Correct Angle for FSW in Fabry-Perot Interferometer

[Blanchdog]

Homework Statement

What are the free spectral range and fringe width in a Fabry-Perot Interferometer made up of two silver coated plates each with reflectance R = 0.9, transmittance T = 0.05, and Absorbance A = 0.05? The plate separation is d = 0.5cm with interior index n1 = 1. Suppose the wavelength being observed near normal incidence is 587nm.

Relevant Equations

FSR = wavelength^2/(2 n d cos (theta1))

Sorry for the lack of formatting, I'm in a rush to write this before I go pick someone up from work. My question is what is theta1? Is it 45 degrees because that's the angle the light hits the boundary of the plate to the n1 medium? Or is it 0? It seems like it should be 45 but there's an example in the textbook that uses 0. Or is cos(theta1) imaginary? Thanks!

 

[BvU]

near normal incidence

For normal incidence this $\theta$ is zero. The central black spot in the first picture here . Make a side view sketch (or look further down ) to see where $\ \theta\ne 0\$ comes in.

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[Blanchdog]

For normal incidence this $\theta$ is zero. The central black spot in the first picture here . Make a side view sketch (or look further down ) to see where $\ \theta\ne 0\$ comes in.

$\$

Ah thank you, I was getting tripped up because I knew $\theta_1$ was the transmission angle but couldn't figure out what it was, but if the light is at normal incidence then the transmission angle will be zero.