- #1
smithg86
- 59
- 0
Homework Statement
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.
Homework 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
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?
Last edited: