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YF_ISM_36

# YF_ISM_36 - DIFFRACTION 36 36.1 IDENTIFY Use y = x tan to...

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36-1 D IFFRACTION 36.1. I DENTIFY : Use tan y x θ = to calculate the angular position θ of the first minimum. The minima are located by Eq.(36.2): sin , m a λ θ = 1, 2, m = ± ± First minimum means 1 m = and 1 sin / a θ λ = and 1 sin . a λ θ = Use this equation to calculate . λ S ET U P : The central maximum is sketched in Figure 36.1. E XECUTE : 1 1 tan y x θ = 1 1 tan y x θ = = 3 3 1.35 10 m 0.675 10 2.00 m × = × 3 1 0.675 10 rad θ = × Figure 36.1 3 3 1 sin (0.750 10 m)sin(0.675 10 rad) 506 nm a λ θ = = × × = E VALUATE : 1 θ is small so the approximation used to obtain Eq.(36.3) is valid and this equation could have been used. 36.2. I DENTIFY : The angle is small, so m m y x a λ = . S ET U P : 1 10.2 mm y = E XECUTE : 7 5 1 3 1 (0.600 m)(5.46 10 m) 3.21 10 m. 10.2 10 m x x y a a y λ λ × = = = × × E VALUATE : The diffraction pattern is observed at a distance of 60.0 cm from the slit. 36.3. I DENTIFY : The dark fringes are located at angles θ that satisfy sin , 1, 2, .... m m a λ θ = = ± ± S ET U P : The largest value of sin θ is 1.00. E XECUTE : (a) Solve for m that corresponds to sin 1 θ = : 3 9 0.0666 10 m 113.8 585 10 m a m λ × = = = × . The largest value m can have is 113. 1 m = ± , 2 ± , …, 113 ± gives 226 dark fringes. (b) For 113 m = ± , 9 3 585 10 m sin 113 0.9926 0.0666 10 m θ × = ± = ± × and 83.0 θ = ± ° . E VALUATE : When the slit width a is decreased, there are fewer dark fringes. When a λ < there are no dark fringes and the central maximum completely fills the screen. 36.4. I DENTIFY and S ET U P : / a λ is very small, so the approximate expression m m y R a λ = is accurate. The distance between the two dark fringes on either side of the central maximum is 1 2 y . E XECUTE : 9 3 1 3 (633 10 m)(3.50 m) 2.95 10 m 2.95 mm 0.750 10 m R y a λ × = = = × = × . 1 2 5.90 mm y = . E VALUATE : When a is decreased, the width 1 2 y of the central maximum increases. 36.5. I DENTIFY : The minima are located by sin m a λ θ = S ET U P : 12.0 cm a = . 40.0 cm x = . 36

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36-2 Chapter 36 E XECUTE : The angle to the first minimum is θ = arcsin a λ = arcsin 9.00 cm 48.6 . 12.00 cm = ° So the distance from the central maximum to the first minimum is just 1 tan y x θ = = (40.0 cm)tan(48.6 ) 45.4 cm. ° = ± E VALUATE : 2 / a λ is greater than 1, so only the 1 m = minimum is seen. 36.6. I DENTIFY : The angle that locates the first diffraction minimum on one side of the central maximum is given by sin a λ θ = . The time between crests is the period T. 1 f T = and v f λ = . S ET U P : The time between crests is the period, so 1.0 h T = . E XECUTE : (a) 1 1 1 1.0 h 1.0 h f T = = = . 1 800 km/h 800 km 1.0 h v f λ = = = . (b) Africa-Antarctica: 800 km sin 4500 km θ = and 10.2 θ = ° . Australia-Antarctica: 800 km sin 3700 km θ = and 12.5 θ = ° . E VALUATE : Diffraction effects are observed when the wavelength is about the same order of magnitude as the dimensions of the opening through which the wave passes. 36.7. I DENTIFY : We can model the hole in the concrete barrier as a single slit that will produce a single-slit diffraction pattern of the water waves on the shore.
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