Homework 5 Solution

Homework 5 Solution - Version PREVIEW Homework 5 Savrasov...

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Unformatted text preview: Version PREVIEW Homework 5 Savrasov (39825) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. Double Slits 02 JMS 001 10.0 points The second-order bright fringe ( m = 2) is 4 . 33 cm from the center line. 4 . 33cm 1 . 17 m . 0292mm S 1 S 2 viewing screen Determine the wavelength of the light. Be sure to use the small angle approximation, sin ( ) Correct answer: 540 . 325 nm. Explanation: Let : y = 4 . 33 cm , L = 1 . 17 m , and d = 0 . 0292 mm , r 2 r 1 y L d S 1 S 2 = ta n 1 parenleftBig y L parenrightBig viewing screen d sin r 2- r 1 O[(182417955]TJ/R14 8.96638 -41 .22391 1830.2 9(6)TjETQq 46 -11 0(i)6 1354.402151 1720 0BI/IM true/W 1/H 1/BPC 1ID �EI QlSq10 0 0 Version PREVIEW Homework 5 Savrasov (39825) 2 glass and from the coating will interfere de- structively.) The wavelength of light inside the coating is n where n is the index of re- fraction of the coating. This implies that the thickness of the coating must be one fourth the wavelength of light inside this medium. Hence (calling the thickness of the material t , the wavelength of light in air , and the index of refraction of the medium n ) t = 4 n = 4 . 38 10 7 m (4) (1 . 43) t = 7 . 65734 10 8 m = . 0765734 m . 003 (part 2 of 2) 10.0 points Now assume that the coatings index of re- fraction is 1 . 64. Assume that the rest of the system (from the previous question) remains the same. What is the minimum thickness of the coat- ing needed to minimize the reflection of this light now? Correct answer: 0 . 133537 m. Explanation: This part is really solved in the same way as Part 1 except now the index of refrac- tion of the material is greater than that for glass. Hence the light reflected from the mate-glass....
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Homework 5 Solution - Version PREVIEW Homework 5 Savrasov...

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