Question 64 of 295

A transparent cube of side $d$, made of a material of refractive index ${\mu }_{2}$, is immersed in a liquid of refractive index $μ1 ( μ1<μ2)$. 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 $BC.$



Then, $\theta$ must satisfy

A$\theta \lt {\sin }^{-1}\sqrt{\dfrac{{\mu }_{2}^{2}}{{\mu }_{1}^{2}}-1}$B$\theta \gt {\sin }^{-1}\dfrac{{\mu }_{1}}{{\mu }_{2}}$C$\theta \lt {\sin }^{-1}\dfrac{{\mu }_{1}}{{\mu }_{2}}$D$\theta \gt {\sin }^{-1}\sqrt{\dfrac{{\mu }_{2}^{2}}{{\mu }_{1}^{2}}-1}$

Show answer and solution▾

Answer: (A) $\theta \lt {\sin }^{-1}\sqrt{\dfrac{{\mu }_{2}^{2}}{{\mu }_{1}^{2}}-1}$

At surface (1)

${\mu }_{1}\sin \theta ={\mu }_{2}\sin r\ldots \left(1\right)$

At surface (2)

${\mu }_{1}={\mu }_{2}\sin \left(90-r\right)$

${\mu }_{1}={\mu }_{2}\cos r...(2)$

From equation $(1)$ and $(2)$

$\Rightarrow \sin ⁡\theta =\dfrac{\sin ⁡r}{\cos ⁡r}$

$\Rightarrow \sin \theta =\dfrac{\dfrac{\left(\sqrt{{\mu }_{2}^{2}-{\mu }_{1}^{2}}\right)}{{\mu }_{2}}}{\left(\dfrac{{\mu }_{1}}{{\mu }_{2}}\right)}$

$\Rightarrow \sin \theta =\sqrt{\dfrac{{\mu }_{2}^{2}}{{\mu }_{1}^{2}}-1}$

$\theta$ must satisfy,

$\Rightarrow \theta \lt {\sin }^{-1}\left(\sqrt{\dfrac{{\mu }_{2}^{2}}{{\mu }_{1}^{2}}-1}\right)$