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Unformatted text preview: Husain, Zeena Homework 14 Due: May 3 2004, 4:00 am Inst: Sonia Paban 1 This print-out should have 13 questions. Multiple-choice 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 half-width of the central maxima of single-slit diffraction pattern to that of the half-width of double- slit interference pattern? Correct answer: 10 . 3448 . Explanation: Basic Concepts: For a single-slit diffrac- tion pattern, the full-width of the central maxima is y single = 2 L a . For a double-slit interference pattern, the full- width of the central maxima is y double = L d . Solution: For the single-slit diffraction pat- tern, the full-width 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 double-slit 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 single-slit central maxima to double-slit central maxima is y s y d = 2 y single 2 y double = (0 . 152593 m) (0 . 0147507 m) = 10 . 3448 , where the full-width is twice the half-width. 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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