Chapter 22 - PHYS 132, WEEK 4, CHAPTER 22 22.2. Model: Two...

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Unformatted text preview: PHYS 132, WEEK 4, CHAPTER 22 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 Dr = r2 - r1 , where Dr = m l = 2l = 2 (500 nm ) = 1000 nm Thus, the position of the m = 2 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 fringe spacing is Dy = lL d fid= lL Dy = ( 589 10 -9 m 150 10 -2 m )( ) 4.0 10 -3 m = 0.221 mm 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 L L lL -3 -9 fi = 3000 Dy = fi 1. 8 10 m = 600 10 m d d d The wavelength is now changed to 400 nm, and L d , being a part of the experimental setup, stays the same. Applying the above equation once again, lL -9 Dy = = 400 10 m (3000 ) = 1.2 mm d ( ) ( ) 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: d sin q m = ml fi d = ml sin q m -6 = ( 2 )(600 10 -9 m ) -6 = 1.89 10 m sin (39.5 ) The number of lines in per millimeter is (1 10 -3 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 d sin q m = ml m = 0, 1, 2, 3, ... The slit spacing is d = 1 mm 500 = 2.00 10 -6 m . The angles of diffraction are -1 (2 ) l ^ q 2 = sin ~ = 30 .66 d -9 m^ -1 (1)l ^ -1 510 10 q 1 = sin ~ = sin ~ = 14.77 -6 d 2.00 10 m Likewise, q 3 = 49 .9 . The next order (m = 4) is not seen because sin q 4 = 4 510 10 -9 m 2 10 -6 m > 1.0 Thus, three diffraction orders are visible. 22.14. 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 yp = p So Dy = y2 - y1 = 2l L a 1lL a = Ll a . -4 lL a fia= lL Dy = ( 633 10 -9 m (1.5 m ) ) 0.00475 m = 2.0 10 m = 0.20 mm 22.15. 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 = 200l is w= 2 lL a = 2 500 10 -9 m (2.0 m ) 0.0005 m ( ) = 0.0040 m = 4.0 mm 22.18. 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 l = 3 10 8 m / s 800 10 6 Hz = 0.375 m To investigate the diffraction of these waves through the spacing between the two buildings, we can use the general condition for complete destructive interference: a sin q p = pl (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 p = 1 and obtain the angular width Dq = 2q1 from -1 l ^ -1 0.375 m ^ a sin q 1 = l fi q 1 = sin ~ = sin = 1.43 15 m a Thus, the angular width of the wave after it emerges from between the buildings is Dq = 2 (1.43) = 2.86 . 22.21. 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 D= 2.44 lL w = 2.44 633 10 -9 m ( 4.0 m ) 2. 5 10 -2 ( ) m = 0.25 mm 22.26. Model: An interferometer produces a new maximum each time L2 increases by increase by l. Visualize: Please refer to the interferometer in Figure 22.20. Solve: From Equation 22.33, the distance the mirror moves is DL 2 = Dml 2 = 1 2 l causing the path-length difference Dr to (33,198)(602 .446 10 -9 m ) 2 = 0.0100000 m = 1.00000 cm Assess: Because the wavelength is known to 6 significant figures and the fringes are counted exactly, we can determine DL to 6 significant figures. 22.30. 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, d sin q m = ml . For the m = 3 bright orange fringe, the interference condition is d sin q 3 = 3 600 10 fringes is the same, ( -9 m . For the m = 4 bright fringe the condition is d sin q 4 = 4l . Because the position of the ) d sin q 3 = d sin q 4 = 4l = 3 600 10 ( -9 m fil = ) 3 4 (600 10 -9 m = 450 nm ) 22.34. Solve: According to Equation 22.7, the fringe spacing between the m fringe and the m + 1 fringe is Dy = lL d . Dy can be obtained from Figure P22.34. 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 Dy = 0.50 10 -2 m= lL d fi L= dDy l = (0.20 10 -3 m 0.50 10 -2 m -9 )( 600 10 m ) = 167 cm Assess: A distance of 167 cm from the slits to the screen is reasonable. 22.38. Model: A diffraction grating produces an interference pattern like the diagram of Figure 22.8. 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 nm700 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 qm at which 400 nm light and 700 nm light are diffracted. We then use ym = L tan q m to find their positions on the screen which is at a distance L = 75 cm. -7 The slit spacing is d = 1 mm 1200 = 8.333 10 m . For m = 1, l = 400 nm: q1 = sin ( 400 nm / d ) = 28 .7 fi y1 = 75 cm tan 28.7 = 41 cm l = 700 nm: q1 = sin ( 700 nm / d ) = 57.1 fi y1 = 75 cm tan 57.1 = 116 cm -1 -1 For m = 2, l = 400 nm: q1 = sin ( 2 400 nm / d ) = 73.8 fi y2 = 75 cm tan 73 .8 = 257 cm -1 -1 -1 For the 700 nm wavelength at m = 2, q 2 = sin (2 700 nm / d ) = sin (1.68) is not defined, so y2 . We see that visible light diffracted at m = 1 will fall in the range 41 cm y 116 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 = 1. (b) To determine the wavelengths, we first find the diffraction angle from the observed position by using q = tan -1 y ^ y^ -1 = tan L 75 cm This angle is then used in the diffraction grating equation for the wavelength with m = 1, l = d sinq 1 . q l Line 56.2 cm 65.9 cm 93.5 cm 36.85 41.30 51.27 500 nm 550 nm 650 nm 22.45. 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 = (1. 0 10 m) 800 = 1.25 10 m . Referring to Figure P22.45, the angle of diffraction is given by y 0.436 m tan q 1 = 1 = = 0.436 fi q 1 = 23 .557 fi sinq 1 = 0.400 L 1.0 m Using the constructive-interference condition d sin q m = ml , l = d sinq 1 1 = 1.25 10 -3 -6 ( -6 m (0.400 ) = 500 nm ) We can obtain the same value of l by using the second-order interference fringe. We first obtain q2: tan q 2 = y2 L = 0.436 m + 0.897 m 1.0 m = 1.333 fi q 2 = 53 .12 fi sin q 2 = 0.800 Using the constructive-interference condition, l = d sinq 2 2 = (1.25 10 -6 m (0.800 ) ) 2 = 500 nm Assess: Calculations with the first-order and second-order fringes of the interference pattern give the same value for the wavelength. 22.51. Model: A narrow slit produces a single-slit diffraction pattern. Solve: The dark fringes in this diffraction pattern are given by Equation 22.21: plL p = 1, 2, 3, ... yp = a We note from Figure P22.51 that the first minimum is 0.50 cm away from the central maximum. Using the above equation, the slit width is a= pl L yp = (1)(500 10 -9 m )(1.0 m ) 0.50 10 -2 m = 0.10 mm Assess: This is a typical slit width for diffraction. 22.57. 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 D 0.0015 m (c) The diameter of the laser beam is the width of the diffraction pattern: w= 2.44 lL D 2.44 lL D = q1 = 1.22 l = 1.22 633 10 -9 m ( ) = 5.15 10 ) ) -4 rad = 0.0295 2.44 633 10 -9 m (3 m ) 0.0015 m ( ( = 0.00309 m = 0.31 cm (d) At L = 1 km = 1000 m, the diameter is w= = 2.44 633 10 -9 m (1000 m 0.0015 m ) = 1.03 m = 103 cm 22.60. 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 qm by d sin q m = m l , where m = 0, 1, 2, 3, ... The space between the slits is d= 1.0 mm 600 = 1.6667 10 -6 m For m = 1, sin q 1 = 633 10 -9 m ^ -1 fi q 1 = sin ~ = 22.3 1.6667 10 -6 m d l (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 l 4.759 10 -7 m -7 l = = = 4.759 10 m fi sin q 1 = = = 0.2855 fi q 1 = 16.6 n water 1.33 d 1.6667 10 -6 m 22.69. Model: A diffraction grating produces a series of constructive-interference fringes at values of qm that are determined by Equation 22.15. Solve: (a) The condition for bright fringes is d sin q = m l . If l changes by a very small amount Dl, such that q changes by Dq, then we can approximate Dl/Dq as the derivative dl/dq: l = d m sin q fi Dl Dq dl dq = d m cosq = d m 1 - sin q = Dl d^ 2 -l m 2 2 ml ^ 1- ~ = d m d 2 d^ 2 ~ -l m 2 fi Dq = (b) We can now obtain the first-order and second-order angular separations for the wavelengths l = 589.0 nm and l + Dl = 589.6 nm. The slit spacing is 1.0 10 -3 m -6 d= = 1.6667 10 m 600 The first-order (m = 1) angular separation is Dq = 0.6 10 -9 m 2.7778 10 -4 -12 m - 0.3476 10 2 -12 m 2 = 0.6 10 -9 m 1.5589 10 -6 m = 3.85 10 rad = 0.022 The second order (m = 2) angular separation is Dq = 0.6 10 -9 m 0.6945 10 -12 m 2 - 0.3476 10 -12 m 2 = 1.02 10 -3 rad = 0.058 22.69. Model: A diffraction grating produces a series of constructive-interference fringes at values of qm that are determined by Equation 22.15. Solve: (a) The condition for bright fringes is d sin q = m l . If l changes by a very small amount Dl, such that q changes by Dq, then we can approximate Dl/Dq as the derivative dl/dq: l = d m sin q fi Dl Dq dl dq = d m cosq = d m 1 - sin q = Dl d^ 2 -l m 2 2 ml ^ 1- ~ = d m d 2 d^ 2 ~ -l m 2 fi Dq = (b) We can now obtain the first-order and second-order angular separations for the wavelengths l = 589.0 nm and l + Dl = 589.6 nm. The slit spacing is 1.0 10 -3 m -6 d= = 1.6667 10 m 600 The first-order (m = 1) angular separation is Dq = 0.6 10 -9 m 2.7778 10 -4 -12 m - 0.3476 10 2 -12 m 2 = 0.6 10 -9 m 1.5589 10 -6 m = 3.85 10 rad = 0.022 The second order (m = 2) angular separation is Dq = 0.6 10 -9 m 0.6945 10 -12 m 2 - 0.3476 10 -12 m 2 = 1.02 10 -3 rad = 0.058 ...
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This note was uploaded on 09/27/2008 for the course PHYS 131-133 taught by Professor All during the Spring '08 term at Cal Poly.

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