- #1

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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