Non-constant index of refraction due to layered material.

[Fraqtive42]

A ray of light travels through a medium with an index of refraction n_{1} and strikes an layered medium such that the index of refraction is n_{2}=ky+1 where y is the depth of the medium and k is a constant. If it hits at an angle of \theta_{1} with respect to the normal, find the angle \theta_{2} at which the light ray refracts as a function of time.

Source: A post that I made on the Art Of Problem Solving forum.

 

[Integral]

As with all homework like questions you must show some work before getting help.

 

[Fraqtive42]

My work:
So far I know that v=\frac{c}{n_{2}} is the speed of the light beam, which is also equal to v=\frac{dy}{dt}. So a differential equation to solve would be \frac{dy}{dt}=\frac{c}{n_{2}}

 

[ehild]

The light ray does not travel along y but at an angle θ₂ with respect to it. θ₂ itself is a function of y.

ehild

 

[Swap]

APhO 2004 problem 2. It is similar to this one. Look at the solution there.

 

[Fraqtive42]

The light ray does not travel along y but at an angle θ₂ with respect to it. θ₂ itself is a function of y.

ehild

But because y is a function of time, that also makes \theta_{2} a function of time.

 

[ehild]

And how are y and θ2 related? ehild

 

[Fraqtive42]

If the material is layered infintesimally so that the index of refraction is proportional to the y, which I stated in the problem, then y is related to \theta₂ because the index of refraction is related to \theta₂

 

[ehild]

What is the relation between the refractive index and θ2?

ehild

 

[Fraqtive42]

The refractive index and \theta₂ are related through Snell's Law.

 

[ehild]

Well. At depth y, the light ray encloses the angle θ2(y) with the y axis. The light travels along a curved path s and ds/dt = c/n2(y). At depth y, θ2 is obtained from Snell's law. Now you can set up the differential equation for θ2 as function of t.

ehild