(adsbygoogle = window.adsbygoogle || []).push({}); 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([tex]\Delta[/tex]/2))}

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

I_t = transmitted intensity

I_0 = original intensity

T = transmittance

R = reflectance

[tex]\Delta[/tex] = [tex]\delta[/tex] + [tex]\delta[/tex]_r

[tex]\delta[/tex] = phase difference per round trip between mirrors

[tex]\delta[/tex]_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 [tex]\Delta[/tex] to an integer multiple of pi, then sin [tex]\Delta[/tex] = 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?

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# Homework Help: Fabry-Perot Interferometer: Energy Conservation

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