The problem involves an equilateral prism, which means the prism angle is $A = 60^\circ$. The refractive index of the prism is given as $n = \sqrt{2}$. A ray of light is incident on one surface, and the emergent ray grazes along the other surface. This specific condition implies that the angle of emergence is $e = 90^\circ$.
Let $r_2$ be the angle of refraction inside the prism at the second surface. According to Snell's Law at the point where the ray exits the prism:
$ n_{prism} \sin(r_2) = n_{air} \sin(e) $
Substituting the known values ($n_{prism} = \sqrt{2}$ and $e = 90^\circ$):
$ \sqrt{2} \sin(r_2) = 1 \times \sin(90^\circ) $
Since $\sin(90^\circ) = 1$:
$ \sqrt{2} \sin(r_2) = 1 $
Solving for $\sin(r_2)$:
$ \sin(r_2) = \frac{1}{\sqrt{2}} $
Therefore, the angle $r_2$ is:
$ r_2 = 45^\circ $
For any prism, the prism angle $A$ is related to the angles of refraction at the first ($r_1$) and second ($r_2$) surfaces by the formula:
$ A = r_1 + r_2 $
We need to find the angle of refraction at the incident surface, which is $r_1$. Using the prism angle formula and the calculated value of $r_2$:
$ 60^\circ = r_1 + 45^\circ $
Solving for $r_1$:
$ r_1 = 60^\circ - 45^\circ $
$ r_1 = 15^\circ $
The angle of refraction ($r_1$) at the incident surface is $15^\circ$.