p36_057 - 57. Let φ1 be the phase difference of the waves...

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Unformatted text preview: 57. Let φ1 be the phase difference of the waves in the two arms when the tube has air in it, and let φ2 be the phase difference when the tube is evacuated. These are different because the wavelength in air is different from the wavelength in vacuum. If λ is the wavelength in vacuum, then the wavelength in air is λ/n, where n is the index of refraction of air. This means φ1 − φ2 = 2L 2πn 2π 4π (n − 1)L − = λ λ λ where L is the length of the tube. The factor 2 arises because the light traverses the tube twice, once on the way to a mirror and once after reflection from the mirror. Each shift by one fringe corresponds to a change in phase of 2π rad, so if the interference pattern shifts by N fringes as the tube is evacuated, 4π (n − 1)L = 2N π λ and n=1+ Nλ 60(500 × 10−9 m) =1+ = 1.00030 . 2L 2(5.0 × 10−2 m) ...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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