= Solution
Let $r=\arcsin(\sin\theta/n)$ be the angle inside the prism relative to the inward normal of face A. For the entry position shown, below the midpoint of A, and $0<\theta<90^\circ$, the transmitted ray first reaches face B. Its incidence angle there is $i_B=|\alpha-r|$. If $n\sin i_B<1$, a transmitted ray leaves B according to <Snell's law>. If $n\sin i_B>1$, it undergoes <total internal reflection> at B and then reaches C. At the equality, transmission is grazing.
On the reflected route through this equilateral prism, the incidence angle at C is $r$, so $n\sin r=\sin\theta<1$ permits emergence from C. The <one-reflection path through an equilateral prism> is drawn alongside the direct-transmission route.
\Image[/past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2018-prism-paths.png]
{title=Principal transmitted ray paths through an <optical prism>: <total internal reflection> at B followed by exit at C, or direct refraction out of B}
Real interfaces also produce partial <Fresnel reflection> below the critical angle; these additional weaker rays are omitted from the principal-path sketch. Which face a ray first reaches also depends on its entrance position, so the route statement uses the entrance geometry illustrated in the source.
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