# Why is the refractive index for light rays travelling in circular paths proportional to $1/r$?

While studying optics, I came across a problem with solution in which the trajectory of light rays was known—circular paths around a fixed point in space, and the question was that of determining the refraction index as a function of the distance $$r$$ from that fixed point (given that the refractive index at $$r = r_0$$ is $$n_0$$. A bit of a baffling question to me, as without at least knowing where the boundaries are Snell's law seemed to be utterly useless. The only thing that made sense to me was that the refractive index would depend solely on $$r$$, and the light rays were staying in the path with the least change in refractive index. Anyway the solution reads
$$n \cdot 2 \pi r = constant$$
$$n = {constant\over r}$$
$$n = {n_0r_0\over r}$$
At first, my intuition justified it as being similar to angular momentum (I understood it as light behaving as a particle), in that $$v$$ would also be proportional to $$1/r$$. But then eventually I realized $$n$$ is $$c/v$$, so in fact here the speed of light is proportional to $$r$$! The only thread I have at this point is Fermat's principle, but when the trajectory is known and not the refraction index I might as well be reconstructing a chess game by only seeing the final position. Can anyone make sense of this? (I don't suppose there is a nice "formula" for the refraction index function, as in like the one for the Lagrangian in mechanics; $${1\over2} mv^2 - V(x)$$, is there?)

Your last comment is on the right track: Fermat's principle of least time can be expressed as a variational principle. Specifically, we have the travel time of light along an arc $$\gamma$$ as $$T = \frac1{c} \int_\gamma n(\vec{r}) \, ds$$ where $$ds$$ is the infinitesimal arc length along the path. If we parametrize the path in polar coordinates as $$r(\theta)$$ this becomes $$T = \frac1{c} \int_{\theta_1}^{\theta_2} n(r) \sqrt{{r'}^2 + r^2} \, d\theta$$ and we can then apply the techniques of calculus of variations to find a differential equation satisfied by $$r(\theta)$$.
Requiring that $$r(\theta) = r_0$$ is a solution for any value of $$r_0$$ will then place a condition on the unknown function $$n(r)$$. (The Beltrami identity might be useful here.)
• Oh, I guess you're right, using the Beltrami identity I immediately find that the refraction index indeed has to be proportional to $1/r$. On another note, I now realize that the solution is indicating that the time travelled by all of these light rays have to be equal. While it makes sense, it's not obvious to me why it has to be true. Thanks for the help anyways! Commented Mar 23 at 6:45