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. IDENTIFY: Use tan yx θ = to calculate the angular position of the first minimum. The minima are located by Eq.(36.2): sin , m a λ = 1, 2, m ± First minimum means 1 m = and 1 sin / a θλ = and 1 sin . a λθ = Use this equation to calculate . SET UP: The central maximum is sketched in Figure 36.1. EXECUTE: 11 tan = 1 1 tan y x == 3 3 1.35 10 m 0.675 10 2.00 m × 3 1 0.675 10 rad Figure 36.1 33 1 sin (0.750 10 m)sin(0.675 10 rad) 506 nm a −− ==× × = EVALUATE: 1 is small so the approximation used to obtain Eq.(36.3) is valid and this equation could have been used. 36.2. IDENTIFY: The angle is small, so m m a = . SET UP: 1 10.2 mm y = EXECUTE: 7 5 1 3 1 (0.600 m)(5.46 10 m) 3.21 10 m. 10.2 10 m xx ya ay λλ × = × × EVALUATE: The diffraction pattern is observed at a distance of 60.0 cm from the slit. 36.3. IDENTIFY: The dark fringes are located at angles that satisfy sin , 1, 2, . ... m m a ± ± SET UP: The largest value of sin is 1.00. EXECUTE: (a) Solve for m that corresponds to sin 1 = : 3 9 0.0666 10 m 113.8 585 10 m a m × = × . The largest value m can have is 113. 1 m , 2 ± , …, 113 ± gives 226 dark fringes. (b) For 113 m , 9 3 sin 113 0.9926 ⎛⎞ × ⎜⎟ × ⎝⎠ and 83.0 ° . EVALUATE: When the slit width a is decreased, there are fewer dark fringes. When a < there are no dark fringes and the central maximum completely fills the screen. 36.4. IDENTIFY and SET UP: / a is very small, so the approximate expression m m yR a = is accurate. The distance between the two dark fringes on either side of the central maximum is 1 2 y . EXECUTE: 9 3 1 3 (633 10 m)(3.50 m) 2.95 10 m 2.95 mm 0.750 10 m R y a × = × = × . 1 25 . 9 0 m m y = . EVALUATE: When a is decreased, the width 1 2 y of the central maximum increases. 36.5. IDENTIFY: The minima are located by sin m a = SET UP: 12.0 cm a = . 40.0 cm x = . 36
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36-2 Chapter 36 EXECUTE: 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 yx == (40.0 cm)tan(48.6 ) 45.4 cm. °=± EVALUATE: 2/ a is greater than 1, so only the 1 m = minimum is seen. 36.6. IDENTIFY: The angle that locates the first diffraction minimum on one side of the central maximum is given by sin a = . The time between crests is the period T. 1 f T = and v f = . SET UP: The time between crests is the period, so 1.0 h T = . EXECUTE: (a) 1 11 1.0 h 1.0 h f T = . 1 800 km/h 800 km 1.0 h v f = . (b) Africa-Antarctica: 800 km sin 4500 km = and 10.2 = ° . Australia-Antarctica: 800 km sin 3700 km = and 12.5 = ° . EVALUATE: 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. IDENTIFY: 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. SET UP: For single-slit diffraction, the angles at which destructive interference occurs are given by sin m = m / a , where m = 1, 2, 3, ….
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YF_ISM_36 - DIFFRACTION 36 36.1. IDENTIFY: Use y = x tan to...

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