# Fabry-Perot Interferometer: Energy Conservation

1. Dec 2, 2007

### smithg86

1. The problem statement, all variables and given/known data

This is regarding a Fabry-Perot Interferometer with identical mirrors:

If you shine a beam of light on a 99% reflecting mirror, 1% goes through and the other 99% is reflected. But if a second mirror is placed behind the first one, where there is only 1% of the light present, somehow 100% of the light appears. Explain this, also explain how energy is conserved.

2. Relevant equations

I_t = I_0 * {T^2 / (1-R^2)} * { 1 / (1 + F * sin^2($$\Delta$$/2))}

F = finesse = 4R / (1-R)^2

I_t = transmitted intensity
I_0 = original intensity

T = transmittance
R = reflectance

$$\Delta$$ = $$\delta$$ + $$\delta$$_r
$$\delta$$ = phase difference per round trip between mirrors
$$\delta$$_r = phase difference due to reflection = 0 or pi

3. The attempt at a solution

I understand mathematically why it's true. If you set $$\Delta$$ to an integer multiple of pi, then sin $$\Delta$$ = 0 and then

I_t/I_0 = {T^2 / (1-R^2)} * { 1 / (1 + F*(0)) }

I_t/I_0 = T^2 / (1 - R^2) = 1

so I_t = I_0, so 100% of the light appears.

I really don't know how to explain this physically. Help?

Last edited: Dec 2, 2007
2. Dec 3, 2007

### Gokul43201

Staff Emeritus
Can you write down the question EXACTLY as it was given to you?

I can't seem to make much sense of it. First of all, the interferometer set-up I'm familiar with has the mirrors facing each other, not one behind the other. Secondly, in this geometry, I expect that only about half the incident power is transmitted through in the forward direction, the rest is transmitted backwards.

What does the question mean by "100% of the light appears"? Appears where?

I may be completely off here, so it would help if someone else takes a look.

3. Feb 7, 2009

### MeChaState

"I really don't know how to explain this physically. Help?"

Has anyone an answer to this question?

That is the question of my life too.