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A modern physics homework problem about interferometry and the binomial approximation. The problem involves calculating the expected fringe shift for a given interferometer setup and discussing the significance of observing or not observing this small shift. The document also mentions the work of michelson and morley and their improvements to the interferometer design.
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A B C
D v
u
u
A u
D C
Figure 1: Geometry for problem #1.
We're given that the wind sp eed v = 30 m/s and that sound propagates in the air with a sp eed of u = 330 m/s:
uB = u + v = 360 m/s (2) uD =
(a) Problem 1.10.
0 : 5 m
= 0 : 017 radians; (4)
where = v =c. This is ab out 1/60 of a complete fringe. The negative result of not observing this small fringe shift was not taken seriously. Michelson and Morley later lengthened the arms of the interferometer by a factor of ab out 20 to 11 m, however, and increased this value to ab out 1/3 of a complete fringe which was easily observable.
, the interferometer will have two arms of length and + . To nd the resulting phase shift, N , caused by the increased one-way length =T , we must compare this extra length traveled by light along the lengthened arm. Light will travel an extra distance 2= 2T when you include its transit down and back the lengthened arm. The phase shift will dep end on the size of this distance relative to the wavelength, , of light:2 (0: 5 C ^1 ) (0: 5 m) (10 ^2 C ) 590 10 ^9 m
= 0 : 2 radians: (5)
Note: There is the p otential for notational confusion here. In this case, N is the phase shift caused by the di erence in arm lengths of the interferometer pro duced by temp erature variations across the interferometer. In the text, the phase shift caused by the earth's motion relative to the aether frame was referred to as N and N was the phase shift up on rotation of the interferometer.