A light ray is incident from a medium of refractive index 2 into a medium of refractive index $\sqrt{3}$.…
A light ray is incident from a medium of refractive index 2 into a medium of refractive index $\sqrt{3}$. The critical angle is
$30^{\circ}$
$45^{\circ}$
$60^{\circ}$
$90^{\circ}$
Solution
Given, absolute refractive index of medium-1 and 2 are given as
$
\mu_1=2 \text { and } \mu_2=\sqrt{3}
$
$\therefore$ Let $i_c$ be the critical angle, then $\sin i_c=\frac{1}{2 \mu_1}$
$
\begin{aligned}
& =\frac{1}{\frac{\mu_1}{\mu_2}}=\frac{1}{2 / \sqrt{3}} \\
& =\frac{\sqrt{3}}{2} \\
\Rightarrow \quad \sin i_c & =\sin 60^{\circ} \\
\Rightarrow \quad i_c & =60^{\circ}
\end{aligned}
$