Solving for Refraction Angle of Light Ray Across Parallel Plate

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Homework Statement

We have a parallel, transparent plate. The plate has an incidence of refraction that is dependent on a transverse coordinate. (i.e. the plate is perpendicular to z direction, and the refractive index changes along x direction)

n(x) = 1.5/(1 - x/0.13)

It is observed that when a light ray is perpendicularly incident on the plate (from the air) at x=0, it emerges on the other side of the plate, making a 30 degree angle from the normal.

Find the index of refraction where the ray exits. (Thickness of the plate is not known)

Homework Equations

I'm not really sure about this. Snell's law tells us that the light ray should stay perpendicular to the normal across both boundaries, so there is something I am clearly missing.

Snell's Law: n1*sin(a1) = n2*sin(a2)

Where n1 and n2 are indices of refraction for the two materials
a1, a2 are the angles the ray makes from the normal of the material

The Attempt at a Solution

Again, using Snell's Law, at the second boundary (going from plate back to air), we know that n1 = n(x), n2 = 1, a1 = ?, a2 = 30 deg

This is one equation with two unknowns, so I am stuck.

If for some reason, the light were to bend at the first boundary, then I would treat the plate as a bunch of thin plates, and integrate (as the light ray would be experiencing a different index of refraction as it traveled through the plate).

In order for the ray to be emerging at a 30 degree angle, something like this must be happening, but I don't know why!

Any help is appreciated!

 

[mark726]

Hello!

I can see why you are stuck on this problem. It is not a straightforward application of Snell's Law, as you have correctly identified. The reason for this is because the index of refraction is changing along the x-direction, so the light ray is experiencing different indices of refraction as it travels through the plate.

To solve this problem, we need to use the concept of Fermat's Principle, which states that light will travel the path that takes the least amount of time. In this case, the light ray will take the path that minimizes the time it takes to travel through the plate.

To find the index of refraction where the ray exits, we need to find the path that takes the least amount of time. This can be done by using the calculus of variations, which involves minimizing a functional. In this case, the functional is the time it takes for the light ray to travel through the plate.

Using the calculus of variations, we can find the path that minimizes the time, and from there, we can determine the index of refraction at that point. This will give us the answer to the problem.

I hope this helps! Let me know if you have any further questions. Good luck with your problem solving!