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Michelson interferometer average power derivation

  1. Sep 21, 2013 #1
    1. The problem statement, all variables and given/known data
    Using the trig product identity, [itex]cosαcosβ=\frac{1}{2}[cos(α+β)+cos(α-β)][/itex], show that the time-average power at the detector can be written as Pavg = 1+cos(δ)

    That = is supposed to be a proportional symbol.


    2. Relevant equations
    Other than the ones given in the problem statement, there are a few:

    E1=E0cos(wt)
    E2=E0cos(wt+δ)

    [tex]δ=\frac{2∏(2x)}{λ}[/tex]
    Etot=E1+E2

    P = Etot2

    That last = is supposed to be a proportional symbol.


    3. The attempt at a solution
    Well, I started off by trying to square Etot, which gives me a long expression:
    E02cos2(wt)+E02[cos(2wt+δ)+cos(δ)]+E02cos2(wt+δ)

    I'm not sure I did that right. I used the trig product rule.

    From here, I can factor out an E02, but I still have a bunch of cosine terms that I don't know what to do with. How in the world could I turn those into 1+cos(δ)?

    Thanks.
     
  2. jcsd
  3. Sep 21, 2013 #2

    haruspex

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    You want average power, so don't you need to integrate wrt t over one cycle?
     
  4. Sep 21, 2013 #3
    I don't know. Actually, the question is asking to derive the relationship Pavg= cos(δ)

    So to derive that expression, I need to integrate? Do I integrate Etot2?

    Thanks
     
  5. Sep 23, 2013 #4
    Anyone with any idea how to do this?
     
  6. Sep 23, 2013 #5

    haruspex

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    Yes, I think it gives the desired answer. Integrate over one period (0 to 2pi/w) and divide by the length of the period.
     
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