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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 θ viewing screen What is the ratio y 2 y 1 ? Correct answer: 1 . 325 . Explanation: The first minimum is at y = λL a For the first light, y 1 = λ 1 L a For the second light, y 2 = λ 2 L a Therefore y 2 y 1 = λ 2 λ 1 = (530 nm) (400 nm) = 1 . 325 . 002 (part 1 of 1) 10 points Hint: Use a small angle approximation; e.g. , sin θ = tan θ = θ . 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 θ 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 λL a . For a doubleslit interference pattern, the full width of the central maxima is y double = λL d . Solution: For the singleslit diffraction pat tern, the fullwidth of the central fringe is y single = 2 λL 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 = λL 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 λL a λL 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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This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
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