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VSayantan
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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?
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