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Homework_20

# Homework_20 - Nguyen Thanh Homework 20 Due 7:00 pm Inst...

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Nguyen, Thanh – Homework 20 – Due: Apr 18 2008, 7:00 pm – Inst: Weathers 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A 1 . 73 m long pole stands vertically in a lake having a depth of 0 . 81 m. When the Sun is 35 . 7 above the horizon, find the length of the pole’s shadow on the bottom of the lake. The index of refraction for water is 1 . 33. Correct answer: 1 . 90482 m. Explanation: θ φ 2 s 2 pole s 1 l φ 1 d Given : = 1 . 73 m , d = 0 . 81 m , n = 1 . 33 , and θ = 35 . 7 . Light just passing the top of the pole has an angle of incidence of φ 1 = 90 - θ = 90 - 35 . 7 = 54 . 3 . This light falls on the water’s surface at dis- tance s 1 = ( - d ) tan φ 1 = (1 . 73 m - 0 . 81 m) tan 54 . 3 = 1 . 28031 m from the pole. As the light enters the water, Snell’s Law becomes sin φ 1 = n sin φ 2 φ 2 = arcsin sin φ 1 n = arcsin sin 54 . 3 1 . 33 = 37 . 6321 . Then distance between the extremity of the pole’s shadow on the water’s surface and the extremity of the pole’s shadow on the bottom of the lake is s 2 = d tan φ 2 = (0 . 81 m) tan 37 . 6321 = 0 . 624507 m . The whole shadow’s length is s = s 1 + s 2 = 1 . 28031 m + 0 . 624507 m = 1 . 90482 m . Note: The shadow’s length without the water would be s = tan φ 1 so its length is shortened by the water by Δ s = s - s = (1 . 73 m) tan 54 . 3 - 1 . 90482 m = 0 . 502726 m . keywords: 002 (part 1 of 4) 10 points Light of wavelength 396 nm is incident on the face of a silica prism at an angle of θ 1 = 56 . 2 (with respect to the normal to the surface). The apex angle of the prism is φ = 72 . 6 . Given: The value of the index of refraction for silica is n = 1 . 455. θ 2 θ 3 θ 1 φ θ 4 Find the angle of refraction at this first surface. Correct answer: 34 . 8286 .

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Nguyen, Thanh – Homework 20 – Due: Apr 18 2008, 7:00 pm – Inst: Weathers 2 Explanation: Basic Concept: Snell’s Law n 1 sin θ 1 = n 2 sin θ 2 . Solution: Applying Snell’s law, the refrac- tion angle at the first surface θ 2 = sin - 1 sin θ 1 n = sin - 1 sin 56 . 2 1 . 455 = 34 . 8286 . 003 (part 2 of 4) 10 points Find the angle of incidence at the second sur- face. Correct answer: 37 . 7714 . Explanation: The incident angle at the second surface is θ 3 = φ - θ 2 = (72 . 6 ) - (34 . 8286 ) = 37 . 7714 , where φ is the apex angle of the prism. 004 (part 3 of 4) 10 points Find the angle of refraction at the second surface. Correct answer: 63 . 0252 . Explanation: The incident angle at the second surface is θ 3 = 37 . 7714 , so the refraction angle is θ 4 = sin - 1 [( n sin θ 3 )] = sin - 1 [(1 . 455) sin(37 . 7714 )] = 63 . 0252 . 005 (part 4 of 4) 10 points Find the angle between the incident and emerging rays.
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Homework_20 - Nguyen Thanh Homework 20 Due 7:00 pm Inst...

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