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Unformatted text preview: Husain, Zeena Homework 14 Due: May 3 2004, 4:00 am Inst: Sonia Paban 1 This printout should have 13 questions. Multiplechoice questions may continue on the next column or page find all choices before making your selection. The due time is Central time. 001 (part 1 of 1) 10 points Consider the setup of a single slit experiment shown in the figure. The first minimum for 400 nm light is at y 1 . The first minimum for 530 nm light is at y 2 . y L a S 1 S 2 viewing screen What is the ratio y 2 y 1 ? Correct answer: 1 . 325 . Explanation: The first minimum is at y = L a For the first light, y 1 = 1 L a For the second light, y 2 = 2 L a Therefore y 2 y 1 = 2 1 = (530 nm) (400 nm) = 1 . 325 . 002 (part 1 of 1) 10 points Hint: Use a small angle approximation; e.g. , sin = tan = . Light of wavelength 598 nm falls on a dou ble slit as shown in the schematic figure below. Note: The dimensions y d and y s are not to scale (for obvious reasons)! . 029mm . 15mm y s y d 3 . 7 m S 1 S 2 What is the ratio y s y d of the halfwidth of the central maxima of singleslit diffraction pattern to that of the halfwidth of double slit interference pattern? Correct answer: 10 . 3448 . Explanation: Basic Concepts: For a singleslit diffrac tion pattern, the fullwidth of the central maxima is y single = 2 L a . For a doubleslit interference pattern, the full width of the central maxima is y double = L d . Solution: For the singleslit diffraction pat tern, the fullwidth of the central fringe is y single = 2 L a = 2 (598 nm)(3 . 7 m) . 029 mm = 2 (5 . 98 10 7 m)(3 . 7 m) 2 . 9 10 5 m = 0 . 152593 m . For the doubleslit interference pattern, the width of the central maxima is y double = L d = (598 nm) (3 . 7 m) (0 . 15 mm) = (5 . 98 10 7 m) (3 . 7 m) (0 . 00015 m) = 0 . 0147507 m . Husain, Zeena Homework 14 Due: May 3 2004, 4:00 am Inst: Sonia Paban 2 The ratio of singleslit central maxima to doubleslit central maxima is y s y d = 2 y single 2 y double = (0 . 152593 m) (0 . 0147507 m) = 10 . 3448 , where the fullwidth is twice the halfwidth. Alternative Solution: y s y d = 2 L a L d = 2 d a = 2 (0 . 15 mm) (0 . 029 mm) = 10 . 3448 . 003 (part 1 of 1) 10 points The resolution of a lens can be estimated by treating the lens as a circular aperture. The resolution is the smallest distance between two point sources that produce distinct im ages. This is similar to the resolution of a single slit, related to the distance from the middle of the central bright band to the first order dark band; however, the aperture is circular instead of a rectangular slit which introduces a scale factor. Suppose the Hubble Space Telescope, 2 . 4 m in diameter, is in orbit 90 . 8 km above Earth and is turned to look at Earth....
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 Spring '08
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