Question

Light ray is incident on a prism of angle A $=$ 60${}^o$ and refractive index $\mu =\sqrt{2}$. The angle of incidence at which the emergent ray grazes the surface is given by :

• A. $si{n}^{-1}\left(\frac{\sqrt{3}-1}{2}\right)$
• B. $si{n}^{-1}\left(\frac{1-\sqrt{3}}{2}\right)$
• C. $si{n}^{-1}\left(\frac{\sqrt{3}}{2}\right)$
• D. $si{n}^{-1}\left(\frac{2}{\sqrt{3}}\right)$

Answer 1

The correct option is A $si{n}^{-1}\left(\frac{\sqrt{3}-1}{2}\right)$

Since the emergent ray grazes the surface, $e={90}^{\circ }$

Applying Snell's law at the second surface gives,

$\mu \sin r_2=1$

$r_2={45}^{\circ }$

Also in a prism $r_1+r_2=A$

Thus $r_1={15}^{\circ }$

Applying Snell's law at the firs surface gives

$\sin i=\mu \sin r_1$

$\sin i=\sqrt{2}\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)=\frac{\sqrt{3}-1}{2}$

Hence $i={\sin }^{-1}\left(\frac{\sqrt{3}-1}{2}\right)$