A transparent cube of side d, made of a material of refractive index $μ_{2},$ is immersed in a liquid of refractive index $μ_{1} (μ_{1} < μ_{2}) .$ A ray is incident on the face AB at an angle $θ$ (shown in the figure). Total internal reflection takes place at point E on the face BC.
Then $θ$ must satisfy:

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$θ < \sin^{- 1} \frac{μ_{1}}{μ_{2}}$

$θ < \sin^{- 1} \sqrt{\frac{μ_{2}^{2}}{μ_{1}^{2}} - 1}$

$θ > \sin^{- 1} \frac{μ_{1}}{μ_{2}}$

$θ > \sin^{- 1} \sqrt{\frac{μ_{2}^{2}}{μ_{1}^{2}} - 1}$

answer is C.

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Detailed Solution

Using, $\sin θ_{\max} = μ_{1} \sqrt{μ_{2}^{2} - μ_{1}^{2}} = \sqrt{\frac{μ_{2}^{2}}{μ_{1}^{2}} - 1}$
or $θ_{\max} = \sin^{- 1} (\sqrt{\frac{μ_{2}^{2}}{μ_{1}^{2}} - 1})$
For $T_{1} R, θ < \sin^{- 1} {(\sqrt{\frac{μ_{2}^{2}}{μ_{1}^{2}} - 1})}_{A}$