# Fermat's principle in exponential refraction [closed]

Determine path light will travel in a index of refraction described as $n(y)=1+be^{-ry}$. Let the light start off at the origin (0,0) and go to the point (L,0). Determine the equation that must be solved to determine the constant in your path equation. Thus far I have used Euler-Lagrange relation applied to time but can't seem to deduce the full equation for the constant.

## closed as off-topic by garyp, Yashas, Kyle Kanos, John Rennie, ZeroTheHeroMay 11 '17 at 16:07

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – garyp, Yashas, Kyle Kanos, John Rennie, ZeroTheHero
If this question can be reworded to fit the rules in the help center, please edit the question.

The general approach of dealing with these kind of situations where the refractive index is varying continuously, is to employ the Fermat's principle. The link contains most of the relevant information, but to summarize the approach, the principle will lead you to an equation of the type $$\delta \int n ds = 0$$ where $ds$ represents the length element along the light ray path. The same can be written in terms of your 2D co-ordinates, in order to segregate out the $x$ and $y$ dependent terms. More generally, substituting for the spatial dependence of $n$, one would arrive at the equation: $$\delta \int n(x,y) \sqrt{(1+(dy/dx)^2)} dx = 0$$
Thereafter, one can use the standard approach of calculus of variations - using Euler-Lagrange equation on this "action", to obtain the equations of motion. If the starting point is the origin, $n_{\rm origin} = 1 + b$ from the above equations, which can act as a boundary condition on $n$. (You need boundary conditions here since, EL equations would turn your problem into a differential equation.)