1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Fringe-Shift in Michelson Interferometer with a Moving Source

  1. Dec 3, 2017 #1
    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 interferometer.jpg
    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. jcsd
  3. Dec 8, 2017 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Fringe-Shift in Michelson Interferometer with a Moving Source
Loading...