A ray of light is incident on one face of an equilateral glass prism having refractive index $\sqrt{2}$. It…

A ray of light is incident on one face of an equilateral glass prism having refractive index $\sqrt{2}$. It produces the emergent ray which just grazes along the adjacent face. The value of angle of incidence is
  1. $\sin ^{-1}\left(\frac{1}{\sqrt{2}} \sin 15^{\circ}\right)$
  2. $\sin ^{-1}\left(\sqrt{2} \sin 30^{\circ}\right)$
  3. $\sin ^{-1}\left(\frac{1}{\sqrt{2}} \sin 45^{\circ}\right)$
  4. $\sin ^{-1}\left(\sqrt{2} \sin 15^{\circ}\right)$

Solution

The emergent ray just grazes the second face. Hence angle of emergence, $\mathrm{e}=90^{\circ}$ $\begin{aligned} & \mu=\frac{\sin \mathrm{e}}{\sin r_2}=\frac{\sin 90^{\circ}}{\sin r_2}=\frac{1}{\sin r_2} \\ & \therefore \frac{1}{\sin r_2}=\sqrt{2} \text { or } \sin r_2=\frac{1}{\sqrt{2}}\end{aligned}$ $\begin{aligned} & \therefore \mathrm{r}_2=45^{\circ} ; \mathrm{A}=\mathrm{r}_1+\mathrm{r}_2 \\ & \therefore \mathrm{r}_1=\mathrm{A}-\mathrm{r}_2=60-45=15^{\circ}\end{aligned}$ Also, $\frac{\sin i}{\sin r_1}=\mu$ $\begin{aligned} & \therefore \sin \mathrm{i}=\mu \sin \mathrm{r}_1=\sqrt{2} \sin 15^{\circ} \\ & \therefore \mathrm{i}=\sin ^{-1}\left(\sqrt{2} \sin 15^{\circ}\right)\end{aligned}$ .

Asked in: MHT CET 2021 (24 Sep Shift 2)

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