Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2023 / Paper 1 / 9

The Fermat principle states that the actual path of a light ray makes its travel time stationary under small admissible variations with fixed endpoints. Equivalently, its optical path length is stationary:where $v$ is the local light speed, $n$ is the refractive index, and the two integrals differ by the constant vacuum light speed. In a uniform medium this reduces to stationarity of geometric path length. The principle is a stationarity statement; for the planar reflection considered next the stationary path is also the unique minimum.

Take the fixed source and receiver to be $A=(a,h_A)$ and $B=(b,h_B)$ on the same side of the mirror, with $h_A,h_B>0$. In the uniform medium, the Fermat principle minimizes the length through a reflection point $P=(s,0)$:Its stationary equation isThe two fractions are the signed tangential components of the ray directions. Equality of their magnitudes gives $\sin {\theta }_i=\sin {\theta }_r$, where both angles are measured from the normal vector to the mirror and lie in $[0,\pi /2)$. Hence the law of reflection isFor uniqueness, differentiate once more:Thus $L$ is a strictly convex function, its derivative increases from $-2$ to $2$, and it has exactly one stationary point, its global minimum. Explicitly,Substitution gives $(s_{\ast }-a)/h_A=(b-s_{\ast })/h_B$, verifying the reflection angles and the existence of that unique solution. Equivalently it is the point where the line from $A$ to the mirror image $(b,-h_B)$ crosses the mirror.

First expand the variational integrand:Its needed partial derivatives are ${\mathcal{L}}_f=3f^{\prime }+4f$ and ${\mathcal{L}}_{f^{\prime }}=2f^{\prime }+3f-x$. The Euler-Lagrange equation is thereforeThe general solution is $f=A\cosh (\sqrt{2}x)+B\sinh (\sqrt{2}x)-1/4$. Subtracting the two endpoint equations gives $B=0$; adding them gives $A=1/[4\cosh \sqrt{2}]$. HenceThis is the unique stationary solution. It is also the unique global minimizer: after discarding the fixed boundary terms, integration by parts rewrites $I$ as ${\int }_{-1}^1[f^{\prime 2}+2f^2+f-2x^2]dx$. For an endpoint-zero variation $\eta$, the linear part vanishes at the displayed solution andwith equality only for $\eta =0$.