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Unformatted text preview: Golshani (ag5873) – Practice Midterm 01 – Yao – (57045) 1 This printout should have 11 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. No credit is assigned to the practice exam so completing it is optional. 001 10.0 points A thin layer of liquid methylene iodide (with index of refraction 1 . 756) is sandwiched be tween two flat parallel plates of glass (of index of refraction 1 . 53). What must be the minimum thickness of the liquid layer if normally incident light of wavelength 480 nm in air is to be strongly reflected? Correct answer: 68 . 3371 nm. Explanation: Let : n film = 1 . 756 , n glass = 1 . 53 , and λ = 480 nm . A phase change of π occurs upon reflection at the first interface, but not at the second. The condition for strong reflection is then 2 t = parenleftbigg m + 1 2 parenrightbigg λ n = 2 m + 1 2 λ n film , m = 0 , 1 , 2 , ... , so for m = 0, t = λ 4 n film = 480 nm 4(1 . 756) = 68 . 3371 nm . 002 10.0 points A ray of light traveling in air strikes the sur face of mineral oil at an angle of 20.2 ◦ with the normal to the surface. If the light travels at 2 . 17 × 10 8 m / s through the oil, what is the angle of refraction? Correct answer: 14 . 4636 ◦ . Explanation: Basic Concepts: n i (sin θ i ) = n r (sin θ r ) n = c v Given: θ i = 20 . 2 ◦ v = 2 . 17 × 10 8 m / s n i = 1 . 00 c = 3 . 00 × 10 8 m / s Solution: n r = c v n i (sin θ i ) = c (sin θ r ) v θ r = sin − 1 bracketleftbigg n i v (sin θ i ) c bracketrightbigg = sin − 1 bracketleftBigg 1 ( 2 . 17 × 10 8 m / s ) (sin20 . 2 ◦ ) 3 × 10 8 m / s bracketrightBigg = 14 . 4636 ◦ 003 10.0 points Consider a double slit experiment as shown in the sketch. y L d S 1 S 2 θ viewing screen θ δ What is the phase angle difference φ and corresponding path difference δ of the 2nd minimum on the screen?...
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This note was uploaded on 09/27/2010 for the course PHY 58650 taught by Professor Yao during the Fall '10 term at University of Texas.
 Fall '10
 YAO

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