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Michelson interferometer average power derivation

  • #1
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Homework Statement


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.


Homework 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.


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.
 

Answers and Replies

  • #2
haruspex
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You want average power, so don't you need to integrate wrt t over one cycle?
 
  • #3
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You want average power, so don't you need to integrate wrt t over one cycle?
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
 
  • #4
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Anyone with any idea how to do this?
 
  • #5
haruspex
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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
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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