= Visibility across a refracting circular tank
For a circular index-$n$ tank, an eye at $(\sqrt2R,0)$, and a swim diameter $x=0$, trace outward single-pass rays through $P=R(\cos\phi,\sin\phi)$, with $|\phi|\le\pi/4$. Backward refraction gives the dimensionless intercept $y=\sin\phi-\cos\phi\tan(\phi-i)$, where $\sin i=\sqrt2\sin\phi/[n\sqrt{3-2\sqrt2\cos\phi}]$. The visible fraction is the largest positive intercept. For $n=1.33$ it is approximately $0.75744$, attained before the grazing outgoing ray. The limiting pair of stationary optical paths merges at a <caustic>.
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