= Solution
Take a horizontal cross-section in units of $R$. The eye is at $E=(\sqrt2,0)$, 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 $P=(\cos\phi,\sin\phi)$. An unobstructed outward ray to the eye has $|\phi|\le\pi/4$.
For $\phi\ge0$, 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> give
$$
\sin e=\frac{\sqrt2\sin\phi}{\sqrt{3-2\sqrt2\cos\phi}},\qquad
i=\sin^{-1}\left(\frac{\sin e}{1.33}\right).
$$
The outgoing internal ray makes angle $\beta=\phi-i$ to $+x$. Its backward continuation intersects the swim diameter at
$$
y(\phi)=\sin\phi-\cos\phi\tan(\phi-i).
$$
The directly visible positive endpoint is the largest such intercept, including all single-pass refracted rays. Maximizing this one-variable expression gives $\phi_*\simeq38.07^\circ$ and $y_*\simeq0.75744$. By symmetry the visible length is $2Ry_*$, so
$$
\boxed{\text{visible fraction}\simeq0.757\quad\text{(about three quarters)}.}
$$
Checking only the grazing exit at $\phi=45^\circ$ gives $\sqrt2/[1+\sqrt{1.33^2-1}]\simeq0.75350$, 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.
\Image[/past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2013-tank-visible-diameter.png]
{title=Single-pass visibility across the fish-tank diameter}
{description=The original ray construction traces rays backward from the eye through a circular n=1.33 interface. The maximum swim-line intercept, not just the grazing outgoing ray, determines the visible interval.}
Back to article page
