# 34 - Kapoor(mk9499 – homework34 – Turner –(60230 1...

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Unformatted text preview: Kapoor (mk9499) – homework34 – Turner – (60230) 1 This print-out should have 13 questions. Multiple-choice 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 . 675 is incident on the glass surface. An unknown liquid condenses on the surface of the glass. Total internal reflection on the glass-liquid interface of 54 . 3 ◦ . As- sume the glass and liquid have perfect planar surfaces. What is the refractive index of the unknown liquid? Correct answer: 1 . 36024. Explanation: Let : n g = 1 . 675 , and θ c = 54 . 3 ◦ . The critical angle at the glass-liquid interface is sin θ c = n l n g , so the refraction index of the liquid is n l = n g sin θ c = 1 . 675 sin 54 . 3 ◦ = 1 . 36024 . 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: 36 . 6564 ◦ . Explanation: Let : n g = 1 . 675 With the liquid removed, the critical angle becomes θ ′ c = sin − 1 parenleftbigg 1 n g parenrightbigg = sin − 1 parenleftbigg 1 1 . 675 parenrightbigg = 36 . 6564 ◦ . 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: 47 . 3211 ◦ . Explanation: Let : n g = 1 . 675 n l = 1 . 36024 , and θ ′ c = 36 . 6564 ◦ . Applying Snell’s law, n g sin θ ′ c = n l sin θ l , so θ l = sin − 1 parenleftbigg n g sin θ ′ c n l parenrightbigg = sin − 1 parenleftbigg 1 . 675sin 36 . 6564 ◦ 1 . 36024 parenrightbigg = 47 . 3211 ◦ . 004 (part 1 of 2) 10.0 points The top of a swimming pool is at ground level. If the pool is 2 . 8 m deep, how far below ground level does the bottom of the pool ap- pear to be located when the pool is completely filled with water? Correct answer: 2 . 10526 m. Explanation: Given : n 1 = 1 . 33 , n 2 = 1 , and p = 2 . 8 m . For a plane surface, n 1 p + n 2 q = n 2- n 1 R be- comes q 1 =- n 2 n 1 p since R → ∞ , so q 1 =- n 2 n 1 p = parenleftbigg- 1 1 . 33 parenrightbigg (2 . 8 m) =- 2 . 10526 m ....
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## This note was uploaded on 06/28/2009 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas.

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34 - Kapoor(mk9499 – homework34 – Turner –(60230 1...

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