Past exam of the mathematics course of the University of Cambridge / 2013 / ib / Paper 2 / 15A / Solution

The Euler-Lagrange equation is $f_y-d(f_{y^{\prime }})/dx=0$. Since $f$ has no explicit $x$ dependence,Thus a stationary curve satisfies the Beltrami identity, $f-y^{\prime }{f}_{y^{\prime }}=C$.

By Fermat principle, the light ray makes the travel time $\int \sqrt{1+y^{\prime 2}}/(y+c_0)dx$ stationary. Its Beltrami identity becomesWriting $R=1/C$ and integrating the resulting first-order equation gives $(x-x_0{)}^2+(y+c_0{)}^2=R^2$. Both endpoints have height zero, so subtracting their equations gives $x_0=0$, and then $R^2=a^2+c_0^2$. The arc lying in $y>0$ isIts maximum height is $\sqrt{a^2+c_0^2}-c_0$. The ray rises into the faster region before returning to the boundary; its circle centre is $(0,-c_0)$. The prescribed boundary endpoints are understood by continuity from the open half-plane. This is the circular light ray in a linear speed profile.