Ray Optics and Optical Instruments - Result Question 62
65. The refractive index of the material of the prism is $\sqrt{3}$; then the angle of minimum deviation of the prism is
(a) $30^{\circ}$
(b) $45^{\circ}$
(c) $60^{\circ}$
(d) $75^{\circ}$
[1999]
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Answer:
Correct Answer: 65. (c)
Solution:
- (c) Angle of minimum deviation
$ \begin{aligned} & \mu=\frac{\sin (\frac{A+\delta_m}{2})}{\sin (\frac{A}{2})} \Rightarrow \sqrt{3}=\frac{\sin (\frac{60^{\circ}+\delta_m}{2})}{\sin (\frac{60^{\circ}}{2})} \\ & \Rightarrow \sin (30^{\circ}+\frac{\delta_m}{2})=\frac{\sqrt{3}}{2} \Rightarrow 30^{\circ}+\frac{\delta_m}{2}=60^{\circ} \\ & \Rightarrow \delta_m=60^{\circ} . \end{aligned} $
