Q. A ray of light passes through an equilateral prism such that the angle of incidence and the angle of emergence are both equal to 3/4th of the angle of prism. The angle of minimum deviation is
A) $15{^\circ}$
B) $30{^\circ}$
C) $45{^\circ}$
D) $60{^\circ}$
Explanation
(b) Given, $A = 60{^\circ}$
$i = i^{'} = \frac{3}{4},A = 45{^\circ}$
$\therefore i + i^{'} = A + \delta$
Or $90{^\circ} = 60{^\circ} + \delta$
$\therefore\delta = 30{^\circ}$
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Note that $i = i^{'}$is the condition for minimum deviation
Hence$,\delta = 30{^\circ} = \delta_{\min}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.