p36_060 - 60. (a) In a reference frame fixed on Earth, the...

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Unformatted text preview: 60. (a) In a reference frame fixed on Earth, the ether travels leftward with speed v . Thus, the speed of the light beam in this reference frame is c − v as the beam travels rightward from M to M1 , and c + v as it travels back from M1 to M . The total time for the round trip is therefore given by t1 = d1 d1 2cd1 . + =2 c−v c+v c − v2 (b) In a reference frame fixed on the ether, the mirrors travel rightward with speed v , while the speed of the light beam remains c. Thus, in this reference frame, the total distance the beam has to travel is given by d2 = 2 d2 + v 2 2 t2 2 [see Fig. 36-37(h)-(j)]. Thus, t2 = d2 2 = c c which we solve for t2 : t2 = √ 2 t2 2 d2 + v 2 , 2d2 . c2 − v 2 (c) We use the binomial expansion (Appendix E) (1 + x)n = 1 + nx + · · · ≈ 1 + nx In our case let x = v/c (|x| 1) . 1, then L1 = 2c2 d1 v = 2d1 1 − 2 − v2 c c and L2 = √ 2 −1 ≈ 2d1 1 + ≈ 2d2 1 + 1 2 2 − 1/ 2 2cd2 v = 2d2 1 − 2 − v2 c c 2 v c , v c 2 . Thus, if d1 = d2 = d then ∆L = L1 − L2 ≈ 2d 1 + v c 2 − 2d 1 + 1 2 v c 2 = dv 2 . c2 (d) In terms of the wavelength, the phase difference is given by dv 2 ∆L = 2. λ λc (e) We now must reverse the indices 1 and 2, so the new phase difference is −∆L dv 2 =− 2 . λ λc The shift in phase difference between these two cases is shift = ∆L λ −− ∆L λ = 2dv 2 . λc2 (f) Assume that v is about the same as the orbital speed of the Earth, so that v ≈ 29.8 km/s (see Appendix C). Thus, shift = 2dv 2 2(10 m)(29.8 × 103 m/s)2 = = 0.40 . λc2 (500 × 10−9 m)(2.998 × 108 m/s)2 ...
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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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