Q. When a glass prism of refracting angle $60{^\circ}$ is immersed in a liquid its angle of minimum deviation is $30{^\circ}$. The critical angle of glass with respect to the liquid medium is
A) $42{^\circ}$
B) $45{^\circ}$
C) $50{^\circ}$
D) $52{^\circ}$
Explanation
(b) $A = 60{^\circ},\ \delta_{m} = 30{^\circ}\ so\ \mu = \frac{\sin\left( \frac{A + \delta_{m}}{2} \right)}{\sin\left( \frac{A}{2} \right)}$
$\mu = \frac{\sin\left( \frac{60{^\circ} + 30{^\circ}}{2} \right)}{\sin\left( \frac{60{^\circ}}{2} \right)} = \frac{\sin{45{^\circ}}}{\sin{30{^\circ}}} = \sqrt{2}$
Also $\mu = \frac{1}{\sin C} \Rightarrow C = \sin^{- 1}\left( \frac{1}{\mu} \right) \Rightarrow C = 45{^\circ}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.