# Homework Help: Fringe-Shift in Michelson Interferometer with a Moving Source

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1. Dec 3, 2017

### VSayantan

1. The problem statement, all variables and given/known data
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?

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

3. 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?

Last edited: Dec 3, 2017
2. Dec 8, 2017

### PF_Help_Bot

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