Consider a ray of light travelling between two points A and B on the $xy$ plane. Using the calculus of variations and Fermat's Principle we can derive equations which give the trajectory of a ray of light through a medium of given refractive index $n(y)$ by minimising,
\begin{equation} T = \int n(y) \sqrt{1+{\Big(\frac{dy}{dx}\Big)}^2}dx \end{equation} and using the Euler-Lagrange equations we arrive quickly at an integral expression for the trajectory as \begin{equation} x = \int \frac{\alpha\> dy}{\sqrt{{n(y)^2-\alpha}^2}} \end{equation} with $\alpha$ as some arbitrary constant based on initial conditions. Now consider the refractive index of the medium to be $\> n(y)= n_0 \cos(ky)$.

At this point I can't seem to progress; Is this integral solvable with this continuously varying index? Is this the correct approach or is there a more convenient way to find the trajectory?


The integral you're trying to solve is $$ x = \int \frac{\beta}{\sqrt{\cos^2 (ky) - \beta^2}} dy = \int \frac{\beta}{\sqrt{(1 - \beta^2) - \sin^2 (ky)}} dy \\= \frac{\beta}{\sqrt{1 - \beta^2}} \int \frac{dy}{\sqrt{1 - \sin^2 (ky)/(1 - \beta^2)}} dy, $$ where $\beta \equiv \alpha/n_0$. This can be recognized to be an incomplete elliptic integral of the first kind: $$ x - x_0 = F \left( ky \left| \frac{1}{1 - \beta^2}\right) \right. $$ where $x_0$ is the $x$-coordinate when $y = 0$. If desired, this can be inverted to find $y(x)$ in terms of the Jacobi elliptic functions: $$ y = \frac{1}{k} \sin^{-1} (\mathrm{sn} (x - x_0)). $$ In this notation, the use of the parameter $1/(1- \beta^2)$ in the definition of the inverse function is understood.

In some sense, I haven't really told you how to evaluate this integral; I've just told you that this integral has a name and has been studied (along with its inverse function.) The problem of plotting $y(x)$ (or $x(y)$) and understanding where the trajectories go is still there; and unless you have a completely intuitive understanding of the elliptic functions (I don't), then this is probably unilluminating. Still, it might give you somewhere to go from here; most mathematical handbooks can tell you about the analytical properties of the elliptic integrals and the Jacobi elliptic functions, and a computer language that can do upper-level math (e.g., Mathematica, Maple, MATLAB, etc.) could plot the trajectories out for you.

  • 1
    $\begingroup$ Indeed. The correct response to getting an elliptic integral is (a very specific flavour of) giving up. Much like getting, say, a Bessel integral, you just reshape it into the canonical form and then that's the end of the line. $\endgroup$ – Emilio Pisanty Feb 6 at 21:10

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