Michelson interferometer average power derivation

[leroyjenkens]

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 P_avg = 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]
E_tot=E₁+E₂

P = E_tot²

That last = is supposed to be a proportional symbol.

The Attempt at a Solution

Well, I started off by trying to square E_tot, 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.

 

[haruspex]

You want average power, so don't you need to integrate wrt t over one cycle?

 

[leroyjenkens]

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 P_avg= cos(δ)

So to derive that expression, I need to integrate? Do I integrate E_tot²?

Thanks

 

[leroyjenkens]

Anyone with any idea how to do this?

 

[haruspex]

I don't know. Actually, the question is asking to derive the relationship P_avg= cos(δ)

So to derive that expression, I need to integrate? Do I integrate E_tot²?

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.