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panda 2 - pokharel(pp7242 Light as a wave walther(16180...

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pokharel (pp7242) – Light as a wave – walther – (16180) 1 This print-out should have 37 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Good Luck 001 10.0 points Will the light from two very close stars pro- duce an interference pattern? 1. Yes; if they are very close to each other, the light has large overlapping area. 2. Yes; strong light from the two stars cre- ates very clear interference patterns. 3. No; light from two very close stars is not likely to have the same frequency. correct 4. No; the light is too strong to produce a discernable interference pattern. Explanation: Light from a pair of stars will not produce an interference pattern because the waves of light from the two separate sources have dif- ferent frequencies and thus are incoherent; when combined they smudge. Interference occurs when light from a single source divides and recombines. 002 10.0 points Plane sound waves of wavelength 0 . 13 m are incident on two narrow slits in a box with nonreflecting walls, as shown in the figure below. At a perpendicular distance of 10 m from the center of slits, a first order maximum occurs at a point which is 1 m from the central maximum. 1 m 10 m sound wavelength 0 . 13 m What is the distance between the two slits? Correct answer: 1 . 30919 m. Explanation: BasicConcept: The rules for determining interference maximum or minimum are the same for sound waves and light waves. Thus, the path length difference is δ = d 2 d 1 = n λ , (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 : λ = 0 . 13 m , L = 10 m , and y = 1 m , The approximation sin θ = δ 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 = radicalBigg L 2 + parenleftbigg y d 2 parenrightbigg 2 and the wave from the lower slit travels a distance d 2 = radicalBigg L 2 + parenleftbigg y + d 2 parenrightbigg 2 [ d 2 + d 1 ] [ d 2 d 1 ] = d 2 2 d 2 1 , so [ d 2 + d 1 ] n λ = L 2 + parenleftbigg y + d 2 parenrightbigg 2 L 2 parenleftbigg y d 2 parenrightbigg 2 = y 2 + y d + d 2 4 y 2 + y d d 2 4 = 2 y d .
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pokharel (pp7242) – Light as a wave – walther – (16180) 2 Since d 1 = radicalBigg (10 m) 2 + parenleftbigg 1 m 1 . 3 m 2 parenrightbigg 2 = 10 . 0061 m and d 2 = radicalBigg (10 m) 2 + parenleftbigg 1 m + 1 . 3 m 2 parenrightbigg 2 = 10 . 1352 m , d = n λ [ d 2 + d 1 ] 2 y = (1) (0 . 13 m) 2 (1 m) × [(10 . 1352 m) + (10 . 0061 m)] = 1 . 30919 m . Alternative Approximate Solution: Since the receiver is at the first maximum, n = 1. From trigonometry, tan θ y L . For constructive interference (using the approxi- mation), n λ = δ = d sin θ . 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 λ sin θ = n λ sin bracketleftBig arctan parenleftBig y L parenrightBigbracketrightBig = (1) (0 . 13 m) sin bracketleftbigg arctan parenleftbigg 1 m 10 m
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