Optics Ques 113
A ray of light is incident at an angle of $60^{\circ}$ on one face of a prism which has an angle of $30^{\circ}$. The ray emerging out of the prism makes an angle of $30^{\circ}$ with the incident ray. Show that the emergent ray is perpendicular to the face through which it emerges and calculate the refractive index of the material of the prism.
(1978)
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Answer:
Correct Answer: 113.$(\mu=\sqrt{3})$
Solution:
Formula:
- Given $i _1=60^{\circ}, A=30^{\circ}, \delta=30^{\circ}$
From the relation
$ \begin{array}{ll} & \delta=\left(i_1+i_2\right)-A \\ \text { we have, } & i_2=0^{\circ} \end{array} $
i.e. the ray is perpendicular to the face from which it emerges.
Further, i_2 =0^{\circ}
$\therefore \quad r _2 =0^{\circ} $
$r_1 + r_2 = A$
$r_1 = A = 30^{\circ}$
$\mu=\frac{\sin i _1}{\sin r _1} =\frac{\sin 60^{\circ}}{\sin 30^{\circ}}=\sqrt{3}$
BETA
