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I Michelson-Morley experiment

  1. Oct 14, 2016 #1
    In Michelson-Morley experiment, the two light beams arrive in phase at the detector, so they say this means waves are travelling at the same speeds because they arrive in phase at the detector,
    What if the waves started in phase and their path difference is an integer multiple of the wave lengths (meaning they will arrive in phase at the detector too).
    I am arguing that same speed is not the only condition for constructive interference.
     
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  3. Oct 14, 2016 #2

    Orodruin

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    While this is true, it does nothing for the interpretation of the experiment. The point is that the interference pattern does not change when the device is rotated. There is also no a priori requirement that the interference should be constructive.
     
  4. Oct 14, 2016 #3

    Ibix

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    But then you rotate the experiment and the relative velocities change (or, more precisely, Michelson and Morley expected them to change).

    Also, with a bit of practice it's fairly easy to recognise the zero path difference fringe. Due to finite coherence times, it has the strongest contrast between it and the adjacent minima (black/white rather than dark grey/light grey).
     
  5. Oct 14, 2016 #4
    Okay thanks, I think its becoming clear now. Is it possible for two waves travelling at different speeds to be in phase? Why?
     
  6. Oct 14, 2016 #5

    Ibix

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    Even under the ether model Michelson and Morley assumed, the waves are travelling at the same speed in the output arm because they're travelling in the same direction. They would pick up a phase difference inside the interferometer, but would then maintain a constant phase relation after they were recombined at the beam splitter.
     
  7. Oct 14, 2016 #6
    Now I think my problem is solved! This means getting bright fringe at the detector means there has been no such phase difference even if the interferometer was rotated CONTINUOUSLY in all angles??
     
  8. Oct 14, 2016 #7

    Ibix

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    You would typically zero your interferometer at a bright fringe, then show that you had a bright fringe at all angles and in all seasons, yes. Strictly, the bright fringe isn't necessary - an unchanging fringe is all you need. Practically, though, the brightest thing is the easiest to find and track.
     
  9. Oct 14, 2016 #8
    Thanks a looooooot!!!!!
     
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