Question

A shift in one fringe in the Michelson-Morley experiment corresponds to a change in the round-trip travel time along one arm of the interferometer by one period of vibration of light (about 2.0 x 10-15 s) when the apparatus is rotated by 9〫0. What velocity through the ether would be deduced from a shift of one fringe? (Take the length of the interferometer arm to be 11 m).Hint: You may find the following expansions helpful for this problem and problems you will see later inthe class.

Answer 1

Step-by-step explanation:

When Michelson-Morley apparatus is turned through
`90^(o)` then position of two mirrors will be changed. The resultant path difference will be as follows.

`(lv^(2))/(\lambda c^(2)) - (-(lv^(2))/(\lambda c^(2))) = (2lv^(2))/(\lambda c^(2))`

Formula for change in fringe shift is as follows.

n =
`(2lv^(2))/(\lambda c^(2))`

`v^(2) = (n \lambda c^(2))/(2l)`

v =
`\sqrt{(n \lambda c^(2))/(2l)}`

According to the given data change in fringe is n = 1. The data is Michelson and Morley experiment is as follows.

l = 11 m

`\lambda = 5.9 * 10^(-7) m`

c =
`3.0 * 10^(8)` m/s

Hence, putting the given values into the above formula as follows.

v =
`\sqrt{(n \lambda c^(2))/(2l)}`

=
`\sqrt{(1 * (5.9 * 10^(-7) m) * (3.0 * 10^(8))^(2))/(2 * 11 m)}`

=
`2.41363 * 10^(9) m/s`

Thus, we can conclude that velocity deduced is
`2.41363 * 10^(9) m/s`.