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Unformatted text preview: Chapter 36 1. The condition for a minimum of a single-slit diffraction pattern is a m sin = where a is the slit width, is the wavelength, and m is an integer. The angle is measured from the forward direction, so for the situation described in the problem, it is 0.60 for m = 1. Thus a m = = = -- sin sin . . . 633 10 0 60 6 04 10 9 5 m m 2. (a) = sin 1 (1.50 cm/2.00 m) = 0.430. (b) For the m th diffraction minimum a sin = m . We solve for the slit width: a m = = = sin sin . . . 2 441 0 430 0118 nm mm b g 3. (a) The condition for a minimum in a single-slit diffraction pattern is given by a sin = m , where a is the slit width, is the wavelength, and m is an integer. For = a and m = 1, the angle is the same as for = b and m = 2. Thus a = 2 b = 2(350 nm) = 700 nm. (b) Let m a be the integer associated with a minimum in the pattern produced by light with wavelength a , and let m b be the integer associated with a minimum in the pattern produced by light with wavelength b . A minimum in one pattern coincides with a minimum in the other if they occur at the same angle. This means m a a = m b b . Since a = 2 b , the minima coincide if 2 m a = m b . Consequently, every other minimum of the b pattern coincides with a minimum of the a pattern. With m a = 2, we have m b = 4. (c) With m a = 3, we have m b = 6. 4. (a) Eq. 36-3 and Eq. 36-12 imply smaller angles for diffraction for smaller wavelengths. This suggests that diffraction effects in general would decrease. (b) Using Eq. 36-3 with m = 1 and solving for 2 (the angular width of the central diffraction maximum), we find 1451 CHAPTER 36 2 2 2 050 50 11 1 1 = F H G I K J = F H G I K J = -- sin sin . . . a m m (c) A similar calculation yields 0.23 for = 0.010 m. 5. (a) A plane wave is incident on the lens so it is brought to focus in the focal plane of the lens, a distance of 70 cm from the lens. (b) Waves leaving the lens at an angle to the forward direction interfere to produce an intensity minimum if a sin = m , where a is the slit width, is the wavelength, and m is an integer. The distance on the screen from the center of the pattern to the minimum is given by y = D tan , where D is the distance from the lens to the screen. For the conditions of this problem, sin . . . = = = --- m a 1 590 10 0 40 10 1475 10 9 3 3 bgc h m m This means = 1.475 10 3 rad and y = (70 10 2 m) tan (1.475 10 3 rad) = 1.0 10 3 m. 6. (a) We use Eq. 36-3 to calculate the separation between the first ( m 1 = 1) and fifth 2 ( 5) m = minima: y D D m a D a m D a m m = = F H G I K J = =- sin ....
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