chapter In Chapter 36, we studied light rays passing through a lens or reflecting from a mirror

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1 chapter 37 Wave Optics 37.1 Young s Double-Slit Experiment 37.2 Analysis Moel: Waves in Interference 37.3 Intensity Distribution of the Double-Slit Interference Pattern 37.4 Change of Phase Due to Reflection 37.5 Interference in Thin Films 37.6 The Michelson Interferometer In Chapter 36, we stuie light rays passing through a lens or reflecting from a mirror to escribe the formation of images. This iscussion complete our stuy of ray optics. In this chapter an in Chapter 38, we are concerne with wave optics, sometimes calle physical optics, the stuy of interference, iffraction, an polarization of light. These phenomena cannot be aequately explaine with the ray optics use in Chapters 35 an 36. We now learn how treating light as waves rather than as rays leas to a satisfying escription of such phenomena. The colors in many of a hummingbir s feathers are not ue to pigment. The iriescence that makes the brilliant colors that often appear on the bir s throat an belly is ue to an interference effect cause by structures in the feathers. The colors will vary with the viewing angle. (RO-MA Stock) 37.1 Young s Double-Slit Experiment In Chapter 18, we stuie the waves in interference moel an foun that the superposition of two mechanical waves can be constructive or estructive. In constructive interference, the amplitue of the resultant wave is greater than that of either iniviual wave, whereas in estructive interference, the resultant amplitue 1084

2 37.1 Young s Double-Slit Experiment 1085 S 1 S 2 Barrier A region marke max in a correspons to a bright fringe in b. max min max min max min max min max Viewing screen Photograph from M. Cagnet, M. Françon, J. C. Thrierr, Atlas of Optical Phenomena, Berlin, Springer-Verlag, 1962 ACTIVE FIGURE 37.1 (a) Schematic iagram of Young s ouble-slit experiment. Slits S 1 an S 2 behave as coherent sources of light waves that prouce an interference pattern on the viewing screen (rawing not to scale). (b) An enlargement of the center of a fringe pattern forme on the viewing screen. The waves a constructively at the re ots an estructively at the black ots. a b is less than that of the larger wave. Light waves also interfere with one another. Funamentally, all interference associate with light waves arises when the electromagnetic fiels that constitute the iniviual waves combine. Interference in light waves from two sources was first emonstrate by Thomas Young in A schematic iagram of the apparatus Young use is shown in Active Figure 37.1a. Plane light waves arrive at a barrier that contains two slits S 1 an S 2. The light from S 1 an S 2 prouces on a viewing screen a visible pattern of bright an ark parallel bans calle fringes (Active Fig. 37.1b). When the light from S 1 an that from S 2 both arrive at a point on the screen such that constructive interference occurs at that location, a bright fringe appears. When the light from the two slits combines estructively at any location on the screen, a ark fringe results. Figure 37.2 is a photograph of an interference pattern prouce by two vibrating sources in a water tank. The linear regions of constructive interference, such as at A, an estructive interference, such as at B, raiating from the area between the sources are analogous to the re an black lines in Active Figure Richar Megna/Funamental Photographs, NYC A B Constructive interference occurs along lines like this one. Destructive interference occurs along lines like this one. Figure 37.2 An interference pattern involving water waves is prouce by two vibrating sources at the water s surface.

3 1086 CHAPTER 37 Wave Optics Figure 37.3 Waves leave the slits an combine at various points on the viewing screen. (All figures not to scale.) Constructive interference occurs at point O when the waves combine. Constructive interference also occurs at point P. Destructive interference occurs at point R when the two waves combine because the lower wave falls one-half a wavelength behin the upper wave. S 1 S 2 Bright O fringe S 1 S 2 P O Bright fringe S 1 S 2 P R O Dark fringe Viewing screen a b c Conitions for interference Figure 37.3 shows some of the ways in which two waves can combine at the screen. In Figure 37.3a, the two waves, which leave the two slits in phase, strike the screen at the central point O. Because both waves travel the same istance, they arrive at O in phase. As a result, constructive interference occurs at this location an a bright fringe is observe. In Figure 37.3b, the two waves also start in phase, but here the lower wave has to travel one wavelength farther than the upper wave to reach point P. Because the lower wave falls behin the upper one by exactly one wavelength, they still arrive in phase at P an a secon bright fringe appears at this location. At point R in Figure 37.3c, however, between points O an P, the lower wave has fallen half a wavelength behin the upper wave an a crest of the upper wave overlaps a trough of the lower wave, giving rise to estructive interference at point R. A ark fringe is therefore observe at this location. If two lightbulbs are place sie by sie so that light from both bulbs combines, no interference effects are observe because the light waves from one bulb are emitte inepenently of those from the other bulb. The emissions from the two lightbulbs o not maintain a constant phase relationship with each other over time. Light waves from an orinary source such as a lightbulb unergo ranom phase changes in time intervals of less than a nanosecon. Therefore, the conitions for constructive interference, estructive interference, or some intermeiate state are maintaine only for such short time intervals. Because the eye cannot follow such rapi changes, no interference effects are observe. Such light sources are sai to be incoherent. To observe interference of waves from two sources, the following conitions must be met: The sources must be coherent; that is, they must maintain a constant phase with respect to each other. The sources shoul be monochromatic; that is, they shoul be of a single wavelength. As an example, single-frequency soun waves emitte by two sie-by-sie louspeakers riven by a single amplifier can interfere with each other because the two speakers are coherent. In other wors, they respon to the amplifier in the same way at the same time. A common metho for proucing two coherent light sources is to use a monochromatic source to illuminate a barrier containing two small openings, usually in the shape of slits, as in the case of Young s experiment illustrate in Active Figure The light emerging from the two slits is coherent because a single source prouces the original light beam an the two slits serve only to separate the original beam into two parts (which, after all, is what is one to the soun signal from two sie-by-sie louspeakers). Any ranom change in the light emitte by the source occurs in both beams at the same time. As a result, interference effects can be observe when the light from the two slits arrives at a viewing screen.

4 37.2 Analysis Moel: Waves in Interference 1087 If the light travele only in its original irection after passing through the slits as shown in Figure 37.4a, the waves woul not overlap an no interference pattern woul be seen. Instea, as we have iscusse in our treatment of Huygens s principle (Section 35.6), the waves sprea out from the slits as shown in Figure 37.4b. In other wors, the light eviates from a straight-line path an enters the region that woul otherwise be shaowe. As note in Section 35.3, this ivergence of light from its initial line of travel is calle iffraction. Light passing through narrow slits oes not behave this way Analysis Moel: Waves in Interference We iscusse the superposition principle for waves on strings in Section 18.1, leaing to a one-imensional version of the waves in interference analysis moel. In Example 18.1 on page 515, we briefly iscusse a two-imensional interference phenomenon for soun from two louspeakers. In walking from point O to point P in Figure 18.5, the listener experience a maximum in soun intensity at O an a minimum at P. This experience is exactly analogous to an observer looking at point O in Figure 37.3 an seeing a bright fringe an then sweeping his eyes upwar to point R, where there is a minimum in light intensity. Let s look in more etail at the two-imensional nature of Young s experiment with the help of Figure The viewing screen is locate a perpenicular istance L from the barrier containing two slits, S 1 an S 2 (Fig. 37.5a). These slits are separate by a istance, an the source is monochromatic. To reach any arbitrary point P in the upper half of the screen, a wave from the lower slit must travel farther than a wave from the upper slit by a istance sin u (Fig. 37.5b). This istance is calle the path ifference (Greek letter elta). If we assume the rays labele r 1 an r 2 are parallel, which is approximately true if L is much greater than, then is given by 5 r 2 2 r 1 5 sin u (37.1) The value of etermines whether the two waves are in phase when they arrive at point P. If is either zero or some integer multiple of the wavelength, the two waves are in phase at point P an constructive interference results. Therefore, the conition for bright fringes, or constructive interference, at point P is sin u bright 5 ml m 5 0, 61, 62,... (37.2) The number m is calle the orer number. For constructive interference, the orer number is the same as the number of wavelengths that represents the path ifference between the waves from the two slits. The central bright fringe at u bright 5 0 is calle the zeroth-orer maximum. The first maximum on either sie, where m 5 61, is calle the first-orer maximum, an so forth. a Light passing through narrow slits iffracts. b Figure 37.4 (a) If light waves i not sprea out after passing through the slits, no interference woul occur. (b) The light waves from the two slits overlap as they sprea out, filling what we expect to be shaowe regions with light an proucing interference fringes on a screen place to the right of the slits. Conition for constructive interference P r 1 r 1 S 1 y S 1 r 2 Q S 2 r 2 u u u O S 2 r 2 r 1 sin u a L Viewing screen When we assume r 1 is parallel to r 2, the path ifference between the two rays is r 2 r 1 sin u. b Figure 37.5 (a) Geometric construction for escribing Young s ouble-slit experiment (not to scale). (b) The slits are represente as sources, an the outgoing light rays are assume to be parallel as they travel to P. To achieve that in practice, it is essential that L...

5 1088 CHAPTER 37 Wave Optics When is an o multiple of l/2, the two waves arriving at point P are 180 out of phase an give rise to estructive interference. Therefore, the conition for ark fringes, or estructive interference, at point P is Conition for estructive interference sin u ark 5 1m l m 5 0, 61, 62, c (37.3) These equations provie the angular positions of the fringes. It is also useful to obtain expressions for the linear positions measure along the screen from O to P. From the triangle OPQ in Figure 37.5a, we see that tan u 5 y L (37.4) Using this result, the linear positions of bright an ark fringes are given by y bright 5 L tan u bright (37.5) y ark 5 L tan u ark (37.6) where u bright an u ark are given by Equations 37.2 an When the angles to the fringes are small, the positions of the fringes are linear near the center of the pattern. That can be verifie by noting that for small angles, tan u < sin u, so Equation 37.5 gives the positions of the bright fringes as y bright 5 L sin u bright. Incorporating Equation 37.2 gives y bright 5 L ml 1small angles2 (37.7) This result shows that y bright is linear in the orer number m, so the fringes are equally space for small angles. Similarly, for ark fringes, y ark 5 L 1m l 1small angles2 (37.8) As emonstrate in Example 37.1, Young s ouble-slit experiment provies a metho for measuring the wavelength of light. In fact, Young use this technique to o precisely that. In aition, his experiment gave the wave moel of light a great eal of creibility. It was inconceivable that particles of light coming through the slits coul cancel one another in a way that woul explain the ark fringes. The principles iscusse in this section are the basis of the waves in interference analysis moel. This moel was applie to mechanical waves in one imension in Chapter 18. Here we see the etails of applying this moel in three imensions to light. Quick Quiz 37.1 Which of the following causes the fringes in a two-slit interference pattern to move farther apart? (a) ecreasing the wavelength of the light (b) ecreasing the screen istance L (c) ecreasing the slit spacing () immersing the entire apparatus in water Example 37.1 Measuring the Wavelength of a Light Source A viewing screen is separate from a ouble slit by 4.80 m. The istance between the two slits is mm. Monochromatic light is irecte towar the ouble slit an forms an interference pattern on the screen. The first ark fringe is 4.50 cm from the center line on the screen. (A) Determine the wavelength of the light.

6 37.2 Analysis Moel: Waves in Interference cont. SOLUTION Conceptualize Stuy Figure 37.5 to be sure you unerstan the phenomenon of interference of light waves. The istance of 4.50 cm is y in Figure Categorize We etermine results using equations evelope in this section, so we categorize this example as a substitution problem. Because L.. y, the angles for the fringes are small. Solve Equation 37.8 for the wavelength an substitute numerical values: (B) Calculate the istance between ajacent bright fringes. SOLUTION Fin the istance between ajacent bright fringes from Equation 37.7 an the results of part (A): y ark l5 1m L m m m m nm 1m 1 12l y m11 2 y m 5 L 2 L ml 5 L l m a m m b m cm For practice, fin the wavelength of the soun in Example 18.1 using the proceure in part (A) of this example. Example 37.2 Separating Double-Slit Fringes of Two Wavelengths A light source emits visible light of two wavelengths: l nm an l nm. The source is use in a ouble-slit interference experiment in which L m an mm. Fin the separation istance between the thir-orer bright fringes for the two wavelengths. SOLUTION Conceptualize In Figure 37.5a, imagine light of two wavelengths incient on the slits an forming two interference patterns on the screen. At some points, the fringes of the two colors might overlap, but at most points, they will not. Categorize We etermine results using equations evelope in this section, so we categorize this example as a substitution problem. Use Equation 37.7 to fin the fringe positions corresponing to these two wavelengths an subtract them: Substitute numerical values: Dy 5 Dy 5 yr bright 2 y bright 5 L mlr m m m m m cm 2 L ml 5 Lm 1lr 2l2 WHAT IF? What if we examine the entire interference pattern ue to the two wavelengths an look for overlapping fringes? Are there any locations on the screen where the bright fringes from the two wavelengths overlap exactly? Answer Fin such a location by setting the location of any bright fringe ue to l equal to one ue to l9, using Equation 37.7: Substitute the wavelengths: L ml 5 L mrlr mr m nm 510 nm S mr m 5 l lr continue

7 1090 CHAPTER 37 Wave Optics 37.2 cont. Therefore, the 51st fringe of the 430-nm light overlaps with the 43r fringe of the 510-nm light. Use Equation 37.7 to fin the value of y for these fringes: y m2 c m m m This value of y is comparable to L, so the small-angle approximation use for Equation 37.7 is not vali. This conclusion suggests we shoul not expect Equation 37.7 to give us the correct result. If you use Equation 37.5, you can show that the bright fringes o inee overlap when the same conition, m9/m 5 l/l9, is met (see Problem 44). Therefore, the 51st fringe of the 430-nm light oes overlap with the 43r fringe of the 510-nm light, but not at the location of 1.32 m. You are aske to fin the correct location as part of Problem Intensity Distribution of the Double-Slit Interference Pattern Notice that the eges of the bright fringes in Active Figure 37.1b are not sharp; rather, there is a graual change from bright to ark. So far, we have iscusse the locations of only the centers of the bright an ark fringes on a istant screen. Let s now irect our attention to the intensity of the light at other points between the positions of maximum constructive an estructive interference. In other wors, we now calculate the istribution of light intensity associate with the ouble-slit interference pattern. Again, suppose the two slits represent coherent sources of sinusoial waves such that the two waves from the slits have the same angular frequency v an are in phase. The total magnitue of the electric fiel at point P on the screen in Figure 37.5 is the superposition of the two waves. Assuming the two waves have the same amplitue E 0, we can write the magnitue of the electric fiel at point P ue to each wave separately as E 1 5 E 0 sin vt an E 2 5 E 0 sin (vt 1 f) (37.9) Although the waves are in phase at the slits, their phase ifference f at P epens on the path ifference 5 r 2 2 r 1 5 sin u. A path ifference of l (for constructive interference) correspons to a phase ifference of 2p ra. A path ifference of is the same fraction of l as the phase ifference f is of 2p. We can escribe this fraction mathematically with the ratio which gives l 5 f 2p Phase ifference f5 2p l 52p l sin u (37.10) This equation shows how the phase ifference f epens on the angle u in Figure Using the superposition principle an Equation 37.9, we obtain the following expression for the magnitue of the resultant electric fiel at point P: E P 5 E 1 1 E 2 5 E 0 [sin vt 1 sin (vt 1 f)] (37.11) We can simplify this expression by using the trigonometric ientity sin A 1 sin B 5 2 sin a A 1 B 2 b cos a A 2 B b 2

8 37.3 Intensity Distribution of the Double-Slit Interference Pattern 1091 Taking A 5 vt 1 f an B 5 vt, Equation becomes E P 5 2E 0 cos a f 2 b sin avt 1 f 2 b (37.12) This result inicates that the electric fiel at point P has the same frequency v as the light at the slits but that the amplitue of the fiel is multiplie by the factor 2 cos (f/2). To check the consistency of this result, note that if f 5 0, 2p, 4p,..., the magnitue of the electric fiel at point P is 2E 0, corresponing to the conition for maximum constructive interference. These values of f are consistent with Equation 37.2 for constructive interference. Likewise, if f 5 p, 3p, 5p,..., the magnitue of the electric fiel at point P is zero, which is consistent with Equation 37.3 for total estructive interference. Finally, to obtain an expression for the light intensity at point P, recall from Section 34.4 that the intensity of a wave is proportional to the square of the resultant electric fiel magnitue at that point (Eq ). Using Equation 37.12, we can therefore express the light intensity at point P as I ~ E P 2 5 4E 0 2 cos 2 a f 2 b sin2 avt 1 f 2 b Most light-etecting instruments measure time-average light intensity, an the timeaverage value of sin 2 (vt 1 f/2) over one cycle is 1 2. (See Fig ) Therefore, we can write the average light intensity at point P as I 5 I max cos 2 a f 2 b (37.13) where I max is the maximum intensity on the screen an the expression represents the time average. Substituting the value for f given by Equation into this expression gives I 5 I max cos 2 a p sin u b (37.14) l Alternatively, because sin u < y/l for small values of u in Figure 37.5, we can write Equation in the form I 5 I max cos 2 a p yb 1small angles2 (37.15) ll Constructive interference, which prouces light intensity maxima, occurs when the quantity py/ll is an integral multiple of p, corresponing to y 5 (ll/)m. This result is consistent with Equation A plot of light intensity versus sin u is given in Figure 37.6 (page 1092). The interference pattern consists of equally space fringes of equal intensity. Figure 37.7 (page 1092) shows similar plots of light intensity versus sin u for light passing through multiple slits. For more than two slits, the pattern contains primary an seconary maxima. For three slits, notice that the primary maxima are nine times more intense than the seconary maxima as measure by the height of the curve because the intensity varies as E 2. For N slits, the intensity of the primary maxima is N 2 times greater than that ue to a single slit. As the number of slits increases, the primary maxima increase in intensity an become narrower, while the seconary maxima ecrease in intensity relative to the primary maxima. Figure 37.7 also shows that as the number of slits increases, the number of seconary maxima also increases. In fact, the number of seconary maxima is always N 2 2, where N is the number of slits. In Section 38.4, we shall investigate the pattern for a very large number of slits in a evice calle a iffraction grating.

9 1092 CHAPTER 37 Wave Optics Figure 37.6 Light intensity versus sin u for a ouble-slit interference pattern when the screen is far from the two slits (L.. ). Photograph from M. Cagnet, M. Françon, J. C. Thrierr, Atlas of Optical Phenomena, Berlin, Springer-Verlag, 1962 I I max 2l l 0 l 2l sin u Figure 37.7 Multiple-slit interference patterns. As N, the number of slits, is increase, the primary maxima (the tallest peaks in each graph) become narrower but remain fixe in position an the number of seconary maxima increases. N 2 I I max Primary maximum Seconary maximum N 3 N 4 N 5 For any value of N, the ecrease in intensity in maxima to the left an right of the central maximum, inicate by the blue ashe arcs, is ue to iffraction patterns from the iniviual slits, which are iscusse in Chapter 38. N 10 2l l 0 l 2l sin u An interference pattern is prouce on the screen as a result of the combination of the irect ray (re) an the reflecte ray (blue). Real source S S Mirror Viewing screen P P Figure 37.8 Lloy s mirror. The reflecte ray unergoes a phase change of 180. Quick Quiz 37.2 Using Figure 37.7 as a moel, sketch the interference pattern from six slits Change of Phase Due to Reflection Young s metho for proucing two coherent light sources involves illuminating a pair of slits with a single source. Another simple, yet ingenious, arrangement for proucing an interference pattern with a single light source is known as Lloy s mirror 1 (Fig. 37.8). A point light source S is place close to a mirror, an a viewing screen is positione some istance away an perpenicular to the mirror. Light waves can reach point P on the screen either irectly from S to P or by the path involving reflection from the mirror. The reflecte ray can be treate as a ray originating from a virtual source S9. As a result, we can think of this arrangement as a oubleslit source where the istance between sources S an S9 in Figure 37.8 is analo- 1 Develope in 1834 by Humphrey Lloy ( ), Professor of Natural an Experimental Philosophy, Trinity College, Dublin.

10 37.5 Interference in Thin Films 1093 For n 1 n 2, a light ray traveling in meium 1 unergoes a 180 phase change when reflecte from meium 2. The same thing occurs when a pulse traveling on a string reflects from a fixe en of the string. For n 1 n 2, a light ray traveling in meium 1 unergoes no phase change when reflecte from meium 2. The same is true of a pulse reflecte from the en of a string that is free to move. 180 phase change No phase change n 1 n 2 n 1 n 2 n 1 n 2 Rigi support n 1 n 2 Free support a b Figure 37.9 Comparisons of reflections of light waves an waves on strings. gous to length in Figure Hence, at observation points far from the source (L.. ), we expect waves from S an S9 to form an interference pattern exactly like the one forme by two real coherent sources. An interference pattern is inee observe. The positions of the ark an bright fringes, however, are reverse relative to the pattern create by two real coherent sources (Young s experiment). Such a reversal can only occur if the coherent sources S an S9 iffer in phase by 180. To illustrate further, consier point P9, the point where the mirror intersects the screen. This point is equiistant from sources S an S9. If path ifference alone were responsible for the phase ifference, we woul see a bright fringe at P9 (because the path ifference is zero for this point), corresponing to the central bright fringe of the two-slit interference pattern. Instea, a ark fringe is observe at P9. We therefore conclue that a 180 phase change must be prouce by reflection from the mirror. In general, an electromagnetic wave unergoes a phase change of 180 upon reflection from a meium that has a higher inex of refraction than the one in which the wave is traveling. It is useful to raw an analogy between reflecte light waves an the reflections of a transverse pulse on a stretche string (Section 16.4). The reflecte pulse on a string unergoes a phase change of 180 when reflecte from the bounary of a enser string or a rigi support, but no phase change occurs when the pulse is reflecte from the bounary of a less ense string or a freely-supporte en. Similarly, an electromagnetic wave unergoes a 180 phase change when reflecte from a bounary leaing to an optically enser meium (efine as a meium with a higher inex of refraction), but no phase change occurs when the wave is reflecte from a bounary leaing to a less ense meium. These rules, summarize in Figure 37.9, can be euce from Maxwell s equations, but the treatment is beyon the scope of this text Interference in Thin Films Interference effects are commonly observe in thin films, such as thin layers of oil on water or the thin surface of a soap bubble. The varie colors observe when white light is incient on such films result from the interference of waves reflecte from the two surfaces of the film. Consier a film of uniform thickness t an inex of refraction n. The wavelength of light l n in the film (see Section 35.5) is l n 5 l n

11 1094 CHAPTER 37 Wave Optics Interference in light reflecte from a thin film is ue to a combination of rays 1 an 2 reflecte from the upper an lower surfaces of the film. Air n 1.00 A Film n B Air n phase change No phase change Rays 3 an 4 lea to interference effects for light transmitte through the film. Figure Light paths through a thin film. t where l is the wavelength of the light in free space an n is the inex of refraction of the film material. Let s assume light rays traveling in air are nearly normal to the two surfaces of the film as shown in Figure Reflecte ray 1, which is reflecte from the upper surface (A) in Figure 37.10, unergoes a phase change of 180 with respect to the incient wave. Reflecte ray 2, which is reflecte from the lower film surface (B), unergoes no phase change because it is reflecte from a meium (air) that has a lower inex of refraction. Therefore, ray 1 is 180 out of phase with ray 2, which is equivalent to a path ifference of l n /2. We must also consier, however, that ray 2 travels an extra istance 2t before the waves recombine in the air above surface A. (Remember that we are consiering light rays that are close to normal to the surface. If the rays are not close to normal, the path ifference is larger than 2t.) If 2t 5 l n /2, rays 1 an 2 recombine in phase an the result is constructive interference. In general, the conition for constructive interference in thin films is 2 2t 5 1m l n m 5 0, 1, 2, c (37.16) This conition takes into account two factors: (1) the ifference in path length for the two rays (the term ml n ) an (2) the 180 phase change upon reflection (the term 1 2l n ). Because l n 5 l/n, we can write Equation as 2nt 5 1m l m 5 0, 1, 2, c (37.17) If the extra istance 2t travele by ray 2 correspons to a multiple of l n, the two waves combine out of phase an the result is estructive interference. The general equation for estructive interference in thin films is 2nt 5 ml m 5 0, 1, 2,... (37.18) Pitfall Prevention 37.1 Be Careful with Thin Films Be sure to inclue both effects path length an phase change when analyzing an interference pattern resulting from a thin film. The possible phase change is a new feature we i not nee to consier for ouble-slit interference. Also think carefully about the material on either sie of the film. If there are ifferent materials on either sie of the film, you may have a situation in which there is a 180 phase change at both surfaces or at neither surface. The foregoing conitions for constructive an estructive interference are vali when the meium above the top surface of the film is the same as the meium below the bottom surface or, if there are ifferent meia above an below the film, the inex of refraction of both is less than n. If the film is place between two ifferent meia, one with n, n film an the other with n. n film, the conitions for constructive an estructive interference are reverse. In that case, either there is a phase change of 180 for both ray 1 reflecting from surface A an ray 2 reflecting from surface B or there is no phase change for either ray; hence, the net change in relative phase ue to the reflections is zero. Rays 3 an 4 in Figure lea to interference effects in the light transmitte through the thin film. The analysis of these effects is similar to that of the reflecte light. You are aske to explore the transmitte light in Problems 33, 34, an 36. Quick Quiz 37.3 One microscope slie is place on top of another with their left eges in contact an a human hair uner the right ege of the upper slie. As a result, a wege of air exists between the slies. An interference pattern results when monochromatic light is incient on the wege. What is at the left eges of the slies? (a) a ark fringe (b) a bright fringe (c) impossible to etermine Newton s Rings Another metho for observing interference in light waves is to place a plano-convex lens on top of a flat glass surface as shown in Figure 37.11a. With this arrangement, 2 The full interference effect in a thin film requires an analysis of an infinite number of reflections back an forth between the top an bottom surfaces of the film. We focus here only on a single reflection from the bottom of the film, which provies the largest contribution to the interference effect.

12 37.5 Interference in Thin Films 1095 P R r O 2 1 Courtesy of Bausch an Lomb Figure (a) The combination of rays reflecte from the flat plate an the curve lens surface gives rise to an interference pattern known as Newton s rings. (b) Photograph of Newton s rings. a b the air film between the glass surfaces varies in thickness from zero at the point of contact to some value t at point P. If the raius of curvature R of the lens is much greater than the istance r an the system is viewe from above, a pattern of light an ark rings is observe as shown in Figure 37.11b. These circular fringes, iscovere by Newton, are calle Newton s rings. The interference effect is ue to the combination of ray 1, reflecte from the flat plate, with ray 2, reflecte from the curve surface of the lens. Ray 1 unergoes a phase change of 180 upon reflection (because it is reflecte from a meium of higher inex of refraction), whereas ray 2 unergoes no phase change (because it is reflecte from a meium of lower inex of refraction). Hence, the conitions for constructive an estructive interference are given by Equations an 37.18, respectively, with n 5 1 because the film is air. Because there is no path ifference an the total phase change is ue only to the 180 phase change upon reflection, the contact point at O is ark as seen in Figure 37.11b. Using the geometry shown in Figure 37.11a, we can obtain expressions for the raii of the bright an ark bans in terms of the raius of curvature R an wavelength l. For example, the ark rings have raii given by the expression r <!mlr/n. The etails are left as a problem (see Problem 62). We can obtain the wavelength of the light causing the interference pattern by measuring the raii of the rings, provie R is known. Conversely, we can use a known wavelength to obtain R. One important use of Newton s rings is in the testing of optical lenses. A circular pattern like that picture in Figure 37.11b is obtaine only when the lens is groun to a perfectly symmetric curvature. Variations from such symmetry prouce a pattern with fringes that vary from a smooth, circular shape. These variations inicate how the lens must be regroun an repolishe to remove imperfections. a Peter Aprahamian/Photo Researchers, Inc. b Dr. Jeremy Burgess/Science Photo Library/Photo Researchers, Inc. (a) A thin film of oil floating on water isplays interference, shown by the pattern of colors when white light is incient on the film. Variations in film thickness prouce the interesting color pattern. The razor blae gives you an iea of the size of the colore bans. (b) Interference in soap bubbles. The colors are ue to interference between light rays reflecte from the front an back surfaces of the thin film of soap making up the bubble. The color epens on the thickness of the film, ranging from black, where the film is thinnest, to magenta, where it is thickest.

13 1096 CHAPTER 37 Wave Optics THIN-FILM INTERFERENCE Problem-Solving Strategy The following features shoul be kept in min when working thin-film interference problems. 1. Conceptualize. Think about what is going on physically in the problem. Ientify the light source an the location of the observer. 2. Categorize. Confirm that you shoul use the techniques for thin-film interference by ientifying the thin film causing the interference. 3. Analyze. The type of interference that occurs is etermine by the phase relationship between the portion of the wave reflecte at the upper surface of the film an the portion reflecte at the lower surface. Phase ifferences between the two portions of the wave have two causes: ifferences in the istances travele by the two portions an phase changes occurring on reflection. Both causes must be consiere when etermining which type of interference occurs. If the meia above an below the film both have inex of refraction larger than that of the film or if both inices are smaller, use Equation for constructive interference an Equation for estructive interference. If the film is locate between two ifferent meia, one with n, n film an the other with n. n film, reverse these two equations for constructive an estructive interference. 4. Finalize. Inspect your final results to see if they make sense physically an are of an appropriate size. Example 37.3 Interference in a Soap Film Calculate the minimum thickness of a soap-bubble film that results in constructive interference in the reflecte light if the film is illuminate with light whose wavelength in free space is l nm. The inex of refraction of the soap film is SOLUTION Conceptualize Imagine that the film in Figure is soap, with air on both sies. Categorize We etermine the result using an equation from this section, so we categorize this example as a substitution problem. The minimum film thickness for constructive interference in the reflecte light correspons to m 5 0 in Equation Solve this equation for t an substitute numerical values: t l 2n 5 l 1600 nm nm 4n WHAT IF? What if the film is twice as thick? Does this situation prouce constructive interference? Answer Using Equation 37.17, we can solve for the thicknesses at which constructive interference occurs: t 5 1m l l 5 12m n 4n m 5 0, 1, 2, c The allowe values of m show that constructive interference occurs for o multiples of the thickness corresponing to m 5 0, t nm. Therefore, constructive interference oes not occur for a film that is twice as thick.

14 37.6 The Michelson Interferometer 1097 Example 37.4 Nonreflective Coatings for Solar Cells Solar cells evices that generate electricity when expose to sunlight are often coate with a transparent, thin film of silicon monoxie (SiO, n ) to minimize reflective losses from the surface. Suppose a silicon solar cell (n 5 3.5) is coate with a thin film of silicon monoxie for this purpose (Fig a). Determine the minimum film thickness that prouces the least reflection at a wavelength of 550 nm, near the center of the visible spectrum. SOLUTION Conceptualize Figure 37.12a helps us visualize the path of the rays in the SiO film that result in interference in the reflecte light. Categorize Base on the geometry of the SiO layer, we categorize this example as a thin-film interference problem. Air n 1 SiO n 1.45 Si n 3.5 a 180 phase change phase change Kristen Brochmann/Funamental Photographs, NYC Figure (Example 37.4) (a) Reflective losses from a silicon solar cell are minimize by coating the surface of the cell with a thin film of silicon monoxie. (b) The reflecte light from a coate camera lens often has a reish-violet appearance. b Analyze The reflecte light is a minimum when rays 1 an 2 in Figure 37.12a meet the conition of estructive interference. In this situation, both rays unergo a 180 phase change upon reflection: ray 1 from the upper SiO surface an ray 2 from the lower SiO surface. The net change in phase ue to reflection is therefore zero, an the conition for a reflection minimum requires a path ifference of l n /2, where l n is the wavelength of the light in SiO. Hence, 2nt 5 l/2, where l is the wavelength in air an n is the inex of refraction of SiO. Solve the equation 2nt 5 l/2 for t an substitute numerical values: t 5 l 550 nm nm 4n Finalize A typical uncoate solar cell has reflective losses as high as 30%, but a coating of SiO can reuce this value to about 10%. This significant ecrease in reflective losses increases the cell s efficiency because less reflection means that more sunlight enters the silicon to create charge carriers in the cell. No coating can ever be mae perfectly nonreflecting because the require thickness is wavelength-epenent an the incient light covers a wie range of wavelengths. Glass lenses use in cameras an other optical instruments are usually coate with a transparent thin film to reuce or eliminate unwante reflection an to enhance the transmission of light through the lenses. The camera lens in Figure 37.12b has several coatings (of ifferent thicknesses) to minimize reflection of light waves having wavelengths near the center of the visible spectrum. As a result, the small amount of light that is reflecte by the lens has a greater proportion of the far ens of the spectrum an often appears reish violet The Michelson Interferometer The interferometer, invente by American physicist A. A. Michelson ( ), splits a light beam into two parts an then recombines the parts to form an interference pattern. The evice can be use to measure wavelengths or other lengths with great precision because a large an precisely measurable isplacement of one of the mirrors is relate to an exactly countable number of wavelengths of light. A schematic iagram of the interferometer is shown in Active Figure (page 1098). A ray of light from a monochromatic source is split into two rays by mirror M 0, which is incline at 45 to the incient light beam. Mirror M 0, calle a beam splitter, transmits half the light incient on it an reflects the rest. One ray is reflecte

15 1098 CHAPTER 37 Wave Optics ACTIVE FIGURE Diagram of the Michelson interferometer. A single ray of light is split into two rays by mirror M 0, which is calle a beam splitter. Light source The path ifference between the two rays is varie with the ajustable mirror M 1. Telescope L 2 M 0 L 1 M 1 As M 1 is move, an interference pattern changes in the fiel of view. M 2 from M 0 to the right towar mirror M 1, an the secon ray is transmitte vertically through M 0 towar mirror M 2. Hence, the two rays travel separate paths L 1 an L 2. After reflecting from M 1 an M 2, the two rays eventually recombine at M 0 to prouce an interference pattern, which can be viewe through a telescope. The interference conition for the two rays is etermine by the ifference in their path length. When the two mirrors are exactly perpenicular to each other, the interference pattern is a target pattern of bright an ark circular fringes. As M 1 is move, the fringe pattern collapses or expans, epening on the irection in which M 1 is move. For example, if a ark circle appears at the center of the target pattern (corresponing to estructive interference) an M 1 is then move a istance l/4 towar M 0, the path ifference changes by l/2. What was a ark circle at the center now becomes a bright circle. As M 1 is move an aitional istance l/4 towar M 0, the bright circle becomes a ark circle again. Therefore, the fringe pattern shifts by one-half fringe each time M 1 is move a istance l/4. The wavelength of light is then measure by counting the number of fringe shifts for a given isplacement of M 1. If the wavelength is accurately known, mirror isplacements can be measure to within a fraction of the wavelength. We will see an important historical use of the Michelson interferometer in our iscussion of relativity in Chapter 39. Moern uses inclue the following two applications, Fourier transform infrare spectroscopy an the laser interferometer gravitational-wave observatory. Fourier Transform Infrare Spectroscopy Spectroscopy is the stuy of the wavelength istribution of raiation from a sample that can be use to ientify the characteristics of atoms or molecules in the sample. Infrare spectroscopy is particularly important to organic chemists when analyzing organic molecules. Traitional spectroscopy involves the use of an optical element, such as a prism (Section 35.5) or a iffraction grating (Section 38.4), which spreas out various wavelengths in a complex optical signal from the sample into ifferent angles. In this way, the various wavelengths of raiation an their intensities in the signal can be etermine. These types of evices are limite in their resolution an effectiveness because they must be scanne through the various angular eviations of the raiation. The technique of Fourier transform infrare (FTIR) spectroscopy is use to create a higher-resolution spectrum in a time interval of 1 secon that may have require 30 minutes with a stanar spectrometer. In this technique, the raiation from a sample enters a Michelson interferometer. The movable mirror is swept through the zero-path-ifference conition, an the intensity of raiation at the viewing position is recore. The result is a complex set of ata relating light intensity as a

16 37.6 The Michelson Interferometer 1099 function of mirror position, calle an interferogram. Because there is a relationship between mirror position an light intensity for a given wavelength, the interferogram contains information about all wavelengths in the signal. In Section 18.8, we iscusse Fourier analysis of a waveform. The waveform is a function that contains information about all the iniviual frequency components that make up the waveform. 3 Equation shows how the waveform is generate from the iniviual frequency components. Similarly, the interferogram can be analyze by computer, in a process calle a Fourier transform, to provie all the wavelength components. This information is the same as that generate by traitional spectroscopy, but the resolution of FTIR spectroscopy is much higher. Laser Interferometer Gravitational-Wave Observatory Einstein s general theory of relativity (Section 39.10) preicts the existence of gravitational waves. These waves propagate from the site of any gravitational isturbance, which coul be perioic an preictable, such as the rotation of a ouble star aroun a center of mass, or unpreictable, such as the supernova explosion of a massive star. In Einstein s theory, gravitation is equivalent to a istortion of space. Therefore, a gravitational isturbance causes an aitional istortion that propagates through space in a manner similar to mechanical or electromagnetic waves. When gravitational waves from a isturbance pass by the Earth, they create a istortion of the local space. The laser interferometer gravitational-wave observatory (LIGO) apparatus is esigne to etect this istortion. The apparatus employs a Michelson interferometer that uses laser beams with an effective path length of several kilometers. At the en of an arm of the interferometer, a mirror is mounte on a massive penulum. When a gravitational wave passes by, the penulum an the attache mirror move an the interference pattern ue to the laser beams from the two arms changes. Two sites for interferometers have been evelope in the Unite States in Richlan, Washington, an in Livingston, Louisiana to allow coincience stuies of gravitational waves. Figure shows the Washington site. The two arms of the Michelson interferometer are evient in the photograph. Five ata runs have been performe as of 2009, an a sixth is in operation with a projecte completion ate of These runs have been coorinate with other gravitational wave etectors, such as GEO in Hannover, Germany, TAMA in Mitaka, Japan, an VIRGO in Cascina, Italy. So far, gravitational waves have not yet been etecte, but the ata runs have provie critical information for moifications an esign features for the next generation of etectors. Funing has been approve for Avance LIGO, an upgrae that shoul increase the sensitivity of the observatory by a factor of 10. The target ate for the beginning of scientific operation of Avance LIGO is Figure The Laser Interferometer Gravitational-Wave Observatory (LIGO) near Richlan, Washington. Notice the two perpenicular arms of the Michelson interferometer. LIGO Hanfor Observatory 3 In acoustics, it is common to talk about the components of a complex signal in terms of frequency. In optics, it is more common to ientify the components by wavelength.

17 1100 CHAPTER 37 Wave Optics Concepts an Principles Summary Interference in light waves occurs whenever two or more waves overlap at a given point. An interference pattern is observe if (1) the sources are coherent an (2) the sources have ientical wavelengths. The intensity at a point in a ouble-slit interference pattern is p sin u I 5 I max cos 2 a b (37.14) l where I max is the maximum intensity on the screen an the expression represents the time average. A wave traveling from a meium of inex of refraction n 1 towar a meium of inex of refraction n 2 unergoes a 180 phase change upon reflection when n 2. n 1 an unergoes no phase change when n 2, n 1. The conition for constructive interference in a film of thickness t an inex of refraction n surroune by air is 2nt 5 1m l m 5 0, 1, 2, c (37.17) where l is the wavelength of the light in free space. Similarly, the conition for estructive interference in a thin film surroune by air is 2nt 5 ml m 5 0, 1, 2,... (37.18) Analysis Moel for Problem Solving Waves in Interference. Young s ouble-slit experiment serves as a prototype for interference phenomena involving electromagnetic raiation. In this experiment, two slits separate by a istance are illuminate by a single-wavelength light source. The conition for bright fringes (constructive interference) is sin u bright 5 ml m 5 0, 61, 62,... (37.2) The conition for ark fringes (estructive interference) is sin u ark 5 1m l m 5 0, 61, 62, c (37.3) u sin u The number m is calle the orer number of the fringe. Objective Questions 1. A plane monochromatic light wave is incient on a oubleslit as illustrate in Active Figure (i) As the viewing screen is move away from the ouble slit, what happens to the separation between the interference fringes on the screen? (a) It increases. (b) It ecreases. (c) It remains the same. () It may increase or ecrease, epening on the wavelength of the light. (e) More information is require. (ii) As the slit separation increases, what happens enotes answer available in Stuent Solutions Manual/Stuy Guie to the separation between the interference fringes on the screen? Select from the same choices. 2. Four trials of Young s ouble-slit experiment are conucte. (a) In the first trial, blue light passes through two fine slits 400 mm apart an forms an interference pattern on a screen 4 m away. (b) In a secon trial, re light passes through the same slits an falls on the same screen. (c) A

18 Conceptual Questions 1101 thir trial is performe with re light an the same screen, but with slits 800 mm apart. () A final trial is performe with re light, slits 800 mm apart, an a screen 8 m away. (i) Rank the trials (a) through () from the largest to the smallest value of the angle between the central maximum an the first-orer sie maximum. In your ranking, note any cases of equality. (ii) Rank the same trials accoring to the istance between the central maximum an the firstorer sie maximum on the screen. 3. Suppose Young s ouble-slit experiment is performe in air using re light an then the apparatus is immerse in water. What happens to the interference pattern on the screen? (a) It isappears. (b) The bright an ark fringes stay in the same locations, but the contrast is reuce. (c) The bright fringes are closer together. () The bright fringes are farther apart. (e) No change happens in the interference pattern. 4. Suppose you perform Young s ouble-slit experiment with the slit separation slightly smaller than the wavelength of the light. As a screen, you use a large half-cyliner with its axis along the miline between the slits. What interference pattern will you see on the interior surface of the cyliner? (a) bright an ark fringes so closely space as to be inistinguishable (b) one central bright fringe an two ark fringes only (c) a completely bright screen with no ark fringes () one central ark fringe an two bright fringes only (e) a completely ark screen with no bright fringes 5. A thin layer of oil (n ) is floating on water (n ). What is the minimum nonzero thickness of the oil in the region that strongly reflects green light (l nm)? (a) 500 nm (b) 313 nm (c) 404 nm () 212 nm (e) 285 nm 6. A monochromatic beam of light of wavelength 500 nm illuminates a ouble slit having a slit separation of m. What is the angle of the secon-orer bright fringe? (a) ra (b) ra (c) ra () ra (e) ra 7. Accoring to Table 35.1, the inex of refraction of flint glass is 1.66 an the inex of refraction of crown glass is (i) A film forme by one rop of sassafras oil, on a horizontal surface of a flint glass block, is viewe by reflecte light. The film appears brightest at its outer margin, where it is thinnest. A film of the same oil on crown glass appears ark at its outer margin. What can you say about the inex of refraction of the oil? (a) It must be less than (b) It must be between 1.52 an (c) It must be greater than () None of those statements is necessarily true. (ii) Coul a very thin film of some other liqui appear bright by reflecte light on both of the glass blocks? (iii) Coul it appear ark on both? (iv) Coul it appear ark on crown glass an bright on flint glass? Experiments escribe by Thomas Young suggeste this question. 8. Green light has a wavelength of 500 nm in air. (i) Assume green light is reflecte from a mirror with angle of incience 0. The incient an reflecte waves together constitute a staning wave with what istance from one noe to the next noe? (a) nm (b) 500 nm (c) 250 nm () 125 nm (e) 62.5 nm (ii). The green light is sent into a Michelson interferometer that is ajuste to prouce a central bright circle. How far must the interferometer s moving mirror be shifte to change the center of the pattern into a ark circle? Choose from the same possibilities as in part (i). (iii). The green light is reflecte perpenicularly from a thin film of a plastic with an inex of refraction The film appears bright in the reflecte light. How much aitional thickness woul make the film appear ark? 9. While using a Michelson interferometer (shown in Active Fig ), you see a ark circle at the center of the interference pattern. (i) As you graually move the light source towar the central mirror M 0, through a istance l/2, what o you see? (a) There is no change in the pattern. (b) The ark circle changes into a bright circle. (c) The ark circle changes into a bright circle an then back into a ark circle. () The ark circle changes into a bright circle, then into a ark circle, an then into a bright circle. (ii) As you graually move the moving mirror towar the central mirror M 0, through a istance l/2, what o you see? Choose from the same possibilities. 10. A film of oil on a pule in a parking lot shows a variety of bright colors in swirle patches. What can you say about the thickness of the oil film? (a) It is much less than the wavelength of visible light. (b) It is on the same orer of magnitue as the wavelength of visible light. (c) It is much greater than the wavelength of visible light. () It might have any relationship to the wavelength of visible light. Conceptual Questions 1. What is the necessary conition on the path length ifference between two waves that interfere (a) constructively an (b) estructively? 2. A soap film is hel vertically in air an is viewe in reflecte light as in Figure CQ37.2. Explain why the film appears to be ark at the top. 3. Explain why two flashlights hel close together o not prouce an interference pattern on a istant screen. enotes answer available in Stuent Solutions Manual/Stuy Guie Richar Megna/Funamental Photographs, NYC Figure CQ37.2 Conceptual Question 2 an Problem 66.