Q. A ray of light is incident normally on one of the face of a prism of angle $30{^\circ}$ and refractive index $\sqrt{2}$. The angle of deviation will be
A) $26{^\circ}$
B) $0{^\circ}$
C) $23{^\circ}$
D) $15{^\circ}$
Explanation
(d) $A = 30{^\circ},\ \mu = \sqrt{2}$. As we know
$A = r_{1} + r_{2} = 0 + r_{2} \Rightarrow A = r_{2}$
Applying Snell’s law for the surface $AC$
$\frac{1}{\mu} = \frac{\sin r_{2}}{\sin e} = \frac{\sin A}{\sin e}$
$\Rightarrow \frac{1}{\sqrt{2}} = \frac{\sin{30{^\circ}}}{\sin e} \Rightarrow e = 45{^\circ}$
$\delta = e - r_{2} = 45{^\circ} - 30{^\circ} = 15{^\circ}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.