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Unformatted text preview: Version PREVIEW – Homework 5 – Savrasov – (39825) 1 This printout should have 10 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. Double Slits 02 JMS 001 10.0 points The secondorder bright fringe ( m = 2) is 4 . 33 cm from the center line. 4 . 33cm 1 . 17 m . 0292mm S 1 S 2 θ viewing screen Determine the wavelength of the light. Be sure to use the small angle approximation, sin ( θ ) ≈ θ 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 θ = ta n − 1 parenleftBig y L parenrightBig viewing screen δ ≈ d sin θ ≈ 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 λ 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 λ , and the index of refraction of the medium n ) t = λ 4 n = 4 . 38 × 10 − 7 m (4) (1 . 43) t = 7 . 65734 × 10 − 8 m = . 0765734 μ m . 003 (part 2 of 2) 10.0 points Now assume that the coating’s 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 μ 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 mateglass....
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This note was uploaded on 01/10/2010 for the course PHY na taught by Professor Richardscalettar during the Fall '09 term at UC Davis.
 Fall '09
 RichardScalettar
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