Light through a medium with variable refraction index

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Homework Statement



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

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

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

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

We need to find \alpha

The Attempt at a Solution



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

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

So I am stuck at this point.
 
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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.
 
I know mathematic gives an analytical expression in terms of \alpha but the equation seems to have no solution. (the integral is negative and the left hand side is positive)
 
Are you sure you're not supposed to assume alpha is small and approximate e^(-2ay) as 1-2ay?
 
actually I have a typo in the original integral it should be \left(\frac{n[y]}{n_0}\right)^2 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.
 
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