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Nightmarish michelson interferometer question

  1. May 31, 2006 #1
    We have to "closely spaced" frequencies f1 and f2. "The moving mirror moves at a constant speed v. Explain how the Doppler effect on the light reflected from the moving mirror affects the recorded signal as a function of time. Estimate the value of v required to resolve the freq. difference of 500 MHz between two modes of a laswe within a time of 0.1s."

    I'm very confused about this. Under normal circumstances I'm pretty sure there is no moving mirror! I can only assume that the "closely" spaced frequencies are so closely spaced that they can't be resolved by the instrument, therefore we have to artificially introduce a doppler freq. shift in one of the components.
    I'm using the doppler shift formula from spec. rel. that shift in freq. = (v/c) * original frequency. Therefore surely v = 500MHz *c/f1
    or we could have used f2, as they are almost equal. I can't see how the time 0.1s comes into it. Thanks very much
  2. jcsd
  3. Jun 1, 2006 #2

    Andrew Mason

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    Can you give us the whole problem? You are assuming we know what you are talking about here.

    There is a doppler shift only if the mirror is moving relative to the source of the light. How do you get such motion in a Michelson interferometer? You need to clearly explain what it is you are talking about.

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