A ray OP of monochromatic light is incident on the face AB of prism ABCD near vertex B at an incident angle of 60º as shown in figure. If the refractive index of the material of the prism is $\sqrt{3}$ , which of the following is(are) correct ?

detailed solution

Correct option is A

Applying Snell’s law at face AB, we get
$\begin{array}{l}1\times \sin {60}^{\circ }=\sqrt{3}\times \sin \mathrm{r}\\ \frac{\sqrt{3}}{2}=\sqrt{3}\sin \mathrm{r}\text{ or }\sin \mathrm{r}=\frac{1}{2}\\ \mathrm{r}={\sin }^{-1}\left(\frac{1}{2}\right)={30}^{\circ }\\ \sin {\mathrm{\theta }}_{\mathrm{c}}=\frac{1}{\sqrt{3}}\end{array}$
(where, ${\mathrm{\theta }}_{\mathrm{c}}$ is the critical angle)
${\mathrm{\theta }}_{\mathrm{c}}={\sin }^{-1}\left(\frac{1}{\sqrt{3}}\right)\text{ or }{\mathrm{\theta }}_{\mathrm{c}}\approx {35}^{\circ }$
From geometry, angle of incidence at Q is 45º. At face CD angle of incidence is greater than ${\mathrm{\theta }}_{\mathrm{c}}$. Therefore, the rays gets totally internally reflected at face CD. Hence, option (a) is correct. From geometry, angle of incidence at R is 30º. At face AD angle of incidence is less than ${\mathrm{\theta }}_{\mathrm{c}}$So, the ray comes through face AD. Hence option (b) is correct. Applying Snell’s law at face AD, we get$\mathrm{e}={\sin }^{-1}\left(\frac{\sqrt{3}}{2}\right)={60}^{\circ }$
The angle between the incident ray and the emergent ray is 90º.
Hence, option (c) is correct and option (d) is incorrect