Question

A transparent cube of side $d$, made of a material of refractive index $\mu_{2}$, is immersed in a liquid of refractive index $\mu_{1} \quad\left(\mu_{1}<\mu_{2}\right)$. A ray is incident on the face $AB$ at an angle $\theta$ (shown in the figure). Total internal reflection takes place at the point $E$ on the face $BC$.

• A. $\theta < sin^{- 1}\sqrt{\frac{\mu _{2}^{2}}{\mu _{1}^{2}} - 1}$
• B. $\theta >sin^{- 1}\frac{\mu _{1}}{\mu _{2}}$
• C. $\theta < sin^{- 1}\frac{\mu _{1}}{\mu _{2}}$
• D. $\theta >sin^{- 1}\sqrt{\frac{\mu _{2}^{2}}{\mu _{1}^{2}} - 1}$

Answer 1

Answer: (A)

At surface (1)
$\mu_{1} \sin \theta=\mu_{2} \sin r$
At surface (2)
$ \begin{array}{l} \mu_{1}=\mu_{2} \sin (90-r) \\ \mu_{1}=\mu_{2} \cos r \ldots \text { (2) } \end{array} $
From equation (1) and (2)
$ \begin{array}{l} \Rightarrow \sin \theta=\frac{\sin r}{\cos r} \\ \Rightarrow \sin \theta=\frac{\frac{\left(\sqrt{\mu_{2}^{2}-\mu_{1}^{2}}\right)}{\mu_{2}}}{\left(\frac{\mu_{1}}{\mu_{2}}\right)} \\ \Rightarrow \sin \theta=\sqrt{\frac{\mu_{2}^{2}}{\mu_{1}^{2}}-1} \end{array} $
$\theta$ must satisfy,
$ \Rightarrow \theta<\sin ^{-1}\left(\sqrt{\frac{\mu_{2}^{2}}{\mu_{1}^{2}}-1}\right) $