= Solution
Square the stationary condition and eliminate the <angle of incidence> using <Snell's law>:
$$
1-n^2\sin^2\beta=\cos^2\alpha=\frac{n^2}{4}\cos^2\beta=\frac{n^2}{4}(1-\sin^2\beta).
$$
Rearranging yields $3n^2\sin^2\beta=4-n^2$. Taking the positive root for a physical incident <light ray>,
$$
\boxed{\sin\beta=\sqrt{\frac{4-n^2}{3n^2}}.}
$$
An interior minimum exists for $1<n<2$, which includes water at visible <wavelengths>. For illustration, $n=4/3$ gives $\beta=40.20^\circ$, $\alpha=59.39^\circ$ and $\delta_{\min}=137.97^\circ$. The corresponding angle from the antisolar direction is $180^\circ-\delta_{\min}=42.03^\circ$. At $n=2$ the stationary solution reaches the endpoint, and for $n>2$ the displayed square root is not real; the water result is not a formula for every possible <refractive index>.
\Image[/past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2015-primary-rainbow-deviation.png]
{title=Stationary deviation for the one-reflection rainbow branch and its dispersion}
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