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A cubical block of glass of refractive index $n_{1}$ , is in contact with the surface of water of refractive index $n_{2}$ . A beam of light is incident on vertical face of the block (see figure). After refraction, a total internal reflection at the base and refraction at the opposite vertical face, the ray emerges out at an angle $θ$ . The value of $θ$ is given by

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

$\tan θ < \sqrt{n_{1}^{2} - n_{2}^{2}}$

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

$\tan θ < \frac{1}{\sqrt{n_{1}^{2} - n_{2}^{2}}}$

answer is A.

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Detailed Solution

$n_{1} \sin θ_{C} = n_{2} \Rightarrow \sin θ_{C} = \frac{n_{2}}{n_{1}}$

$\begin{array}{l} 1 \times \sin θ = n_{1} \sin (90^{\circ} - θ_{C}) = n_{1} \cos θ_{C} = n_{1} \sqrt{1 - \frac{n_{2}^{2}}{n_{1}^{2}}} \\ θ = \sin^{- 1} (\sqrt{n_{1}^{2} - n_{2}^{2}}); \sin θ < \sqrt{n_{1}^{2} - n_{2}^{2}} \end{array}$

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