Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2023 / Paper 1 / 20 / b

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.

Put $s=\sin \theta$. By part (i), $\sin \varphi =\sqrt{2}s$. The acute-angle geometry requires $0<s<1/\sqrt{2}$. Expressing the tangents in terms of their sines in the geometric constraint gives the required implicit equationThe hint suggests $s=1/m$ for an integer $m\ge 2$. Testing $m=2$ givesThus $\boxed{\sin \theta =1/2}$. Each term on the left is strictly increasing in $s$ over the allowed interval, so this solution is unique.

The acute angles follow from part (ii) and the Snell's law relation:The path lengths below and above the interface are respectivelyHence the total distance and minimum travel time areFor a sketch to scale, choose the interface as $y=0$, with $A=(0,-\sqrt{3})$ and $B=(2,1)$. The crossing is $C=(1,0)$ because $d_1\tan \theta =d_2\tan \varphi =1$. The figure uses equal axis scales and shows the two normal-relative angles.