In the figure,a ray of light $PQ$ is incident normally on one face $AB$ of an equilateral glass prism. What are the angles of incidence at faces $AB$ and $AC$?

(N/A) We have the angle of incidence at the face $AB = 0^{\circ}$ as the ray $PQ$ falls normally on this face.
The normally incident ray $PQ$ passes through the prism undeviated and strikes the face $AC$ at the point $R$.
We draw the normal $MN$ to the face $AC$ at the point $R$.
The angle of incidence at the point $R$ is the angle $NRQ$.
From the geometry of the equilateral triangle $ABC$,the angle at vertex $A$ is $60^{\circ}$. In the triangle formed by the ray $PQ$ and the sides of the prism,the angle at $Q$ is $90^{\circ}$ and the angle at $A$ is $60^{\circ}$,so the angle at $R$ inside the triangle is $180^{\circ} - (90^{\circ} + 60^{\circ}) = 30^{\circ}$.
Since $MN$ is normal to $AC$,the angle of incidence $i = 90^{\circ} - 30^{\circ} = 60^{\circ}$.
Hence,the angle of incidence at face $AB = 0^{\circ}$ and the angle of incidence at face $AC = 60^{\circ}$.