known to be uncharged. If the cathode rays carried charge,they were definitely matter and not light, and they were presum-ably being made to jump the gap by the simultaneous repulsion ofthe negative charge in the cathode and attraction of the positivecharge in the anode. The rays would overshoot the anode becauseof their momentum. (Although electrically charged particles do notnormally leap across a gap of vacuum, very large amounts of chargewere being used, so the forces were unusually intense.)

13.2.3 Thomson’s experiments

Physicist J.J. Thomson at Cambridge carried out a series ofdefinitive experiments on cathode rays around the year 1897. Byturning them slightly off course with electrical forces, c, he showedthat they were indeed electrically charged, which was strong evi-dence that they were material. Not only that, but he proved thatthey had mass, and measured the ratio of their mass to their charge,m/q. Since their mass was not zero, he concluded that they werea form of matter, and presumably made up of a stream of micro-scopic, negatively charged particles. When Millikan published hisresults fourteen years later, it was reasonable to assume that thecharge of one such particle equaled minus one fundamental charge,q = −e, and from the combination of Thomson’s and Millikan’s re-sults one could therefore determine the mass of a single cathode rayparticle.

c / Thomson’s experiment provingcathode rays had electric charge(redrawn from his original paper).The cathode, C, and anode, A,are as in any cathode ray tube.The rays pass through a slit inthe anode, and a second slit, B,is interposed in order to makethe beam thinner and eliminaterays that were not going straight.Charging plates D and E showsthat cathode rays have charge:they are attracted toward the pos-itive plate D and repelled by thenegative plate E.

The basic technique for determining m/q was simply to measurethe angle through which the charged plates bent the beam. Theelectric force acting on a cathode ray particle while it was betweenthe plates would be proportional to its charge,

Felec = Eq,

where E is the electric field.

Application of Newton’s second law, a = F/m, would allow m/qto be determined:

m

q=E

a

There was just one catch. Thomson needed to know the cathode rayparticles’ velocity in order to figure out their acceleration. At that


point, however, nobody had even an educated guess as to the speedof the cathode rays produced in a given vacuum tube. The beamappeared to leap across the vacuum tube practically instantaneously,so it was no simple matter of timing it with a stopwatch!

Thomson’s clever solution was to observe the effect of both elec-tric and magnetic forces on the beam. The magnetic force exertedby a particular magnet would depend on both the cathode ray’scharge and its velocity:

Fmag = Bqv

Thomson played with the electric and magnetic forces until eitherone would produce an equal effect on the beam, allowing him tosolve for the velocity,

v =E

B.

Knowing the velocity (which was on the order of 10% of the speedof light for his setup), he was able to find the acceleration and thusthe mass-to-charge ratio m/q. Thomson’s techniques were relativelycrude (or perhaps more charitably we could say that they stretchedthe state of the art of the time), so with various methods he cameup with m/q values that ranged over about a factor of two, evenfor cathode rays extracted from a cathode made of a single mate-rial. The best modern value is m/q = 5.69 × 10−12 kg/C, which isconsistent with the low end of Thomson’s range.

13.2.4 The cathode ray as a subatomic particle: the electron

What was significant about Thomson’s experiment was not theactual numerical value of m/q, however, so much as the fact that,combined with Millikan’s value of the fundamental charge, it gavea mass for the cathode ray particles that was thousands of timessmaller than the mass of even the lightest atoms. Even withoutMillikan’s results, which were 14 years in the future, Thomson rec-ognized that the cathode rays’ m/q was thousands of times smallerthan the m/q ratios that had been measured for electrically chargedatoms in chemical solutions. He correctly interpreted this as evi-dence that the cathode rays were smaller building blocks — he calledthem electrons — out of which atoms themselves were formed. Thiswas an extremely radical claim, coming at a time when atoms hadnot yet been proven to exist! Even those who used the word “atom”often considered them no more than mathematical abstractions, notliteral objects. The idea of searching for structure inside of “un-splittable” atoms was seen by some as lunacy, but within ten yearsThomson’s ideas had been amply verified by many more detailedexperiments.

Section 13.2 The electron 291

d / The raisin cookie modelof the atom with four units ofcharge, which we now know to beberyllium.

Discussion questions

A Thomson started to become convinced during his experiments thatthe “cathode rays” observed coming from the cathodes of vacuum tubeswere building blocks of atoms — what we now call electrons. He thencarried out observations with cathodes made of a variety of metals, andfound that m/q was roughly the same in every case, considering his lim-ited accuracy. Given his suspicion, why did it make sense to try differentmetals? How would the consistent values of m/q serve to test his hypoth-esis?

B My students have frequently asked whether the m/q that Thomsonmeasured was the value for a single electron, or for the whole beam. Canyou answer this question?

C Thomson found that the m/q of an electron was thousands of timessmaller than that of charged atoms in chemical solutions. Would this implythat the electrons had more charge? Less mass? Would there be no wayto tell? Explain. Remember that Millikan’s results were still many years inthe future, so q was unknown.

D Can you guess any practical reason why Thomson couldn’t justlet one electron fly across the gap before disconnecting the battery andturning off the beam, and then measure the amount of charge depositedon the anode, thus allowing him to measure the charge of a single electrondirectly?

E Why is it not possible to determine m and q themselves, rather thanjust their ratio, by observing electrons’ motion in electric and magneticfields?

13.3 The raisin cookie model of the atom

Based on his experiments, Thomson proposed a picture of theatom which became known as the raisin cookie model. In the neutralatom, d, there are four electrons with a total charge of −4e, sittingin a sphere (the “cookie”) with a charge of +4e spread throughout it.It was known that chemical reactions could not change one elementinto another, so in Thomson’s scenario, each element’s cookie spherehad a permanently fixed radius, mass, and positive charge, differentfrom those of other elements. The electrons, however, were not apermanent feature of the atom, and could be tacked on or pulled outto make charged ions. Although we now know, for instance, that aneutral atom with four electrons is the element beryllium, scientistsat the time did not know how many electrons the various neutralatoms possessed.

This model is clearly different from the one you’ve learned ingrade school or through popular culture, where the positive chargeis concentrated in a tiny nucleus at the atom’s center. An equallyimportant change in ideas about the atom has been the realizationthat atoms and their constituent subatomic particles behave entirelydifferently from objects on the human scale. For instance, we’ll seelater that an electron can be in more than one place at one time.


The raisin cookie model was part of a long tradition of attemptsto make mechanical models of phenomena, and Thomson and hiscontemporaries never questioned the appropriateness of building amental model of an atom as a machine with little parts inside. To-day, mechanical models of atoms are still used (for instance thetinker-toy-style molecular modeling kits like the ones used by Wat-son and Crick to figure out the double helix structure of DNA), butscientists realize that the physical objects are only aids to help ourbrains’ symbolic and visual processes think about atoms.

Although there was no clear-cut experimental evidence for manyof the details of the raisin cookie model, physicists went ahead andstarted working out its implications. For instance, suppose you hada four-electron atom. All four electrons would be repelling eachother, but they would also all be attracted toward the center of the“cookie” sphere. The result should be some kind of stable, sym-metric arrangement in which all the forces canceled out. Peoplesufficiently clever with math soon showed that the electrons in afour-electron atom should settle down at the vertices of a pyramidwith one less side than the Egyptian kind, i.e., a regular tetrahe-dron. This deduction turns out to be wrong because it was basedon incorrect features of the model, but the model also had manysuccesses, a few of which we will now discuss.

Flow of electrical charge in wires example 1One of my former students was the son of an electrician, andhad become an electrician himself. He related to me how hisfather had remained refused to believe all his life that electronsreally flowed through wires. If they had, he reasoned, the metalwould have gradually become more and more damaged, eventu-ally crumbling to dust.

His opinion is not at all unreasonable based on the fact that elec-trons are material particles, and that matter cannot normally passthrough matter without making a hole through it. Nineteenth-century physicists would have shared his objection to a charged-particle model of the flow of electrical charge. In the raisin-cookiemodel, however, the electrons are very low in mass, and there-fore presumably very small in size as well. It is not surprising thatthey can slip between the atoms without damaging them.

Flow of electrical charge across cell membranes example 2Your nervous system is based on signals carried by charge mov-ing from nerve cell to nerve cell. Your body is essentially all liquid,and atoms in a liquid are mobile. This means that, unlike the caseof charge flowing in a solid wire, entire charged atoms can flow inyour nervous system

Emission of electrons in a cathode ray tube example 3Why do electrons detach themselves from the cathode of a vac-uum tube? Certainly they are encouraged to do so by the re-

Section 13.3 The raisin cookie model of the atom 293

pulsion of the negative charge placed on the cathode and theattraction from the net positive charge of the anode, but these arenot strong enough to rip electrons out of atoms by main force —if they were, then the entire apparatus would have been instantlyvaporized as every atom was simultaneously ripped apart!

The raisin cookie model leads to a simple explanation. We knowthat heat is the energy of random motion of atoms. The atoms inany object are therefore violently jostling each other all the time,and a few of these collisions are violent enough to knock electronsout of atoms. If this occurs near the surface of a solid object, theelectron may can come loose. Ordinarily, however, this loss ofelectrons is a self-limiting process; the loss of electrons leavesthe object with a net positive charge, which attracts the lost sheephome to the fold. (For objects immersed in air rather than vacuum,there will also be a balanced exchange of electrons between theair and the object.)

This interpretation explains the warm and friendly yellow glow ofthe vacuum tubes in an antique radio. To encourage the emissionof electrons from the vacuum tubes’ cathodes, the cathodes areintentionally warmed up with little heater coils.

Discussion questions

A Today many people would define an ion as an atom (or molecule)with missing electrons or extra electrons added on. How would peoplehave defined the word “ion” before the discovery of the electron?

B Since electrically neutral atoms were known to exist, there had to bepositively charged subatomic stuff to cancel out the negatively chargedelectrons in an atom. Based on the state of knowledge immediately afterthe Millikan and Thomson experiments, was it possible that the positivelycharged stuff had an unquantized amount of charge? Could it be quan-tized in units of +e? In units of +2e? In units of +5/7e?

13.4 The nucleus13.4.1 Radioactivity

Becquerel’s discovery of radioactivity

How did physicists figure out that the raisin cookie model wasincorrect, and that the atom’s positive charge was concentrated ina tiny, central nucleus? The story begins with the discovery of ra-dioactivity by the French chemist Becquerel. Up until radioactivitywas discovered, all the processes of nature were thought to be basedon chemical reactions, which were rearrangements of combinationsof atoms. Atoms exert forces on each other when they are close to-gether, so sticking or unsticking them would either release or storeelectrical energy. That energy could be converted to and from otherforms, as when a plant uses the energy in sunlight to make sugarsand carbohydrates, or when a child eats sugar, releasing the energy


e / Henri Becquerel (1852-1908).

f / Becquerel’s photographicplate. In the exposure at thebottom of the image, he hasfound that he could absorb theradiations, casting the shadowof a Maltese cross that wasplaced between the plate and theuranium salts.

in the form of kinetic energy.

Becquerel discovered a process that seemed to release energyfrom an unknown new source that was not chemical. Becquerel,whose father and grandfather had also been physicists, spent thefirst twenty years of his professional life as a successful civil engi-neer, teaching physics on a part-time basis. He was awarded thechair of physics at the Musee d’Histoire Naturelle in Paris after thedeath of his father, who had previously occupied it. Having now asignificant amount of time to devote to physics, he began studyingthe interaction of light and matter. He became interested in the phe-nomenon of phosphorescence, in which a substance absorbs energyfrom light, then releases the energy via a glow that only graduallygoes away. One of the substances he investigated was a uraniumcompound, the salt UKSO5. One day in 1896, cloudy weather in-terfered with his plan to expose this substance to sunlight in orderto observe its fluorescence. He stuck it in a drawer, coincidentally ontop of a blank photographic plate — the old-fashioned glass-backedcounterpart of the modern plastic roll of film. The plate had beencarefully wrapped, but several days later when Becquerel checked itin the darkroom before using it, he found that it was ruined, as if ithad been completely exposed to light.

History provides many examples of scientific discoveries thathappened this way: an alert and inquisitive mind decides to in-vestigate a phenomenon that most people would not have worriedabout explaining. Becquerel first determined by further experimentsthat the effect was produced by the uranium salt, despite a thickwrapping of paper around the plate that blocked out all light. Hetried a variety of compounds, and found that it was the uraniumthat did it: the effect was produced by any uranium compound, butnot by any compound that didn’t include uranium atoms. The effectcould be at least partially blocked by a sufficient thickness of metal,and he was able to produce silhouettes of coins by interposing thembetween the uranium and the plate. This indicated that the effecttraveled in a straight line., so that it must have been some kind ofray rather than, e.g., the seepage of chemicals through the paper.He used the word “radiations,” since the effect radiated out fromthe uranium salt.

At this point Becquerel still believed that the uranium atomswere absorbing energy from light and then gradually releasing theenergy in the form of the mysterious rays, and this was how hepresented it in his first published lecture describing his experiments.Interesting, but not earth-shattering. But he then tried to determinehow long it took for the uranium to use up all the energy that hadsupposedly been stored in it by light, and he found that it neverseemed to become inactive, no matter how long he waited. Not onlythat, but a sample that had been exposed to intense sunlight for awhole afternoon was no more or less effective than a sample that

Section 13.4 The nucleus 295

had always been kept inside. Was this a violation of conservationof energy? If the energy didn’t come from exposure to light, wheredid it come from?

Three kinds of “radiations”

Unable to determine the source of the energy directly, turn-of-the-century physicists instead studied the behavior of the “radia-tions” once they had been emitted. Becquerel had already shownthat the radioactivity could penetrate through cloth and paper, sothe first obvious thing to do was to investigate in more detail whatthickness of material the radioactivity could get through. They soonlearned that a certain fraction of the radioactivity’s intensity wouldbe eliminated by even a few inches of air, but the remainder wasnot eliminated by passing through more air. Apparently, then, theradioactivity was a mixture of more than one type, of which one wasblocked by air. They then found that of the part that could pene-trate air, a further fraction could be eliminated by a piece of paperor a very thin metal foil. What was left after that, however, wasa third, extremely penetrating type, some of whose intensity wouldstill remain even after passing through a brick wall. They decidedthat this showed there were three types of radioactivity, and with-out having the faintest idea of what they really were, they made upnames for them. The least penetrating type was arbitrarily labeledα (alpha), the first letter of the Greek alphabet, and so on throughβ (beta) and finally G (gamma) for the most penetrating type.

Radium: a more intense source of radioactivity

The measuring devices used to detect radioactivity were crude:photographic plates or even human eyeballs (radioactivity makesflashes of light in the jelly-like fluid inside the eye, which can beseen by the eyeball’s owner if it is otherwise very dark). Becausethe ways of detecting radioactivity were so crude and insensitive,further progress was hindered by the fact that the amount of ra-dioactivity emitted by uranium was not really very great. The vi-tal contribution of physicist/chemist Marie Curie and her husbandPierre was to discover the element radium, and to purify and iso-late significant quantities of it. Radium emits about a million timesmore radioactivity per unit mass than uranium, making it possibleto do the experiments that were needed to learn the true natureof radioactivity. The dangers of radioactivity to human health werethen unknown, and Marie died of leukemia thirty years later. (Pierrewas run over and killed by a horsecart.)

Tracking down the nature of alphas, betas, and gammas

As radium was becoming available, an apprentice scientist namedErnest Rutherford arrived in England from his native New Zealandand began studying radioactivity at the Cavendish Laboratory. Theyoung colonial’s first success was to measure the mass-to-charge ra-


g / A simplified version ofRutherford’s 1908 experiment,showing that alpha particles weredoubly ionized helium atoms.

h / These pellets of uraniumfuel will be inserted into the metalfuel rod and used in a nuclearreactor. The pellets emit alphaand beta radiation, which thegloves are thick enough to stop.

tio of beta rays. The technique was essentially the same as the oneThomson had used to measure the mass-to-charge ratio of cathoderays by measuring their deflections in electric and magnetic fields.The only difference was that instead of the cathode of a vacuumtube, a nugget of radium was used to supply the beta rays. Notonly was the technique the same, but so was the result. Beta rayshad the same m/q ratio as cathode rays, which suggested they wereone and the same. Nowadays, it would make sense simply to usethe term “electron,” and avoid the archaic “cathode ray” and “betaparticle,” but the old labels are still widely used, and it is unfortu-nately necessary for physics students to memorize all three namesfor the same thing.

At first, it seemed that neither alphas or gammas could be de-flected in electric or magnetic fields, making it appear that neitherwas electrically charged. But soon Rutherford obtained a much morepowerful magnet, and was able to use it to deflect the alphas butnot the gammas. The alphas had a much larger value of m/q thanthe betas (about 4000 times greater), which was why they had beenso hard to deflect. Gammas are uncharged, and were later found tobe a form of light.

The m/q ratio of alpha particles turned out to be the sameas those of two different types of ions, He++ (a helium atom withtwo missing electrons) and H+

2 (two hydrogen atoms bonded into amolecule, with one electron missing), so it seemed likely that theywere one or the other of those. The diagram shows a simplified ver-sion of Rutherford’s ingenious experiment proving that they wereHe++ ions. The gaseous element radon, an alpha emitter, was in-troduced into one half of a double glass chamber. The glass walldividing the chamber was made extremely thin, so that some of therapidly moving alpha particles were able to penetrate it. The otherchamber, which was initially evacuated, gradually began to accu-mulate a population of alpha particles (which would quickly pick upelectrons from their surroundings and become electrically neutral).Rutherford then determined that it was helium gas that had ap-peared in the second chamber. Thus alpha particles were proved tobe He++ ions. The nucleus was yet to be discovered, but in modernterms, we would describe a He++ ion as the nucleus of a He atom.

To summarize, here are the three types of radiation emitted byradioactive elements, and their descriptions in modern terms:

α particle stopped by a few inches of air He nucleus

β particle stopped by a piece of paper electron

G ray penetrates thick shielding a type of light

Discussion question

A Most sources of radioactivity emit alphas, betas, and gammas, notjust one of the three. In the radon experiment, how did Rutherford knowthat he was studying the alphas?


i / Ernest Rutherford (1871-1937).

j / Marsden and Rutherford’sapparatus.

13.4.2 The planetary model

The stage was now set for the unexpected discovery that thepositively charged part of the atom was a tiny, dense lump at theatom’s center rather than the “cookie dough” of the raisin cookiemodel. By 1909, Rutherford was an established professor, and hadstudents working under him. For a raw undergraduate named Mars-den, he picked a research project he thought would be tedious butstraightforward.

It was already known that although alpha particles would bestopped completely by a sheet of paper, they could pass through asufficiently thin metal foil. Marsden was to work with a gold foilonly 1000 atoms thick. (The foil was probably made by evaporatinga little gold in a vacuum chamber so that a thin layer would bedeposited on a glass microscope slide. The foil would then be liftedoff the slide by submerging the slide in water.)

Rutherford had already determined in his previous experimentsthe speed of the alpha particles emitted by radium, a fantastic 1.5×107 m/s. The experimenters in Rutherford’s group visualized themas very small, very fast cannonballs penetrating the “cookie dough”part of the big gold atoms. A piece of paper has a thickness of ahundred thousand atoms or so, which would be sufficient to stopthem completely, but crashing through a thousand would only slowthem a little and turn them slightly off of their original paths.

Marsden’s supposedly ho-hum assignment was to use the appa-ratus shown in figure j to measure how often alpha particles weredeflected at various angles. A tiny lump of radium in a box emit-ted alpha particles, and a thin beam was created by blocking allthe alphas except those that happened to pass out through a tube.Typically deflected in the gold by only a small amount, they wouldreach a screen very much like the screen of a TV’s picture tube,which would make a flash of light when it was hit. Here is the firstexample we have encountered of an experiment in which a beam ofparticles is detected one at a time. This was possible because eachalpha particle carried so much kinetic energy; they were moving atabout the same speed as the electrons in the Thomson experiment,but had ten thousand times more mass.

Marsden sat in a dark room, watching the apparatus hour afterhour and recording the number of flashes with the screen moved tovarious angles. The rate of the flashes was highest when he set thescreen at an angle close to the line of the alphas’ original path, but ifhe watched an area farther off to the side, he would also occasionallysee an alpha that had been deflected through a larger angle. Afterseeing a few of these, he got the crazy idea of moving the screen tosee if even larger angles ever occurred, perhaps even angles largerthan 90 degrees.


l / The planetary model ofthe atom.

k / Alpha particles being scattered by a gold nucleus. On this scale,the gold atom is the size of a car, so all the alpha particles shown hereare ones that just happened to come unusually close to the nucleus.For these exceptional alpha particles, the forces from the electrons areunimportant, because they are so much more distant than the nucleus.

The crazy idea worked: a few alpha particles were deflectedthrough angles of up to 180 degrees, and the routine experimenthad become an epoch-making one. Rutherford said, “We have beenable to get some of the alpha particles coming backwards. It wasalmost as incredible as if you fired a 15-inch shell at a piece of tissuepaper and it came back and hit you.” Explanations were hard tocome by in the raisin cookie model. What intense electrical forcescould have caused some of the alpha particles, moving at such astro-nomical speeds, to change direction so drastically? Since each goldatom was electrically neutral, it would not exert much force on analpha particle outside it. True, if the alpha particle was very near toor inside of a particular atom, then the forces would not necessarilycancel out perfectly; if the alpha particle happened to come veryclose to a particular electron, the 1/r2 form of the Coulomb forcelaw would make for a very strong force. But Marsden and Ruther-ford knew that an alpha particle was 8000 times more massive thanan electron, and it is simply not possible for a more massive objectto rebound backwards from a collision with a less massive objectwhile conserving momentum and energy. It might be possible inprinciple for a particular alpha to follow a path that took it veryclose to one electron, and then very close to another electron, and soon, with the net result of a large deflection, but careful calculationsshowed that such multiple “close encounters” with electrons wouldbe millions of times too rare to explain what was actually observed.

At this point, Rutherford and Marsden dusted off an unpop-ular and neglected model of the atom, in which all the electronsorbited around a small, positively charged core or “nucleus,” justlike the planets orbiting around the sun. All the positive charge


and nearly all the mass of the atom would be concentrated in thenucleus, rather than spread throughout the atom as in the raisincookie model. The positively charged alpha particles would be re-pelled by the gold atom’s nucleus, but most of the alphas would notcome close enough to any nucleus to have their paths drasticallyaltered. The few that did come close to a nucleus, however, couldrebound backwards from a single such encounter, since the nucleus ofa heavy gold atom would be fifty times more massive than an alphaparticle. It turned out that it was not even too difficult to derive aformula giving the relative frequency of deflections through variousangles, and this calculation agreed with the data well enough (towithin 15%), considering the difficulty in getting good experimentalstatistics on the rare, very large angles.

What had started out as a tedious exercise to get a studentstarted in science had ended as a revolution in our understandingof nature. Indeed, the whole thing may sound a little too muchlike a moralistic fable of the scientific method with overtones ofthe Horatio Alger genre. The skeptical reader may wonder whythe planetary model was ignored so thoroughly until Marsden andRutherford’s discovery. Is science really more of a sociological enter-prise, in which certain ideas become accepted by the establishment,and other, equally plausible explanations are arbitrarily discarded?Some social scientists are currently ruffling a lot of scientists’ feath-ers with critiques very much like this, but in this particular case,there were very sound reasons for rejecting the planetary model. Asyou’ll learn in more detail later in this course, any charged particlethat undergoes an acceleration dissipate energy in the form of light.In the planetary model, the electrons were orbiting the nucleus incircles or ellipses, which meant they were undergoing acceleration,just like the acceleration you feel in a car going around a curve. Theyshould have dissipated energy as light, and eventually they shouldhave lost all their energy. Atoms don’t spontaneously collapse likethat, which was why the raisin cookie model, with its stationaryelectrons, was originally preferred. There were other problems aswell. In the planetary model, the one-electron atom would haveto be flat, which would be inconsistent with the success of molecu-lar modeling with spherical balls representing hydrogen and atoms.These molecular models also seemed to work best if specific sizeswere used for different atoms, but there is no obvious reason in theplanetary model why the radius of an electron’s orbit should be afixed number. In view of the conclusive Marsden-Rutherford results,however, these became fresh puzzles in atomic physics, not reasonsfor disbelieving the planetary model.

Some phenomena explained with the planetary model

The planetary model may not be the ultimate, perfect model ofthe atom, but don’t underestimate its power. It already allows us


7 (a) find a case where the magni cation of a curved ... · 8 as discussed in question 7, there are...

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