How to Prove the Refractive Index Equation in a Michelson Interferometer?

[t1mbro]

I've been doing some working with a michelson interferometer, specifically finding the refractive index of a glass block. I am given this eqn:


n = (2t-ML)(1-cos)/ (2t (1 - cos) - mL)

Where n is the refractive index of the glass
M is number of fringes passing
I have left the thetas out because they mess up how it looks but u can assume cos = cos theta1.
t is thickness of the glass
L is Lambda the wavelength of the light.

I am trying to prove this eqn.

Here is what I have.

I know that the optical path length is the distance the light travels in the glass * n.

Where B is the distance traveled in the glass B = t cos theta2

Theta1 is angle of incidence, theta 2 is angle of refraction.

From elsewhere L = 2d/M

In this case B will be my d as this is the distance that the beam travel in excess of its normal path.

Now I have tried just shoving everything in and hoping I can simplify it but to no avail. How should I be going about this?

 

[rush_mart]

interferometer

hi

You know what, we have the same problem. I'm also trying to simplify it. I'm using a backward approach. But I still can't find it. But anyway, if I'm going to be able to derive it totally, I'll just send it to you.

 

[JeffKoch]

Not entirely sure I follow (sketch would help), but where do you get L=2d/M? B=tcos(theta2), but if I understand your problem then this is the path within the glass, not the distance the beam travels in excess of it's "normal" path.