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Unformatted text preview: Chapter 36 1. The condition for a minimum of a singleslit diffraction pattern is a m sin Î¸ = Î» where a is the slit width, Î» is the wavelength, and m is an integer. The angle Î¸ is measured from the forward direction, so for the situation described in the problem, it is 0.60Â° for m = 1. Thus a m = = Ã— Â° = Ã— Î» sin sin . . . Î¸ 633 10 0 60 6 04 10 9 5 m m 2. (a) Î¸ = sin â€“1 (1.50 cm/2.00 m) = 0.430Â°. (b) For the m th diffraction minimum a sin Î¸ = m Î» . We solve for the slit width: a m = = Â° = Î» sin sin . . . Î¸ 2 441 0 430 0118 nm mm b g 3. (a) The condition for a minimum in a singleslit diffraction pattern is given by a sin Î¸ = m Î» , where a is the slit width, Î» is the wavelength, and m is an integer. For Î» = Î» a and m = 1, the angle Î¸ is the same as for Î» = Î» b and m = 2. Thus Î» a = 2 Î» 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 Î» a , and let m b be the integer associated with a minimum in the pattern produced by light with wavelength Î» b . A minimum in one pattern coincides with a minimum in the other if they occur at the same angle. This means m a Î» a = m b Î» b . Since Î» a = 2 Î» b , the minima coincide if 2 m a = m b . Consequently, every other minimum of the Î» b pattern coincides with a minimum of the Î» a pattern. With m a = 2, we have m b = 4. (c) With m a = 3, we have m b = 6. 4. (a) Eq. 363 and Eq. 3612 imply smaller angles for diffraction for smaller wavelengths. This suggests that diffraction effects in general would decrease. (b) Using Eq. 363 with m = 1 and solving for 2 Î¸ (the angular width of the central diffraction maximum), we find 1451 CHAPTER 36 2 2 2 050 50 11 1 1 Î¸ = F H G I K J = F H G I K J = Â° sin sin . . . Î» a m m (c) A similar calculation yields 0.23Â° for Î» = 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 Î¸ to the forward direction interfere to produce an intensity minimum if a sin Î¸ = m Î» , where a is the slit width, Î» 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 Î¸ , where D is the distance from the lens to the screen. For the conditions of this problem, sin . . . Î¸ = = Ã— Ã— = Ã— m a Î» 1 590 10 0 40 10 1475 10 9 3 3 bgc h m m This means Î¸ = 1.475 Ã— 10 â€“3 rad and y = (70 Ã— 10 â€“2 m) tan (1.475 Ã— 10 â€“3 rad) = 1.0 Ã— 10 â€“3 m. 6. (a) We use Eq. 363 to calculate the separation between the first ( m 1 = 1) and fifth 2 ( 5) m = minima: âˆ† âˆ† âˆ† âˆ† y D D m a D a m D a m m = = F H G I K J = = sin ....
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This note was uploaded on 04/19/2008 for the course PHYS 1100, 1200 taught by Professor Many during the Fall '08 term at Rensselaer Polytechnic Institute.
 Fall '08
 Many
 Physics, Diffraction

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