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

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.

      $\frac{lv^{2}}{\lambda c^{2}} - (-\frac{lv^{2}}{\lambda c^{2}}) = \frac{2lv^{2}}{\lambda c^{2}}$

Formula for change in fringe shift is as follows.

          n = $\frac{2lv^{2}}{\lambda c^{2}}$

       $v^{2} = \frac{n \lambda c^{2}}{2l}$

             v = $\sqrt{\frac{n \lambda c^{2}}{2l}}$

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

             l = 11 m

    $\lambda = 5.9 \times 10^{-7} m$

           c = $3.0 \times 10^{8}$ m/s

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

            v = $\sqrt{\frac{n \lambda c^{2}}{2l}}$

               = $\sqrt{\frac{1 \times (5.9 \times 10^{-7} m) \times (3.0 \times 10^{8})^{2}}{2 \times 11 m}}$

               = $2.41363 \times 10^{9} m/s$

Thus, we can conclude that velocity deduced is $2.41363 \times 10^{9} m/s$.