Optics Ques 108
- A ray of light is incident on a prism $A B C$ of refractive index $\sqrt{3}$ as shown in figure.
$(2005,4\ M)$
(a) Find the angle of incidence for which the deviation of light ray by the prism $A B C$ is minimum.
(b) By what angle should the second identical prism be rotated, so that the final ray suffers net minimum deviation.
Show Answer
Answer:
Correct Answer: 108.(a) $60^{\circ}$ (b) $ 60^{\circ}$
Solution:
Formula:
- (a) At minimum deviation, $r_1=r_2=30^{\circ}$
$\therefore$ From Snell’s law
$ \begin{aligned} & \mu=\frac{\sin i _1}{\sin r _1} \text { or } \sqrt{3}=\frac{\sin i _1}{\sin 30^{\circ}} \\ \therefore \quad & \sin i _1=\frac{\sqrt{3}}{2} \text { or } i _1=60^{\circ} \end{aligned} $
(b) In the position shown net deviation suffered by the ray of light should be minimum. Therefore, the second prism should be rotated by $60^{\circ}$ (anti-clockwise).
BETA
