Michelson Interferometer

[Melawrghk]

Homework Statement

Basically, there is an interferometer with one perfect mirror and a mirror with varying thickness across it. You shine parallel coherent light, wavelength 633nm, and see an interference pattern. The interference batter is wavy dark and light fringes of varying thickness. Two neighbouring fringes have their centers marked, A and B and those are separated by 3mm.
Find the slope of the line between A and B on the uneven mirror's surface wrt plane of constant phase of the wave.

Homework Equations

y=m(lambda)/2

The Attempt at a Solution

I understand why thick fringes are formed - the source isn't a point source, so it is a beam of light. That beam can be broken up into infinitely many point sources, each of which results in an interference pattern.
But, I have absolutely no idea how to approach this question. I guess I want to find the wavelength that would have resulted in fringe center separation of 3mm and then compare it to 633nm (maybe get slope out of that). But I don't know how. :S Any hint would be greatly appreciated.

 

[Jhero]

First, let's define some variables:
- y is the distance from the mirror's surface to the plane of constant phase
- m is the order of the fringe (1, 2, 3, etc.)
- λ is the wavelength of the light (633nm in this case)
- d is the distance between neighboring fringes (3mm in this case)
- θ is the angle between the line connecting A and B and the surface of the mirror

We can then use the equation y=m(λ)/2 to calculate the distance between the plane of constant phase and each fringe. Since the distance between neighboring fringes is d, we can set up the following equation:

y(m+1) - ym = d

Substituting in the equation for y, we get:

m(λ)/2 + (λ)/2 - m(λ)/2 = d

Simplifying, we get:

(λ)/2 = d

Now, we can use the small angle approximation to say that θ is approximately equal to y/d. We can then rearrange the equation to solve for y:

y = θd

Since we want to find the slope of the line between A and B with respect to the plane of constant phase, we can take the derivative of y with respect to θ:

dy/dθ = d

This means that the slope of the line between A and B on the uneven mirror's surface with respect to the plane of constant phase is simply equal to the distance between neighboring fringes (3mm in this case).

I hope this helps! Let me know if you have any further questions.