PSE 3E Chapter 22 EOC Conceptual Questions Larry Smith

22 WAVE OPTICS

Conceptual Questions

22.1.

The initial light pattern is a double-slit interference pattern. It is centered behind the midpoint of the slits. The slight decrease in intensity going outward from the middle indicates that the light from each of the individual slits is not uniform but slowly decreases toward the edges of the screen. If the right slit is covered, light comes through only the left slit. Without a second slit, there is no interference. Instead, we get simply the spread-out pattern of light diffracting through a single slit, such as in the center of the photograph of Figure 22.2. The intensity is a maximum directly behind the left slit, and—as we discerned from the intensities in the double-slit pattern—the single-slit intensity fades gradually toward the edges of the screen.

mL 22.2. Because y  increasing  and L increases the fringe spacing. Increasing d decreases the fringe spacing. m d Submerging the experiment in water decreases and decreases the fringe spacing. So the answers are (a) and (c).

22.3. The fringe separation for the light intensity pattern of a double-slit is determined by y  L/d. (a) If  increases, y will decrease. (b) If d decreases, y will increase. (c) If L decreases, y will decrease. (d) Since the dot is in the m  2 bright fringe, the path length difference from the two slits is 2  1000 nm

22.4. (a) The equation for gratings does not contain the number of slits, so increasing the number of slits can’t affect the angles at which the bright fringes appear as long as d is the same. So the number of fringes on the screen stays the same. (b) The number of slits does not appear in the equation for the fringe spacing, so the spacing stays the same. (c) Decreases; the fringes become narrower. (d) The equation for intensity does contain the number of slits, so each 2 fringe becomes brighter: INImax 1

22.5.   a. Several secondary maxima appear. For apsinp  , the first minima from the central maximum require asin1  , which must be less than 1.

1 22 1 22 22.6. (a) The width increases because   . (b) The width decreases because   . (c) Almost 1 D 1 D uniformly gray with no minima.

22.7. When the experiment is immersed in water the frequency of the light stays the same. But the speed is slower, L so the wavelength is smaller. When the wavelength is smaller the fringes get closer together because y   d

22.8. (a) Because the hole is so large we do not use the wave model of light. The ray model says we treat the light as if it travels in straight lines, so increasing the hole diameter will increase the size of the spot on the screen. (b) When the hole is smaller than 1 mm we need to consider diffraction effects; making the hole 20% larger will decrease the diameter of the spot on the screen.

22.9. Moving a mirror by 200 nm increases or decreases the path length by 400 nm. Since one mirror is moved in and the other out each increases the path length difference by 400 nm. The total is a path length difference of 800 nm, so the spot will still be bright.

22.10. If the wavelength is changed to /2 there will still be crests everywhere there were crests before (plus new crests halfway between, where there used to be troughs). If the interferometer was set up to display constructive interference originally, then crests were arriving from each arm of the interferometer together at the detector. That will still be true when the wavelength is halved. So the central spot remains bright.

Exercises and Problems

Section 22.2 The Interference of Light

22.1. Model: Two closely spaced slits produce a double-slit interference pattern. Visualize: The interference pattern looks like the photograph of Figure 22.3(b). It is symmetrical with the m  3 fringes on both sides of and equally distant from the central maximum. Solve: The bright fringes occur at angles m such that

dsinm  m m  0, 1, 2, 3,  3(600 109 m) 180 sin33   0 0225 0 023 rad  0 023 rad   1 3 (80 106 m)  rad

22.2. Model: Two closely spaced slits produce a double-slit interference pattern. Visualize: The interference pattern looks like the photograph of Figure 22.3(b).

Solve: The bright fringes are located at positions given by Equation 22.4, dmsinm  . For the m  3 bright orange 9 fringe, the interference condition is d sin 3  3(600 10 m). For the m  4 bright fringe the condition is d sin4  4 Because the position of the fringes is the same, ddsin sin   4   3(600  1099 m)   3 (600  10 m)  450 nm 34 4

22.3. Model: Two closely spaced slits produce a double-slit interference pattern. Visualize: The interference pattern looks like the photograph of Figure 22.3(b). It is symmetrical, with the m  2 fringes on both sides of and equally distant from the central maximum.

Solve: The two paths from the two slits to the bright fringe differ by r  r21  r , where rm  2  2(500 nm)  1000 nm Thus, the position of the bright fringe is 1000 nm farther away from the more distant slit than from the nearer slit.

22.4. Model: Two closely spaced slits produce a double-slit interference pattern. Visualize: The interference pattern looks like the photograph of Figure 22.3(b). Solve: The formula for fringe spacing is LLL y  1  8  1039 m  (600  10 m)   3000 d d d The wavelength is now changed to 400 nm, and Ld/, being a part of the experimental setup, stays the same. Applying the above equation once again, L y  (400  109 m)(3000)  1  2 mm d

L 22.5. Visualize: The fringe spacing for a double slit pattern is y   We are given L  2.0 m and  600 nm d We also see from the figure that y 1 cm  3 Solve: Solve the equation for d. L (600 109 m)(2 0 m) d   0  36 mm y 1 2 3 10 m Assess: 0.36 mm is a typical slit spacing. 22.6. Model: Two closely spaced slits produce a double-slit interference pattern. Visualize: The interference pattern looks like the photograph of Figure 22.3(b). Solve: The fringe spacing is LL(589 1092 m)(150 10 m) yd     0  22 mm dy 4 0 103 m

22.7. Model: Two closely spaced slits produce a double-slit interference pattern. Visualize: The interference fringes are equally spaced on both sides of the central maximum. The interference pattern looks like Figure 22.3(b). Solve: In the small-angle approximation         (mm  1)   mm1 d d d Since d  200 , we have  1    rad  0  286  d 200

22.8. Model: Two closely spaced slits produce a double-slit interference pattern. Visualize: The interference pattern looks like the photograph of Figure 22.3(b). Solve: The dark fringes are located at positions given by Equation 22.9: L y  m 1 m 0, 1, 2, 3,  m  2  d 2 11LL 3 4  (60 10 m) yy515  1  6 0 10 m   500 nm 2dd 20 20 10 3 m Section 22.3 The Diffraction Grating 22.9. Model: A diffraction grating produces an interference pattern. Visualize: The interference pattern looks like the diagram in Figure 22.8. Solve: The bright constructive-interference fringes are given by Equation 22.15: dsinm  m m  0, 1, 2,  (1)(550 109 m) sin11   0  0275   1  6  (4 0 102 m)/2000

Likewise, sin22 0  055 and  3  2  .

22.10. Model: A diffraction grating produces a series of constructive-interference fringes at values of  m determined by Equation 22.15. Solve: We have

dsinm  m  m  0, 1, 2, 3,  d sin20  0   1  and d sin 2  2  Dividing these two equations,

sin22 2sin20  0   0  6840   43  2 

22.11. Model: A diffraction grating produces an interference pattern. Visualize: The interference pattern looks like the diagram in Figure 22.8. Solve: The bright constructive-interference fringes are given by Equation 22.15: 9 m (2)(600 10 m) 6 dsinm  m  d    1  89  10 m sin m sin(39 5 )

The number of lines in per millimeter is (1 1036 m)/(1  89  10 m)  530.

22.12. Model: A diffraction grating produces an interference pattern. Visualize: The interference pattern looks like the diagram of Figure 22.8. Solve: The bright interference fringes are given by

dmsinm  m  0, 1, 2, 3, 

The slit spacing is d 1 mm/500  2  00  106 m and m 1. For the red and blue light, 656 1099 m   486 10 m  sin11   19  15  sin    14  06  1 red66  1 blue   2 00  10 m   2  00  10 m 

The distance between the fringes, then, is y  y1 red  y 1 blue where y (1  5 m)tan19  15   0  521 m 1 red y1 blue (1  5 m)tan14  06   0  376 m So, y 0  145 m  14  5 cm 

22.13. Model: A diffraction grating produces an interference pattern. Visualize: The interference pattern looks like the diagram in Figure 22.8.

Solve: The bright fringes are given by Equation 22.15: dsinm  m  m  0, 1, 2, 3,  d sin 11  (1)   d   /sin 

The angle 1 can be obtained from geometry as follows: (0 32 m)/2 tan  0  080   tan1 (0.080) 4.57  112 0 m

Using sin1  sin4  57   0  07968, 633 109 m d  7  9  m 0 07968

22.14. Model: A diffraction grating produces a series of constructive-interference fringes at values of  m that are determined by Equation 22.15. Solve: For the m  3 maximum of the red light and the m  5 maximum of the unknown wavelength, Equation 22.15 gives 9 ddsin3 3(660  10 m) sin  5  5  unknown

The m  5 fringe and the m  3 fringe have the same angular positions. This means 53 . Dividing the two equations, 9 5unknown 3(660  10 m)  unknown  396 nm

Section 22.4 Single-Slit Diffraction

22.15. Model: A narrow single slit produces a single-slit diffraction pattern. Visualize: The intensity pattern for single-slit diffraction will look like Figure 22.14. Solve: The minima occur at positions L yp p a 2LLLL 1    (633 109 m)(1 5 m) So y y y a  2 0 104 m  0 20 mm 21 a a a y 0 00475 m

22.16. Model: A narrow slit produces a single-slit diffraction pattern. Visualize: The intensity pattern for single-slit diffraction will look like Figure 22.14. Solve: The width of the central maximum for a slit of width a  200 is 2L 2 (2 0 m) w    20 mm a 200

22.17. Visualize: We are given L 1 0 m and w  4000 . 2L Solve: Solve w  for a. a 2L 2 (1 0 m) 1 0 m a    0  50 mm w 4000 2000 Assess: This is a typical slit width.

2L 22.18. Visualize: Use Equation 22.22: w  We are given   600 nm and L  2.0 m. We see from the figure a that w 1  0 cm  Solve: Solve the equation for a. 2L 2(600 109 m)(2 0 m) a   0  24 mm w 0 010 m Assess: 0.24 mm is in the range of typical slit widths.

22.19. Model: A narrow slit produces a single-slit diffraction pattern. Visualize: The intensity pattern for single-slit diffraction will look like Figure 22.14. Solve: The width of the central maximum for a slit of width a  200 is 2L 2(500 109 m)(2 0 m) w   0  0040 m  4  0 mm a 0 0005 m

22.20. Model: The spacing between the two buildings is like a single slit and will cause the radio waves to be diffracted. Solve: Radio waves are electromagnetic waves that travel with the speed of light. The wavelength of the 800 MHz waves is 3 108 m/s   0  375 m 800 106 Hz To investigate the diffraction of these waves through the spacing between the two buildings, we can use the general condition for complete destructive interference: asinp  p  p  1, 2, 3,  where a is the spacing between the buildings. Because the width of the central maximum is defined as the distance between the two p 1 minima on either side of the central maximum, we will use and obtain the angular width  2 1 from

11  0 375 m  asin11     sin   sin    1  43  a  15 m  Thus, the angular width of the wave after it emerges from between the buildings is  2(1  43  ) 2 86  2 9

22.21. Model: The crack in the cave is like a single slit that causes the ultrasonic sound beam to diffract. Visualize:

Solve: The wavelength of the ultrasound wave is 340 m/s   0  0113 m 30 kHz Using the condition for complete destructive interference with p 1,

10 0113 m asin11     sin  2  165  0 30 m From the geometry of the diagram, the width of the sound beam is wy211  2(100 m  tan )  200 m  tan2  165   7  6 m Assess: The small-angle approximation is almost always valid for the diffraction of light, but may not be valid for the diffraction of sound waves, which have a much larger wavelength.

Section 22.5 Circular-Aperture Diffraction

22.22. Model: Light passing through a circular aperture leads to a diffraction pattern that has a circular central maximum surrounded by a series of secondary bright fringes.

Solve: The width of the central maximum for a circular aperture of diameter D is 2 44L (2  44)(500  109 m)(2  0 m) w    4.9 mm D 0 50 103 m

22.23. Model: Light passing through a circular aperture leads to a diffraction pattern that has a circular central maximum surrounded by a series of secondary bright fringes. Visualize: The intensity pattern will look like Figure 22.15. Solve: According to Equation 22.23, the angle that locates the first minimum in intensity is 1 22 (1  22)(2.5  106 m) 1   0.01525 rad  0.874  D 0 20 103 m These should be rounded to 0.015 rad 0.87 .

22.24. Model: Light passing through a circular aperture leads to a diffraction pattern that has a circular central maximum surrounded by a series of secondary bright fringes. Visualize: The intensity pattern will look like Figure 22.15. Solve: From Equation 22.24, Dw (0 12  1032 m)(1  0  10 m) L    78 cm 2 44 2 44(633 109 m)

22.25. Model: Light passing through a circular aperture leads to a diffraction pattern that has a circular central maximum surrounded by a series of secondary bright fringes. Visualize: The intensity pattern will look like Figure 22.15. Solve: From Equation 22.24, the diameter of the circular aperture is 2 44L 2  44(633  109 m)(4  0 m) D   0  25 mm w 2 5 102 m

Section 22.6 Interferometers

Model: An interferometer produces a new maximum each time L increases by 1  causing the path-length 22.26. 2 2 difference r to increase by . Visualize: Please refer to the interferometer in Figure 22.20. Solve: From Equation 22.33, the number of fringe shifts is 2L 2(1  00  102 m) m 2   30,467  656 45 109 m

22.27. Model: An interferometer produces a new maximum each time increases by causing the path-length difference to increase by . Visualize: Please refer to the interferometer in Figure 22.20. Solve: From Equation 22.33, the wavelength is 2L 2(100 106 m)  2  4  0  107 m  400 nm m 500

22.28. Model: An interferometer produces a new maximum each time increases by causing the path-length difference to increase by . Visualize: Please refer to the interferometer in Figure 22.20. Solve: From Equation 22.33, the distance the mirror moves is m (33,198)(602  446  109 m) L   0  0100000 m  1  00000 cm 2 22 Assess: Because the wavelength is known to 6 significant figures and the fringes are counted exactly, we can determine L to 6 significant figures.

Model: An interferometer produces a new maximum each time L increases by 1  causing the path-length 22.29. 2 2 difference r to increase by . Visualize: Please refer to the interferometer in Figure 22.20.

Solve: For sodium light of the longer wavelength ()1 and of the shorter wavelength (2 ),  L  m12  L ( m  1) 22

We want the same path difference 2(LL21 ) to correspond to one extra wavelength for the sodium light of shorter wavelength (2 ). Thus, we combine the two equations to obtain:

1  2  2 589 0 nm m( m 1) m (1  2 )  2 m   981 67 982 2 212 589  6 nm  589  0 nm

Thus, the distance by which M2 is to be moved is 1 589 6 nm Lm  982  0  2895 mm 22

22.30. Model: Two closely spaced slits produce a double-slit interference pattern with the intensity graph looking like Figure 22.3(b). The intensity pattern due to a single slit diffraction looks like Figure 22.14. Both the spectra consist of a central maximum flanked by a series of secondary maxima and dark fringes. Solve: (a) The light intensity shown in Figure P22.30 corresponds to a double-slit aperture. This is because the fringes are equally spaced and the decrease in intensity with increasing fringe order occurs slowly. (b) From Figure P22.30, the fringe spacing is y 1  0 cm  1  0  102 m  Therefore, LL(6 00  109 m)(2  5 m) yd     0  15 mm dy0 010 m

22.31. Model: Two closely spaced slits produce a double-slit interference pattern with the intensity graph looking like Figure 22.3(b). The intensity pattern due to a single slit diffraction looks like Figure 22.14. Both the spectra consist of a central maximum flanked by a series of secondary maxima and dark fringes. Solve: (a) The light intensity shown in Figure P22.31 corresponds to a single slit aperture. This is because the central maximum is twice the width and much brighter than the secondary maximum. (b) From Figure P22.31, the separation between the central maximum and the first minimum is 2 y1 1.0 cm  1.0  10 m. Therefore, using the small-angle approximation, Equation 22.21 gives the condition for the dark minimum: pL L (2 5 m)(600 109 m) ya    0  15 mm p 2 dy1 1 0 10 m

22.32. Model: Two closely spaced slits produce a double-slit interference pattern. Visualize: The interference pattern looks like the photograph of Figure 22.3(b). Solve: In a span of 12 fringes, there are 11 gaps between them. The formula for the fringe spacing is L 52 1039 m (633  10 m)(3  0 m) y    d  0.40 mm  dd11 Assess: This is a reasonable distance between the slits, ensuring dL/ 1  34  104 1.

22.33. Solve: According to Equation 22.7, the fringe spacing between the m fringe and the m  1 fringe is y L/. d y can be obtained from Figure P22.33. The separation between the m  2 fringes is 2.0 cm, implying that the separation between the two consecutive fringes is 1 (2.0 cm) 0.50 cm. Thus, 4 L d y (0  20  1032 m)(0  50  10 m) yL 0  50  102 m      167 cm d  600 109 m Assess: A distance of 167 cm from the slits to the screen is reasonable.

22.34. Solve: According to Equation 22.7, the fringe spacing between the m fringe and the m  1 fringe is y L/. d y can be obtained from Figure P22.33. Because the separation between the m  2 fringes is 2.0 cm, two consecutive fringes are y 1 (2  0 cm)  0  50 cm apart. Thus, 4 L d y (0  20  1032 m)(0  50  10 m) y 0  50  102 m      500 nm dL 2 0 m

22.35. Solve: The intensity of light of a double-slit interference pattern at a position y on the screen is

2 d Idouble 4 I 1 cos  y L where is the intensity of the light from each slit alone. At the center of the screen, that is, at II 1 . I1 y  0 m, 14 double 2 From Figure P22.33, Idouble at the central maximum is 12 mW/m . So, the intensity due to a single slit 2 is I1 3 mW/m

22.36. Prepare: The path length difference is affected by this rotation. See accompanying figure.

We are given  1  The new part of the path length difference is rd1 sin while the old piece is still

r2 sin , so the total path length difference must obey dm(sin sin  )  for constructive interference. However, looking at the central maximum, m  0, so sin  sin  This is independent of  and d. Solve: y Ltan  L tan[sin11 (  sin )]  (1.00 m)tan[sin (  sin1.0  )]   1 75 cm So the central maximum moves by 1.75 cm on the screen Assess: This seems like a large shift for a 1 rotation, but is probably reasonable for L 1m.

22.37. Model: Two closely spaced slits produce a double-slit interference pattern. The interference pattern is symmetrical on both sides of the central maximum. Visualize:

In figure (a), the interference of laser light from the two slits forms a constructive-interference central (m  0) fringe at P. The paths 1P and 2P are equal. When a glass piece of thickness t is inserted in the path 1P, the interference between the two waves yields the 10th dark fringe at P. Note that the glass piece is not present in figure (a). Solve: The number of wavelengths in the air-segment of thickness t is t m  1  The number of wavelengths in the glass piece of thickness t is t t nt m2    glass /n  The path length has thus increased by m wavelengths, where t m  m  m ( n  1) 21  From the 10th dark fringe to the 1st dark fringe is 9 fringes and from the dark fringe to the 0th bright fringe is one-half of a fringe. Hence, 1 19 19t 19 19 (633 109 m) m9 ( n 1) t   12 0 m 22 2 2(1)21510n    

22.38. Model: A diffraction grating produces an interference pattern. Visualize: The interference pattern looks like the diagram in Figure 22.8. Solve: 500 lines per mm on the diffraction grating gives a spacing between the two lines of d 1 mm/500 36 (1 10 m)/500  2  0  10 m. The wavelength diffracted at angle  m 30 in order m is d sin (2 0  106 m)sin30  1000 nm  m   m m m We’re told it is visible light that is diffracted at 30 , and the wavelength range for visible light is 400–700 nm. Only m  2 gives a visible light wavelength, so   500 nm.

22.39. Model: Each wavelength of light is diffracted at a different angle by a diffraction grating. Solve: Light with a wavelength of 501.5 nm creates a first-order fringe at y  21.90 cm. This light is diffracted at angle

121 90 cm 1 tan  23  65  50 00 cm

We can then use the diffraction equation dmsinm  , with m 1, to find the slit spacing:  501 5 nm d   1250 nm sin1 sin(23 65 ) The unknown wavelength creates a first order fringe at y  31.60 cm, or at angle

131 60 cm 1 tan  32  29  50 00 cm With the split spacing now known, we find that the wavelength is

d sin1  (1250 nm)sin(32  29  )  667  8 nm Assess: The distances to the fringes and the first wavelength were given to 4 significant figures. Consequently, we can determine the unknown wavelength to 4 significant figures.

22.40. Model: Assume the incident light is coherent and monochromatic. Solve: (a) Where the path length difference is  between adjacent slits there will be constructive interference from all three 2 slits, and INImax 1 applies. 2 IIImax39 1 1

(b) For the case where the path length difference is /2 adjacent slits will produce destructive interference, so the third slit will just give a total intensity of I1.

22.41. Model: We will assume that the listeners did not hear any other loud spots between the center and 1.4 m on each side, m 1. We’ll use the diffraction grating equations. First solve Equation 22.16 for 1 and insert it into Equation 22.15. We need the wavelength of the sound waves, and we’ll use the fundamental relationship for periodic waves to get it. v 340 m/s    3  4 cm  0  034 m f 10000 Hz

We are also given L 10 m and y1 1  4 m  Solve:

11y1 1 4 m 1 tan  tan  0  14 rad L 10 m (This may be small enough to use the small-angle approximation, but we are almost finished with the problem without it, so maybe we can save the approximation and try it in the assess step.) (1) 0 034 m d   0  25 m  25 cm sin1 sin(0 14 rad) Assess: These numbers seem reasonable given the size (wavelength) of sound waves. With the small-angle approximation,sin  tan  , m 1(0 034 m) d    24 cm yLm/ 1 4 m/10 m

This is almost the same, but rounded down to two significant figures rather than up. The 1 we computed seemed almost small enough to use the approximation, and it would probably be OK for many applications when the angle is this small.

22.42. Model: A diffraction grating produces an interference pattern. Visualize: The interference pattern looks like the diagram of Figure 22.8. Solve: (a) A grating diffracts light at angles sinm  md / . The distance between adjacent slits is d 1 mm/600 1 667  106 m  1667 nm  The angle of the m 1fringe is

1500 nm 1 sin  17  46  1667 nm The distance from the central maximum to the bright fringe on a screen at distance L is o yL11tan  (2 m)tan17  46  0  629 m (Note that the small angle approximation is not valid for the maxima of diffraction gratings, which almost always have angles 10 . ) There are two bright fringes, one on either side of the central maximum. The distance between them is yy 21  1  258 m  1  3 m 

(b) The maximum number of fringes is determined by the maximum value of m for which sinm does not exceed 1 because there are no physical angles for which sin  1. In this case, mm (500 nm) sin  m d 1667 nm We can see by inspection that m  1, m  2, and m  3 are acceptable, but m  4 would require a physically impossible sin 4  1 Thus, there are three bright fringes on either side of the central maximum plus the central maximum itself for a total of seven bright fringes.

22.43. Model: Assume the screen is centered behind the slit. We actually want to solve for m, but given the other data, it is unlikely that we will get an integer from the equations for the edge of the screen, so we will have to truncate our answer to get the largest order fringe on the screen.

Visualize: Refer to Figure 22.7. Use Equation 22.15: dmsinm  , and Equation 22.16: yLmmtan We are given   510 nm, L 2 0 m, and d 1 mm As mentioned above, we are not guaranteed that a bright fringe will 500 occur exactly at the edge of the screen, but we will kind of assume that one does and set ym 1 0 m; if we do not get an integer for m then the fringe was not quite at the edge of the screen and we will truncate our answer to get an integer m.

Solve: Solve Equation 22.16 for m and insert it in Equation 22.15. y   tan1 m m L Solve Equation 22.15 for m 1 d d11 ym 500 mm 1 0 m  m sin m  sin tan   sin  tan   1  8 L 510 nm  2 0 m  Indeed, we did not get an integer, so truncate 1.8 to get m 1 This means we will see three fringes, one for m  0, and one on each side for m  1  Assess: Our answer fits with the statement in the text: “Practical gratings, with very small values for d, display only a few orders.”

22.44. Model: A diffraction grating produces an interference pattern that is determined by both the slit spacing and the wavelength used. The visible spectrum spans the wavelengths 400 nm to 700 nm.

Solve: According to Equation 22.16, the distance ym from the center to the mth maximum is yLmm tan . The angle of diffraction is determined by the constructive-interference condition dmsinm  , where m 0, 1, 2, 3,

The width of the rainbow for a given fringe order is thus w yred y violet. The slit spacing is 1 mm 1 0 103 m d   1  6667  106 m 600 600 For the red wavelength and for the m 1order, 9 11 700 10 m d sin11 (1)     sin  sin  24  83  d 1 6667 106 m From the equation for the distance of the fringe,

yLredtan 1  (2  0 m)tan(24  83  )  92  56 cm Likewise for the violet wavelength, 400 109 m  sin1  13 88y (2 0 m)tan(13  88 ) 49 42 cm 16 violet 1 6667 10 m The width of the rainbow is thus 92.56 cm 49.42 cm  43.14 cm  43 cm.

22.45. Model: A diffraction grating produces an interference pattern that is determined by both the slit spacing and the wavelength used. Solve: (a) If blue light (the shortest wavelengths) is diffracted at angle  , then red light (the longest wavelengths) is diffracted at angle   30°. In the first order, the equations for the blue and red wavelengths are  sinb d sin(   30  )   d r Combining the two equations we get for the red wavelength, 2 bb d(sin  cos30  cos  sin30  ) d (0 8660sin   0 50cos  )  d 0  8660 d (0 50) 1 r d 2  d 2 b 2 2 2 2 (0 50)dd 1 r 0 8660  b  (0 50) (   b ) (  r 0 8660  b ) d 2

2 rb0 8660 2 d   b 0 50 9 9 7 Using b 400 10 m and r 700 10 m, we get d 8  125  10 m  This value of d corresponds to 1 mm 1 0 103 m 1230 lines/mm d 8 125 107 m (b) Using the value of d from part (a) and  589 109 m, we can calculate the angle of diffraction as follows: 79 d sin1 (1)   (8  125  10 m)sin  1  589  10 m   1  46  5 

22.46. Model: A diffraction grating produces an interference pattern that is determined by both the slit spacing and the wavelength used. Solve: An 800 line/mm diffraction grating has a slit spacing d (1  0  1036 m)/800  1  25  10 m  Referring to Figure P22.46, the angle of diffraction is given by y 0 436 m tan1  0 436  23 557 sin   0 400 1L 1 0 m 1 1

Using the constructive-interference condition dmsinm  , d sin  1 (1  25  106 m)(0  400)  500 nm 1

We can obtain the same value of  by using the second-order interference fringe. We first obtain 2: y 0 436 m  0  897 m tan2  1 333  53 12 sin   0 800 2L 1 0 m 2 2 Using the constructive-interference condition, d sin (1 25  106 m)(0  800)  2   500 nm 22 Assess: Calculations with the first-order and second-order fringes of the interference pattern give the same value for the wavelength.

22.47. Model: A diffraction grating produces an interference pattern that is determined by both the slit spacing and the wavelength used. Solve: From Figure P22.46, 0 436 m tan  0  436    23  557   sin   0  400 11 0 m 1 1

Using the constructive-interference condition dmsinm  , 600 109 m ddsin23 557   (1)(600  1096 m)    1  50  10 m sin(23 557 ) Thus, the number of lines per millimeter is 1 0 103 m  670 lines/mm 1 50 106 m Assess: The same answer is obtained if we perform calculations using information about the second-order bright constructive-interference fringe.

22.48. Model: A narrow slit produces a single-slit diffraction pattern. Visualize: The diffraction-intensity pattern from a single slit will look like Figure 22.14. Solve: The dark fringes are located at pL y p = 0, 1, 2, 3, p a

The locations of the first and third dark fringes are LL3 yy 13aa Subtracting the two equations, 2LL 2 2(589 109 m)(0 75 m) (y y )   a    0.12 mm 31 3 a y31 y 7 5 10 m

22.49. Visualize: The relationship between the diffraction grating spacing d, the angle at which a particular order of constructive interference occurs  m, the wavelength of the light, and the order of the constructive interference m is dmsinm  . Also note Nd1/ Solve: The first order diffraction angle for green light is 1  1  7  6  1 1 sin ( /d )  sin (5  5 10 m/2  0 10 m)  sin (0  275) 0 278 rad  16 Assess: This is a reasonable angle for a first order maximum.

22.50. Model: The peacock feather is acting as a reflection grating, so we may use Equation 22.15: dmsinm  , with m 1(we are told it is first-order diffraction),   470 nm, and1 15   0  262 rad  Solve: Solve Equation 22.15 for d m (1)(470 109 m) d   1  8 m sinm sin(0.262 rad) Assess: The answer is small, but plausible for the bands on the barbules.

22.51. Model: Assume the incident light is coherent and monochromatic.

Visualize: The distances given in the table are 2y1 since the measurement is between the two first order fringes. Also note that m 1 in this problem.

Solve: The equation for diffraction gratings is dmsinm  where d is the distance (in mm) between slits; we seek 1 , the number of lines per mm. d 1 sin 1 d From yL tan and y  distance , where “distance” is the distance between first order fringes as given in the mm1 2 table, we arrive at

1distance 1 sin tan   2Ld

1distance 1 This leads us to believe that a graph of sin tan vs.  will produce a straight line whose slope is and 2L d whose intercept is zero.

We see from the spreadsheet that the linear fit is excellent and that the slope is 799.84 mm1 and the intercept is very small. Because the number of lines per mm is always reported as an integer we round this answer to 800 lines/mm. Assess: This number of lines per mm is a typical number for a decent grating.

22.52. Model: Small angle approximations would not apply here. Visualize: We see from the figure that if yL1  then 1 45 .

Solve: From dmsinm  we get  sin  1 d 2 but  45 , and sin45 . 1 2 2   d 2 Assess: This large diffraction angle corresponds to a wavelength that is about the same size as the slit spacing.

22.53. Model: A narrow slit produces a single-slit diffraction pattern. Visualize: The diffraction-intensity pattern from a single slit will look like Figure 22.14. Solve: These are not small angles, so we can’t use equations based on the small-angle approximation. As given by

Equation 22.19, the dark fringes in the pattern are located at apsinp  , where p 1, 2, 3, For the first minimum of the pattern, p 1. Thus, ap 1  sin p sin 1 For the three given angles the slit width to wavelength ratios are a 1 a 1 a 1 (a):  2, (b):   1  15, (c):  1  30 sin30  60 sin60  90 sin90 Assess: It is clear that the smaller the a/ ratio, the wider the diffraction pattern. This is a conclusion that is contrary to what one might expect.

22.54. Model: A narrow slit produces a single-slit diffraction pattern. Visualize: The diffraction-intensity pattern from a single slit will look like Figure 22.14.

Solve: As given by Equation 22.19, the dark fringes in the pattern are located at apsinp  , where

For the diffraction pattern to have no minima, the first minimum must be located at least at 1 90 . From the constructive-interference condition apsinp  , we have p aa     633 nm sin p sin90

22.55. Model: A narrow slit produces a single-slit diffraction pattern. Visualize: The dark fringes in this diffraction pattern are given by Equation 22.21: pL yp 1,  2,  3,  p a We note that the first minimum in the figure is 0.50 cm away from the central maximum. We are given a  0.02 nm and L 1  5 m  Solve: Solve the above equation for  ya (0 50  1023 m)(0  20  10 m)  p   670 nm pL (1)(1 5 m) Assess: 670 nm is in the visible range.

22.56. Model: A narrow slit produces a single-slit diffraction pattern. Solve: The dark fringes in this diffraction pattern are given by Equation 22.21:

We note from Figure P22.55 that the first minimum is 0.50 cm away from the central maximum. Thus, ay (0 15  1032 m)(0  50  10 m) L p  1.3 nm p (1)(600 109 m) Assess: This is a typical slit to screen separation.

22.57. Model: Light passing through a circular aperture leads to a diffraction pattern that has a circular central maximum surrounded by a series of secondary bright fringes. Solve: Within the small angle approximation, which is almost always valid for the diffraction of light, the width of the central maximum is L wy2  2  44 1 D

From Figure P22.55, w 1.0 cm, so 2 44L 2.44(500 109 m)(1.0 m) D    0.12 mm w (1.0 102 m) Assess: This is a typical size for an aperture to show diffraction.

22.58. Model: Light passing through a circular aperture leads to a diffraction pattern that has a circular central maximum surrounded by a series of secondary bright fringes. Solve: Within the small-angle approximation, the width of the central maximum is L w 2 44 D Because wD , we have L DDLD2 44  2 44  (2 44)(633  109 m)(0  50 m)  0 88 mm D

22.59. Model: Light passing through a circular aperture leads to a diffraction pattern that has a circular central maximum surrounded by a series of secondary bright fringes. Solve: (a) Because the visible spectrum spans wavelengths from 400 nm to 700 nm, we take the average wavelength of sunlight to be 550 nm. (b) Within the small-angle approximation, the width of the central maximum is L (550 109 m)(3 m) wD2 44 (1 1024 m)  (2 44)  4 03 10 m  0 40 mm DD

22.60. Model: The antenna is a circular aperture through which the microwaves diffract. Solve: (a) Within the small-angle approximation, the width of the central maximum of the diffraction pattern is w 2.44 L / D . The wavelength of the radiation is c 3 1083 m/s 2  44(0  025 m)(30  10 m)    0  025 m w   920 m f 12 109 Hz 2 0 m That is, the diameter of the beam has increased from 2.0 m to 915 m, a factor of 458. (b) The average microwave intensity is P 100 103 W I/  0  15 W m2 2 area  1 (915 m) 2

22.61. Model: The laser light is diffracted by the circular opening of the laser from which the beam emerges. Solve: The diameter of the laser beam is the width of the central maximum. We have 2 44LL 2  44 2  44(532  1098 m)(3  84  10 m) wD    0  50 m Dw 1000 m In other words, the laser beam must emerge from a laser of diameter 50 cm.

22.62. Model: Two closely spaced slits produce a double-slit interference pattern.

Visualize:

Solve: (a) The m 1 bright fringes are separated from the m  0 central maximum by L (600 109 m)(1 0 m) y   0  0030 m  3  0 mm d 0 0002 m (b) The light’s frequency is fc/  5.00  1014 Hz. Thus, the period is Tf1/  2.00  10 15 s. A delay of 5.0 1016 s  0.50  10 15 s is 1 T . 4 (c) The wave passing through the glass is delayed by 1 of a cycle. Consequently, the two waves are not in phase as 4 they emerge from the slits. The slits are the sources of the waves, so there is now a phase difference 0 between the two sources. A delay of a full cycle ()tT would have no effect at all on the interference because it corresponds to a phase difference  2 rad  Thus a delay of 1 of a cycle introduces a phase difference  0 4 0 11(2 ) rad 42 (d) The text’s analysis of the double-slit interference experiment assumed that the waves emerging from the two slits were in phase, with 0 0 rad. Thus, there is a central maximum at a point on the screen exactly halfway between the two slits, where  0 rad and r 0 m. Now that there is a phase difference between the sources, the central maximum—still defined as the point of constructive interference where  0 rad—will shift to one side. The wave leaving the slit with the glass was delayed by 1 of a period. If it travels a shorter distance to the screen, taking 1 of 4 4 a period less than the wave coming from the other slit, it will make up for the previous delay and the two waves will arrive in phase for constructive interference. Thus, the central maximum will shift toward the slit with the glass. How far? A phase difference 0 2 would shift the fringe pattern by y 3.0 mm, making the central maximum fall exactly where the m 1 bright fringe had been previously. This is the point where r (1) , exactly compensating for a phase shift of 2π at the slits. Thus, a phase shift of  11   (2  ) will shift the fringe pattern 0 24 by 1 (3 mm) 0.75 mm. The net effect of placing the glass in the slit is that the central maximum (and the entire 4 fringe pattern) will shift 0.75 mm toward the slit with the glass.

22.63. Model: A diffraction grating produces an interference pattern, which looks like the diagram of Figure 22.8. Solve: (a) Nothing has changed while the aquarium is empty. The order of a bright (constructive interference) fringe is related to the diffraction angle m by dmsinm  , where m 0, 1, 2, 3, The space between the slits is 1 0 mm d  1  6667  106 m 600 For m 1,  633 109 m sin   sin1  22  3  116 d 1 6667 10 m

(b) The path-difference between the waves that leads to constructive interference is an integral multiple of the wavelength in the medium in which the waves are traveling, that is, water. Thus, 633 nm 633 nm 4 759 107 m   4 759 107 m  sin   0 2855  16 6 116 ndwater 1 33 1 6667 10 m

Model: An interferometer produces a new maximum each time L increases by 1  causing the path- 22.64. 2 2 length difference r to increase by . Visualize: Please refer to the interferometer in Figure 22.20.

Solve: The path-length difference between the two waves is r 2 L21  2 L . The condition for constructive interference is rm, hence constructive interference occurs when 2(L L )  m  L  L 11 m   1200   600  2 1 2 122 2   where   632.8 nm is the wavelength of the helium-neon laser. When the mirror M2 is moved back and a hydrogen discharge lamp is used, 1200 fringes shift again. Thus, LL 12001    600  21  2  where  656  5 nm  Subtracting the two equations, 99 (LLLL2 1 ) 2 1 600(  ) 600(632  8 10 m  656 5 10 m) 6 LL22  14  2  10 m That is, is now 14.2  m closer to the beam splitter.

22.65. Model: The gas increases the number of wavelengths in one arm of the interferometer. Each additional wavelength causes one bright-dark-bright fringe shift. Solve: From the equation in Challenge Example 22.9, the number of fringe shifts is 2d (2)(2 00 102 m) m  m  m ( n  1)  (1  00028  1)  19 21 9 vac 600 10 m

22.66. Model: The piece of glass increases the number of wavelengths in one arm of the interferometer. Each additional wavelength causes one bright-dark-bright fringe shift. Solve: We can rearrange the equation in Challenge Example 22.9 to find that the index of refraction of glass is  m (500 109 nm)(200) n 1 vac  1   1.50 2d 2(0 10 103 m)

22.67. Model: The arms of the interferometer are of equal length, so without the crystal the output would be bright. Visualize: We need to consider how many more wavelengths fit in the electro-optic crystal than would have occupied that space (6.70  m) without the crystal; if it is an integer then the interferometer will produce a bright output; if it is a half-integer then the interferometer will produce a dark output. But the wavelength we need to consider is the wavelength inside the crystal, not the wavelength in air.    n n We are told the initial n with no applied voltage is 1.522, and the wavelength in air is 1  000  m  Solve: The number of wavelengths that would have been in that space without the crystal is 6 70 m 6 70 1 000 m (a) With the crystal in place (andn  1.522) the number of wavelengths in the crystal is 6.70 m 10.20 1.000 m/1.522 10.20 6.70 3.50

© Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 22-20 Chapter 22 which shows there are a half-integer number more wavelengths with the crystal in place than if it weren’t there. Consequently the output is dark with the crystal in place but no applied voltage. (b) Since the output was dark in the previous part, we want it to be bright in the new case with the voltage on. That means we want to have just one-half more extra wavelengths in the crystal (than if it weren’t there) than we did in the previous part. That is, we want 4.00 extra wavelengths in the crystal instead of 3.5, so we want 6.70 4.00 10.70 wavelengths in the crystal. 6 70  m 10  70(1  000  m) 10  70 n   1  597 1 000  m/n 6  70  m Assess: It seems reasonable to be able to change the index of refraction of a crystal from 1.522 to 1.597 by applying a voltage.

22.68. Model: A diffraction grating produces an interference pattern like the one shown in Figure 22.8. We also assume that the small-angle approximation is valid for this grating. Solve: (a) The general condition for constructive-interference fringes is

dsinm  m m  0, 1, 2, 3, 

When this happens, we say that the light is diffracted at an angle  m. Since it is usually easier to measure distances rather than angles, we will consider the distance ym from the center to the mth maximum. This distance is

yLmm tan . In the small-angle approximation, sinmm tan , so we can write the condition for constructive interference as y m L dm  m  y  Ldm The fringe separation is L y  y   y  mm1 d (b) Now the laser light falls on a film that has a series of “slits” (i.e., bright and dark stripes), with spacing L d  d Applying once again the condition for constructive interference: y m L m L dsin m   d m  m   y     md mmL d  L/ d

The fringe separation is ymm1  y    y   d. That is, using the film as a diffraction grating produces a diffraction pattern whose fringe spacing is d, the spacing of the original slits.

22.69. Model: The laser beam is diffracted through a circular aperture. Visualize:

Solve: (a) No. The laser light emerges through a circular aperture at the end of the laser. This aperture causes diffraction, hence the laser beam must gradually spread out. The diffraction angle is small enough that the laser beam appears to be parallel over short distances. But if you observe the laser beam at a large distance it is easy to see that the diameter of the beam is slowly increasing.

(b) The position of the first minimum in the diffraction pattern is more or less the “edge” of the laser beam. For diffraction through a circular aperture, the first minimum is at an angle 1 22 1  22(633  109 m)    7  72  104 rad  0  044  1 D 0 0010 m (c) The diameter of the laser beam is the width of the diffraction pattern: 2 44L 2  44(633  109 m)(3  0 m) w   4  6 mm D 0 0010 m (d) At L 1 km 1000 m, the diameter is 2 44L 2  44(633  109 m)(1000 m) w   1  5 m D 0 0010 m

2 d 22.70. Visualize: To find the location y where the intensity is I1 use Equation 22.14: Idouble 4 I 1 cos y . L 1 L Then divide by the distance to the first minimum y  to get the fraction desired. 0 2 d

Solve: First set IIdouble 1.

2 d Idouble 4 I 1 cos  y L

2 d I11 4 I cos  y L

1 2 d  cos y 4 L 1 d  cosy 2 L

L 11 y  cos  d 2 Now set up the ratio that will give the desired fraction. L 11 cos  y d 2 21 1 2 2  cos    y0 1 L 2 3 3 2 d Assess: The fraction must be less than 1, and 2 seems reasonable. 3

22.71. Model: A diffraction grating produces a series of constructive-interference fringes at values of m that are determined by Equation 22.15. Solve: (a) The condition for bright fringes is dmsin . If  changes by a very small amount , such that  changes by , then we can approximate / as the derivative d/d  22 d d  d d22 d m    d  sin     cos   1  sin   1        m d m m m d   m      2 d 2   m (b) We can now obtain the first-order and second-order angular separations for the wavelengths   589.0 nm and    589.6 nm. The slit spacing is 1 0 103 m d  1  6667  106 m 600

The first-order (m  1) angular separation is

0 6  1099 m 0  6  10 m    6 2 7778  1012 m 2  0  3476  10 12 m 2 1 5589 10 m 3  85  104 rad  0  022  The second order (m  2) angular separation is 0 6 109 m   1  02  103 rad  0  058  0 6945  1012 m 2  0  3476  10 12 m 2

22.72. Model: The intensity in a double-slit interference pattern is determined by diffraction effects from the slits. Solve: (a) For the two wavelengths  and  passing simultaneously through the grating, their first-order peaks are at LL()   yy 11dd Subtracting the two equations gives an expression for the separation of the peaks: L y  y  y  11 d (b) For a double-slit, the intensity pattern is

2 d Idouble 4 I 1 cos  y L

The intensity oscillates between zero and 4,I1 so the maximum intensity is 4.I1 The width is measured at the point where the intensity is half of its maximum value. For the intensity to be 1 II 2 for the m 1 peak: 2 max 1

22d    d  1  d   L 2I1 4 I 1 cos y half   cos  y half    y half   y half  L    L  2  L 4 4 d

The width of the fringe is twice y half . This means L wy2 half 2d But the location of the peak is so we get wy 1 . y1   L/, d 2 1

(c) We can extend the result obtained in (b) for two slits to w y1/ N The condition for barely resolving two diffraction fringes or peaks is wy min. From part (a) we have an expression for the separation of the first-order peaks and from part (b) we have an expression for the width. Thus, combining these two pieces of information,

y11minL y d L d  ymin    min    N d LN d NL N (d) Using the result of part (c),  656 27 109 m N    3646 lines 9 min 0 18 10 m

22.73. Model: A diffraction grating produces an interference pattern like the one shown in Figure 22.8, when the incident light is normal to the grating. The equation for constructive interference will change when the light is incident at a nonzero angle.

Solve: (a) The path difference between waves 1 and 2 on the incident side is rd1 sin  The path difference between the waves 1 and 2 on the diffracted side, however, is rd2 sin  The total path difference between waves 1 and 2 is thus r12   r  d(sin  sin ). Because the path difference for constructive interference must be equal to m, d(sin sin  )  m  m  0,  1,  2,  (b) For  30 the angles of diffraction are (1)(500 109 m) sin11  sin30  0 20 11 5 (1 0 103 m)/600 ( 1)(500 109 m) sin11  sin30  0 80 53 1 (1 0 103 m)/600

22.74. Solve: (a)

We have two incoming and two diffracted light rays at angles  and  and two wave fronts perpendicular to the rays. We can see from the figure that the wave 1 travels an extra distance rdsin to reach the reflection spot. Wave 2 travels an extra distance rdsin from the reflection spot to the outgoing wave front. The path difference between the two waves is

r   r12   r  d(sin  sin ) (b) The condition for diffraction, with all the waves in phase, is still rm. Using the results from part (a), the diffraction condition is

dsinm  m   d sin  m  2,  1, 0, 1, 2,  Negative values of m will give a different diffraction angle than the corresponding positive values. (c) The “zeroth order” diffraction from the reflection grating is m   From the diffraction condition of part (b), this implies ddsin0  sin and hence 0  That is, the zeroth order diffraction obeys the law of reflection—the angle of reflection equals the angle of incidence. (d) A 700 lines per millimeter grating has spacing d 1 mm  1  429  106 m  1429 nm  The diffraction angles 700 are given by

11mm  (500 nm)  m sin  sin   sin   sin40   d  1429 nm 

M  m   not defined         0  1   2 not defined

(e)

22.75. Model: Diffraction patterns from two objects can just barely be resolved if the central maximum of one image falls on the first dark fringe of the other image. Solve: (a) Using Equation 22.24 with the width equal to the aperture diameter, 2 44L w D   D 2  44 L  (2  44)(550  107 m)(0.20 m)  0  52 mm D (b) We can now use Equation 22.23 to find the angle between two distant sources that can be resolved. The angle  1 22(550 109 m)  1.22   1.29  103 rad  0.074  D 0 52 103 m (c) The distance that can be resolved is (1000 m)  (1000 m)(1  29  103 rad)  1  3 m