A ray is incident normally on the face of an equilateral prism of refracting angle of \( 60^\circ \). The refractive index of the prism is \( \displaystyle\frac{2}{\sqrt{3}} \). The angle of deviation will be:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 30^\circ \)
For a prism the angles of refraction inside satisfy \( r_1 + r_2 = A \), and the deviation is \( \delta = i + e - A \). At normal incidence the ray enters without bending, so \( i = 0 \) and \( r_1 = 0 \).
\( i = 0,\ r_1 = 0 \)
\( r_2 = A - r_1 = 60^\circ \)
At the second face, Snell's law (glass to air) gives the angle of emergence \( e \).
\( \displaystyle \sin e = \mu \sin r_2 = \frac{2}{\sqrt{3}} \times \frac{\sqrt{3}}{2} = 1 \)
\( \displaystyle e = 90^\circ \)
\( \displaystyle \delta = i + e - A = 0 + 90^\circ - 60^\circ = 30^\circ \)
The ray emerges grazing the second face, since \( r_2 \) is the critical angle here.