 Solution

The angles in the PDF are measured from the vertical normal to the interface. The horizontal displacements of the two path segments are $d_1\tan \theta$ and $d_2\tan \varphi$, while their lengths are $d_1\sec \theta$ and $d_2\sec \varphi$. Thus the geometric constraint and travel time areThe Lagrange multiplier equations $\nabla T=\lambda \nabla g$ giveCancelling the positive geometric factors gives $\lambda =2\sin \theta /c=\sqrt{2}\sin \varphi /c$, henceThis is Snell's law, $\sin \theta /(c/2)=\sin \varphi /(c/\sqrt{2})$. The relation depends only on the ratio of the two speeds, so changing $D,d_1,d_2$ leaves the relation unchanged; it changes which angles satisfy it together with the geometric constraint.

For a minimum check, let $u$ be the horizontal distance to the crossing point. ThenandThus the travel time is a strictly convex function of $u$, and its stationary point is the unique global minimum. For positive $D,d_1,d_2$, $T^{\prime }(0)<0<T^{\prime }(D)$, placing the crossing between the two endpoint projections.