Q. The exit surface of a prism with refractive index \(n\) is coated with a material
having refractive index \( \dfrac{n}{2} \). When this prism is set for minimum angle
of deviation, it exactly meets the condition of critical angle. The prism angle
is _____.
Step-by-Step Solution
For a prism at minimum deviation, the refraction angles inside the prism are equal.
\[
r_1 = r_2 = \frac{A}{2}
\]
At the exit surface, the ray meets the critical angle condition.
Critical angle \(C\) is given by:
\[
\sin C = \frac{\text{Refractive index of rarer medium}}
{\text{Refractive index of denser medium}}
\]
Here,
\[
\sin C = \frac{(n/2)}{n} = \frac{1}{2}
\]
\[
C = 30^\circ
\]
Since the ray inside the prism makes angle \( \dfrac{A}{2} \) with the normal at the
exit face and it just satisfies the critical angle condition,
\[
\frac{A}{2} = C = 30^\circ
\]
\[
A = 60^\circ
\]
Therefore, the prism angle is
\[
\boxed{60^\circ}
\]