Wave Optics. Why is the sky blue? What causes the beautiful colors in a soap bubble or an oil

Transcription

1 HAPTER26 C. Return to Table of Contents Wave Optics Colors produced by a thin layer of oil on the surface of water result from constructive and destructive interference of light. Why is the sky blue? What causes the beautiful colors in a soap bubble or an oil film? Why are clouds and ocean surf white, though both are formed by tiny drops of clear, colorless water? Why do Polaroid sunglasses reduce reflected glare? In this chapter we shall answer these questions. Additionally, we shall see how two light beams can combine to produce darkness, we shall show how to measure the wavelength of light using a meter stick, and we shall see why the magnification of any optical microscope is limited by the wave properties of light. We begin by giving a brief qualitative introduction to the wave phenomena of polarization, diffraction, and interference before returning to a more quantitative discussion of each. In our description of various experiments we shall often use a laser as our light source because of its wonderfully simple properties. 731

2 732 CHAPTER 26 Wave Optics Fig The electric field in a linearly polarized, plane, monochromatic wave. Fig A polarized laser beam. Fig Cross section of an unpolarized light beam Wave Properties of Light Polarization The simplest kind of light wave is a plane, monochromatic wave, which is linearly polarized (Fig. 26 1). Linear polarization means that the electric field vector is always directed parallel to a single line (the y-axis in the figure). Fig shows only the electric field. An associated magnetic field is parallel to the z-axis and oscillates in phase with the electric field. Values of the electric field are shown at a particular instant of time for various points along the x-axis. In a plane wave the value of the electric field is the same along any plane perpendicular to the direction of the wave s motion. The figure shows plane wavefronts, along which the electric field is maximum. Rays show the direction of motion of the wave along the x-axis. The plane wave described in Fig is approximated by a section of a polarized laser beam (Fig. 26 2). Within the beam, rays are approximately parallel, and wavefronts are approximately cross sections of the beam.* By turning the laser on its side, rotating it 90, we can produce a wave linearly polarized in the horizontal direction, rather than in the vertical direction. By rotating the laser through some other angle we can get polarization in any direction perpendicular to the beam. Most natural light sources and many lasers have random polarization. This means that at a given instant the electric field at any point in the wave is just as likely to be directed along any line perpendicular to the direction of motion. The light is then said to be unpolarized, and we indicate this state as shown in Fig Frequency Bandwidths Any real source of light is not exactly monochromatic; that is, there is never just one precise value of frequency. Instead there is a range or band of frequencies, which may be wide or narrow. The narrower the band, the more nearly the wave approximates a monochromatic wave. Laser light is nearly monochromatic. A common helium-neon laser emits light at a frequency of Hz with a bandwidth of about 10 8 Hz. This means that the frequency range is less than 1 part in More expensive, frequency-stabilized lasers have bandwidths as low as 10 4 Hz. Some lasers have even achieved a stabilized frequency range of less than 100 Hz. *There is some slight spreading of the beam, and hence the beam is not exactly a plane wave. A plane wave is also approximated by a small section of a spherical wave from a distant point source.

3 26 1 Wave Properties of Light By way of comparison, spectral lines emitted by various gas discharge tubes typically have bandwidths of roughly 10 9 Hz, and white light, ranging in frequency from Hz to Hz, has a bandwidth of Hz. The order of magnitude of these bandwidths is summarized in Table Coherence It is often important to be able to predict the relationship between the phase of a light wave at two different times at the same point in space. For example, suppose that at some instant t 0, at one point in a laser beam, the electric field vector has its maximum value; that is, you are at a peak in the wave. At a time t later, will the wave again have its peak value or will it have some other value (Fig. 26 4)? If the laser light were truly monochromatic, the solution would be easy. We could simply determine the exact number of cycles elapsed in a given time interval t by multiplying the frequency (the number of cycles per second) by the time interval t. (If the result were a whole number, the wave would again be at a peak. Or if the result were a whole number plus 1, the electric field would be zero at that instant.) However, even laser light is not 4 exactly monochromatic. There is always some frequency range f to f f. The number of cycles per second during any particular time interval can be anywhere in this range. So the number of cycles completed during a time interval t is somewhere in the range f t to ( f f ) t. If the time interval t is small enough, the product f t will be much less than 1 cycle, and there will be little uncertainty in the number of cycles completed, or in the final phase of the cycle. We can then predict the final phase from the initial phase, and we say that the wave is coherent over the time interval t. This means that there is a definite, predictable phase relationship. The condition for coherence then is that the time interval be small enough that or that f t 1 Table 26 1 Frequency band widths typical of various kinds of light Light Stabilized He-Ne laser Common He-Ne laser Spectral line White light f (Hz) Fig What phase of the cycle occurs at a time t after t 0? t (condition for coherence) (26 1) f EXAMPLE 1 Coherence of Light Sources Will you have coherence over a time interval of 10 6 s for light from (a) a gas discharge tube; (b) a stabilized He-Ne laser? SOLUTION (a) For a single line from a gas discharge tube, we see from Table 26 1 that even for a single spectral line f 10 9 Hz. Thus 1/ f 10 9 s, and the time interval t 10 6 s is much too long to satisfy Eq. 26 1, since 10 6 s is certainly not less than 10 9 s. Thus this kind of light is not coher ent over such a time interval. The number of cycles completed is uncertain by f t (10 9 s 1 )(10 6 s) 10 3 cycles, and so there is no ability to predict the phase over such a time interval. (b) For light from a stabilized He-Ne laser, f 10 4 Hz. Thus the condition that must be satisfied for such light is or t 1 f t 10 4 s A time interval of 10 6 s satisfies this condition. Thus the laser light is coherent over this time interval. Notice that f t (10 4 s 1 )(10 6 s) 10 2, and so we know the number of elapsed cycles with an uncertainty of only one hundredth of a cycle.

4 734 CHAPTER 26 Wave Optics Fig During a time interval t a wave peak advances a distance x c t. Thus com par - ing the phases at two points a distance x apart at a fixed time is the same as comparing the phases at the same point in space over a time interval t. Each wavefront in a light wave advances at the speed of light. Therefore, we can relate the coherence of light at a fixed point in a plane wave at two different times to coherence at two different points in a plane wave at the same time. As illustrated in Fig. 26 5, during a time interval t a wavefront advances a distance x c t, and so comparison of phases at two points x apart is equivalent to comparing the phases at a fixed point over a time interval t. Since t 1/ f is the condition for coherence, two points in a wavefront will be coherent if x c f The distance c/ f is called the coherence length, denoted by x c. c x c (26 2) f The condition for coherence may be expressed in terms of x c : x x c (coherence condition) (26 3) For a stabilized He-Ne laser with a frequency range of 10 4 Hz, we find c 3 10 x c 8 m/s m f 104 Hz Two points in the laser beam have a predictable phase relationship, as long as they are much less than 30,000 m apart! For a common laboratory He-Ne laser, the bandwidth is of the order of 10 8 Hz. Thus c 3 10 x c 8 m/s 3 m f 108 Hz The two points must be much closer than 3 m. Certainly two points a few cm apart are coherent. For white light, c 3 10 x c 8 m/s m f Hz Thus points in a beam of white light must be considerably less than a thousandth of a millimeter apart to be coherent.

5 26 1 Wave Properties of Light 735 Diffraction Suppose you are standing behind an open doorway, listening to a conversation in the next room. You can easily hear the voices from the room because the sound waves bend around the doorway. This phenomenon is called diffraction. It is a property common to all waves to bend or diffract around an obstacle. However, the amount of bending depends on the wavelength of the wave and the dimensions of the obstacle. In general, the longer the wavelength, the greater is the diffraction. Light, with its relatively short wavelength, bends or diffracts very little around an open doorway, but sound waves with their much longer wavelengths, diffract a great deal. Thus you can hear the conversation though you cannot see those who are talking. Seeing diffraction of light requires careful observation. Suppose we pass an intense beam of light through a narrow slit in an opaque screen and project it onto a white screen (Fig. 26 6). If the slit is relatively wide, (say, at least a millimeter), we get an image of the slit on the screen (Fig. 26 6a). As predicted by geometrical optics, the rays passing through the slit travel straight to the screen. But if we make the slit very narrow (say, less than about 0.1 mm), the image on the screen actually gets wider (Fig. 26 6b), violating the prediction of geometrical optics. We find that the narrower we make the slit, the more the light bends outward. Of course if we make the slit much less than 0.1 mm, there will be too little light to be seen, even if the light source illuminating the slit is very intense. But if we could make a slit with a width much less than the wavelength of light and still have enough light intensity to see the small amount of light passing through, we would see the light spread out in all directions, forming a cylindrical wave. Or if we replaced the slit by a tiny circular hole, with a diameter much less than the wavelength of light, we would produce a spherical wave. This result, illustrated in Fig. 26 7, is the essence of the Huygens-Fresnel principle, according to which each section of wavefront in the diffracting aperture is the source of a spherical wave. Fig shows a photograph, illustrating this principle for water waves. The waves in Fig are incident on an aperture much smaller than the wavelength. Diffraction of light was first observed and recorded by the Jesuit priest Francesco Grimaldi, a contemporary of Newton. Grimaldi observed the spreading of a narrow beam of sunlight entering a darkened room. Geometrical optics, which was then based on a picture of light consisting of particles,* could not account for this phenomenon, and so Grimaldi proposed that light is a wave. Grimaldi s idea was rejected by Newton, who believed that if light were a wave, the diffraction effect would be much greater than observed. It must have seemed unlikely to Newton that light could have the incredibly small wavelength necessary to explain such a small amount of diffraction. Newton s authority was so great that the wave theory was not accepted for another 200 years. (a) (b) Fig (a) Little diffraction is produced by a slit wider than 1 mm. (b) Considerable diffraction is produced by a slit less than 0.1 mm wide. Fig Cross section of a spherical wave, resulting from diffraction of light by a circular hole with a diameter much smaller than the wavelength of the light. Interference Like sound waves or waves on a string, light waves can interfere constructively or destructively. Constructive interference occurs when two light waves are in phase, and destructive interference occurs when two light waves are 180 out of phase. The colors in the photograph of the oil slick at the beginning of this chapter result from interference of the light reflected from the top and bottom surfaces of the thin film of oil. The thick - ness of the film varies, and, as a result, different colors of light interfere constructively. *In the twentieth century it was discovered that light does indeed have some particle-like properties. However this modern idea of a photon as a particle of light refers to emission or absorption of light, not to the way it propagates. Light travels as a wave, not as a bunch of particles, contrary to Newton s belief. The dual character of light as wavelike (in transmission) and particle-like (in absorption and emission) is at the heart of modern quantum physics, to be discussed in Chapter 28. Fig Diffraction of water waves.

6 736 CHAPTER 26 Wave Optics For example, where the film appears blue the thickness is such that blue light reflected from the two surfaces interferes constructively, while red light interferes destructively. In the sections that follow we shall investigate interference and diffraction phenomena quantitatively. (a) Constructive interference 26 2 Interference Fig. 26 9a illustrates constructive interference of light, which occurs when two light waves are in phase at a certain point in space over a period of time. Fig. 26 9b shows destructive interference, which occurs at a point in space where the waves are 180 out of phase over a period of time. If the two waves are of equal amplitude and 180 out of phase, the presence of two sources of light actually produces darkness! Fig illustrates the intensity of light that is seen at a given point in space where two waves of equal amplitude interfere either constructively or destructively. Since the intensity of a wave is proportional to the square of its amplitude, constructive interference of two equal-amplitude waves produces in the resultant wave 2 times the amplitude or 4 times the intensity of the individual waves. Whether the interference is constructive or destructive at a given point in space depends on the position of the point relative to the sources of light. Interference of light typically produces a pattern of light and dark areas (Fig ). (a) (b) (c) Fig Light seen on a screen at a point in space where: (a) There is one light wave of intensity I; (b) There are two light waves, each of intensity I, interfering constructively and producing a total intensity 4I; (c) There are two light waves, each of intensity I, interfering destructively and producing no light. (b) Destructive interference Fig Two light waves can interfere either (a) constructively or (b) destructively. In order for the eye to perceive interference of light, there must be a definite phase relationship between the two waves over a time interval that the eye can detect. The eye 1 has a response time on the order of of a second. Thus interference effects must be 20 stable for at least this long to be visible. This is longer than the coherence time of even the most monochromatic sources available today. The relative phases of two independent light sources (say, two different lasers) will vary randomly over time intervals greater than the coherence time. Thus, if we illuminate an area with two different sources, the interference of their light waves at a given point will rapidly oscillate from constructive to destructive, and so no interference pattern is visible. All one sees is a uniform illumination equal to the sum of the two intensities. For example, two equalamplitude waves from separate sources produce instantaneous intensities rapidly oscillating between 0 and 4 times each wave s intensity I, at each point in space. Thus one sees only the time average of the instantaneous intensity, which at all points is the same: twice the intensity of each source s wave, since the average of 0 and 4I equals 2I. With the present state of technology, it is impossible to see interference of light from two independent light sources.* Fig An interference pattern. *It is possible to detect electronically interference of two independent sources, as demonstrated by Brown and Twiss in 1952.

7 26 2 Interference 737 Interference effects are easily observable when a single wavefront is divided into two separate parts, which then follow separate paths to the point where interference is observed. Since the two waves arise from a common wavefront, changes in phase are common to each, and the interference pattern remains constant. Young s Double Slit One of the simplest ways to produce interfering light is to use a double-slit arrangement like the one studied by the English physician Thomas Young in 1801.* If a monochromatic plane wave is incident on a pair of thin, closely spaced slits, the two slits serve as sources of coherent light. The slits must be narrow enough and close enough that there is a significant amount of diffraction and overlap of the two wavefronts. As illustrated in Fig , what is seen on a screen in front of the slits is a pattern of alternating light and dark fringes. Fig shows how the location of the fringes relates to the distance to each of the slits. Point P is equidistant from the two slits, and so the two waves are exactly in phase at this point, and therefore point P is at the center of an interference maximum a bright fringe. The first dark fringe above the central bright fringe is at a point Q, which is one-half wavelength farther from slit 2 than from slit 1. The next bright fringe is at point R, which is 1 wavelength farther from slit 2 than from slit 1. (a) (b) Fig (a) Overlapping wavefronts interfere constructively at certain points and destructively at others. (Figure is not drawn to scale.) (b) Photograph of interference fringes from double slits illuminated with a He-Ne laser. Fig Constructive interference occurs at P and R. Destructive interference occurs at Q. (Figures are not drawn to scale.) *In 1801 Young used his double-slit experiment to measure the wavelength of light and to provide support for his belief that light is a wave. His ideas gained wide acceptance only after many years.

8 738 CHAPTER 26 Wave Optics Fig A point S on a screen is located a distance r 1 from slit 1 and a distance r 2 from slit 2. (This figure is not drawn to scale.) The location of the fringes can be determined with the aid of Fig The figure shows an arbitrary point S some distance y from the center of the interference pattern at point P. The angular displacement of point S is measured by the angle. The difference in the path lengths from S to each of the slits is related to the same angle. As shown in the figure, this path-length difference is d sin. Constructive interference occurs when this distance equals a whole number of wavelengths: d sin m m 0, 1, 2, (constructive interference) (26 4) Destructive interference occurs when the difference in path lengths equals a whole number of wavelengths plus wavelength: d sin (m 2 ) m 0, 1, 2, (destructive interference) (26 5) EXAMPLE 2 Measuring Light s Wavelength With a Meter Stick Light from a He-Ne laser illuminates two narrow slits, 0.20 mm apart, producing interference fringes on a wall 6.67 m from the slits (Fig ). The centers of the bright fringes are 2.1 cm apart. (a) Determine the wavelength of the laser light. (b) What would the fringe separation be if the slits were illuminated with violet light of wavelength 400 nm? SOLUTION (a) From Fig , we see that the distance y from the center of the interference pattern to any point S is related to the angle and the distance from slits to screen: tan y Any point in the interference pattern is at a very small angle, for which sin and tan are very nearly identical. Thus y sin Inserting this equation into Eq yields an expression for the distance y m to the mth fringe or y m d m m 0, 1, 2, y m m d

9 26 2 Interference 739 EXAMPLE 2 Measuring Light s Wavelength With a Meter Stick Continued Thus Thus, using measurements in cm and m, we indirectly measure 2 the wavelength of the laser light, a length less than a thousandth y 0 0, y 1 y 2, d d of a mm. The fringes are equally spaced, separated by a distance (b) Applying Eq. 26 6, we find that for violet light of wavelength 400 nm, the fringe spacing changes to y (26 6) d ( Solving for, we find 9 m)(6.67 m) y m d m d y ( m)( m) 6.67 m 1.3 cm m Because of its shorter wavelength, violet light produces interference fringes that are closer together. 630 nm Thin Films There is another simple way to split a single light wave into separate, coherent waves, which can then interfere. When light is incident on a partially reflecting surface (for example, a glass surface), part of the incident light is reflected and part is transmitted into the second medium. If the two light waves again come together, after having followed paths of somewhat different length, they will interfere. The difference in path lengths, however, must be less than the coherence length of the light, or the two waves will be totally incoherent. The Michelson interferometer, described in Problem 21, uses a half-silvered mirror to equally divide the incident wavefront from a monochromatic source. A thin transparent film can also serve to produce two coherent light waves, one reflected from the top and one from the bottom of the film (Fig ). If the film is very thin, you can see the interference, even with a white light source, as indicated by the photo of the colored oil film at the beginning of this chapter. In Fig the second light wave travels a distance greater than the first. The difference in the path lengths may cause a phase difference between the two waves. In addition, there may also be phase changes produced by the reflections. When light is incident on the surface of a medium with a higher refractive index than that of the incident medium, reflected light experiences a 180 phase change. If the second medium has a lower index than the first, reflection causes no phase change. The situation is the same as that of a wave pulse on a rope, partially reflected because of a change in density of the rope. As illustrated in Chapter 16, Fig , the reflected pulse is inverted if the second section of rope has higher density than the first. If the second section is of lower density than the first, there is no inversion of the reflected pulse, that is, no change in phase. The microscope slides in Fig show a pattern of interference fringes produced by light reflected on either side of the thin film of air trapped between the slides. The air film varies in thickness, and hence the interference alternates between constructive (bright) and destructive (dark). Fig The top and bottom sur - faces of a thin film reflect light upward. Fig An interference pattern is formed by reflection from the thin film of air between two microscope slides.

10 740 CHAPTER 26 Wave Optics EXAMPLE 3 Colors On An Oil Slick What color will be brightest when white light is reflected at normal incidence from a film of oil 250 nm thick on the surface of a puddle of water? The oil has a refractive index of 1.4. SOLUTION The light reflected from the upper surface of the oil film undergoes a 180 phase change, since oil s refractive index is higher than air s (Fig ). The light reflected from the lower surface of the oil experiences no phase change, since water s index is lower than oil s. Thus, if light reflected from the lower surface is to arrive at the upper surface in phase with the light reflected from the top surface, the extra distance traveled must give it a net 180 phase change. This means that the additional path length, equal to twice the film s thickness t for normal incidence, must equal a whole number of wavelengths plus wavelength t (m ) n where m 0, 1, 2, (26 7) 2 The wavelength n is of course the wavelength in the oil. This is related to the vacuum wavelength 0 by Eq : 0 n n where n is the refractive index of the oil. Inserting this expression into Eq and solving for 0, we find 2tn 0 for m 0, 1, 2, (26 8) m 1 2 Fig Trying possible values of m, we find 2tn 2(250 nm)(1.4) m 0: nm 1 2 2tn 1400 nm m 1: nm tn 1400 nm m 2: nm Only for m 1 do we find a wavelength in the visible range (400 to 700 nm). Blue light of wavelength 470 nm will interfere con - structively, and therefore blue will be the color most strongly re - flected by the film. The film will appear blue. One can show that red light will experience destructive interference (Problem 14). 1 2 EXAMPLE 4 Nonreflective Glass Coating Uncoated glass reflects 4% of the light incident on its surface at normal incidence. Sometimes glass is coated with a thin layer of a transparent material so that the intensity of the reflected light is reduced. Find the minimum thickness of a coating of magnesium fluoride, MgF 2 (n 1.38), which will produce destructive interference at a wavelength in the middle of the visible spectrum (550 nm). SOLUTION Both reflected waves experience a 180 phase change, since both are reflected from a medium with a higher index than that of the incident medium (Fig ). The only relative change in phase results from a difference in path length. Destructive interference occurs for a minimum path difference of wavelength. 1 2 Fig or 0 2t 1 2 n 0 2n 550 nm t 4n 99.6 nm 4(1.38)

11 26 3 Diffraction 741 EXAMPLE 4 Nonreflective Glass Coating Continued This is the minimum thickness of MgF 2 that will produce destructive interference for nm. Such a layer will not completely eliminate reflection at this wavelength, since the amplitudes of the two interfering waves are not equal, and so there is only partial cancellation of the waves, as in Fig. 26 9b. One might wonder whether this coating will produce a reduction in reflected intensity at other wavelengths or whether it could perhaps enhance reflection by constructive interference at some wavelengths. The condition for constructive interference here is that the path-length difference equals a whole number of wavelengths. or 0 2t m n m n interferes constructively. A more detailed analysis, which takes into account the intensities of the interfering waves, shows that the effect of the coating is to reduce reflection fairly uniformly across the visible spectrum to an average of about 1% of incident intensity, compared to a 4% reflection for uncoated glass. How - ever, there is a slight enhancement of reflected intensity in the blue part of the spectrum. Coating glass with several thin layers of different materials, carefully selected for index and thickness, can provide a further reduction in intensity. One of the most important applications of such coatings is for lenses in optical instruments, which use a large number of lenses that would otherwise produce much unwanted reflected light. Fig shows eyeglasses that have a nonreflective coating on one lens. 0 2tn m where m 1, 2, 3, m 1: 0 2tn 2(99.6 nm)(1.38) 275 nm 2tn m 2: nm 2 Other values of m yield smaller wavelengths. Thus no value of m gives a wavelength in the visible range. No visible light Fig Diffraction Historical Background The quantitative study of diffraction was of great historical importance in establishing the wave nature of light. Although Young s double-slit experiment supported the wave theory, many nineteenth-century scientists clung to Newton s particle theory of light. Full acceptance of the wave theory followed careful quantitative studies of diffraction by various scientists, especially Fresnel and Arago. In evaluating the wave theory of diffraction proposed by Fresnel, Poisson objected that it led to a rather strange and, to Poisson, an obviously false prediction: that at the center of the shadow of a round object would be a bright spot. Poisson argued that waves diffracted around the edges would travel an equal distance to the center and therefore interfere constructively there, if the wave theory were correct. Poisson presented this argument as a proof that the wave theory was wrong. Arago promptly performed the crucial experiment and found the predicted bright spot at the center of the shadow (Fig ). Based on such results, the wave theory of light was strongly established by Fig Diffraction pattern of a penny. Constructive interference of light diffracted around the edge of the penny produces a bright spot at the center of the shadow.

12 742 CHAPTER 26 Wave Optics Fraunhofer and Fresnel Diffraction When an object producing diffraction is illuminated by a plane wave and the resulting diffraction pattern is viewed on a screen at a large enough distance from the object, detailed analysis of the diffraction pattern is greatly simplified. This case is referred to as Fraunhofer diffraction. When the illuminating source is not a plane wave or the screen is not far enough away, the diffraction is called Fresnel diffraction. Analysis of Fresnel diffraction is more complicated than analysis of Fraunhofer diffraction, and the Fresnel diffraction pattern itself looks quite different from the Fraunhofer diffraction pattern of the same object. Figs and are examples of Fresnel dif - fraction. We shall analyze only Fraunhofer diffraction because of its relative simplicity. Fig Diffraction by a straight edge and the corresponding graph of intensity versus position. The dashed line indicates the intensity predicted by geometrical optics, implying a sharp-edged shadow. Single Slit A particularly simple case, Fraunhofer diffraction by a single slit, is shown in Fig In Section 26 1 we described diffraction of light around the edges of the slit (Fig. 26 6b). Actually the phenomenon is somewhat more complicated because light coming from various parts of the slit interferes to form a series of light and dark bands, as shown in Fig Fig Single-slit diffraction. Light diffracted by the slit forms a series of light and dark bands on a distant screen. Fig Rays emanate from points in the opening, a thin slit of width a. All rays in the direction strike the screen at point P. (This figure is not drawn to scale.)

13 26 3 Diffraction 743 We can use the Huygens-Fresnel principle to analyze single-slit diffraction. According to this principle, each section of a wavefront in the diffracting aperture is the source of a spherical wavelet. The amplitude of the light wave at any point beyond the aperture is the superposition of all these wavelets. Fig shows a plane wave incident on a narrow slit and a distant screen for viewing the resulting diffraction pattern. An enlarged section of the figure shows the spherical wavelets and associated rays emanating from several points in the opening. Rays striking the screen at point P interfere constructively or destructively, depending on the relative phases of the waves. Since P is at a great distance, the rays reaching P are nearly parallel. These rays form an angle with a line drawn to point O, which is directly opposite the slit. Fig shows rays directed at the center of the diffraction pattern ( 0). Since these parallel rays travel equal distances to the screen, interference is constructive, and the center of the pattern is therefore a diffraction maximum. (a) Fig Rays in the 0 direction strike the screen at point O, directly in front of the slit. Next we shall locate the diffraction minima produced by a single slit. We can use Fig to find the values of corresponding to diffraction minima. The first minimum is indicated in Fig a, where the slit of width a has been divided into two halves. We compare any ray from the top half with a ray a distance a/2 below it. Let the difference in path length of two such rays be /2, so that they interfere destructively. As shown in the figure this occurs for rays at an angle, where a sin Since all rays from the top half of the slit can be paired with a cancelling ray from the bottom half, this angle corresponds to a diffraction minimum. A similar argument can be applied to find other diffraction minima. In Fig b the slit is divided into 4 seg - ments, and it is shown that another diffraction minimum occurs at an angle, such that a sin 2 In Fig c, the slit is divided into 6 segments, and a diffraction minimum is found at an angle, where a sin 3 In general, single-slit diffraction minima occur at any angle satisfying the equation a sin m m 1, 2, 3, (diffraction minima) (26 9) (b) (c) Fig Locating the (a) first, (b) second, and (c) third diffraction minima for a single slit of width a.

14 744 CHAPTER 26 Wave Optics EXAMPLE 5 A Wide Image of a Thin Slit Find the width of the central diffraction maximum of a slit of width mm, as seen on a screen 2.00 m from the slit. The slit is illuminated by a He-Ne laser beam ( 633 nm). SOLUTION The edge of the central diffraction maximum will correspond to the first minimum (m 1). Applying Eq. 26 9, we find the angle corresponding to this point. or a sin m 633 nm sin a m rad (or ) Next we find the linear distance y from the center of the pattern to the point on the screen corresponding to this angle. Since the angle is quite small, we may approximate sin by y/, where y and are shown in Fig Thus or y sin y ( ) ( )(2.00 m) m 1.27 cm Since the central maximum extends an equal distance below the midpoint, the width of the maximum is double this value, or 2.54 cm. Thus the central maximum has a width 254 times the slit width ( cm). In other words, the width of the slit s central image is 254 times greater than the width of the image predicted by geometrical optics. Circular Aperture Fraunhofer diffraction by a circular aperture is of particular importance because of its application to the eye and to optical instruments, which generally have circular apertures. However, quantitative analysis of the circular aperture is considerably more complicated than analysis of the slit. Therefore we shall simply state without proof the one most important result of that analysis: the first diffraction minimum of an aperture of diameter D is at an angle, where D sin 1.22 Notice that this equation is similar to the equation for the first minimum of a slit of width a (Eq with m 1: a sin ). Since sin is very nearly equal to in radians, for the angles we normally encounter, we can express this result 1.22 (first minimum; circular aperture) (26 10) D Fig Fraunhofer diffraction by a circular aperture. The central maximum is called the Airy disk, which contains 84% of the light in the pattern. Since the diameter of the Airy disk is inversely proportional to the aperture diameter D, as the diameter of the aperture decreases, the disk gets bigger. Fig shows the Fraunhofer diffraction pattern of a circular aperture and the corresponding graph of intensity versus. The circular aperture was illuminated by a plane wave. One can think of this wave as originating from a distant point source. Since most of the light energy is concentrated in the central disk, called the Airy disk, for simplicity we can regard this disk as the image of the point.

15 26 3 Diffraction Rayleigh Criterion for Resolution When a point source of light is imaged by an optical system with a circular aperture, the image is an Airy disk. For example, the image of a star formed by a telescope is such a disk. If two points are very close, their Airy disks will overlap, and you may not be able to distinguish separate images. Fig shows the image of two points that are (a) clearly resolved, (b) barely resolved, or (c) unresolved. As a quantitative measure of the resolution of two points, Lord Rayleigh proposed the following criterion the Rayleigh criterion: two points are barely resolved when the center of one s Airy disk is at the edge of the other s Airy disk. Fig illustrates the formation of the image of two points that are barely resolved according to Rayleigh s criterion. Notice that the angular separation min of the two points P and Q is the angle from the center of an Airy disk to the first minimum, expressed by Eq Thus two points are resolved only if they subtend an angle at least as big as this minimum value min, where min 1.22 (Rayleigh s criterion for resolution) (26 11) D Any image formed by an optical system consists of a set of Airy disks, each of which is the image of a single point on the object. The size of these disks determines the resolution. 745 (a) Points are clearly resolved (b) Points are barely resolved (c) Points unresolved Fig Resolution of two point sources of light diffracted by a circular aperture. Fig Rayleigh s criterion for resolution. EXAMPLE 6 Diffraction of Light by the Eye (a) Find the prediction of diffraction theory for the minimum angle subtended by two points that are barely resolved by the eye. Assume a pupil diameter of 2.0 mm, and use a wavelength at the center of the visible spectrum. (b) Find the distance between the two points if they are 25 cm from the eye, at its near point. SOLUTION (a) We apply Eq , using for the wavelength inside the eye, where the refractive index n At the center of the visible spectrum the vacuum wavelength nm, and Thus 0 n (b) From Fig we see that this angle equals the separation d between the points divided by the distance of 25 cm. d rad 25 cm d (25 cm)( rad) cm The eye should be unable to resolve points closer than about 0.06 mm. The minimum angle we have calculated is fairly close to the measured minimum angle between points barely resolved by the normal eye. The factors other than diffraction that affect this minimum angle are discussed in Chapter 25, Section 25 5 (Factors Limiting Visual Acuity) min m nd 1.22 D (1.34)( m) rad

16 746 CHAPTER 26 Wave Optics EXAMPLE 7 Diffraction Limit of a Microscope Find an expression for the minimum separation between two points that are barely resolved by a microscope with an object - ive of diameter D and focal length f. SOLUTION find Using Fig and applying Eq , we f d f min 1.22 (26 12) D To make d as small as possible we need to minimize the ratio f/d. But it is not possible to make f less than about D/2, the radius of the lens.* Setting f D/2 in the expression above, we obtain d 0.61 *For a simple symmetrical lens made of glass with a refractive index of 1.5, the focal length equals the radius of curvature of the first surface, as shown in Problem But the radius of curvature can be no smaller than the radius of the lens itself. Therefore the focal length is always less than the radius D/2. Fig Thus the minimum distance between two points that can be resolved by any microscope equals roughly half the wavelength of the light used to illuminate the points. For example, using the value 550 nm for the center of the visible spectrum, we find d (0.61)( m) m Diffraction Gratings A diffraction grating consists of thousands of very narrow, closely spaced slits, made by etching precisely spaced grooves on a glass plate (Fig ). The slits are the transparent spaces between the grooves. Typically a grating consists of thousands or tens of thousands of transparent lines per cm. It follows from our discussion of singleslit diffraction that each line, because of its extremely narrow width, produces diffracted light spread over a considerable angle perhaps 20 or 30. Of course, the diffraction pattern of one such line alone would not produce enough intensity to be seen by itself. But when diffracted light from thousands of lines interfere, bright, sharp diffraction maxima are produced, (Figs and 26 32). Fig A diffraction grating consists of thousands of narrow, closely spaced slits. Fig The diffraction pattern produced by a grating illuminated by a He-Ne laser. Fig The diffraction pattern produced by two per - pendicular diffraction gratings illuminated by a He-Ne laser.

17 26 3 Diffraction 747 Fig can be used to locate the diffraction maxima of a grating with spacing d between adjacent slits. As indicated in the figure, for light in the direction, the difference in path length of rays from adjacent slits is d sin. These adjacent rays will interfere constructively if the difference in path length is an integral multiple of, that is, 0,, 2, 3, and so forth. Indeed rays from all the slits will interfere constructively if they are directed at an angle, such that d sin m m 0, 1, 2, (26 13) Unlike the maxima produced by a double slit, diffraction grating maxima are very narrow and sharp, as shown in Fig This can be understood when we consider how a slight change in the angle away from a value satisfying Eq will affect the intensity. Suppose that the change in angle is so slight that light from adjacent slits is still nearly in phase. If there were only two slits interfering, such an angle would still give an intensity close to the maximum value. However, with the thousands of slits in a grating, there can be destructive interference in many ways. For example, light from slits 100 spacings apart might interfere destructively. If all pairs of slits 100 spacings apart interfere destructively, there will be no light in that particular direction. Diffraction gratings can be used to measure the wavelength of light, as illustrated in the following example. Fig Rays from a diffraction grating. EXAMPLE 8 Separating the Sodium Doublet Find the first-order (m 1) diffraction angles for the sodium doublet, using a grating with 10 6 lines/m. The sodium doublet consists of two yellow lines in the spectrum of sodium, with nearly identical wavelengths: nm and nm. SOLUTION Applying Eq to each of the wavelengths, with d 10 6 m and m 1, we find sin sin 1 m d (1)( m) 10 6 m Fig (1)( sin 2 9 m) 10 6 m Thus the angular separation between the lines is 0.04, or rad. Viewed at a distance of 1 m, the lines are 0.7 mm apart. X-ray diffraction is a technique that utilizes the small spacing between the atoms in a crystal as a three-dimensional diffraction grating. The atomic spacing is on the same order as wavelengths in the X-ray portion of the electromagnetic spectrum. So X rays, rather than visible light, are diffracted by a crystal. And the resulting diffraction pattern can be used to discover the crystal structure. X-ray diffraction of DNA was used by Watson and Crick in discovering the structure of DNA in 1951 (Fig ). Fig X-ray diffraction pattern of DNA. The double helix structure of DNA was revealed by this historic photograph taken by Rosalind Franklin.

18 748 CHAPTER 26 Wave Optics 26 4 Polarization Polarization by Absorption Most light sources produce unpolarized light (Fig. 26 3), as opposed to the polarized light produced by some lasers (Figs and 26 2). However, there are ways to polarize light that is initially unpolarized, or to change the direction of polarization of polarized light. One way is to pass the light through a Polaroid sheet, a synthetic material first produced by Edwin Land in 1928 when he was an undergraduate. There is a direction along each Polaroid sheet called its transmission axis. Light linearly polarized along this axis passes through the sheet (Fig a), whereas light polarized in the perpendicular direction is completely absorbed (Fig b). If the incident light is linearly polarized at some angle relative to the transmission axis, the light will be partially absorbed and partially transmitted. As illustrated in Fig c, the component of the electric field parallel to the axis is transmitted. The light that emerges is thus polarized along the direction of the transmission axis and has an amplitude E related to the incident amplitude E 0 by the equation E E 0 cos (a) Incident light polar ized along transmission axis (b) Incident light polarized perpendicu - lar to transmission axis (c) Incident light polarized at an angle with transmission axis Fig The effect of a Polaroid sheet on initially polarized light depends on the direction of initial polarization relative to the direction of the sheet s transmission axis. The intensity of light is proportional to the square of its amplitude (Eq. 23 5: I av 1 0 ce 02 ). Squaring the equation above, we obtain 2 E 2 E 02 cos 2 1 Multiplying both sides of this equation by the appropriate constant ( 2 0 c), we obtain a re - lationship between the average transmitted intensity I and the average incident intensity I 0 : I I 0 cos 2 (26 14)

19 26 4 Polarization 749 This result is known as the law of Malus. The intensity of the transmitted light has its maximum value, I I 0, when 0, and has its minimum value, I 0, when 90. Unpolarized light consists of a superposition of linearly polarized waves, with vary - ing directions of polarization, as illustrated in Fig a. The electric field associated with each of these waves can be resolved into x and y components, relative to an arbitrary coordinate system. Since the direction of polarization is random, the resultant x and y components are equal. We can replace the many randomly directed linearly polarized waves by just two linearly polarized waves of equal intensity, with mutually perpendicular polarization directions, as illustrated in Figs b and 26 37c. When unpolarized light is incident on a Polaroid sheet, only the component along the transmission axis is transmitted. Since the two components have equal intensity in unpolarized light, this means that the intensity of the transmitted light is half the intensity of the incident light (Fig ). 1 I I 0 (initially unpolarized light) (26 15) 2 (a) (b) (c) Fig Equivalent representations of unpolarized light. Fig A Polaroid sheet polarizes initially unpolarized light. EXAMPLE 9 Light Passing Through Two Polarizers An unpolarized laser beam of intensity 1000 W/m 2 is incident on a Polaroid sheet with a vertical transmission axis. The light passing through this sheet strikes a second Polaroid sheet, with a transmission axis at an angle of 30.0 from the vertical (Fig ). Find the polarization and the intensity of the light emerging from the second sheet. SOLUTION The light transmitted by the first Polaroid sheet is vertically polarized and, according to Eq , has an intensity equal to half the incident intensity. 1 1 I I 0 (1000 W/m 2 ) 500 W/m Fig The light incident on the second sheet is polarized at an angle of 30 relative to this sheet s transmission axis and, according to the law of Malus (Eq ), has intensity I I 0 cos 2 (500 W/m 2 )(cos ) 375 W/m 2

20 750 CHAPTER 26 Wave Optics (a) (b) Fig (a) Light polarized parallel to the reflecting surface is more strongly reflected than (b) light polarized in a perpendicular direction. Polarization by Reflection When light is reflected from the surface of a dielectric, such as water or glass, the intensity of the reflected light depends on the angle of incidence and on the polarization of the incident light. Light polarized parallel to the reflecting surface (Fig a) is always more strongly reflected than light polarized in a perpendicular direction (Fig b). Unpolarized light can be thought of as consisting of two equalintensity polarized waves one polarized parallel to the surface and a second polarized perpendicular to the first. After reflection, the component polarized parallel to the surface is more intense than the other component. Fig shows the intensities of the two components in a beam of initially unpolarized light reflected by water, for several angles of incidence. The intensities are predicted by Maxwell s equations. Notice that both components are more strongly reflected for very large angles of incidence. Any smooth dielectric surface becomes mirror-like as the angle of incidence approaches 90. You can observe this effect by holding up a smooth sheet of paper so that rays from a light source are reflected at glancing incidence. The angle of incidence at which the reflected light is 100% polarized is known as Brewster s angle, denoted by B. Brewster s angle has a value of 53 for reflection by water, as indicated in Fig c. Maxwell s equations can be used to derive an ex pres - sion for Brewster s angle, relating it to the refractive index n of the incident medium and the index n of the reflecting medium. We present the result here without proof: n tan B (Brewster s angle) (26 16) n (a) (b) (c) Fig Initially unpolarized light of total intensity 200 W/m 2 is reflected by water at various angles of incidence. (d) (e)

21 26 4 Polarization 751 EXAMPLE 10 Brewster s Angle for Glass Calculate Brewster s angle for light incident from air onto a glass surface if the glass has a refractive index of 1.5. SOLUTION Applying Eq , using the refractive indices for glass and air, we find n 1.5 tan B 1.5 n 1.0 B 56 Polaroid sunglasses are effective at reducing reflected glare from the surface of a body of water or from other surfaces (Fig ). The lenses are made of Polaroid sheets with vertical transmission axes. The reflected light consists mainly of horizontally polarized light, and such light is completely absorbed by the lenses. Fig An early ad for Polaroid sunglasses. EXAMPLE 11 A Sunset Seen Through Polaroid Sunglasses The setting sun is reflected from the surface of a lake at an angle of incidence of 75. The intensity of the incident light is 200 W/m 2, as in Fig e. Find the intensity of the reflected light reaching the eye of an observer wearing Polaroid sun - glasses (Fig ). SOLUTION From Fig e we see that the intensity of the reflected light having a polarization along the transmission axis of the sunglasses is 11 W/m 2. Therefore this is the intensity of the reflected light reaching the eye. Without sunglasses, the observer would see reflected light of both polarizations. From Fig e, the intensity of reflected light seen by the observer would be 11 W/m 2 31 W/m 2 42 W/m 2. Fig Non-Polaroid sunglasses that produce the same darkening as these Polaroid sunglasses would absorb 50% of all incident light, and so they would transmit to the observer reflected light of intensity 21 W/m 2.

22 752 CHAPTER 26 Wave Optics Polarization by Scattering When an electromagnetic wave is incident on an atom, the atom s electrons oscillate in response to the oscillating electric field. The electrons behave like tiny antennas; they emit their own radiation with the same frequency as the incident electromagnetic wave, but scatter the radiation in various directions. The intensity of this scattered radiation depends on the light s frequency. Blue light is scattered much more effectively than red light. Fig shows how scattering of sunlight by the earth s atmosphere gives us blue skies and red sunsets. Observers A and B both see blue sky as a result of the blue part of the sun s spectrum being scattered toward their eyes by the atmosphere. Meanwhile, observer C sees an orange or red sun as a result of the blue part of the spectrum s having been scattered out of the beam of direct sunlight. This kind of scattering, called Rayleigh scattering, is the result of independent, incoherent radiation by many atoms. Fig Scattering of sunlight by the earth s atmosphere results in blue skies (seen by A and B) and red sunsets (seen by C). Scattering is also the basic mechanism at the heart of reflection and refraction by a solid or a liquid. But the higher density and relative immobility of the atoms in the liquid or solid mean that light scattered by neighboring atoms is coherent and can therefore interfere constructively or destructively. The result of this interference is remarkably simple. The scattered waves interfere destructively in all directions, except those corresponding to the reflected and refracted waves, for which the interference is constructive. So we see only a reflected wave and a refracted wave. The particles of water in a cloud or in ocean surf also scatter sunlight, but in this case the scattered light is white, in contrast to the blue sky. The tiny water droplets simply reflect and refract incident light. The result of multiple reflections and refractions by a very large number of droplets is to redirect or scatter the incident white light in all directions.

23 26 4 Polarization 753 Scattering of sunlight by the atmosphere tends to polarize the light. Fig shows how this polarization arises. A beam of unpolarized sunlight, incident on the atmosphere, travels along the x-axis. This transverse electromagnetic wave has an electric field that oscillates in the yz plane; there is no component along the x-axis, the direction of propagation of the wave. Electrons within atoms in the atmosphere oscillate in the yz plane, in response to the incident wave, and scatter light in various directions. The nature of radiation produced by any source of radiation is that there can be a component of the electric field in a given direction only if there is a component of motion of the radiating source parallel to that direction. Therefore there can be no x- component of the electric field in scattered radiation, since there is none in the incident wave. This implies that radiation scattered along the yz plane, perpendicular to the incident beam, must be polarized. For radiation scattered in any such direction there is only a single line perpendicular to the direction of propagation, along which the electric field vector can oscillate, as indicated in Fig Light scattered in other directions is partially polarized. Fig Polarization by scattering.

24 ACloser Look Magic in the Sky Mirage! Looming! Mountain specter! What do you think of when you hear these terms? Illusions or ghosts, perhaps. The names are suggestive of the fear and be - wilderment that these phenomena have aroused over the centuries. It may surprise you to learn, however, that the names describe real optical effects in the atmosphere. You already know enough about the basic principles of optics to understand such phenomena. Mirages You have heard of people lost in the desert who imagined that they saw tempting pools of blue water just beyond the next sand dune and who dragged themselves forward in hopeless pursuit of them. You may think that such mirages are simply hallucinations caused by heat and thirst. Actually, they are almost always real phenomena real light is behaving in a way that creates an illusion in which anyone, hot and thirsty or not, can share (Fig. 26 A). Mirages are refraction phenomena. Light rays are bent by layers of air at dif ferent temperatures. Warm air has lower density than cool air and has a lower refractive index. As a ray coming from a cooler layer enters a warmer layer which is what happens when the ray is moving downward Fig. 26 A A fairly common kind of mirage: the dry surface of a road appears to be wet. Fig. 26 B Formation of an inferior mirage. toward a searingly hot desert surface the ray is refracted away from the normal. The ray can be bent so much that it curves back upward (Fig. 26 B). When it reaches the viewer s eye, it is automatically traced back by the brain as if it came from a source directly in line with its final segment. That is, it is seen as if it were ahead and below, rather than ahead and above. Light coming from a clear blue sky produces the illusion of a bright blue pool on the ground ahead. A mirage of the sort seen in a hot desert is called an inferior mirage because it appears below the light source (the sky). There is another kind of mirage, called a superior mirage, which appears above the source of light (Fig. 26 C). A superior mirage is typically caused when light moves upward from a layer of cool, dense air into warmer, less dense layers. The rays are bent away from the normal, as before, but this time they turn down rather than upward (Fig. 26 D). When they strike the eye, they are traced up to a mirage seen at their apparent point of origin. Under such conditions, a ship moving on the water below the horizon can look like a ghost ship sail - ing through the sky! The appearance of a superior mirage is sometimes called looming, for obvious reasons: the mirage looms above its source. Under special conditions, looming can produce truly uncanny effects. One par - ticular type of looming is called fata morgana, after Morgan le Fay, the fairyenchantress of the King Arthur legends who lived on a magical island. Fata morgana is most often seen in the Strait of Messina, a waterway that separates Sicily from Italy and that was long dreaded for its deadly currents, rocks, and whirlpools. The mirage is caused by irregular layerings of air of various densities, which produce multiple refractions and multiple overlapping images. The result is an apparent vertical elongation of the source object, sometimes to enormous proportions. For example, when

25 ACloser Look Fig. 26 C A superior mirage of a ferry, which appears to be vertically elongated. Fig. 26 D Formation of a superior mirage. seen from a ship in the strait, objects such as trees or hills on the shore can look like huge, weirdly shaped figures that can be disorienting and dangerous to unsuspecting mariners (Fig. 26 E). Coronas and anticoronas Interference or diffraction effects can occur when light rays pass near the edges of tiny objects in the atmosphere. If the diffraction appears around the light source, the effect is usually a ring, called a corona. Coronas can appear around the sun or moon when light rays pass near the edges of water droplets in the atmosphere. Rings of different colors can be seen because different wavelengths are diffracted to varying degrees. Fig. 26 E What appears in this picture to be icy castles in the sky is actually an extremely rare, mirage-like effect known as fata morgana.

26 ACloser Look A related phenomenon caused by the presence of hexagonal ice crystals is called a halo (Fig. 26 F). If the ice crystals line up in just the right way on a sunny day, they can produce two separate bright spots, one to either side of the sun. These ghost suns are called parhelia or more popularly, sundogs. When diffraction occurs around a shadow area, the effect is called an anticorona. Anticoronas are also known by the names glory (Fig. 26 G) and mountain specter. These are rare and awesome phenomena. The glory is seen as a halo surrounding the shadow of the observer s head. Be - cause of the extreme directional dependence of this effect, no one else sees the halo around the observer s head. Two observers might each see a halo around his or her own head, but each observer will not be able to see the other s halo. Fig. 26 F A street light blocks the sun s direct rays, allowing a halo around the sun to be clearly seen. One particularly famous example of an anticorona occurs occasionally near the Brocken, a mountain in the Harz range of central Germany. Because it is known in German legend as the site of the Walpurgis Night witchcraft rituals, the Brocken is an Fig. 26 G The bright circle around the plane s shadow is called a glory. appropriate location for the anticorona, which is there given the name Brocken specter. The specter can be seen at twi - light on sunny days by observers who stand near the foot of the mountain when there are misty banks of fog or cloud just above them that do not reach as high as the mountain top. The low-lying sun then casts a huge shadow of the mountain peak onto the upper surface of the mist. This silhouette of the peak appears surrounded by rings of colored light, as rays passing around the edge of the peak are bent and sep - arated out according to their wavelength (Fig. 26 H). Fig. 26 H Viewing the Brocken Specter.

27 C 26 HAPTER SUMMARY Light propagates as an electromagnetic wave. An ideal monochromatic source produces a wave with a definite frequency and wavelength. If the wave is also linearly polarized, the electric field oscillates parallel to a line perpendicular to the direction of wave propagation. If the wave is unpolarized, it can be thought of as a superposition of two linearly polarized waves, with perpendicular lines of polarization. Any real source is not perfectly monochromatic. There is always some uncertainty in frequency, f. Light is said to be coherent over a time interval t if there is a predictable phase relationship between the phases of the wave at the beginning and end of the interval. The condition for coherence is that the time interval be much less than the inverse of the uncertainty in frequency: 1 t (coherence condition) f Two points in a wave separated by a distance x are co - herent, that is, have a predictable phase relationship, if they are much closer than the wave s coherence length x c, where x c c f x x c (coherence condition) When two light waves are in phase at a given point in space, they are said to interfere constructively. The intensity of the light is maximum at such a point. When two light waves are 180 out of phase at a point, they are said to interfere destructively. At such a point the intensity of the light is minimum. If the two interfering waves have equal intensity, this minimum intensity equals zero; that is, the superposition of two light waves produces darkness. An interference pattern is a pattern of light and dark areas, corresponding to constructive and destructive interference by overlapping waves. In order for interference patterns to be visible, the phase relationship between the interfering waves must remain stable; that is, the waves must be coherent over a sufficiently long time interval. In practice this means that the two waves must be produced by splitting of a single wave into two separate parts, which travel over different paths. Then, if the difference in path lengths of the two waves is less than the coherence length, an interference pattern is visible. If a plane light wave of wavelength is incident on two narrow slits separated by a distance d, a pattern of equalwidth light and dark fringes is seen on a distant screen. An angle locating the center of a bright fringe satisfies the equation d sin m where m 0, 1, 2, and an angle locating the center of a dark fringe satisfies the equation 1 d sin (m ) where m 0, 1, 2, Light reflected from the top and bottom surfaces of a thin film will interfere if the film s thickness is much less than the coherence length of the light. The phase difference in the two interfering waves arises from a difference in path length and may also arise from a 180 phase change of light reflected from a medium with a higher refractive index than that of the incident medium. When there is no relative phase change between the two waves caused by reflection, constructive interference by a film of thickness t and index n occurs for light with vacuum wavelengths 2tn 0 where m 1, 2, 3, m When there is a 180 relative phase change caused by reflection, constructive interference occurs for wavelengths 2tn 0 where m 1, 2, 3, m 1 2 Light bends or diffracts around obstacles placed in its path. The Huygens-Fresnel principle states that each section of a wavefront in a diffracting aperture is the source of a spherical wave. The superposition of these waves produces a diffraction pattern a pattern of light and dark areas, corresponding to constructive and destructive interference of the waves. The simplest kind of diffraction, known as Fraunhofer diffraction, occurs when the diffracting object is illuminated by a plane wave and the diffraction pattern is viewed on a distant screen. Fraunhofer diffraction by a single slit of width a produces a broad central maximum and weaker successive maxima. The diffracting minima occur at angles satisfying the equation a sin m where m 1, 2, 3, Fraunhofer diffraction by a circular aperture of diameter D produces a circular diffraction maximum, called an Airy disk. The first diffraction minimum occurs at an angle given by the following approximate expression: 1.22 D The image of a point source of light produced by an optical system with a circular aperture is an Airy disk. Two object points may not be resolvable as separate points if they are too close, because their Airy disks overlap

28 758 CHAPTER 26 Wave Optics According to the Rayleigh criterion, two points are barely resolved if the center of one s Airy disk is at the edge of the other s Airy disk. The minimum angle subtended by points that are barely resolved by a circular aperture of diameter D is given by min 1.22 D A diffraction grating, which consists of thousands of very narrow, closely spaced slits, produces bright, sharp diffraction maxima at angles satisfying the equation d sin m where m 0, 1, 2, where d is the slit spacing. Light that is initially unpolarized becomes polarized as it passes through a Polaroid sheet. The light that emerges is polarized along the transmission axis of the sheet and has an intensity I equal to half the intensity I 0 of the unpolarized light. 1 I I 0 If polarized light is incident on a Polaroid sheet, the light that emerges is polarized along the sheet s transmission 2 axis and has an intensity I related to the initial intensity I 0 by the law of Malus: I I 0 cos 2 where is the angle between the incident light s polarization direction and the transmission axis. Light reflected from the surface of a dielectric is at least partially polarized, with the component of the electric field parallel to the surface more strongly reflected than the component in the perpendicular direction. For an angle of incidence known as Brewster s angle, the polarization is complete; that is, the reflected light is polarized parallel to the reflecting surface. Brewster s angle B is related to the refracting indices n and n of the incident medium and the reflecting medium by the equation n tan B n Scattering of sunlight by atoms in the earth s atmosphere produces complete polarization of the scattered light at a scattering angle of 90, or partial polarization at other scattering angles, as indicated in Fig Questions 1 (a) If visible light had a much longer wavelength on the order of meters would it be possible to see around a corner? (b) Would you be able to see much detail? 2 Linearly polarized light travels along the y-axis. Is it possible for there to be a component of the electric field (a) in the x direction; (b) in the y direction; (c) in the z direction; (d) in both the x and the y directions; (e) in any direction perpendicular to the y-axis? 3 Would you expect to be able to see an interference pat - tern produced by sunlight reflected from the surfaces of a glass windowpane? Explain. 4 Suppose you have just invented a new laser with an in - credible bandwidth of only 1 Hz. Would it be possible to produce a visible interference pattern by allowing light from two such lasers to overlap? 5 When two equally intense, coherent light waves are in phase at a point in space, the total intensity is four times the intensity of each of the waves alone more than the sum of the intensities of the two waves. Explain why this does not violate the principle of conservation of energy. 6 Suppose that Young s double-slit experiment is per - formed underwater. Would the interference fringes be closer, farther apart, or spaced the same as for the experiment performed in air? 7 A high-quality, diffraction-limited microscope will have better resolution if illuminated with (a) white light; (b) blue light; (c) red light; (d) orange laser light. 8 The soap bubble in Fig is colored because of inter ference of light reflected from its surfaces. Just before a bubble bursts, its thickness is much less than the wavelength of light. What color will the bubble then appear? Explain. Fig

29 Problems A diffraction pattern is produced by a rectangular aperture with a width of 0.1 mm and a height of 0.2 mm. Will the central maximum have greater height or width? 10 What happens to the width of the central diffraction max - imum produced by a thin slit as the slit width is decreased? 11 When unpolarized light is normally incident on a plane surface, is there any possibility that the reflected light will be polarized? Explain. 12 Two Polaroid sheets are oriented with perpendicular trans mission axes, as shown in Fig Unpolarized light is incident from the left side. (a) Will any light be transmitted through the sheet on the right? (b) Suppose a third Polaroid sheet is placed between the two sheets shown in the figure. Could any light possibly be transmitted through the sheets? (c) What are the orientations of the middle sheet s transmission axis that will result in no light being transmitted through all three sheets? 13 Suppose you are swimming underwater. Is the light that reaches your eyes partially polarized? Explain. Fig A beam of sunlight is reflected by a store window. The angle of incidence is 56. Will polarized sunglasses reduce the glare? 15 Suppose you are driving a car in the late afternoon, look - ing directly into the sun. Would Polaroid sunglasses be more effective than ordinary sunglasses in reducing the intensity of direct sunlight? 16 An intense unpolarized laser beam is incident on a horizontal glass surface at Brewster s angle. The reflected beam then strikes a Polaroid sheet. The sheet will heat up most if the transmission axis is (a) horizontal; (b) ver - tical; (c) at an angle of 45 relative to the vertical. 17 (a) What is the color of the sky on the moon? (b) What is the color of the setting sun seen from the moon? 18 Is light scattered by a cloud polarized? Answers to Odd-Numbered Questions 1 (a) yes; (b) no; 3 No. The coherence length of white light is much less than the thickness of the glass; 5 The total energy equals the sum of the energy in the separate waves, but energy is redistributed, greater where interference is con structive and less where it is destructive; 7 b; 9 width; 11 No. All directions perpendicular to the direction of wave motion are physically equivalent. There is nothing to distinguish a polarization direction; 13 Yes, slightly. The component of polarization parallel to the surface is somewhat weaker, since this component is reflected more strongly; 15 no; 17 (a) black; (b) white Problems (listed by section) 26 1 Wave Properties of Light 1 TV channel 2 broadcasts over a frequency range from 54 MHz to 60 MHz. Suppose that at t 0 at a certain point in space the electric field of the station s signal is 0.1 N/C, directed upward, its peak value. Can you estimate the electric field at that point at (a) t 1 s; (b) t 1 ns? 2 VHF TV channels have a bandwidth of 6 MHz. (a) What is the coherence length of a VHF signal? (b) Would it be possible for TV signals to form an interference pattern by superposition of a direct signal and one reflected from a hill, with the reflected wave traveling an additional 1000 m? *3 A CO 2 laser produces infrared radiation with a wavelength of 10.6 m and a bandwidth of 10 5 nm. What is the coherence length of the radiation? 4 The coherence length of an ordinary white light source can be increased if we place a color filter in front of the source, so that the light that passes through the filter is somewhat monochromatic. The minimum wavelength of the emerging light is 600 nm. What is the maximum wavelength in order for the coherence length to be mm? 5 A laser beam of intensity W/m 2 with a wavelength of 600 nm is incident on an opaque screen with a tiny circular aperture of radius m. Apply the Huygens-Fresnel principle to find the intensity of the diffracted light at a distance of 1.00 cm from the center of the aperture.

30 760 CHAPTER 26 Wave Optics 26 2 Interference 6 Two thin slits separated by mm are illuminated by light from a He-Ne laser ( 633 nm), producing interference fringes on a distant screen. Find the angle between the centers of the central bright fringe and the next bright fringe. 7 Two thin slits separated by 0.15 mm are illuminated by a monochromatic plane wave, producing interference fringes on a distant screen. If the angle between adjacent fringes is rad, what is the color of the fringes? 8 Blue interference fringes are formed on a screen 2.0 m away from a double slit illuminated by monochromatic light of wavelength 480 nm. The distance between the centers of adjacent fringes is 4.0 mm. Find the separation between the two slits. 9 Two narrow slits separated by 0.40 mm are illuminated by monochromatic light of wavelength 500 nm. How many bright fringes can be seen on a screen 1.0 cm wide placed 1.0 m in front of the slits? 10 How far is the second dark fringe to the right of the cen - tral bright fringe in the double-slit interference pattern seen on a screen 3 m from the slits? The slits are separated by 0.3 mm and are illuminated by monochromatic light of wavelength 600 nm. *11 The experiment described in Example 2a is performed underwater. Find the separation between the centers of the bright fringes. What color are the fringes? *12 Two narrow slits, separated by 0.10 mm, are illuminated by white light. Is it possible to see interference fringes? Explain. 13 Two narrow slits, illuminated by light consisting of two distinct wavelengths, produce two overlapping colored interference patterns on a distant screen. The center of the fourth bright fringe in one pattern coincides with the center of the third bright fringe in the other pattern. What is the ratio of the two wavelengths? 14 White light is reflected at normal incidence from a 250 nm thick oil film of index 1.40 (Fig ). Light of which vacuum wavelength will experience destructive interference? 15 A layer of oil, with a refractive index of , exactly 1 mm thick, floats on water. (a) Light of vacuum wavelength nm, emitted by a stabilized He-Ne laser, is reflected at normal incidence by the oil. How many wavelengths are con tained in the light wave passing back and forth through the oil? (b) Does the light reflected from the two surfaces of the oil interfere constructively or destructively? *16 White light is reflected at normal incidence by a thin film of water on glass. Both 500 nm and 700 nm light reflected from the two surfaces interferes constructively. Find the minimum thickness of the film. 17 White light is normally incident on two glass microscope slides with a thin layer of air of variable thickness between the slides, as seen in Fig Which wavelengths interfere constructively when reflected at a point where the air layer is 800 nm thick? **18 An interference pattern, consisting of alternating bright and dark rings and known as Newton s rings, is formed by interference of monochromatic light reflected from the two air-glass surfaces between a convex lens and a glass plate (Fig ). (a) Show that the radius of the mth bright ring is given by r R(m 1 ) 1 (m 1 2 )2 2 2 where R is the radius of curvature of the convex surface. (b) Find the radius of the tenth ring for light of wavelength 550 nm if R 200 cm. Fig Newton s rings. 19 A human hair is placed between the ends of two microscope slides, forming a thin wedge of air between the slides, as indicated in Fig An expanded He-Ne laser beam of wavelength 633 nm is incident on the slides from above. An observer viewing the slides from above sees a series of thin, straight, alternating light and dark fringes, with 120 bright fringes spread over the 4

31 Problems cm length of the slides. Find the diameter of the hair. Fig *20 The colors seen in the photo in Fig are the result of interference of light reflected from the surfaces of a thin film produced when a wire frame is dipped into soapy water. A white light source in front of the film illuminates it at normal incidence. The liquid tends to drain toward the bottom, and so there is a gradual increase in thickness from top to bottom. At the instant the photo was made, the film thickness at the top was much less than the wavelength of light, and the film was about to break. (c) Find a film thickness such that yellow light of wavelength 575 nm interferes constructively and light near either end of the spectrum interferes destructively. *21 A Michelson interferometer is shown in Fig a. Fig b shows the main elements of the interferometer. A light wave incident from the left is split into two separate waves by a half-silvered mirror M h inclined at an angle of 45 relative to the incident beam. Half of the incident light is transmitted, while the other half is reflected by M h, producing two perpendicular beams of equal intensity. Mirrors M 1 and M 2 reflect the respective beams back to M h, which then reflects half and transmits half of each beam. Half of the light reflected from M 1 is transmitted through M h and falls on the screen, while half the light reflected from M 2 is reflected by M h and falls on the screen. Thus the two beams come together again, after having followed different paths. Mirror M 2 is fixed, but one may move mirror M 1 by turning a micro - meter screw, so that the path length 1 can be varied precisely. Do the light waves interfere constructively or destructively if the value of 1 2 is (a) 275 nm (using white light); (b) mm (using white light); (c) mm (using He-Ne laser light)? (a) Fig Reflection of white light by a thin soap film. (a) Why is the top of the film black? (b) Just beneath the dark section of the film is a section that appears white, indicating constructive interference for all wavelengths. Show that if the film is 100 nm thick the difference in path lengths of the two reflected waves is such that all visible wavelengths come closer to satisfying the condition for constructive interference than the condition for destructive interference. (b) Fig Michelson interferometer.

32 762 CHAPTER 26 Wave Optics **22 In the Michelson interferometer, described in the last problem, suppose that mirror M 1 is rotated by a small angle, as indicated in Fig a. This increases the path length traveled by the left side of the beam reflected from M 1. Assume that the right side of the beam interferes constructively, so that a bright fringe appears at point P on the screen. (a) Find an equation expressing the condition for con - structive interference at point Q at the opposite edge of the beam. (b) Find the number of fringes appearing on the screen, using an expanded He-Ne laser beam of width 10.0 cm, when rad (Fig b). (a) (b) Fig Interference fringes produced by a Michelson interferometer Diffraction 23 (a) Find the angle locating the first minimum in the Fraunhofer diffraction pattern of a single slit of width mm, using light of wavelength 400 nm. (b) Find the angle locating the second minimum. 24 How many diffraction maxima are contained in a region of the Fraunhofer single-slit pattern, subtending an angle of 2.00, for a slit width of mm, using light of wavelength 580 nm? 25 A Fraunhofer diffraction pattern of a single slit of width mm is produced using light of wavelength 600 nm. Find the angle subtended by the central diffraction maximum. 26 At what distance from a single slit of width mm should a screen be placed so that the width of the Fraunhofer central diffraction maximum is 15.0 mm (100 times the slit width), using light of wavelength 500 nm? 27 A tall narrow doorway of width 0.8 m opens into a large empty room. Plane sound waves of wavelength are normally incident on the doorway from the outside. In some directions within the room you may hear no sound. Find the angles specifying these directions (measured relative to the perpendicular to the doorway) for a wavelength of (a) 20.0 cm ( f 1720 Hz); (b) 78.0 cm ( f 440 Hz); (c) 1.00 m ( f 344 Hz)? (Assume that you can ignore the effect of sound reflected by the floor of the room and that you are far enough inside the room that the conditions of Fraunhofer diffraction apply.) 28 Water waves of wavelength 10.0 m are normally incident on a 30.0 m wide entrance to a harbor channel. Find the angle, relative to the direction of motion of the incident waves, corresponding to the first diffraction minimum. *29 Light consisting of two wavelengths (400 nm and 600 nm) is used to produce overlapping Fraunhofer diffraction patterns of a single slit of width 2.00 mm. Find an angle at which diffraction minima for the two wavelengths coincide. 30 Find the angle subtended by the Airy disk in the Fraunhofer diffraction pattern of a circular aperture of diameter mm illuminated by light from a He-Ne laser ( 633 nm). 31 A monochromatic plane wave of wavelength 500 nm is incident on an opaque screen with a pinhole of diameter mm. Find the diameter of the Airy disk on a wall 5.00 m in front of the pinhole.

33 Problems 763 *32 The planet Venus has a diameter of 12,100 km and its distance from the earth is km at its closest approach to the earth. On the basis of diffraction theory, should the human eye, with the pupil fully dilated to a diameter of 6.0 mm, be able to distinguish Venus from a star because of the larger angle subtended by Venus? Use 550 nm for the vacuum wavelength of light and 1.34 for the refractive index of the eye. (Actually there are other limits to the visual acuity of the eye, which normally cannot resolve points with an angular separation of less than about rad. See Section 25 5.) 33 You are looking at starlight of wavelength 600 nm with your pupils fully dilated to a diameter of 6.00 mm. Find the diameter of the corresponding Airy disk on each retina. The distance from pupil to retina is approximately 21.0 mm, and the refractive index within the eye is You are driving at night on a long, straight highway in the desert as another car approaches. What is the max - imum distance at which you can tell that the car is not a motorcycle by seeing its two headlights, which are 1.5 m apart? (a) Assume that your visual acuity is limited only by diffraction. Use 550 nm for the vacuum wavelength, and assume that your pupils are dilated to a 6.0 mm diameter. (b) Assume a more realistic, typical visual acuity with min rad. 35 On the basis of diffraction theory, calculate the mini - mum angle between objects barely resolvable by the human eye for the following values of the pupil diameter D and vacuum wavelength 0 : (a) D 2.00 mm, nm; (b) D 2.00 mm, nm; (c) D 6.00 mm, nm; (d) D 6.00 mm, nm. 36 The diffraction pattern shown in Fig was pro - duced by a grating with 500 lines/cm illuminated by a He-Ne laser ( 633 nm). Find the angle between adjacent maxima. 37 A monochromatic plane wave is normally incident on a diffraction grating with lines/m. For what visible wavelengths would this grating produce a maxi - mum at 10.0? 38 Looking at a white light source through a diffraction grating with lines/m, you see a spectrum of colors (Fig ). Find the wavelength of the light seen in the first-order spectrum at an angle of The surface of a compact disk (CD) behaves as a kind of diffraction grating. When light is reflected from the sur - face, each of the closely spaced ridges on a CD behaves as the source of a spherical light wave. Except for the fact that light is reflected rather than transmitted, the effect is the same as for the transmission gratings described in the text; that is, one sees a spectrum of colors (Fig ). Suppose a beam of sunlight is normally incident on a CD. You see yellow light of wavelength 575 nm reflected at an angle of 30.0 relative to the normal. Find the number of ridges per mm on the CD Polarization 40 You look through a Polaroid sheet at a 100 W light bulb 3.00 m away. (a) Find the intensity of the light you see. (b) Next you place a second Polaroid sheet in front of the first sheet with an angle of 45.0 between the two transmission axes. What intensity do you observe? 41 Unpolarized light is incident on a series of two Polaroid sheets. The intensity of the light emerging from the second sheet is 10.0% of the intensity of the light incident on the first sheet. Find the angle between the transmission axes of the two sheets. **42 Initially light is incident on a series of two Polaroid sheets, with perpendicular transmission axes, so that no light emerges from the second sheet. Without changing these sheets, a third Polaroid sheet is inserted between the other two. Is it possible that light will now emerge from the series of three sheets? If so, what should be the angle between the transmission axes of the inserted sheet and the first sheet in order that the final intensity of the emerging light will be maximized? 43 Suppose you are underwater, wearing Polaroid sun - glasses. Looking upward toward the water s surface, you view the setting sun by seeing light rays from the sun that strike the water at an angle of incidence of 75. The intensity of the incident sunlight is 200 W/m 2. What is the intensity of the light you see? 44 Suppose you are photographing a quiet lake that reflects a mirrorlike image of trees on the opposite shore. You place a Polaroid sheet in front of your camera. What orientation of its transmission axis will increase the relative intensity of the reflected light compared to direct light from objects in the background, thereby making the reflected image appear brighter in the photograph?

34 764 CHAPTER 26 Wave Optics 45 Calculate Brewster s angle for light incident from glass into air if the glass has an index of 1.5. **46 Let denote Brewster s angle for light reflected from a surface when the light is incident from medium A onto medium B. Let denote Brewster s angle for light re - flected from the same surface but with the light incident from medium B onto medium A. Prove that Calculate Brewster s angle for light incident from water onto glass of refractive index 1.5. *48 An unpolarized laser beam enters a container of water. The beam is partially reflected from the water-glass surface, as indicated in Fig For what angle of incidence will this reflected beam be completely polarized? Additional Problems Fig A Michelson interferometer (described in Problem 21) can be used to accurately measure the wavelength of its light source. Mirror M 1 is moved a distance of mm. What is seen at a certain point on the screen changes from bright to dark to bright 542 times as the mirror is being moved. Find the wavelength of the source. *50 A certain diffraction grating, illuminated by a gas dis - charge tube, produces a diffraction pattern in which the higher-order spectra overlap. Both a violet line and an orange line appear side by side at an angle of approximately 40. How many lines per meter does the grating have? **51 In the double-slit interference pattern, what is seen is the interference of the diffracted waves produced by each of the slits. The brightest fringes are those that correspond to the central diffraction maximum of each slit. Suppose that each of the slits is mm wide, the slits centers are mm apart, and they are illuminated by a source of wavelength 600 nm. How many bright fringes will be seen in the central maximum? **52 The width of each opening in a diffraction grating is mm, and the distance between adjacent openings is mm. (a) Find the angle subtended by the central diffraction maximum of each opening when illuminated by a He-Ne laser ( 633 nm). (b) How many bright spots are seen in the central maximum? **53 Suppose you are trying to demonstrate diffraction by a single aperture in such a way that maximizes the angle subtended by the central diffraction maximum so that the effect is as dramatic as possible. Thus you want as small an aperture as possible. If, however, you make the aperture too small, there will be too little light to be seen. So there is a practical limit to how small you can make it. Suppose a circular aperture is illuminated by a He-Ne laser beam ( 633nm), with an intensity of W/m 2. The Airy disk, seen on a screen 10.0 cm in front of the aperture, contains 84.0% of the radiant energy passing through the aperture. In order to be visible, the average intensity of the light forming the Airy disk should be at least W/m 2. Find the minimum diameter of the aperture and the diameter of the corresponding Airy disk. *54 Find the minimum angular magnification of a microscope, such that an eye with normal acuity ( min rad) and a normal near point of 25 cm can distinguish points as close as possible with a light microscope. Assume 550 nm. 55 Estimate the maximum distance at which you should be able to recognize a familiar face, looking through binoculars that provide an angular magnification of 7.0. Sup - pose that to accomplish this you need to be able to resolve points 1.0 cm apart. Assume that in bright sunlight resolution is limited only by diffraction of light by your eye, which then has a pupil diameter of 2.0 mm. Use 550 nm for the vacuum wavelength of light. **56 (a) Estimate the minimum diameter for an aperture of a camera used to detect ground-based missiles from a spy satellite. (b) Estimate the minimum focal length of the camera s lens if the 9 m diameter of the grains in the photographic film are not to limit the resolution.