Tipler Mosca Physics Scientists Engineers Ch 33 Interference and Diffraction PDF

C H A P T E R 33 WHITE LIGHT IS REF...

C H A P T E R 33 WHITE LIGHT IS REFLECTED OFF A SOAP BUBBLE. WHEN LIGHT OF ONE Interference WAVELENGTH IS INCIDENT ON A THIN SOAP-AND-WATER FILM, LIGHT IS and Diffraction REFLECTED FROM BOTH THE FRONT AND THE BACK SURFACES OF THE FILM. IFTHE ORDER OF MAGNITUDE OF THE THICKNESS OF THE FILM IS ONE 33-1 Phase Difference and Coherence WAVELENGTH OF THE LIGHT, THE TWO REFLECTED LIGHT WAVES INTERFERE. 33-2 Interference in Thin Films IFTHE TWO REFLECTED WAVES ARE 180° 33-3 Two-Slit Interference Pattern OUT OF PHASE, THE REFLECTED WAVES INTERFERE DESTRUCTIVELY, SO THE NET 33-4 Diffraction Pattern of a Single Slit RESULT IS THAT NO LIGHT IS REFLECTED. IF WHITE LIGHT, WHICH * 33-5 Using Phasors to Add Harmonic Waves CONTAINS A CONTINUUM OF 33-6 Fraunhofer and Fresnel Diffraction WAVELENGTHS, IS INCIDENT ON THE THIN FILM, THEN THE REFLECTED WAVES 33-7 Diffraction and Resolution WILL INTERFERE DESTRUCTIVELY * 33-8 Diffraction Gratings ONLY FOR CERTAIN WAVELENGTHS, AND FOR OTHER WAVELENGTHS THEY WILL INTERFERE CONSTRUCTIVELY. THIS PROCESS PRODUCES THE nterference and diffraction are the important phenomena that distinguish COLORED FRINGES THAT YOU SEE IN I waves from particles.* Interference is the formation of a lasting intensity pat- THE SOAP BUBBLE. (Aaron Haupt/ tern by two or more waves that superpose in space. Diffraction is the bending Photo Researchers.) of waves around corners that occurs when a portion of a wavefront is cut off by a barrier or obstacle. Have you ever wondered In this chapter, we will see how the pattern of the resulting wave can be cal- ? if the phenomenon that produces the bands that you see in culated by treating each point on the original wavefront as a point source, the light reflected off a soap bubble according to Huygens’s principle, and calculating the interference pattern has any practical applications? resulting from these sources. (See Example 33-2.) * Before you study this chapter, you may wish to review Chapter 15 and Chapter 16, where the general topics of inter- ference and diffraction of waves are first discussed. 1141 1142 | CHAPTER 33 Interference and Diffraction 33-1 PHASE DIFFERENCE AND COHERENCE When two harmonic sinusoidal waves of the same frequency and wavelength but of different phase combine, the resultant wave is a harmonic wave whose amplitude de- pends on the phase difference. If the phase difference is zero, or an integer multiplied by 360°, the waves are in phase and interfere constructively. The resultant amplitude equals the sum of the two individual amplitudes, and the intensity (which is pro- portional to the square of the amplitude) is maximum. (If the amplitudes are equal and the waves are in phase, the intensity is four times that of either individual wave.) If the phase difference is 180° or any odd integer multiplied by 180°, the waves are out of phase and interfere destructively. The resultant amplitude is then the differ- ence between the two individual amplitudes, and the intensity is a minimum. (If the amplitudes are equal and the waves are 180° out of phase, the intensity is zero.) A phase difference between two waves is often the result of a difference in path lengths. When a light wave reflects from a thin transparent film, such as a soap bubble, the reflected light is a superposition of the light reflected from the front surface of the film and the light reflected from the back surface of the film. The additional distance traveled by the light reflected from the back surface is called the path-length difference between the two reflected waves. A path-length difference of one wavelength pro- duces a phase difference of 360°, which is equivalent to no phase difference at all. A path-length difference of one-half wavelength produces a 180° phase difference. In general, a path-length difference of ¢r contributes a phase difference d given by ¢r ¢r d 2p  360° 33-1 l l P H AS E D I F F E R E N C E D U E TO A PAT H - L E N G T H D I F F E R E N C E Example 33-1 Phase Difference (a) What is the minimum path-length difference that will produce a phase difference of 180° for light of wavelength 800 nm? (b) What phase difference will that path-length difference produce in light of wavelength 700 nm? PICTURE The phase difference is to 360° as the path-length difference is to the wavelength. SOLVE d ¢r (a) The phase difference d is to 360° as the path-length difference  360° l ¢r is to the wavelength l. We know that l  800 nm and d  d 180° 180°: ¢r  l (800 nm)  400 nm 360° 360° ¢r 400 nm (b) Set l  700 nm, ¢r  400 nm, and solve for d: d 360°  360°  206°  3.59 rad l 700 nm CHECK The Part (b) result is somewhat larger than 180°. This result is expected because 400 nm is longer than half of the 700-nm wavelength. Another cause of phase difference is the 180° phase change a wave sometimes undergoes upon reflection from a surface. This phase change is analogous to the inversion of a pulse on a string when it reflects from a point where the density suddenly increases, such as when a light string is attached to a heavier string or rope. The inversion of the reflected pulse is equivalent to a phase change of 180° for a sinusoidal wave (which can be thought of as a series of pulses). When light traveling in air strikes the surface of a medium in which light travels more slowly, such as glass or water, there is a 180° phase change in the reflected light.  Interference in Thin Films SECTION 33-2 | 1143 When light is traveling in the liquid wall of a soap bubble, there is no phase change in the light reflected from the surface between the liquid and the air. This situation is analogous to the reflection without inversion of a pulse on a heavy string at a point where the heavy string is attached to a lighter string. If light traveling in one medium strikes the surface of a medium in which light travels more slowly, there is a 180° phase change in the reflected light. P H AS E D I F F E R E N C E D U E TO R E F L E C T I O N As we saw in Chapter 16, interference of waves is observed when two or more co- herent waves overlap. Interference of overlapping waves from two sources is not ob- served unless the sources are coherent. Because the light from each source is usually the result of millions of atoms radiating independently, the phase difference between the waves from such sources fluctuates randomly many times per second, so two light sources are usually not coherent. Coherence in optics is often achieved by splitting the light beam from a single source into two or more beams that can then be combined to produce an interference pattern. The light beam can be split by reflecting the light from the two surfaces of a thin film (Section 33-2), by diffracting the beam through two small openings or slits in an opaque barrier (Section 33-3), or by using a single point source and its image in a plane mirror for the two sources (Section 33-3). Today, lasers are the most important sources of coherent light in the laboratory. Light from an ideal monochromatic source is a sinusoidal wave of infinite dura- tion, and light from certain lasers approaches this ideal. However, light from conven- tional monochromatic sources, such as gas discharge tubes designed for this purpose, consists of packets of sinusoidal light that are only a few million wavelengths long. The light from such a source consists of many such packets, each approximately the same length. The packets have essentially the same wavelength, but the packets differ in phase in a random manner. The length of the individual packets is called the coherence length of the light, and the time it takes one of the packets to pass a point in space is the coherence time. The light emitted by a gas discharge tube designed to produce monochromatic light has a coherence length of only a few millimeters. By comparison, some highly stable lasers produce light that has a coherence length many kilometers long. 33-2 INTERFERENCE IN THIN FILMS You have probably noticed the colored bands in a soap bubble or in the film on the surface of oily water. These bands are due to the interference of light reflected from the top and bottom surfaces of the film. The different colors arise because of varia- tions in the thickness of the film, causing interference for different wavelengths at different points. When waves traveling in a medium cross a surface where the wave speed changes, part of the wave is reflected and part is transmitted. In addition, the re- 1 2 flected wave undergoes a 180° phase shift upon reflection if the transmitted wave travels at a slower speed than do the incident and reflected waves. (This 180° phase shift is established for waves on a string in Section 15-4 of Chapter 15.) The reflected t Water wave does not undergo a phase shift upon reflection if the transmitted wave travels at a faster speed than do the incident and reflected waves. Consider a thin film of water (such as a small section of a soap bubble) of uniform thickness viewed at small angles with the normal, as shown in Figure 33-1. Part of the light is reflected from the upper air–water interface where it undergoes a 180° FIGURE 33-1 Light rays reflected from phase change. Most of the light enters the film and part of it is reflected by the bot- the top and bottom surfaces of a thin film are coherent because both rays come from the same tom water–air interface. There is no phase change in this reflected light. If the light is source. If the light is incident almost normally, nearly perpendicular to the surfaces, both the light reflected from the top surface the two reflected rays will be very close to each and the light reflected from the bottom surface can enter the eye. The path-length other and will produce interference. 1144 | CHAPTER 33 Interference and Diffraction difference between these two rays is 2t, where t is the thickness of the film. This path- Air n = 1.00 length difference produces a phase difference of (2t>l)360°, where l  l>n is the wavelength of the light in the film, n is the index of refraction of the film, and l is Water n = 1.33 the wavelength of the light in vacuum. The total phase difference between the two Glass n = 1.50 rays is thus 180° plus the phase difference due to the path-length difference. Destructive interference occurs when the path-length difference 2t is zero or a whole number of wavelengths l (in the film). Constructive interference occurs when the F I G U R E 3 3 - 2 The interference of light path-length difference is an odd number of half-wavelengths. reflected from a thin film of water resting on a glass surface. In this case, both rays undergo When a thin film of water lies on a glass surface, as in Figure 33-2, the ray that re- a change in phase of 180° upon reflection. flects from the lower water–glass interface also undergoes a 180° phase change, be- cause the index of refraction of glass (approximately 1.50) is greater than that of water (approximately 1.33). Thus, both the rays shown in the figure have undergone a 180° phase change upon reflection. The phase difference d between these rays is due solely to the path-length difference and is given by d  (2t>l)360°. When a thin film of varying thickness is viewed with monochromatic light, such as the yellow light from a sodium lamp, alternating bright and dark bands or lines, called interference fringes, are observed. The distance between a bright fringe and a dark fringe is that distance over which the film’s thickness t changes so that the path- length difference 2t changes by l>2. Figure 33-3a shows the interference pattern ob- served when light is reflected from an air film between a spherical glass surface and a plane glass surface in contact. These circular interference fringes are known as Newton’s rings. Typical rays reflected at the top and bottom of the air film are shown in Figure 33-3b. Near the point of contact of the surfaces, where the path-length difference be- tween the ray reflected from the upper glass–air interface and the ray reflected from the lower air–glass interface is approximately zero (it is small compared with the wave- length of light) the interference is destructive because of the 180° Glass phase shift of the ray reflected from Glass the lower air–glass interface. This (a) (b) central region in Figure 33-3a is Air film Extra path length therefore dark. The first bright fringe occurs at the radius at which F I G U R E 3 3 - 3 (a) Newton’s rings observed when light is reflected from a thin film of air the path-length difference is l>2, between a plane glass surface and a spherical glass surface. At the center, the thickness of the air film which contributes a phase differ- is negligible and the interference is destructive because of the 180° phase change of one of the rays ence of 180°. This adds to the phase upon reflection. (b) Glass surfaces for the observation of Newton’s rings shown in Figure 33-3a. shift due to reflection to produce a The thin film in this case is the film of air between the glass surfaces. (Courtesy of Bausch & Lomb.) total phase difference of 360°, which is equivalent to a zero phase differ- 1 2 ence. The second dark region occurs at the radius at which the path- length difference is l, and so on. Glass Slip of paper θ t Example 33-2 A Wedge of Air Glass A wedge-shaped film of air is made by placing a small slip of paper To edge x between the edges of two flat pieces of glass, as shown in Figure 33-4. Light of wavelength 500 nm is incident normally on the glass, and interference fringes are observed by reflection. If the angle u F I G U R E 3 3 - 4 The angle u, which is less than 0.02°, is made by the plates is 3.0  104 rad (0.017°), how many dark inter- exaggerated. The incoming and outgoing rays are virtually ference fringes per centimeter are observed? perpendicular to all air–glass interfaces.  Two-Slit Interference Pattern SECTION 33-3 | 1145 PICTURE We find the number of fringes per centimeter by finding the horizontal distance x to the mth fringe and solving for m>x. Because the ray reflected from the bottom plate un- dergoes a 180° phase shift, the point of contact (where the path-length difference is zero) will be dark. The mth dark fringe after the contact point occurs when 2t  ml, where l  l is the wavelength in the air film, and t is the plate separation at x, as shown in Figure 33-4. Because the angle u is small, we can use the small-angle approximation u  tan u  t>x. SOLVE 1. The mth dark fringe from the contact point occurs when 2t  ml  ml the path-length difference 2t equals m wavelengths: 2t m l t 2. The thickness t is related to the angle u: u x 2xu 3. Substitute t  xu into the equation for m: m l m 2u 2(3.0  104) 4. Calculate m>x:    1200 m1  12 cm1 x l 5.0  107 m CHECK The expression for the number of dark fringes per unit length in step 4 shows that the number per centimeter would decrease if light of a longer wavelength is used. This result is as expected. TAKING IT FURTHER We observe 12 dark fringes per centimeter. In practice, the number of fringes per centimeter, which is easy to count, can be used to determine the angle. Note that if the angle of the wedge is increased, the fringes become more closely spaced. PRACTICE PROBLEM 33-1 How many dark fringes per centimeter are observed if light of wavelength 650 nm is used? Figure 33-5a shows interference fringes produced by a wedge-shaped air film be- tween two flat glass plates, as in Example 33-2. Plates that produce straight fringes, such as those in Figure 33-5a, are said to be optically flat. To be optically flat, a sur- face must be flat to within a small fraction of a wavelength. A similar wedge-shaped air film formed by two ordinary glass plates yields the irregular fringe pattern in Figure 33-5b, which indicates that these plates are not optically flat. (a) One application of interference effects in thin films is in nonreflecting lenses, which are made by coating the surface of a lens with a thin film of a material that has an index of refraction equal to approximately 1.38, which is between the index of refraction of glass and that of air. The intensities of the light reflected from the top and bottom surfaces of the film are approximately equal, and because the re- flected rays undergo a 180° phase change at both surfaces there is no phase differ- ence due to reflection between the two rays. The thickness of the film is chosen to be 14 l  14 ln, where l is the wavelength, in vacuum, that is in the middle of the visible spectrum, so that there is a phase change of 180° due to the path-length dif- ference of l>2 for light of normal incidence. Reflection from the coated surface is thus minimized, which means that transmission through the surface is maximized. (b) 33-3 TWO-SLIT INTERFERENCE PATTERN FIGURE 33-5 (a) Straight-line fringes Interference patterns of light from two or more sources can be observed only if the from a wedge-shaped film of air, like that sources are coherent. The interference in thin films discussed previously can be shown in Figure 33-4. The straightness of the fringes indicates that the glass plates are observed because the two beams come from the same light source but are sepa- optically flat. (b) Fringes from a wedge-shaped rated by reflection. In Thomas Young’s famous 1801 experiment, in which he film of air between glass plates that are not demonstrated the wave nature of light, two coherent light sources are produced optically flat. (Courtesy T. A. Wiggins.) 1146 | CHAPTER 33 Interference and Diffraction by illuminating two very narrow parallel slits using a single light source. We saw in Chapter 15 that when a wave encounters a barrier that has a very small open- ing, the opening acts as a point source of waves (Figure 33-6). During Young’s experiment, diffraction causes each slit to act as a line source (which is equivalent to a point source in two dimensions). The interference pattern is observed on a screen far from the slits (Figure 33-7a). At very large distances from the slits, the lines from the two slits to some point P on the screen are approximately parallel, and the path-length difference is approximately d sin u, where d is the separation of the slits, as shown in Figure 33-7b. When the path- length difference is equal to an integral number of wavelengths, the interference is constructive. We thus have interference maxima at an angle um given by d sin um  ml m  0, 1, 2, Á 33-2 T WO - S L IT I N T E R F E R E N C E M A X I M A where m is called the order number. The interference minima occur at F I G U R E 3 3 - 6 Plane water waves in a d sin um  A m  Bl ripple tank encountering a barrier that has a 1 2 m  1, 2, 3, Á 33-3 small opening. The waves to the right of the barrier are circular waves that are concentric T WO - S L IT I N T E R F E R E N C E M I N I M A about the opening, just as if there were a point source at the opening. (From PSSC Physics, The phase difference d at a point P is related to the path-length difference d sin u by 2nd Edition, 1965. D. C. Heath & Co. and Education Development Center, Newton MA.) ¢r d sin u d 2p  2p 33-4 l l We can relate the distance ym measured along the screen from the central point to the mth bright fringe (see Figure 33-7b) to the distance L from the slits to the screen: ym tan um  L For small angles, tan u  sin u. Substituting ym >L for sin um in Equation 33-2 and solving for ym gives lL ym  m 33-5 d From this result, we see that for small angles the fringes are equally spaced on the screen. P y S1 θ d L S2 Interference Screen maxima S1 d θ d θ S2 d sin θ (a) (b) F I G U R E 3 3 - 7 (a) Two slits act as coherent sources of light for the observation of interference in Young’s experiment. Cylindrical waves from the slits overlap and produce an interference pattern on a screen. (b) Geometry for relating the distance y measured along the screen to L and u. When the screen is very far away compared with the slit separation, the rays from the slits to a point on the screen are approximately parallel, and the path-length difference between the two rays is d sin u.  Two-Slit Interference Pattern SECTION 33-3 | 1147 Example 33-3 Fringe Spacing from Slit Spacing Try It Yourself Two narrow slits separated by 1.50 mm are illuminated by yellow light from a sodium lamp that has a wavelength equal to 589 nm. Find the spacing of the bright fringes observed on a screen 3.00 m away. PICTURE The distance ym measured along the screen to the mth bright fringe is given by m Equation 33-5, where L  3.00 m, d  1.50 mm, and l  589 nm. Bright fringes 3 SOLVE θ 3 d  (589 nm)> (1.50 mm)  0.0004. Trigonometry Example 33-4 How Many Fringes? Conceptual Example Two narrow slits are illuminated by monochromatic light. If the distance between the slits is equal to 2.75 wavelengths, what is the maximum number of bright fringes that can be seen on a screen? (a) 1, (b) 2, (c) 3, (d) 4, (e) 5, (f) 6 or more PICTURE A bright fringe (constructive interference) exists at points on the screen for which the distance to the two slits differs by an integer multiplied by the wavelength. However, the maximum difference in distance possible is equal to the distance between the two slits. SOLVE 1. Find the maximum difference in At all points on the screen, the difference distance from points on the screen to in distance from the two slits is 2.75 the two slits: wavelengths or less. 2. A bright fringe (constructive Bright fringes exist on the screen at interference) exists at points on the screen for which the distance to the two slits differs by an integer multiplied by the wavelength: places where the difference in distance to the slits is 2 wavelengths, 1 wavelength, or zero wavelengths. ✓ CONCEPT CHECK 33-1 What is the maximum number 3. Count up the bright fringes. There is (e) 5 of dark fringes that can be seen the central maximum and two on on a screen? either side of the central maximum: 1148 | CHAPTER 33 Interference and Diffraction CALCULATION OF INTENSITY To calculate the intensity of the light on the screen at a general point P, we need to add two harmonic wave functions that differ in phase.* The wave functions for elec- tromagnetic waves are the electric field vectors. Let E1 be the electric field at some point P on the screen due to the waves from slit 1, and let E2 be the electric field at that point due to waves from slit 2. Because the angles of interest are small, we can treat the fields as though they are parallel. Both electric fields oscillate with the same frequency (they result from a single source that illuminates both slits) and they have the same amplitude. (The path-length difference is only of the order of a few wave- lengths of light at most.) They have a phase difference d given by Equation 33-4. If we represent the wave functions by E1  A 0 sin vt and E2  A 0 sin(vt  d) the resultant wave function is E  E1  E2  A 0 sin vt  A 0 sin(vt  d) 33-6 By making use of the identity sin a  sin b  2 cos 12(a  b) sin 12(a  b) the resultant wave function is given by E  C 2A 0 cos 12 d D sin A vt  12 d B 33-7 The amplitude of the resultant wave is thus 2A 0 cos 12 d. It has its maximum value of 2A 0 when the waves are in phase and is zero when they are 180° out of phase. Because the intensity is proportional to the square of the amplitude, the intensity at any point P is I  4I0 cos2 12 d 33-8 I N T E N S IT Y I N T E R M S O F P H AS E D I F F E R E N C E where I0 is the intensity of the light reaching the screen from either slit separately. The phase angle d is related to the position on the screen by d  (d sin u>l)2p (Equation 33-4). Figure 33-9a shows the intensity pattern as seen on a screen. A graph of the in- tensity as a function of sin u is shown in Figure 33-9b. For small u, this graph is equiv- alent to a plot of intensity versus y (because y  L tan u  L sin u). The intensity I0 is (a) the intensity from each slit separately. The dashed line in Figure 33-9b shows the av- erage intensity 2I0 , which is the result of averaging over a distance containing many Intensity interference maxima and minima. This is the intensity that would arise from the two sources if they acted independently without interference, that is, if they were not co- 4I0 herent. Then, the phase difference between the two sources would fluctuate ran- domly, so that only the average intensity would be observed. Iav = 2I0 Figure 33-10 shows another method of producing the two-slit interference pattern, an arrangement known as Lloyd’s mirror. A monochromatic horizontal line source is λ 2λ sin θ placed at a distance 12 d above the plane of a mirror. Light striking the screen directly d d from the source interferes with the light that is reflected from the mirror. The reflected light can be considered to come from the virtual image of the line source formed by (b) the mirror. Because of the 180° change in phase upon reflection at the mirror, the in- terference pattern is that of two coherent line sources that differ in phase by 180°. The FIGURE 33-9 (a) The interference pattern pattern is the same as that shown in Figure 33-9 for two slits, except that the maxima observed on a screen far away from the two slits shown in Figure 33-7. (b) Plot of intensity versus and minima are interchanged. Constructive interference occurs at points for which the sin u. The maximum intensity is 4I0 , where I0 is path-length difference is a half-wavelength or any odd number of half-wavelengths. the intensity due to each slit separately. The At those points, the 180° phase difference due to the path-length difference combines average intensity (dashed line) is 2I0. with the 180° phase difference of the sources to produce constructive interference. (Courtesy of Michael Cagnet.) * We did this in Chapter 16 where we first discussed the superposition of two waves.  Diffraction Pattern of a Single Slit SECTION 33-4 | 1149 Light Screen source d Mirror F I G U R E 3 3 - 1 0 Lloyd’s mirror for Image producing a two-slit interference pattern. The of two sources (the source and its virtual image) source are coherent and are 180° out of phase. PRACTICE PROBLEM 33-2 A point source of light (l  589 nm) is placed 0.40 mm above the surface of a glass mirror. Interference fringes are observed on a screen 6.0 m away, and the interference is between the light reflected from the front surface of the glass and the light traveling from the source directly to the screen. Find the spacing of the fringes on the screen. The physics of Lloyd’s mirror was used in the early days of radio astronomy to determine the location of distant radio sources on the celestial sphere. A radio-wave receiver was placed on a cliff overlooking the sea, and the surface of the sea served as the mirror. 33-4 DIFFRACTION PATTERN OF A SINGLE SLIT In our discussion of the interference patterns produced by two or more slits, we as- sumed that the slits were very narrow so that we could consider the slits to be line sources of cylindrical waves, which in our two-dimensional diagrams are point sources of circular waves. We could therefore assume that the value of the intensity due to one slit acting alone was the same (I0) at any point P on the screen, indepen- dent of the angle u made between the ray to point P and the normal line between the slit and the screen. When the slit is not narrow, the intensity on a screen far away is not independent of angle but decreases as the angle increases. Consider a slit of width a. Figure 33-11 shows the intensity pattern on a screen far away from the slit of width a as a function of sin u. We can see that the intensity is maximum in the forward direction (sin u  0) and de- creases to zero at an angle that depends on the slit width a and the wavelength l. Most of the light intensity is concen- trated in the broad central diffraction maximum, although there are minor sec- (a) ondary maxima bands on either side of Intensity the central maximum. The first zeroes in the intensity occur at angles specified by sin u1  l>a 33-9 Note that for a given wavelength l, Equation 33-9 describes how variations in the slit width result in variations in the angular width of the central maximum. If we increase the slit width a, the angle u1 at which the intensity first becomes zero decreases, giving a more narrow central diffraction maximum. Conversely, if we (b) λ 2λ sin θ decrease the slit width, the angle of the a a first zero increases, giving a wider central diffraction maximum. When a is smaller FIGURE 33-11 (a) Diffraction pattern of a single slit as observed on a screen far away. than l, then sin u1 would have to exceed 1 (b) Plot of intensity versus sin u for the pattern in Figure 33-11a. (Courtesy of Michael Cagnet.) 1150 | CHAPTER 33 Interference and Diffraction to satisfy Equation 33-9. Thus, for a less than l, there are no points of zero intensity in the pattern, and the slit acts as a line source (a point source in two dimensions) ra- diating light energy essentially equal in all directions. Multiplying both sides of Equation 33-9 by a>2 gives 1 2a sin u1  12 l 33-10 1 2a 1 θ The quantity 2a sin u1 is the path-length difference between a light ray leaving the middle of the upper half of the slit and one leaving the middle of the lower half of θ the slit. We see that the first diffraction minimum occurs when these two rays are 180° out of phase, that is, when their path-length difference equals a half-wavelength. We can understand this result by considering each point on a wavefront to be a point source of light in accordance with Huygens’s principle. In Figure 33-12, we have placed a line of dots on the wavefront at the slit to represent these point sources schematically. Suppose, for example, that we have 100 such dots and that we look at an angle u1 for which a sin u1  l. Let us consider the slit to be divided into two halves, with sources 1 through 50 in the upper half and sources 51 through 100 in the F I G U R E 3 3 - 1 2 A single slit is represented lower half. When the path-length difference between the middle of the upper half by a large number of point sources of equal and the middle of the lower half of the slit equals a half-wavelength, the path-length amplitude. At the first diffraction minimum of a single slit, the waves from each point source in difference between source 1 (the first source in the upper half) and source 51 (the first the upper half of the slit are 180° out of phase source in the lower half) is also 12 l. The waves from those two sources will be out of with the wave from the point source a distance phase by 180° and will thus cancel. Similarly, waves from the second source in each a>2 lower in the slit. Thus, the interference from region (source 2 and source 52) will cancel. Continuing this argument, we can see each such pair of point sources is destructive. that the waves from each pair of sources separated by a>2 will cancel. Thus, there will be no light energy at that angle. We can extend this argument to the second and third minima in the diffraction pattern of Figure 33-11. At an angle u2 where a sin u2  2l, we can divide the slit into four regions, two regions for the top half and two regions for the bottom half. Using this same argument, the light intensity from the top half is zero because of the cancellation of pairs of sources; similarly, the light intensity from the bottom half is zero. The general expression for the points of zero intensity in the diffraction pattern of a single slit is thus y1 θ1 a sin um  ml m  1, 2, 3, Á 33-11 a L POINTS OF ZERO INTENSITY FOR A SINGLE-SLIT DIFFRACTION PATTERN Usually, we are just interested in the first occurrence of a minimum in the light in- tensity because nearly all of the light energy is contained in the central diffraction Screen maximum. In Figure 33-13, the distance y1 from the central maximum to the first diffraction F I G U R E 3 3 - 1 3 The distance y1 minimum is related to the angle u1 and the distance L from the slit to the screen by measured along the screen from the central maximum to the first diffraction minimum is related to the angle u1 by tan u1  y1 >L, y1 tan u1  L where L is the distance to the screen. Example 33-5 Width of the Central Diffraction Maximum During a lecture demonstration of single-slit diffraction, a laser beam that has a wavelength equal to 700 nm passes through a vertical slit 0.20 mm wide and hits a screen 6.0 m away. Find the width of the central diffraction maximum on the screen; that is, find the distance between the first minimum on the left and the first minimum on the right of the central maximum. PICTURE Referring to Figure 33-13, the width of the central diffraction maximum is 2y1. SOLVE y1 1. The half-width of the central maximum y1 is related to the tan u1  L angle u1 by:  Diffraction Pattern of a Single Slit SECTION 33-4 | 1151 2. The angle u1 is related to the slit width a by Equation 33-11: sin u1  l>a 2y1  2L tan u1  2L tana sin1 b l 3. Solve the step-2 result for u1 , substitute into the step-1 result, a and solve for 2y1: 700  109 m  2(6.0 m) tanasin1 b 0.00020 m  4.2  102 m  4.2 cm CHECK Because sin u1  l>a  (700 nm)>(0.20 mm)  0.0035, we can use the small-angle approximation to evaluate 2y1. In this approximation, sin u1  tan u1 , so l>a  y1 >L and 2y1  2Ll>a  2(6.0 m)(700 nm)>(0.20 mm)  4.2 cm. (This approximate value is in agree- ment with the exact value to within 0.0006 percent.) INTERFERENCE–DIFFRACTION PATTERN OF TWO SLITS When there are two or more slits, the intensity pattern on a screen far away (a) is a combination of the single-slit diffraction pattern of the individual slits and the multiple-slit interference pattern we have studied. Figure 33-14 shows the intensity pattern on a screen far from two slits whose separation d is 10a, where a is the width of each slit. The pattern is the same as the two- slit pattern that has very narrow slits (Figure 33-11) except that it is modu- I lated by the single-slit diffraction pattern; that is, the intensity due to each 4I0 slit separately is now not constant but decreases with angle, as shown in Figure 33-14b. Note that the central diffraction maximum in Figure 33-14 has 19 interfer- ence maxima—the central interference maximum and 9 maxima on either (b) side. The tenth interference maximum on either side of the central one is at 0 2λ 4λ 6λ 8λ 10λ sin θ d d d d d the angle u10 , given by sin u10  10l>d  l>a, because d  10a. This coincides with the position of the first diffraction minimum, so this interference maxi- F I G U R E 3 3 - 1 4 (a) Interference–diffraction mum is not seen. At these points, the light from the two slits would be in pattern for two slits whose separation d is equal phase and would interfere constructively, but there is no light coming from to 10 times their width a. The tenth interference either slit because the points are at diffraction minima of each slit. In general, maximum on either side of the central we can see that if m  d>a, the mth interference maximum will fall at the first interference maximum is missing because it falls diffraction minimum. Because the mth fringe is not seen, there will be m  1 at the first diffraction minimum. (b) Plot of fringes on each side of the central fringe for a total of N fringes in the central intensity versus sin u for the central band of the pattern in Figure 33-14a. (Courtesy of Michael Cagnet.) maximum, where N is given by N  2(m  1)  1  2m  1 33-12 Example 33-6 Interference and Diffraction Two slits that each have a width a  0.015 mm are separated by a distance d  0.060 mm and are illuminated by light of wavelength l  650 nm. How many bright fringes are seen in the central diffraction maximum? PICTURE We need to find the value of m for which the mth interference maximum coincides with the first diffraction minimum. Then there will be N  2m  1 fringes in the central maximum. SOLVE l 1. Relate the angle u1 of the first diffraction minimum to the slit sin u1  (first diffraction minimum) a width a: ml 2. Relate the angle um of the mth interference maxima to the slit sin um  (mth interference maxima) d separation d:  1152 | CHAPTER 33 Interference and Diffraction ml l 3. Set the angles equal and solve for m:  d a d 0.060 mm m   4.0 a 0.015 mm 4. The first diffraction minimum coincides with the fourth bright N 7 bright fringes fringe. Therefore, there are 3 bright fringes visible on either side of the central diffraction maximum. These 6 maxima, plus the central interference maximum, combine for a total of 7 bright fringes in the central diffraction maximum: * 33-5 USING PHASORS TO ADD HARMONIC WAVES To calculate the interference pattern produced by three, four, or more coherent light sources and to calculate the diffraction pattern of a single slit, we need to combine several harmonic waves of the same frequency that differ in phase. A sim- ple geometric interpretation of harmonic wave functions leads to a method of adding harmonic waves of the same frequency by geometric construction. Let the wave functions for two waves at some point be E1  A 1 sin a and E2  A 2 sin(a  d), where a  vt. Our problem is then to find the sum: E1  E2  A 1 sin a  A 2 sin(a  d) We can represent each wave function by the y component of a two-dimensional y vector, as shown in Figure 33-15. The geometric method of addition is based on the fact that the y component of the sum of two or more vectors equals the sum of the y components of the vectors, as illustrated in the figure. The wave function E1 S is represented by the y component of the vector A 1. As the time continues on, A sin (α + δ’ ) S A2 sin (α +δ ) this vector rotates in the xy plane with angular frequency v. The vector A 1 is called A a phasor. (We encountered phasors in our study of ac circuits in Section 29-5.) A2 α +δ The wave function E2 is the y component of a phasor of magnitude A 2 that makes A1 an angle a  d with the x axis. By the laws of vector addition, the sum of the y δ’ components of the individual phasors equals the y component of the resultant pha- α A1 sin α S x sor A , as shown in Figure 33-15. The y component of the resultant phasor, α+δ ’ A sin(a  d), is a harmonic wave function that is the sum of the two original wave functions. That is, Phasor representation of A 1 sin a  A 2 sin(a  d)  A sin(a  d) 33-13 FIGURE 33-15 wave functions. where A (the amplitude of the resultant wave) and d (the phase of the resultant wave relative to the phase of the first wave) are found by adding the phasors rep- resenting the waves. As time varies, a varies. The phasors representing the two wave functions and the resultant phasor representing the resultant wave function rotate in space, but their relative positions do not change because they all rotate with the same angular velocity v. Example 33-7 Wave Superposition Using Phasors Try It Yourself Use the phasor method of addition to derive E  C 2A 0 cos 12 d D sin A vt  12 d B (Equation 33-7) for the superposition of two waves of the same amplitude. PICTURE Represent the waves y1  A 0 sin a and y2  A 0 sin(a  d) by vectors (phasors) of length A 0 making an angle d with one another. The resultant wave yr  A sin(a  d) is represented by the sum of these vectors, which form an isosceles triangle, as shown in Figure 33-16.  Using Phasors to Add Harmonic Waves SECTION 33-5 | 1153 SOLVE y 1 Cover the column to the right and try these on your own before looking at the answers. 2A δ’ Steps Answers A0 1. Relate d and d using the theorem: “An external d  d  d angle to a triangle is equal to the sum of the two non- 1 δ adjacent internal angles.” 2A α 2. Solve for d. d  12 d δ’ A0 1 α 2A 3. Write cos d in terms of A and A 0. cos d  x A0 4. Solve for A in terms of d. A  2A 0 cos d  2A 0 cos 12 d FIGURE 33-16 5. Use your results for A and d to write the resultant yr  A sin(a  d) wave function.  C 2A 0 cos 12 d D sin A a  12 d B CHECK The step-5 result is identical to Equation 33-7 (see Problem statement). PRACTICE PROBLEM 33-3 Find the amplitude and phase constant of the resultant wave function produced by the superposition of the two waves E1  (4.0 V>m) sin(vt) and E2  (3.0 V>m) sin(vt  90°). *THE INTERFERENCE PATTERN OF THREE OR MORE EQUALLY SPACED SOURCES We can apply the phasor method of addition to calculate the interference pattern of three or more coherent sources that are equally spaced and in phase. We are most interested in the location of the interference maxima and minima. Figure 33- 17 illustrates the case of three such sources. The geometry is the same as for two sources. At a great distance from the sources, the rays from the sources to a point P on the screen are approximately parallel. The path- length difference between the first and second source is then d sin u, as before, and the path-length difference be- (a) P tween the first and third source is 2d sin u. The wave at point P is the sum of the three waves. Let a  vt be the S1 y d θ phase of the first wave at point P. We thus have the prob- S2 L lem of adding three waves of the form d S3 E1  A 0 sin a Screen E2  A 0 sin(a  d) 33-14 E3  A 0 sin(a  2d) (b) where 2p 2p yd d d sin u  33-15 l l L d θ as in the two-slit problem. θ At u  0, d  0, so all the waves are in phase. The am- d sin θ plitude of the resultant wave is 3 times that of each indi- d vidual wave and the intensity is 9 times that due to each 2d sin θ source acting separately. As the angle u increases from u  0, the phase angle d increases and the intensity de- creases. The position u  0 is thus a position of maximum F I G U R E 3 3 - 1 7 Geometry for calculating the intensity pattern far away from three equally spaced, coherent sources that are in phase. intensity. 1154 | CHAPTER 33 Interference and Diffraction Figure 33-18 shows the phasor addition of three waves for a phase angle d  30°  p>6 rad. This corresponds to a point P on the screen for which u is given δ by sin u  ld>(2pd)  l>(12d). The resultant amplitude A is considerably less than A0 3 times the amplitude A 0 of each source. As d increases, the resultant amplitude decreases until the amplitude is zero at d  120°. For this value of d, the three pha- sors form an equilateral triangle (Figure 33-19). This first interference minimum for A A0 three sources occurs at a smaller value of d (and therefore at a smaller angle u) than δ it does for only two sources (for which the first interference minimum occurs at d  180°). As d increases from 120°, the resultant amplitude increases, reaching a A0 secondary maximum at d  180°. At the phase angle d  180°, the amplitude is the same as that from a single source, because the waves from the first two sources α cancel each other, leaving only the third. The intensity of the secondary maximum is one-ninth that of the maximum at u  0. As d increases beyond 180°, the ampli- F I G U R E 3 3 - 1 8 Phasor diagram for tude again decreases and is zero at d  180°  60°  240°. For d greater than 240°, determining the resultant amplitude A due the amplitude increases and is again 3 times that of each source when d  360°. to three waves, each of amplitude A 0 , that This phase angle corresponds to a path-length difference of 1 wavelength for the have phase differences of d and 2d due to waves from the first two sources and 2 wavelengths for the waves from the first path-length differences of d sin u and 2d sin u. and third sources. Hence, the three waves are in phase at this point. The largest The angle a  vt varies with time, but this does not affect the calculation of A. maxima, called the principal maxima, are at the same positions as for just two sources, which are those points corresponding to the angles u given by d sin um  ml m  0, 1, 2, Á 33-16 δ = 120° These maxima are stronger and narrower than those for two sources. They occur A0 δ = 120° at points for which the path-length difference between adjacent sources is zero or A0 an integral number of wavelengths. These results can be generalized to more than three sources. For four coherent A0 sources that are equally spaced and in phase, the principal interference maxima α are again given by Equation 33-16, but the maxima are even more intense, they are narrower, and there are two small secondary maxima between each pair of prin- F I G U R E 3 3 - 1 9 The resultant amplitude cipal maxima. At u  0, the intensity is 16 times that due to a single source. The for the waves from three sources is zero when first interference minimum occurs when d is 90°, as can be seen from the phasor d is 120°. This interference minimum occurs at diagram of Figure 33-20. The first secondary maximum is near d  132°. The in- a smaller angle u than does the first minimum tensity of the secondary maximum is about one-fourteenth that of the central for two sources, which occurs when d is 180°. maximum. There is another minimum at d  180°, another secondary maximum near d  228°, and another minimum at d  270° before the next principal maxi- δ = 90° mum at d  360°. Figure 33-21 shows the intensity patterns for two, three, and four equally A0 spaced coherent sources. Figure 33-22 shows a graph of I>I0 , where I0 is the in- A0 δ = 90° tensity due to each source acting separately. For three sources, there is a very small secondary maximum between each pair of principal maxima, and the prin- δ = 90° A0 A0 cipal maxima are sharper and more intense than those due to just two sources. α For four sources, there are two small secondary maxima between each pair of principal maxima, and the principal maxima are even more narrow and intense. F I G U R E 3 3 - 2 0 Phasor diagram for the From this discussion, we can see that as we in- first minimum for four coherent sources that crease the number of sources, the intensity be- are equally spaced and in phase. The amplitude is zero when the phase difference comes more and more concentrated in the princi- of the waves from adjacent sources is 90°. pal maxima given by Equation 33-16, and these maxima become narrower. For N sources, the in- tensity of the principal maxima is N 2 times that F I G U R E 3 3 - 2 1 Intensity patterns for due to a single source. The first minimum occurs two, three, and four coherent sources that are at a phase angle of d  360°>N, for which the N equally spaced and in phase. There is a secondary maximum between each pair of phasors form a closed polygon of N sides. There principal maxima for three sources, and two are N  2 secondary maxima between each pair secondary maxima between each pair of of principal maxima. These secondary maxima are principal maxima for four sources. (Courtesy of very weak compared with the principal maxima. Michael Cagnet.)  Using Phasors to Add Harmonic Waves SECTION 33-5 | 1155 I/I0 Four sources Three sources Two sources F I G U R E 3 3 - 2 2 Plot of relative intensity 0 sin θ – λ λ versus sin u for two, three, and four coherent d d sources that are equally spaced and in phase. As the number of sources is increased, the principal maxima become sharper and more intense, and the intensities of the secondary maxima become negligible compared to those of the principal maxima. *CALCULATING THE SINGLE-SLIT DIFFRACTION PATTERN We now use the phasor method for the addition of harmonic waves to calculate the intensity pattern shown in Figure 33-11. We assume that the slit of width a is di- vided into N equal intervals and that there is a point source of waves at the mid- point of each interval (Figure 33-23). If d is the distance between two adjacent a sources and a is the width of the opening, we have d  a>N. Because the screen on θ which we are calculating the intensity is far from the sources, the rays from the sources to a point P on the screen are approximately parallel. The path-length dif- ference between any two adjacent sources is d sin u, and the phase difference d is related to the path-length difference by F I G U R E 3 3 - 2 3 Diagram for calculating the diffraction pattern far away from a narrow d sin u d 2p slit. The slit width a is assumed to contain a l large number of in-phase, equally spaced If A 0 is the amplitude due to a single source, the amplitude at the central maxi- point sources separated by a distance d. The rays from the sources to a point far away are mum, where u  0 and all the waves are in phase, is A max  NA 0 (Figure 33-24). approximately parallel. The path-length We can find the amplitude at some other point at an angle u by using the pha- difference for the waves from adjacent sources sor method for the addition of harmonic waves. As in the addition of two, three, or is d sin u. four waves, the intensity is zero at any point where the phasors representing the waves form a closed polygon. In this case, the polygon has N sides (Figure 33-25). At the first minimum, the wave from the first source just below the top of the open- ing and the wave from the source just below the middle of the opening are 180° out of phase. In this case, the waves from the source near the top of the opening differ 360° δ = N Amax = NA0 N F I G U R E 3 3 - 2 5 Phasor diagram for sources A0 calculating the first minimum in the single-slit diffraction pattern. When the waves from Screen the N sources completely cancel, the N phasors form a closed polygon. The phase difference between the waves from adjacent F I G U R E 3 3 - 2 4 A single slit is represented by N sources, each of amplitude A 0. At the sources is then d  360°>N. When N is very central maximum point, where u  0, the waves from the sources add in phase, giving a resultant large, the waves from the first and last sources amplitude A max  NA 0. are approximately in phase. 1156 | CHAPTER 33 Interference and Diffraction from those from the bottom of the opening by nearly 360°. [The phase difference A0 is, in fact, 360°  (360°>N). ] Thus, if the number of sources is very large, 360°>N is r A0 negligible and we get complete cancellation if the waves from the first and last sources are out of phase by 360°, corresponding to a path-length difference of one φ 2 A0 wavelength, in agreement with Equation 33-11. φ A A0 We will now calculate the amplitude at a general point at which the waves from 2 two adjacent sources differ in phase by d. Figure 33-26 shows the phasor diagram A0 r for the addition of N waves, where the subsequent waves differ in phase from A0 the first wave by d, 2d, Á , (N  1)d. When N is very large and d is very small, the A0 φ phasor diagram approximates the arc of a circle. The resultant amplitude A is A0 A0 A 0 the length of the chord of this arc. We will calculate this resultant amplitude in terms of the phase difference f between the first wave and the last wave. From F I G U R E 3 3 - 2 6 Phasor diagram for Figure 33-26, we have calculating the resultant amplitude due to A>2 the waves from N sources in terms of the sin 12 f  phase difference f between the wave from r the first source just below the top of the slit or and the wave from the last source just above the bottom of the slit. When N is very A  2r sin 12 f 33-17 large, the resultant amplitude A is the chord where r is the radius of the arc. Because the length of the arc is A max  NA 0 and of a circular arc of length NA 0  A max. the angle subtended is f, we have A max f 33-18 r or A max r f Substituting this into Equation 33-17 gives 2A max sin 12 f A sin 12 f  A max 1 f 2f Because the amplitude at the center of the central maximum (u  0) is A max , the ratio of the intensity at any other point to that at the center of the central maximum is 1  2  a 1 b I A2 sin 2 f 2 I0 A max 2f or sin 12 f I  I0 a b 2 1 33-19 2f I N T E N S IT Y F O R A S I N G L E - S L IT D I F F R AC T I O N PAT T E R N The phase difference f between the first and last waves is related to the path- length difference a sin u between the top and bottom of the opening by a sin u f 2p 33-20 l Equation 33-19 and Equation 33-20 describe the intensity pattern shown in Figure 33-11. The first minimum occurs at a sin u  l, which is the point where the waves from the middle of the upper half and the middle of the lower half of the slit have a path-length difference of l>2 and are 180° out of phase. The second minimum oc- curs at a sin u  2l, where the waves from the upper half of the upper half of the slit and those from the lower half of the upper half of the slit have a path-length difference of l>2 and are 180° out of phase.  Using Phasors to Add Harmonic Waves SECTION 33-5 | 1157 There is a secondary maximum approximately midway between the first and sec- 2 Circumference C = NA0 ond minima at a sin u  32 l. Figure 33-27 shows the phasor diagram for determining 3 2 the approximate intensity of this secondary maximum. The phase difference f = 3 Amax = π A between the first and last waves is approximately 2p  p. The phasors thus com- plete 1 12 circles. The resultant amplitude is the diameter of a circle that has a circum- A = 2 Amax A 3π ference which is two-thirds the total length A max. If C  23 A max is the circumference, 2 4 the diameter A is A = A2 9π 2 max 2 C 3 A max 2 A   A F I G U R E 3 3 - 2 7 Phasor diagram for p p 3p max calculating the approximate amplitude of the and first secondary maximum of the single-slit 4 2 diffraction pattern. The secondary maximum A2  A occurs near the midpoint between the first 9p2 max and second minima when the N phasors The intensity at this point is complete 1 12 circles. 4 1 I I  I 33-21 9p2 0 22.2 0 *CALCULATING THE INTERFERENCE–DIFFRACTION PATTERN OF MULTIPLE SLITS The intensity of the two-slit interference–diffraction pattern can be calculated from the two-slit pattern (Equation 33-8) where the intensity of each slit (I0 in that equation) is replaced by the diffraction pattern intensity due to each slit, I, given by Equation 33-19. The intensity for the two-slit interference–diffraction pattern is thus sin 12 f I  4I0 a b cos2 12 d 2 1 33-22 2f I N T E R F E R E N C E – D I F F R AC T I O N I N T E N S IT Y F O R T WO S L IT S where f is the difference in phase between rays from the top and bottom of each slit, which is related to the width of each slit by a sin u f 2p l and d is the difference in phase between rays from the centers of two adjacent slits, which is related to the slit separation by d sin u d 2p l In Equation 33-22, the intensity I0 is the intensity at u  0 due to one slit alone. Example 33-8 Five-Slit Interference–Diffraction Pattern Find the interference–diffraction intensity pattern for five equally spaced slits, where a is the width of each slit and d is the distance between adjacent slits. PICTURE First, find the interference intensity pattern for the five slits, assuming no angular variations in the intensity due to diffraction. To do this, first construct a phasor diagram to find the amplitude of the resultant wave in an arbitrary direction u. Intensity is proportional to the square of the amplitude. Next, correct for the variation of intensity with u by using the single-slit diffraction pattern intensity relation (Equation 33-20 and Equation 33-20). 1158 | CHAPTER 33 Interference and Diffraction SOLVE sin 12 f I  I0 a b 2 1. The diffraction pattern intensity I due to a slit of width a is 1 given by Equation 33-19 and Equation 33-20: 2f where 2p f a sin u l 2. The interference pattern intensity I is proportional to the square I  A2 of the amplitude A of the superposition of the wave functions where for the light from the five slits: A sin(a  d)  A 0 sin a  A 0 sin(a  d)  A 0 sin(a  2d)  A 0 sin(a  3d)  A 0 sin(a  4d) d sin u and where a  vt and d 2p l π − δ' 3. To solve for A, we construct a phasor diagram (Figure 33-28). d  b  d The amplitude A equals the sum of the projections of the so A0 individual phasors onto the resultant phasor: δ' b  d  d  2d  d  d δ A0 A β δ A0 δ FIGURE 33-28 β δ' A0 δ A0 π − δ' 4. To find d, we add the exterior angles. The sum of the exterior 2(p  d)  4d  2p ⇒ d  2d angles equals 2p (if you walk the perimeter of any polygon you rotate through the sum of the exterior angles, and you rotate through 2p radians): 5. Solve for A from the figure: A  2A 0 cos d  2A 0 cos b  A 0 6. Substitute for d using the step-4 result, and substitute for b using A  A 0(2 cos 2d  2 cos d  1) the relation b  d. (That b and d are equal follows from the theorem “If two parallel lines are cut by a transversal, the interior and exterior angles on the same side of the transversal are equal.”): 7. Square both sides to relate the intensities. Recall, I is A2  A 20(2 cos 2d  2 cos d  1)2 the intensity from a single slit, and A 0 is the amplitude from so a single slit: I  I(2 cos 2d  2 cos d  1)2 sin 12 f I  I0 a b (2 cos 2d  2 cos d  1)2 2 8. Substitute for I using the step-1 result: 1 2f a sin u d sin u where f  2p and d 2p l l CHECK If u  0, both f  0 and d  0. So, for u  0, step 5 becomes A  5A 0 and step 8 becomes I  52I0  25I0 as expected.  Fraunhofer and Fresnel Diffraction SECTION 33-6 | 1159 33-6 FRAUNHOFER AND As the screen is moved closer, FRESNEL DIFFRACTION the Fraunhofer pattern observed Diffraction patterns, like the single-slit pattern shown in Figure 33-11, that are ob- far from the slit... served at points for which the rays from an aperture or an obstacle are nearly paral- lel are called Fraunhofer diffraction patterns. Fraunhofer patterns can be observed at great distances from the obstacle or the aperture so that the rays reaching any point are approximately parallel, or they can be observed using a lens to focus par- allel rays on a viewing screen placed in the focal plane of the lens. The diffraction pattern observed near an aperture or an obstacle is called a Fresnel diffraction pattern. Because the rays from an aperture or an obstacle close to a screen cannot be considered parallel, Fresnel diffraction is much more difficult gradually changes to analyze. Figure 33-29 illustrates the difference between the Fresnel and the into... Fraunhofer patterns for a single slit.* Figure 33-30a shows the Fresnel diffraction pattern of an opaque disk. Note the bright spot at the center of the pattern caused by the constructive interference of the light waves diffracted from the edge of the disk. This pattern is of some historical in- terest. In an attempt to discredit Augustin Fresnel’s wave theory of light, Siméon Poisson pointed out that it predicted a bright spot at the center of the shadow, which he assumed was a ridiculous contradiction of fact. However, Fresnel immediately the Fresnel pattern demonstrated experimentally that such a spot does, in fact, exist. This demonstration observed near convinced many doubters of the validity of the wave theory of light. The Fresnel dif- the slit. fraction pattern of a circular aperture is shown in Figure 33-30b. Comparing this with the pattern of the opaque disk in Figure 33-30a, we can see that the two patterns are F I G U R E 3 3 - 2 9 Diffraction patterns for a complements of each other. single slit at various screen distances. Figure 33-31a shows the Fresnel diffraction pattern of a straightedge illuminated by light from a point source. A graph of the intensity versus distance (measured along a line (a) perpendicular to the edge) is shown in Figure 33-31b. The light intensity does not fall abruptly to zero in the geometric shadow, but it decreases rapidly and is negligible within a few wavelengths of the edge. The Fresnel diffraction pattern (b) Intensity Geometric