Past exam of the physics 2 paper of the University of Oxford / 2013 / 9 / i / Solution

Take a horizontal cross-section in units of $R$. The eye is at $E=(\sqrt{2},0)$, and the tadpole swims along the perpendicular diameter $T=(0,y)$ shown in the source. Since glass and water have the same red-light index, their internal interface causes no red-light refraction. Write an exit point on the circular wall as $P=(\cos \varphi ,\sin \varphi )$. An unobstructed outward ray to the eye has $|\varphi |\le \pi /4$.

For $\varphi \ge 0$, let $e$ be the magnitude of the emerging ray's angle to the outward normal and $i$ its internal angle. Geometry and Snell's law giveThe outgoing internal ray makes angle $\beta =\varphi -i$ to $+x$. Its backward continuation intersects the swim diameter atThe directly visible positive endpoint is the largest such intercept, including all single-pass refracted rays. Maximizing this one-variable expression gives ${\varphi }_{\ast }\simeq 38.07^{\circ }$ and $y_{\ast }\simeq 0.75744$. By symmetry the visible length is $2R{y}_{\ast }$, soChecking only the grazing exit at $\varphi =45^{\circ }$ gives $\sqrt{2}/[1+\sqrt{1.33^2-1}]\simeq 0.75350$, a close estimate but a slightly smaller fraction. The exact limiting ray occurs before that grazing ray, where two refracted paths merge at a caustic. This is a geometrical, single-pass visibility boundary; finite pupil size and detection thresholds would smooth the transition.