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Unformatted text preview: nguyen (jmn727) oldhomework 35 Turner (59070) 1 This printout should have 10 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 3) 10.0 points A light ray in dense flint glass with an index of refraction of 1 . 635 is incident on the glass surface. An unknown liquid condenses on the surface of the glass. Total internal reflection on the glassliquid interface of 54 . 6 . As sume the glass and liquid have perfect planar surfaces. What is the refractive index of the unknown liquid? Correct answer: 1 . 33273. Explanation: Let : n g = 1 . 635 , and c = 54 . 6 . The critical angle at the glassliquid interface is sin c = n l n g , so the refraction index of the liquid is n l = n g sin c = 1 . 635 sin 54 . 6 = 1 . 33273 . 002 (part 2 of 3) 10.0 points If the liquid is removed, what is the angle of incidence for total internal reflection? Correct answer: 37 . 7068 . Explanation: Let : n g = 1 . 635 With the liquid removed, the critical angle becomes c = sin 1 parenleftbigg 1 n g parenrightbigg = sin 1 parenleftbigg 1 1 . 635 parenrightbigg = 37 . 7068 . 003 (part 3 of 3) 10.0 points For the angle of incidence found in previous question, what is the angle of refraction of the ray into the liquid film? Correct answer: 48 . 6196 . Explanation: Let : n g = 1 . 635 n l = 1 . 33273 , and c = 37 . 7068 . Applying Snells law, n g sin c = n l sin l , so l = sin 1 parenleftbigg n g sin c n l parenrightbigg = sin 1 parenleftbigg 1 . 635sin 37 . 7068 1 . 33273...
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 Spring '08
 Turner
 Work, Light

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