Optics 4 Question 19
20. 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 the second identical prism must be rotated, so that the final ray suffers net minimum deviation.
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Solution:
- (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).
