Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 9

The Fermat principle makes the optical path length stationary. For a graph $y=\xi (x)$ the variational integrand isThe Euler-Lagrange equation isUse the chain rule $dn(x,\xi (x))/dx=n_x+n_y{\xi }^{\prime }$ and differentiate ${\xi }^{\prime }/\sqrt{1+{\xi }^{\prime 2}}$ to obtainMultiplying through and combining the $n_y$ terms gives the required second-order ordinary differential equation with no derivatives in denominators:All partial derivatives of the refractive index are evaluated along the ray.

Introduce a constant Lagrange multiplier $\mu$ for this isoperimetric constraint and make $A-\mu (C-B)$ stationary. Its integrand is $(1-\mu )y-\mu y^{\prime 2}$, so the Euler-Lagrange equation givesThe endpoint conditions therefore force a parabola $y=Kx(1-x)$. Its area and budget areSolving the quadratic and choosing positive area givesThis requires $B>0$; no positive-area barrier is possible for a nonpositive budget. The corresponding multiplier is $\mu =1/(1+4K)>0$.

To verify a global maximum rather than just stationarity, use the sharp area-cost bound for a barrier with fixed endpoints. For $w=x(1-x)$, integration by parts gives $2A={\int }_0^1{y}^{\prime }{w}^{\prime }dx$, and ${\int }_0^1{w}^{\prime 2}dx=1/3$. The Cauchy-Schwarz inequality impliesFor $A>0$ the last expression increases strictly, yielding exactly the boxed upper bound. Equality requires $y^{\prime }$ proportional to $w^{\prime }$, hence $y=Kw$ after imposing the endpoints. Thus the displayed optimum is unique and lies above the baseline throughout the interior.