Light through a medium with variable refraction index

[WackStr]

Homework Statement

This is from hand and finch. We proved in the previous problem that (using euler lagrange equation):

[tex]x=\int_0^y\frac{dy}{\sqrt{\left(\frac{n[y]}{n_0}\right)-1}}[/tex]

where [tex]n_0[/tex] is the refractive index at y=0 and x=0. The ray enters horizontally.

As an actual computation the book says that assume [tex]n[y]=n_0e^{-\alpha y}[/tex] and [tex]n_0=1.5[/tex]. Aslo y(30)=-1.

We need to find [tex]\alpha[/tex]

The Attempt at a Solution

From the information given it seems like the equation we need to solve is

[tex]30=\int_0^{-1}\frac{dy}{\sqrt{e^{-2\alpha y}-1}}[/tex] for [tex]\alpha[/tex] but it seems like this equation has no solution.

So I am stuck at this point.

 

[omoplata]

Mathematica solves that integral.

But I don't know how to actually solve it.

I can post the solution to the integral if you want.

 

[WackStr]

I know mathematic gives an analytical expression in terms of [tex]\alpha[/tex] but the equation seems to have no solution. (the integral is negative and the left hand side is positive)

 

[ideasrule]

Are you sure you're not supposed to assume alpha is small and approximate e^(-2ay) as 1-2ay?

 

[WackStr]

actually I have a typo in the original integral it should be [tex]\left(\frac{n[y]}{n_0}\right)^2[/tex] and I figured out what the problem was. To get the expression we had to take a square root. So there should be a +/- sign. If we use the + sign we don't get a solution because RHS < 0 and LHS > 0, but we do get a solution if we put a - sign next to the initial integral and it gives a value for alpha and a nice trajectory for the light ray in the medium.