CH-36 - Chapter 36 1 The condition for a minimum of a...

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Chapter 36 1. The condition for a minimum of a single-slit diffraction pattern is a m sin θ = λ where a is the slit width, λ is the wavelength, and m is an integer. The angle θ is measured from the forward direction, so for the situation described in the problem, it is 0.60° for m = 1. Thus a m = = × ° = × - - λ sin sin . . . θ 633 10 0 60 6 04 10 9 5 m m 2. (a) θ = sin –1 (1.50 cm/2.00 m) = 0.430°. (b) For the m th diffraction minimum a sin θ = m λ . We solve for the slit width: a m = = ° = λ sin sin . . . θ 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 θ = m λ , where a is the slit width, λ is the wavelength, and m is an integer. For λ = λ a and m = 1, the angle θ is the same as for λ = λ b and m = 2. Thus λ a = 2 λ 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 λ a , and let m b be the integer associated with a minimum in the pattern produced by light with wavelength λ b . A minimum in one pattern coincides with a minimum in the other if they occur at the same angle. This means m a λ a = m b λ b . Since λ a = 2 λ b , the minima coincide if 2 m a = m b . Consequently, every other minimum of the λ b pattern coincides with a minimum of the λ 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 θ (the angular width of the central diffraction maximum), we find 1451

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CHAPTER 36 2 2 2 050 50 11 1 1 θ = F H G I K J = F H G I K J = ° - - sin sin . . . λ a m m (c) A similar calculation yields 0.23° for λ = 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 θ to the forward direction interfere to produce an intensity minimum if a sin θ = m λ , where a is the slit width, λ 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 θ , where D is the distance from the lens to the screen. For the conditions of this problem, sin . . . θ = = × × = × - - - m a λ 1 590 10 0 40 10 1475 10 9 3 3 bgc h m m This means θ = 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 . θ λ λ λ 2 1 b g Solving for the slit width, we obtain a D m m y = - = × - = - λ 2 1 6 400 550 10 5 1 0 35 2 5 b g b gc hb g mm mm mm mm . . . (b) For m = 1, sin . . . θ = = × = × - - m a λ 1 550 10 2 5 2 2 10 6 4 bgc h mm mm The angle is θ = sin –1 (2.2 × 10 –4 ) = 2.2 × 10 –4 rad. 7. The condition for a minimum of intensity in a single-slit diffraction pattern is a sin θ = m λ , where a is the slit width, λ is the wavelength, and m is an integer. To find the angular position of the first minimum to one side of the central maximum, we set m = 1: 172
θ 1 1 1 9 3 4 589 10 100 10 589 10 = F H G I K J = × × F H G I

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