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 – fnd 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 nonre±ecting walls, as shown in the fgure below. At a perpendicular distance oF 10 m From the center oF slits, a frst 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: Basic Concept: The rules For determining interFerence maximum or minimum are the same For sound waves and light waves. Thus, the path length di²erence is δ = d 2 d 1 = n λ , (1) where n = 1 For the frst maximum. Solution: Double Slit interFerence. y d 2 y y + d 2 L d 1 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 defnitions and basic concepts oF constructive and destruc- tive interFerence. ³rom the picture and using the Pythagorean Theorem, the wave From the upper slit travels a distance d 1 = r L 2 + p y d 2 P 2 and the wave From the lower slit travels a distance d 2 = r L 2 + p y + d 2 P 2 [ d 2 + d 1 ] [ d 2 d 1 ] = d 2 2 d 2 1 , so [ d 2 + d 1 ] n λ = L 2 + p y + d 2 P 2 L 2 p y d 2 P 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 = r (10 m) 2 + p 1 m 1 . 3 m 2 P 2 = 10 . 0061 m and d 2 = r (10 m) 2 + p 1 m + 1 . 3 m 2 P 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 frst maximum, n = 1. From trigonometry, tan θ y L . For constructive inter±erence (using the approxi- mation), n λ = δ = d sin θ . This approxima- tion assumes that L d , which is only good to a ±ew percent in this case. Solving ±or d , we have d = n λ sin θ = n λ sin b arctan ± y L ²B = (1) (0 . 13 m) sin
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This note was uploaded on 06/07/2010 for the course PHYSICS 58930 taught by Professor Kleinman during the Fall '09 term at University of Texas at Austin.

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

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