22.1. Visualize: Please refer to Figure Ex22.1. Solve: (a)

Transcription

1 22.. Visualize: Please refer to Figure Ex22.. Solve: (a) (b) 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 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.

2 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). 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 m = 2 bright fringe differ by r = r2 r, where r = mλ = 2λ = 2( 500 nm)= 000 nm Thus, the position of the m = 2 bright fringe is 000 nm farther away from the more distant slit than from the nearer slit.

3 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 bright fringes occur at angles θ m such that dsinθ ( ) ( ) m = mλ m = 0,, 2, 3, m 80 sin θ 2 = = 002. θ 6 2 = rad = rad = m π rad

4 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 fringe spacing is 9 2 λl λl ( m) ( 50 0 m) y = d = = = 0.22 mm 3 d y m

5 22.5. 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: λl ym = ( m+ 2 ) m = 0,, 2, 3, d 2 λl λl 4λ 60 0 y5 y = ( 5+ 2 ) ( + 2 ) ( m ). m = λ = 500 nm 3 d d m

6 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 formula for fringe spacing is λl 3 9 y =. 8 0 m = ( m) L d d L d = 3000 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, λl y = = ( m)( 3000)=.2 mm d

7 22.7. 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 2 fringes, there are gaps between them. The formula for the fringe spacing is L y = λ m ( m)( 3.0 m) = d = 0.40 mm d d 4 Assess: This is a reasonable distance between the slits, ensuring dl= <<.

8 22.8. Model: A diffraction grating produces an interference pattern. Visualize: The interference pattern looks like the diagram in Figure Solve: The bright constructive-interference fringes are given by Equation 22.5: dsinθ = mλ m = 0,, 2, 3, = () m sinθ = θ = (. 0 0 m) 000 Likewise, sin θ = 0. 0 and θ = m ( )

9 22.9. Model: A diffraction grating produces an interference pattern. Visualize: The interference pattern looks like the diagram in Figure Solve: The bright constructive-interference fringes are given by Equation 22.5: m dsinθm = mλ d = = ( ) ( 9 λ m) = sinθ sin m ( ) 3 6 The number of lines in per millimeter is ( 0 m) ( m ) = m

10 22.0. Model: A diffraction grating produces an interference pattern. Visualize: The interference pattern looks like the diagram in Figure Solve: The bright fringes are given by Equation 22.5: dsinθ m = mλ m = 0,, 2, 3, d sinθ λ The angle θ can be obtained from geometry as follows: = () d = λ θ tan θ = ( 0.32 m ) 2 = θ = tan (0.080) m Using sinθ = sin = , m d = = 7.94 µ m sin

11 22.. Model: A diffraction grating produces an interference pattern. Visualize: The interference pattern looks like the diagram of Figure Solve: (a) A grating diffracts light at angles sinθm = mλ d. The distance between adjacent slits is d = mm 600 = m = 667 nm. The angle of the m = fringe is 500 nm θ = sin 7 46 =. 667 nm The distance from the central maximum to the m = bright fringe on a screen at distance L is y = Ltanθ = ( 2 m) tan = m (Note that the small angle approximation is not valid for the maxima of diffraction gratings, which almost always have angles >0.) There are two m = bright fringes, one on either side of the central maximum. The distance between them is y = 2y =.258 m. (b) The maximum number of fringes is determined by the maximum value of m for which sinθ m does not exceed because there are no physical angles for which sinθ >. In this case, mλ m( 500 nm) sinθ m = = d 667 nm We can see by inspection that m =, m = 2, and m = 3 are acceptable, but m = 4 would require a physically impossible sinθ 4 >. Thus, there are three bright fringes on either side of the central maximum plus the central maximum itself for a total of 7 bright fringes.

12 22.2. Model: A diffraction grating produces an interference pattern. Visualize: The interference pattern looks like the diagram of Figure Solve: The bright interference fringes are given by dsinθ m = mλ m = 0,, 2, 3, The slit spacing is d = mm 500 = m. The angles of diffraction are 9 () λ 50 0 m ( 2) λ θ = sin = sin =. θ 2 = sin = d m d Likewise, θ 3 = The next order (m = 4) is not seen because m sin θ 4 = > m Thus, three diffraction orders are visible.

13 22.3. Model: A diffraction grating produces an interference pattern. Visualize: The interference pattern looks like the diagram of Figure Solve: The bright interference fringes are given by dsinθm = mλ m = 0,, 2, 3, The slit spacing is d = mm 500 = m and m =. For the red and blue light, m m θ red = sin = 9. 5 θ 6 blue = sin = m m The distance between the fringes, then, is y = y y where red blue y red = (. 5 m ) tan 9. 5 = m y blue = (. 5 m ) tan = m So, y = 0.45 m = 45. cm.

14 22.4. Model: A narrow single slit produces a single-slit diffraction pattern. Visualize: The intensity pattern for single-slit diffraction will look like Figure Solve: The minima occur at positions y = p L λ p. a 2λL λl λl λl So y = y2 y = = a = = a a a y 9 ( m)(.5 m) = m = m 0.20 mm

15 22.5. Model: A narrow slit produces a single-slit diffraction pattern. Visualize: The intensity pattern for single-slit diffraction will look like Figure Solve: The width of the central maximum for a slit of width a = 200λ is ( )( ) = = 2λ L m 2.0 m w = = a m m 4.0 mm

16 22.6. Model: A narrow slit produces a single-slit diffraction pattern. Visualize: The intensity pattern for single-slit diffraction will look like Figure Solve: The width of the central maximum for a slit of width a = 200λ is 2λL 2λ( 2.0 m) w = = = 2.0 cm a 200λ

17 22.7. Model: A narrow slit produces a single-slit diffraction pattern. Visualize: The intensity pattern for single-slit diffraction will look like Figure Solve: Angle θ = = rad is a small angle (<< rad). Thus we use Equation to find the wavelength of light. The angles of the minima of intensity are λ θ p = p p =, 2, 3, K a 3 aθ ( m)( rad p ) λ = = = 6 nm p 2

18 22.8. Model: The spacing between the two building 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 m/s λ = = m 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: asinθp = pλ (p =, 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 = minima on either side of the central maximum, we will use p = and obtain the angular width θ = 2θ from λ m asinθ = λ θ = sin = sin 43. a 5 m = Thus, the angular width of the wave after it emerges from between the buildings is θ = 2(. 43 )=

19 22.9. 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 λ = = 30 khz 0.03 m Using the condition for complete destructive interference with p =, asinθ = λ θ = 0.03 m sin 2 65 = m From the geometry of the diagram, the width of the sound beam is w = 2y = 2( 00 m tanθ)= 200 m tan = 7.56 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.

20 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 L w = 244. λ = ( ) D ( m)( 2.0 m) 3 m = 4.88 mm.

21 22.2. 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 Solve: From Equation 22.24, the diameter of the circular aperture is λ L 2. 44( m)( 4.0 m) D = = = 0.25 mm 2 w m

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. Visualize: The intensity pattern will look like Figure Solve: From Equation 22.24, 3 2 Dw ( m )(. 0 0 m) L = = = 77.7 cm λ m ( )

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 Solve: According to Equation 22.23, the angle that locates the first minimum in intensity is λ (. 0 m) θ = = = rad = D m

24 Model: An interferometer produces a new maximum each time L 2 increases by 2 λ causing the pathlength difference r to increase by λ. Visualize: Please refer to the interferometer in Figure Solve: From Equation 22.33, the wavelength is 2 L λ = = m ( m) = m = 400 nm

25 Model: An interferometer produces a new maximum each time L 2 increases by 2 λ causing the pathlength difference r to increase by λ. Visualize: Please refer to the interferometer in Figure Solve: From Equation 22.33, the number of fringe shifts is 2 2 L 2( m) 2 m = = = 30, λ m

26 Model: An interferometer produces a new maximum each time L 2 increases by 2 λ causing the pathlength difference r to increase by λ. Visualize: Please refer to the interferometer in Figure Solve: From Equation 22.33, the distance the mirror moves is L 2 ( ) = 9 mλ 33, m m = cm 2 2 = = ( ) Assess: Because the wavelength is known to 6 significant figures and the fringes are counted exactly, we can determine L to 6 significant figures.

27 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 Equation 22.36, the number of fringe shifts is d m = m2 m = ( n ) 2 = λ vac ( ) ( ) 2 (. m) m = 9

28 Model: Two closely spaced slits produce a double-slit interference pattern with the intensity graph looking like Fig. 22.3(b). The intensity pattern due to a single slit diffraction looks like Figure 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 P22.28 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 P22.28, the fringe spacing is y =.0 cm = m. Therefore, λl y = d λ L ( m )( 25. m) d = = = 05. mm y m

29 Model: Two closely spaced slits produce a double-slit interference pattern with the intensity graph looking like Fig. 22.3(b). The intensity pattern due to a single slit diffraction looks like Figure 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 P22.29 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 P22.29, the separation between the central maximum and the first minimum is y =.0 cm = m. Therefore, using the small-angle approximation, Equation 22.2 gives the condition for the dark minimum: pl L ( ) ( y = p a d = y = 9 λ λ 2. 5 m m) = 0.5 mm m

30 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, dsinθ = mλ. For the m = 3 bright ( ) 9 orange fringe, the interference condition is d sinθ 3 = m. For the m = 4 bright fringe the condition is d sinθ4 = 4λ. Because the position of the fringes is the same, dsinθ3 = dsinθ4 = 4λ = 3( m ) λ = 4 ( m ) = 450 nm m

31 22.3. 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 Since d = 200 λ, we have λ λ λ θ = θm+ θm = ( m + ) m = d d d λ θ = = rad = d 200

32 Model: Two closely spaced slits produce a double-slit interference pattern. Solve: The light intensity of a double-slit interference pattern at position y is I = 2 d I π L y double 4 cos λ where I is the intensity due to each wave. The intensity of a bright fringe is 4I. For the central maximum, m = 0. So, y central = 0 m. The first minimum occurs at y = λ L/ 2d. Thus, the halfway position between the maximum and the minimum is y = λl 4 d. Substituting into the intensity equation, the intensity at the halfway position is I = 2 πd λl 2 double I = I π 4 cos 4 cos = 2I λl 4d 4

33 Model: Two closely spaced slits produce a double-slit interference pattern. Solve: The light intensity of a double-slit interference pattern at a position y on the screen is I double = 2 d I π L y = I 2 4 cos 4 cos π λ yy where y = λ L/ d = 4. 0 mm is the fringe spacing. Using this value for λl d, we can find the position on the interference pattern where I double = I as follows: 2 4I cos π y I = π m π = cos m y 2 = 3 rad y = m =.33 mm 3

34 Solve: According to Equation 22.7, the fringe spacing between the m fringe and the m + fringe is y = λ L d. y can be obtained from Figure P The separation between the m = 2 fringes is 2.0 cm implying that the separation between the two consecutive fringes is 4 (2.0 cm) = 0.50 cm. Thus, L y = m = λ 3 2 d y ( m )( m) L = = = 67 cm 9 d λ m Assess: A distance of 67 cm from the slits to the screen is reasonable.

35 Solve: According to Equation 22.7, the fringe spacing between the m fringe and the m + fringe is y = λ L d. y can be obtained from Figure P Because the separation between the m = 2 fringes is 2.0 cm, two consecutive fringes are y = 4 ( 2.0 cm)= 0.50 cm apart. Thus, ( )( ) = L y = m = λ 3 2 d y m m λ = = d L 2.0 m 500 nm

36 Solve: The intensity of light of a double-slit interference pattern at a position y on the screen is I = 2 d I π L y double 4 cos λ where I is the intensity of the light from each slit alone. At the center of the screen, that is, at y = 0 m, I = 4 I. double From Figure P22.34, I double at the central maximum is 2 mw/m 2. So, the intensity due to a single slit is I = 3 mw / m 2.

37 Model: A diffraction grating produces an interference pattern. Visualize: The interference pattern looks like the diagram in Figure Solve: 500 lines per mm on the diffraction grating gives a spacing between the two lines of d = mm 500 = 3 6 ( 0 m) 500 = m. The wavelength diffracted at angle θ m = 30 in order m is ( ) = d sin θ m sin30 m 000 nm λ = = m m m We re told it is visible light that is diffracted at 30, and the wavelength range for visible light is nm. Only m = 2 gives a visible light wavelength, so λ = 500 nm.

38 Model: A diffraction grating produces an interference pattern like the diagram of Figure Visualize: Solve: (a) A key statement is that the lines are seen on the screen. This means that the light is visible light, in the range 400 nm 700 nm. We can determine where the entire visible spectrum falls on the screen for different values of m. We do this by finding the angles θ m at which 400 nm light and 700 nm light are diffracted. We then use ym = Ltanθ m to find their positions on the screen which is at a distance L = 75 cm. The slit spacing is d = mm 200 = m. For m =, For m = 2, λ = 400 nm: θ = sin ( 400 nm / d)= y = 75 cm tan = 4 cm λ = 700 nm: θ = sin ( 700 nm / d)= 57. y = 75 cm tan 57. = 6 cm λ = 400 nm: θ = sin ( nm / d)= y = 75 cm tan = 257 cm ( )= ( ) For the 700 nm wavelength at m = 2, θ 2 = sin nm / d sin. 68 is not defined, so y 2. We see that visible light diffracted at m = will fall in the range 4 cm y 6 and that visible light diffracted at m = 2 will fall in the range y 257 cm. These ranges do not overlap, so we can conclude with certainty that the observed diffraction lines are all m =. (b) To determine the wavelengths, we first find the diffraction angle from the observed position by using θ = y = tan tan y L 75 cm This angle is then used in the diffraction grating equation for the wavelength with m =, λ = d sinθ. Line θ λ 56.2 cm nm 65.9 cm nm 93.5 cm nm 2

39 Model: Each wavelength of light is diffracted at a different angle by a diffraction grating. Solve: Light with a wavelength of 50.5 nm creates a first-order fringe at y = 2.90 cm. This light is diffracted at angle 2 90 θ =. cm tan = cm We can then use the diffraction equation dsinθ m = mλ, with m =, to find the slit spacing: λ nm d = = sinθ sin( ) = 250 nm The unknown wavelength creates a first order fringe at y = 3.60 cm, or at angle 3 θ =.60 cm tan = cm With the split spacing now known, we find that the wavelength is λ = d sin θ = ( 250 nm) sin( ) = 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.

40 Model: A diffraction grating produces a series of constructive-interference fringes at values of θ m determined by Equation Solve: We have Dividing these two equations, dsin θ = mλ m = 0,, 2, 3, K d sin = λ and d sinθ = λ m sinθ 2 = 2sin = θ 2 =

41 22.4. Model: A diffraction grating produces a series of constructive-interference fringes at values of θ m that are determined by Equation Solve: For the m = 3 maximum of the red light and the m = 5 maximum of the unknown wavelength, Equation 22.5 gives ( ) 9 d sinθ 3 = m d sinθ λ 5 = 5 unknown The m = 5 fringe and the m = 3 fringe have the same angular positions. This means θ 5 = θ 3. Dividing the two equations, 5λ unknown = ( m) λunknown = 396 nm

42 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 22.6, the distance y m from the center to the mth maximum is ym = Ltanθ m. The angle of diffraction is determined by the constructive-interference condition dsinθm = mλ, where m = 0,, 2, 3, The width of the rainbow for a given fringe order is thus w = y red y violet. The slit spacing is mm. 0 0 d = = m = For the red wavelength and for the m = order, 9 λ m d sinθ = () λ θ = sin = sin = d m From the equation for the distance of the fringe, yred = Ltanθ = ( 2.0 m) tan ( )= cm Likewise for the violet wavelength, θ sin = The width of the rainbow is thus cm cm = 43. cm. 9 m = y violet = ( 2.0 m) tan ( )= cm m 6 6 m

43 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: The slit spacing is m 6 d = = m 300 The constructive-interference condition is dsinθm = mλ, where m = 0,, 2, 3, For the red and violet light in the third-order rainbow, m m θ3 red = sin = θ violet = sin = m m That is, the angular spread of the third-order rainbow is from 2.0 to For the fourth-order rainbow, m m θ4 red = sin 57 4 θ =. 4 violet = sin 6 = m m We clearly see that the overlap of wavelengths is from an angle of to an angle of Since we are asked to find wavelengths in the third-order rainbow that overlap the wavelengths in the fourth-order rainbow, we will calculate the wavelengths in the third-order that correspond to an angle of Using the constructiveinterference condition, 6 d sin θ = 3λ λ = m sin ( ) ( )= 533 nm Thus, the wavelengths in the range nm overlap with some of the wavelengths of the fourth order.

44 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 θ, then red light (the longest wavelengths) is diffracted at angle θ In the first order, the equations for the blue and red wavelengths are λ sinθ = b d Combining the two equations we get for the red wavelength, d sin( θ + 30 )= λr 2 λb λb λr = ( θ + θ )= ( θ + θ)= d sin cos30 cos sin 30 d sin 0. 50cos d d( 0. 50) d 2 d 2 λb 2 ( 0. 50) d = λ λ 2 r b ( 0. 50) ( d λb )= λr λb d d = λr λb 2 + λb 050. ( ) Using λ b = m and λ r = m, we get d = m. This value of d corresponds to 3 mm 0. 0 m = = 23 lines / mm 7 d m (b) Using the value of d from part (a) and λ = m, we can calculate the angle of diffraction as follows: 2 d sinθ = () λ m sinθ m θ = ( ) =

45 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 = 3 6 (. 0 0 m) 800 = m. Referring to Figure P22.45, the angle of diffraction is given by y m tan θ = = = θ = sin θ = L.0 m Using the constructive-interference condition dsinθm = mλ, d sinθ λ = = ( m)( )= 500 nm We can obtain the same value of λ by using the second-order interference fringe. We first obtain θ 2 : y m m tanθ 2 = = =. 333 L.0 m θ 2 = sin θ 2 = Using the constructive-interference condition, 6 d sin θ ( m)( ) 2 λ = = = 500 nm 2 2 Assess: Calculations with the first-order and second-order fringes of the interference pattern give the same value for the wavelength.

46 Model: A diffraction grating produces an interference pattern that is determined by both the slit spacing and the wavelength used. Solve: From Figure P22.45, m tan θ = = θ = sin θ = m Using the constructive-interference condition dsinθ = mλ, Thus, the number of lines per millimeter is m dsin =() ( m) d = sin m 6 m m = 667 lines / mm ( ) = Assess: The same answer is obtained if we perform calculations using information about the second-order bright constructive-interference fringe. 6 m

47 Model: A narrow slit produces a single-slit diffraction pattern. Visualize: The diffraction-intensity pattern from a single slit will look like Figure Solve: The dark fringes are located at p L y = λ p p = 0,, 2, 3, a The locations of the first and third dark fringes are L y = λ 3 L y3 = λ a a Subtracting the two equations, 3 ( )( ) 2 L ( y3 y)= λ 9 2λ L m 0.75 m a = = = 0.8 mm 3 a y y m

48 Model: A narrow slit produces a single-slit diffraction pattern. Visualize: The diffraction-intensity pattern from a single slit will look like Figure Solve: 45 is not a small angle, so we can t use equations based on the small-angle approximation. As given by Equation 22.9, the dark fringes in the pattern are located at asinθp = pλ, where p =, 2, 3, For the first minimum to be at 45, p = and the width of the slit is pλ m a = = = 895 nm sinθ sin 45 p

49 Model: A narrow slit produces a single-slit diffraction pattern. Visualize: The diffraction-intensity pattern from a single slit will look like Figure Solve: These are not small angles, so we can t use equations based on the small-angle approximation. As given by Equation 22.9, the dark fringes in the pattern are located at asinθp = pλ, where p =, 2, 3, For the first minimum of the pattern, p =. Thus, a p = = λ sinθp sinθ For the three given angles the slit width to wavelength ratios are a (a): = λ 30 = 2, (b): a = sin λ 60 = 5., (c): a = sin λ 90 = sin Assess: It is clear that the smaller the a/λ ratio, the wider the diffraction pattern. This is a conclusion that is contrary to what one might expect. 90

50 Model: A narrow slit produces a single-slit diffraction pattern. Visualize: The diffraction-intensity pattern from a single slit will look like Figure Solve: As given by Equation 22.9, the dark fringes in the pattern are located at asinθp = pλ, where p =, 2, 3, For the diffraction pattern to have no minima, the first minimum must be located at least at θ = 90. From the constructive-interference condition asinθ = pλ, we have p pλ λ a = a = θ = λ sin sin 90 = 633 nm p

51 22.5. Model: A narrow slit produces a single-slit diffraction pattern. Solve: The dark fringes in this diffraction pattern are given by Equation 22.2: p L y = λ p p =, 2, 3, a We note from Figure P22.5 that the first minimum is 0.50 cm away from the central maximum. Using the above equation, the slit width is p L a = = () 9 λ ( m)(.0 m) = 0.0 mm 2 y p m Assess: This is a typical slit width for diffraction.

52 Model: A narrow slit produces a single-slit diffraction pattern. Solve: The dark fringes in this diffraction pattern are given by Equation 22.2: p L y = λ p p =, 2, 3, a We note from Figure P22.5 that the first minimum is 0.50 cm away from the central maximum. Thus, 3 2 ay p ( m )( m) L = = =.25 m 9 pλ () ( m) Assess: This is a typical slit to screen separation.

53 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 λl w = 2y = 244. D From Figure P22.5, w =.0 cm, so λ L 2. 44( m)(.0 m) D = = = 0.22 mm 2 w ( 0. 0 m) Assess: This is a typical size for an aperture to show diffraction.

54 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 Because w = D, we have λl D = 244. D = 244. L D w λl = 244. D λ D = ( 2 44) ( )( ) =. m 0.50 m mm

55 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 w ( m)( 3 m) L = 244. λ ( m)= ( ) D D D = m = 0.40 mm

56 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 = 2.44λL/D. The wavelength of the radiation is ( ) ( ) = λ = = 8 3 c 3 0 m/s m 30 0 m = m w = 9 f 2 0 Hz 2.0 m That is, the diameter of the beam has increased from 2.0 m to 95 m, a factor of 458. (b) The average microwave intensity is P W I = = = 0.52 W / m 2 2 area π 95 m [ 2 ( )] 95 m

57 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 λ. 22( m) 4 θ = = = rad = D m (c) The diameter of the laser beam is the width of the diffraction pattern: ( )( ) = = 244. λ L m 3 m w = = D m (d) At L = km = 000 m, the diameter is 244. λ L m 000 m w = = D m m 0.3 cm ( )( ) = =.03 m 03 cm

58 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 ( )( ) = = L w = 244. λ λ L m m D = = D w 000 m In other words, the laser beam must emerge from a laser of diameter 49.8 cm m 49.8 cm

59 Model: Two closely spaced slits produce a double-slit interference pattern. Visualize: Solve: (a) The m = bright fringes are separated from the m = 0 central maximum by ( )( ) = = λl m.0 m y = = m 3.0 mm d m (b) The light s frequency is f = c/λ = Hz. Thus, the period is T = /f = s. A delay of s = s is 4 T. (c) The wave passing through the glass is delayed by 4 of a cycle. Consequently, the two waves are not in phase as they emerge from the slits. The slits are the sources of the waves, so there is now a phase difference φ 0 between the two sources. A delay of a full cycle ( t = T) would have no effect at all on the interference because it corresponds to a phase difference φ0 = 2πrad. Thus a delay of 4 of a cycle introduces a phase difference φ 0 = 4 ( 2π)= 2 πrad. (d) The text s analysis of the double-slit interference experiment assumed that the waves emerging from the two slits were in phase, with φ 0 = 0 rad. Thus, there is a central maximum at a point on the screen exactly halfway between the two slits, where φ = 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 φ = 0 rad will shift to one side. The wave leaving the slit with the glass was delayed by 4 of a period. If it travels a shorter distance to the screen, taking 4 of 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 φ0 = 2π would shift the fringe pattern by y = 3.0 mm, making the central maximum fall exactly where the m = bright fringe had been previously. This is the point where r = ()λ, exactly compensating for a phase shift of 2π at the slits. Thus, a phase shift of φ0 = 2 π = 4 ( 2π) will shift the fringe pattern by 4 ( 3 mm )= mm. The net effect of placing the glass in the slit is that the central maximum (and the entire fringe pattern) will shift 0.75 mm toward the slit with the glass.

60 Model: A diffraction grating produces an interference pattern, which looks like the diagram of Figure Solve: (a) Nothing has changed while the aquarium is empty. The order of a bright (constructive interference) fringe is related to the diffraction angle θ m by d sinθ = mλ, where m = 0,, 2, 3, The space between the slits is For m =, m.0 mm d = = m 600 λ sinθ = d θ = sin m = 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, nm 633 nm λ m λ = = = m sinθ = = = θ = d m n water 6

61 22.6. Model: An interferometer produces a new maximum each time L 2 increases by 2 λ causing the pathlength difference r to increase by λ. Visualize: Please refer to the interferometer in Figure Solve: The path-length difference between the two waves is r = 2L 2 2L. The condition for constructive interference is r = mλ, hence constructive interference occurs when ( 2 )= 2 = 2 2 = ( 2 )= 2 L L mλ L L mλ 200 λ 600λ where λ = nm is the wavelength of the helium-neon laser. When the mirror M 2 is moved back and a hydrogen discharge lamp is used, 200 fringes shift again. Thus, L L = 200( 2 λ )= 600λ where λ = nm. Subtracting the two equations, 9 9 ( L2 L) ( L2 L)= 600( λ λ )= m m 6 L = L m 2 2 That is, M 2 is now 4.2 µm closer to the beam splitter. ( )

62 Model: An interferometer produces a new maximum each time L 2 increases by 2 λ causing the pathlength difference r to increase by λ. Visualize: Please refer to the interferometer in Figure Solve: For sodium light of the longer wavelength (λ ) and of the shorter wavelength (λ 2 ), λ λ2 L = m L = ( m+ ) 2 2 We want the same path difference 2(L 2 L ) to correspond to one extra wavelength for the sodium light of shorter wavelength (λ 2 ). Thus, we combine the two equations to obtain: λ λ2 λ nm m = ( m+) m λ λ2 λ2 2 2 ( )= m = = = λ λ nm nm Thus, the distance by which M 2 is to be moved is λ nm L = m = = mm

63 Model: The water increases the number of wavelengths in one arm of the interferometer. Solve: (a) The incident light has λ vac = 500 nm and f = c/ λ vac = Hz. Water doesn t affect the frequency, which is still fw = f = Hz. The wavelength changes to λw = λvac / n w = 376 nm. (b) Light travels in a layer of thickness L twice once to the right and then, after reflecting, once to the left for a total distance 2L. The number of wavelengths in distance 2L is N = 2L/λ. If the mm layer is a vacuum, the number of wavelengths is L N = 2 = 2( 0.00 m) vac = m 4000 λ If the vacuum is replaced by mm of water, the number of wavelengths is L N = 2 = 2( 0.00 m) w = m 539 λ vac w So with the water added, the light travels 39 extra wavelengths. (c) Each additional wavelength of travel shifts the pattern by fringe. So the addition of the water shifts the interference pattern by 39 fringes.

64 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 Equation to find that the index of refraction of glass is n = + ( )( ) λ m nm 200 vac = + = d m ( )

65 Model: The number of wavelengths in one arm of the interferometer increases as the refractive index of the electro-optic crystal increases. Each bright-dark-bright fringe shift corresponds to an additional path-difference increase of one wavelength. A path-difference of one-half wavelength thus will cause the bright fringe to shift to a dark fringe. Visualize: Please refer to the interferometer in Figure Solve: We will follow the treatment of Section 22.6 in the textbook. For an initial index of refraction n of the electro-optic crystal, the number of wavelengths inside the crystal is (including motion in both directions) 2d m = λvac n Similarly, the number of wavelengths inside the crystal when its refractive index increases to n is 2d m2 = n The increase in the number of wavelengths due to a change in refractive index from n to n is λ vac 2d m = m m = n n 2 λ vac ( ) Because the external voltage causes a bright fringe to shift to a dark one, m = 2. Thus, 6 2d λ = ( n n ) n = n+ vac m = n + n = n n = λ vac 4d m Thus, n = = ( )

66 Model: A diffraction grating produces an interference pattern like the one shown in Figure We also assume that the small-angle approximation is valid for this grating. Solve: (a) The general condition for constructive-interference fringes is dsinθm = mλ m = 0,, 2, 3, When this happens, we say that the light is diffracted at an angle θ m. Since it is usually easier to measure distances rather than angles, we will consider the distance y m from the center to the mth maximum. This distance is ym = Ltanθ m. In the small-angle approximation, sinθ tanθ, so we can write the condition for constructive interference as The fringe separation is m m d y m L = mλ y = m mλl d λl ym+ ym = y = d (b) Now the laser light falls on a film that has a series of slits (i.e., bright and dark stripes), with spacing λl d = d Applying once again the condition for constructive interference: d sinθm = mλ d ym = m L λ = mλl mλl ym = = md d λld The fringe separation is ym + ym = 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.

67 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 P and 2P are equal. When a glass piece of thickness t is inserted in the path P, the interference between the two waves yields the 0 th 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 = λ The number of wavelengths in the glass piece of thickness t is t t nt m2 = = = λ λ n λ glass The path length has thus increased by m wavelengths, where t m = m2 m = ( n ) λ From the 0 th dark fringe to the st dark fringe is 9 fringes and from the first dark fringe to the zeroth bright fringe is one-half of a fringe. Hence, t 9 λ m m = 9 + = = ( n ) t = λ 2 n ( ) = ( ) = 2. 0 µm

68 Model: Two closely spaced slits produce a double-slit interference pattern. The fringe intensity decreases because the intensity of the two individual slits is not uniform. Visualize: Solve: To a good approximation, each slit illuminates an area on the screen that is the width of a single-slit diffraction pattern. (A single slit diffraction pattern does have secondary maxima, but they are very weak in comparison with the central maximum.) The width of the single-slit pattern is 9 2λ L 2( m)(2.0 m) w = = = 60. cm 5 a m Thus the fringes of the double-slit pattern can be seen over a range of 6.0 cm, from y = 3.0 cm to y = +3.0 cm. How many fringes fit within this distance? The fringe spacing of the double slit pattern is 9 λ L ( m)(2.0 m) yfringes = = = 060. mm d m The central maximum is at y = 0, then on both sides there are fringes at y = 0.6 cm,.2 cm,.8 cm, and 2.4 cm. The next fringe, which would be at 3.0 cm, won t be seen because that is where the intensity has dropped to zero. Thus a total of 9 fringes are seen. Assess: In practice, with a sufficiently strong light source such as a laser, a few fringes may be seen farther away from the center where the secondary maxima of the single-slit pattern restore some intensity, but they will be very dim in comparison with these 9 fringes.

69 Model: A diffraction grating produces a series of constructive-interference fringes at values of θ m that are determined by Equation Solve: (a) The condition for bright fringes is dsinθ = mλ. If λ changes by a very small amount λ, such that θ changes by θ, then we can approximate λ/ θ as the derivative dλ/dθ: 2 2 λ = d θ m sin λ dλ d d 2 d = cosθ = sin θ = θ dθ m m mλ = d m d m λ θ = 2 d 2 λ m (b) We can now obtain the first-order and second-order angular separations for the wavelengths λ = nm and λ + λ = nm. The slit spacing is m 6 d = = m 600 The first-order (m = ) angular separation is m θ = m m 4 = rad = The second order (m = 2) angular separation is θ = m m m m = m 3 = rad = λ

70 Model: The intensity in a double-slit interference pattern is determined by diffraction effects from the slits. Visualize: Please refer to Figure CP Solve: (a) For the two wavelengths λ and λ + λ passing simultaneously through the grating, their first-order peaks are at L y = λ ( λ + λ) L y = d d Subtracting the two equations gives an expression for the separation of the peaks: λl y = y y = d (b) For a double-slit, the intensity pattern is I = 2 d I π L y double 4 cos λ The intensity oscillates between zero and 4I, so the maximum intensity is 4I. The width is measured at the point where the intensity is half of its maximum value. For the intensity to be 2 Imax = 2I for the m = peak: 2 d 2 2I 4I L y d L y d 2 L y 4 y L = cos π cos π = π = π λ λ = λ half half half half λ 4d The width of the fringe is twice y half. This means λl w = 2yhalf = 2d But the location of the m = peak is y = λ L d, so we get w = 2 y. (c) We can extend the result obtained in (b) for two slits to w = y N. The condition for barely resolving two diffraction fringes or peaks is w = y 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, (d) Using the result of part (c), y λ L yd L d = y = = = min N d LN λ λ min λmin = d NL N 9 λ m N = = = 3646 lines 9 λ min m

71 22.7. 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. Visualize: Please refer to Figure CP22.7. Solve: (a) The path difference between waves and 2 on the incident side is r = dsinφ. The path difference between the waves and 2 on the diffracted side, however, is r2 = dsinθ. The total path difference between waves and 2 is thus r + r2 = d( sinθ + sinφ ). Because the path difference for constructive interference must be equal to mλ, d( sinθ + sinφ)= mλ m = 0, ±, ±2, (b) For φ = 30 the angles of diffraction are sinθ = () ( m) sin 30 = 020. θ. 3 = m 600 sinθ ( ) ( ) 9 ( m) ( 0. 0 m) 600 = sin 30 = 080. θ = 53. 3

72 Visualize: Please refer to Figure CP Solve: (a) We have two incoming and two diffracted light rays at angles φ and θ and two wave fronts perpendicular to the rays. We can see from the figure that the wave travels an extra distance r = dsinφ to reach the reflection spot. Wave 2 travels an extra distance r = dsinθ from the reflection spot to the outgoing wave front. The path difference between the two waves is ( ) r = r r2 = d sinθ sinφ (b) The condition for diffraction, with all the waves in phase, is still r = mλ. Using the results from part (a), the diffraction condition is dsinθm = mλ + dsinφ m = 2,, 0,, 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 = 0. From the diffraction condition of part (b), this implies dsinθ0 = dsinφ and hence θ0 = φ. That is, the zeroth order diffraction obeys the law of reflection the angle of reflection equals the angle of incidence. 6 (d) A 700 lines per millimeter grating has spacing d = 700 mm = m = 429 nm. The diffraction angles are given by (e) θ m mλ m( 500 nm) = sin + sinφ = sin + d sin nm M θ m not defined not defined

73 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. Visualize: Please refer to Figure CP Solve: (a) Using Equation with the width be equal to the aperture diameter, 244. λl w = D = D = 2. 44λ L = ( 2. 44) ( m)( 0.20 m ) = 0.52 mm D (b) We can now use Equation to find the angle between two distant sources that can be resolved. The angle is (c) The distance that can be resolved is ( ) 9 λ m α = 22. = 3 D m ( ) = ( ) = ( ) 3 rad= m α 000 m rad =.3 m