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Chapter 24

# Chapter 24 - CHAPTER 24 Quick Quizzes 1 2 n(b The outer...

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C HAPTER 24 Quick Quizzes 1. (b). The outer edges of the central maximum occur where sin a θ λ = ± . Thus, as a , the width of the slit, becomes smaller, the width of the central maximum will increase. 2. The compact disc. The tracks of information on a compact disc are much closer together than on a phonograph record. As a result, the diffraction maxima from the compact disc will be farther apart than those from the record. 269

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CHAPTER 24 Problem Solutions 24.1 () 1 1 bright m m LL yyy m m dd L d λ λλ + ∆=− = + − = ( ) 9 2 3 632. 81 0 m5 . 00 m 1.58 10 m 0.200 m × == × 1.58 cm = × 24.2 (a) For a bright fringe of order m , the path difference is m δ = , where At the location of the third order bright fringe, 0,1,2, m =… 3 m = and ( ) 3 3 3 589 nm 1.77 10 nm δλ = × = 1.77 m µ (b) For a dark fringe, the path difference is 1 2 m ⎛⎞ =+ ⎜⎟ ⎝⎠ , where At the third dark fringe, m 2 m = and 3 15 2 589 nm 1.47 22 = = × = 1.47 m 24.3 (a) The distance between the central maximum and the first order bright fringe is mm L yy y d ∆= = , or ( ) ( ) 9 3 3 546. 11 m1 . 20 m 2.62 m 0.250 m L y d × = = × = × 2.62 (b) The distance between the first and second dark bands is dark L y d = = as in (a) above. 24.4 From 1 2 L ym d , the spacing between the first and second dark fringes is 31 L y L = . Thus, the required distance to the screen is ( ) ( )( ) 3- 3 9 4.00 m0 . 300 10 m 460 m yd L ×× = 2.61 m × 270
CHAPTER 24 24.5 (a) From sin dm θ λ = , the angle for the 1 m = maximum for the sound waves is 11 1 354 m s 1 si ns i i n 0.300 m 2000 Hz sound v mm dd f θλ −− ⎡⎤ ⎛⎞ == = = ⎢⎥ ⎜⎟ ⎝⎠ ⎣⎦ 36.2 ° (b) For 3.00-cm microwaves, the required slit spacing is ( )( ) 13 .00 cm i n36.2 m d = 5.08 cm ° (c) The wavelength is d m = ; and if this is light, the frequency is ( ) ( ) ( ) 8 -6 .00 10 m s 1.00 10 ms in 36.2 mc c f d λθ × = = × ° 14 5.08 10 Hz × 24.6 The position of the first order bright fringe for wavelength is 1 L y d = . Thus, () ( ) ( ) 9 3 1 3 700 400 10 m1 . 5 m 1. 51 0 m 0.30 m L y d −× ∆= = = × = × 1.5 24.7 Note, with the conditions given, the small angle approximation does not work well . That is, n, tan , and are significantly different. The approach to be used is outlined below. (a) At the 2 m = maximum, n2 d δ = = , or 22 y Ly + or ( ) 300 m 400 m 2 1000 m4 0 0 m + 55.7 m 1000 m 300 m 400 m 271

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CHAPTER 24 (b) The next minimum encountered is the 2 m = minimum; and at that point, 15 sin 22 dm δ θλ ⎛⎞ == + = ⎜⎟ ⎝⎠ λ , or () 11 5 55.7 m 5 si ns i n2 7 2 2 300 m d θ −− = 7 . ° Then, , so the car must travel an additional 1000 mtan 27.7 524 m y = 124 m 24.8 For bright fringes, the path difference is = = . Thus, ()( ) 9 6 1 575 10 m 2.02 m i n 16.5 m d × = 2.02 m ° µ 24.9 The path difference in the two waves received at the home is 2 d = , where d is the distance from the home to the mountain. Neglecting any phase change upon reflection, the condition for destructive interference is 1 2 m =+ with 0,1,2, m = , so 13 0 0 m 0 22 4 4 min d = 75.0 m 24.10 Since the screen distance from the central maximum to the maximum of order m is bright L y d m = , the wavelength is given by ( )( ) ( ) 33 7 3.40 m0 .
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Chapter 24 - CHAPTER 24 Quick Quizzes 1 2 n(b The outer...

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