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Unformatted text preview: Wong, Lemuel Homework 3 Due: Feb 9 2007, 11:00 pm Inst: Matt Mackie 1 This printout should have 23 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. Note: HW is due 11pm central time, which is midnight out time. 001 (part 1 of 2) 10 points Plane sound waves of wavelength 0 . 17 m are incident on two narrow slits in a box with nonreflecting walls, as shown in the figure below. At a perpendicular distance of 12 m from the center of slits, a first order maximum occurs at point which is 7 m from the central maximum. 7m 12 m sound wavelength . 17 m Using the small angle approximation, what is the distance between the two slits? Correct answer: 0 . 291429 m. Explanation: Basic Concept: The rules for determining interference maximum or minimum are the same for sound waves and light waves, so the path length difference is = d 2 d 1 = n, (1) where n = 1 for the first maximum. Solution: Double Slit interference. y d 2 y y + d 2 L d 1 d 2 d Let : = 0 . 17 m , L = 12 m , and y = 7 m , The approximation sin = d requires L d , which does NOT apply here; the sig nals are NOT traveling nearly parallel to each other. We must go back to the definitions and basic concepts of constructive and destruc tive interference. From the picture and using the Pythagorean Theorem, the wave from the upper slit travels a distance d 1 = s L 2 + y d 2 2 and the wave from the lower slit travels a distance d 2 = s L 2 + y + d 2 2 [ d 2 + d 1 ][ d 2 d 1 ] = d 2 2 d 2 1 [ d 2 + d 1 ] n = L 2 + y + d 2 2 L 2 y d 2 2 = y 2 + y d 2 + d 2 4 y 2 + y d 2 d 2 4 = y d. Since d 1 = s (12 m) 2 + 7 m . 291429 m 2 2 = 13 . 8196 m and d 2 = s (12 m) 2 + 7 m + . 291429 m 2 2 = 13 . 9664 m , d = n [ d 2 + d 1 ] y = (1)(0 . 17 m) 7 m [(13 . 9664 m) + (13 . 8196 m)] = . 337402 m . Wong, Lemuel Homework 3 Due: Feb 9 2007, 11:00 pm Inst: Matt Mackie 2 Since the receiver is at the first maximum, n = 1. From trigonometry, tan y L . For constructive interference (using the approxi mation), n = = d sin . This approxima tion assumes that L d , which is only good to a few percent in this case. Solving for d , we have d = n sin = n sin h arctan y L i = (1)(0 . 17 m) sin arctan 7 m 12 m = 0 . 337388 m . Such an estimate, if L > d , is usually fairly good, but may not be close enough to give one percent accuracy. 002 (part 2 of 2) 0 points What is the percent error in the determination of the distance between the sits when using the small angle approximation. Correct answer: 13 . 6257 %....
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This note was uploaded on 02/17/2011 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Physics, Work

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