Fringe-Shift in Michelson Interferometer with a Moving Source

[VSayantan]

Homework Statement

The Michelson interferometer in the figure below can be used to study properties of light emitted by distant sources

A source ##S_1##, when at rest, is known to emit light at wavelength ##632.8~ \rm nm##. In this case, if the movable mirror is translated through a distance ##d##, it is seen that ##99,565## interference fringes pass across the photo-detector. For another source ##S_2##, moving at an uniform speed ##1.5\times {10}^7~ \rm {ms^{-1}}## towards the interferometer along the straight line joining it to the beam splitter, one sees ##100,068## interference fringes pass across the photo-detector for the same displacement ##d## of the movable mirror. What is the wavelength of light emitted by the source ##S_2##, in its own rest frame?

Homework Equations

Fringe shift $$n=\frac {2 d v^2}{\lambda c^2}$$

The Attempt at a Solution

When the whole set up is at rest (in the laboratory frame) there should be no fringe shift. But there will be a shift if the apparatus moves, as is the case in the frame of ##S_2##.

For the first source ##S_1## the total path difference is $$\Delta = 2\cdot d \cdot \cos \theta~ + ~\frac {\lambda}{2}$$
So, for one fringe to appear or disappear $$d= \frac {n \lambda}{2}$$

When the source ##S_2## moves towards the interferometer, along the specified direction, in its own reference frame the interferometer moves away from the source ##S_2##.
The distance ##d## through which the movable mirror moves also moves away from the source, with speed ##v = 1.5\times {10}^7~\rm {ms^{-1}}## - along the direction of motion.
Therefore this distance ##d## is contracted by a factor of ##\sqrt {1-{(\frac {v}{c})}^2}##.
So, for the second source $$d\cdot \sqrt {1-{(\frac {v}{c})}^2}=\frac {n' {\lambda}'}{2}$$

Eliminating ##d## from the two expressions, one obtains
$$n\lambda=\frac {n' {\lambda}'}{\sqrt {1-{(\frac {v}{c})}^2}}$$
Simplifying
$${\lambda}'=\frac {n {\lambda}{\sqrt {1-{(\frac {v}{c})}^2}}}{n'}$$
Substituting values,
$${\lambda}'= \frac {99,565 \times 632.8 ~\rm {nm} \times {\sqrt {1-{(\frac {1.5\times {10}^7~\rm {ms^{-1}}}{3\times {10}^8~\rm {ms^{-1}} } )}^2}}}{100,068}$$
Which gives a value $${\lambda}'\approx 628.83~\rm {nm}$$.

Is this right?

 

[mark726]

Yes, your solution is correct. You have correctly used the formula for fringe shift and accounted for the Lorentz contraction of distance in the frame of the moving source. Your final answer for the wavelength of light emitted by source ##S_2## in its own rest frame is also consistent with the given data. Well done!