Non-constant index of refraction due to layered material.

Fraqtive42

A ray of light travels through a medium with an index of refraction [tex]n_{1}[/tex] and strikes an layered medium such that the index of refraction is [tex]n_{2}=ky+1[/tex] where [tex]y[/tex] is the depth of the medium and [tex]k[/tex] is a constant. If it hits at an angle of [tex]\theta_{1}[/tex] with respect to the normal, find the angle [tex]\theta_{2}[/tex] 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 [tex]v=\frac{c}{n_{2}}[/tex] is the speed of the light beam, which is also equal to [tex]v=\frac{dy}{dt}[/tex]. So a differential equation to solve would be [tex]\frac{dy}{dt}=\frac{c}{n_{2}}[/tex]

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

But because [tex]y[/tex] is a function of time, that also makes [tex]\theta_{2}[/tex] 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 [tex]\theta[/tex]₂ because the index of refraction is related to [tex]\theta[/tex]₂

ehild

What is the relation between the refractive index and θ2?

ehild

Fraqtive42

The refractive index and [tex]\theta[/tex]₂ 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