# hw21b - aljabr(faa335 Hw21 Ross(89251 This print-out should...

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aljabr (faa335) – Hw21 – Ross – (89251) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – Fnd all choices before answering. 001 10.0 points In the Fgure, a ray of light enters the liquid from air and is bent toward the normal as shown. 10 13 . 3 13 . 3 10 n What is the index of refraction n for the liquid? Correct answer: 1 . 33. Explanation: Let : a = 13 . 3 , and b = 10 . b a a b c c θ 1 θ 2 Basic Concept: Snell’s Law: n 1 sin θ 1 = n 2 sin θ 2 Solution: Since the light is incident in air, which has an index of refraction of n 1 = 1 , and sin θ 1 = a c sin θ 2 = b c , so 1 a c = n b c n 2 = a c c b = a b = 13 . 3 10 = 1 . 33 , and θ 1 = arctan p a b P = arctan ± 13 . 3 10 ² = 53 . 0612 θ 2 = arctan ± b a ² = arctan ± 10 13 . 3 ² = 36 . 9388 . 002 (part 1 of 3) 10.0 points The wavelength of red light in air is 376 nm . What is its frequency? Correct answer: 7 . 9732 × 10 14 Hz. Explanation: The relationship of speed of light, its wave- length and its frequency is c = f λ , so the frequency is f = c λ = (2 . 99792 × 10 8 m / s) (376 nm) (1 × 10 9 m / nm) = 7 . 9732 × 10 14 Hz .

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aljabr (faa335) – Hw21 – Ross – (89251) 2 003 (part 2 of 3) 10.0 points What is its wavelength in glass that has an index of refraction of 1 . 5? Correct answer: 250
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## This note was uploaded on 01/27/2011 for the course PHYS 251 taught by Professor Gavrine during the Spring '10 term at Purdue.

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hw21b - aljabr(faa335 Hw21 Ross(89251 This print-out should...

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