Q. A ray of light is incident at polarising angle such that its deviation is $24{^\circ},$ then angle of incidence is
A) $24{^\circ}$
B) $57{^\circ}$
C) $66{^\circ}$
D) $90{^\circ}$
Explanation
(b) Here, $i_{p} = r + \delta$
[image:10_image39.emf]$i_{p} = r + 24\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
Moreover, $i_{p} + r = 90{^\circ}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
From Eqs. (i) and (ii), we get
$i_{p} + \left( i_{p} - 24 \right) = 90{^\circ}$
$\Rightarrow i_{p} = 57{^\circ}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.