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Unformatted text preview: Wong, Lemuel Homework 3 Due: Feb 9 2007, 11:00 pm Inst: Matt Mackie 1 This print-out should have 23 questions. Multiple-choice 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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