1031_PartUniversity Physics Solution

1031_PartUniversity Physics Solution - 36-14 Chapter 36 1 =...

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36-14 Chapter 36 (iii) For 0 0 0 1 10, sin ; 5.74 ; 2 11.5 10 a θ θ θ λ = = = ° = ° E VALUATE : Either definition of the width shows that the central maximum gets narrower as the slit gets wider. 36.54. I DENTIFY : The two holes behave like double slits and cause the sound waves to interfere after they pass through the holes. The motion of the speakers causes a Doppler shift in the wavelength of the sound. S ET U P : The wavelength of the sound that strikes the wall is λ = λ 0 v s T s , and destructive interference first occurs where sin θ = λ /2. E XECUTE : (a) First find the wavelength of the sound that strikes the openings in the wall. λ = λ 0 v s T s = v / f s – v s / f s = ( v – v s )/ f s = (344 m/s – 80.0 m/s)/(1250 Hz) = 0.211 m Destructive interference first occurs where d sin θ = λ /2, which gives d = λ /(2 sin θ ) = (0.211 m)/(2 sin 12.7°) = 0.480 m (b) λ = v/f = (344 m/s)/(1250 Hz) = 0.275 m sin θ = λ /2 d = (0.275 m)/[2(0.480 m)] θ = ±16.7° E VALUATE : The moving source produces sound of shorter wavelength than the stationary source, so the angles at which destructive interference occurs are smaller for the moving source than for the stationary source. 36.55. I DENTIFY and S ET U P : sin / a θ λ = locates the first dark band. In the liquid the wavelength changes and this changes the angular position of the first diffraction minimum. E XECUTE : liquid air air liquid sin ; sin a a λ λ θ θ = = liquid liquid air air sin 0.4836 sin θ λ λ θ = = air / n λ λ = (Eq.33.5), so air liquid / 1/0.4836 2.07 n λ λ = = = E VALUATE : Light travels faster in air and n must be 1.00. > The smaller λ in the liquid reduces θ that located the first dark band. 36.56. I DENTIFY : 1 d N = , so the bright fringes are located by 1 sin N θ λ = S ET U P : Red: R 1 sin 700 nm N λ = . Violet: V 1 sin 400 nm N λ = . E XECUTE : R V sin 7 sin 4 θ θ = . R V R V 15 15 θ θ θ θ = ° → = + ° . V V sin( 15 ) 7 . sin 4 θ θ + ° = Using a trig identify from Appendix B, V V V sin cos15 cos sin15 7 4 sin θ θ θ ° + ° = . V cos15 cot sin15 7 4 θ ° + ° = . V V tan 0.330 18.3 θ θ = = ° and R V 15 18.3 15 33.3 θ θ = + ° = ° + ° = ° . Then R 1 sin 700 nm N θ = gives 5 R 9 sin sin 33.3 7.84 10 lines m 7840 lines cm 700 nm 700 10 m N θ = = = × = × D . The spectrum begins at 18 .3 D and ends at 33.3 D . E VALUATE : As N is increased, the angular range of the visible spectrum increases. 36.57. (a) I DENTIFY and S ET U P : The angular position of the first minimum is given by sin a m θ λ = (Eq.36.2), with 1. m = The distance of the minimum from the center of the pattern is given by tan . y x θ = 9 3 3 3 540 10 m sin 1.50 10 ; 1.50 10 rad 0.360 10 m a λ θ θ × = = = × = × × 3 3 1 tan (1.20 m)tan(1.50 10 rad) 1.80 10 m 1.80 mm. y x θ = = × = × = (Note that θ is small enough for sin tan , θ θ θ and Eq.(36.3) applies.) (b) I DENTIFY and S ET U P : Find the phase angle β where 0 / 2. I I = Then use Eq.(36.6) to solve for θ and tan y x θ = to find the distance. E XECUTE : From part (a) of Problem 36.53, 0 1 2 I I = when 2.78 rad. β = 2 sin a π β θ λ = (Eq.(36.6)), so sin .
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