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$\text{Hint: } \sin \theta_c = \frac{1}{\mu}$ $\text{Step: Find the refractive index of the prism.}$ $\text{Let the refractive index of the prism is } \mu.$ $\text{The condition of total internal refraction takes place at point } P \text{ as shown in}$ $\text{the figure below as the ray goes from a denser medium to a rarer medium.}$ $\text{By applying Snell's law at point } P \text{ we get;} \mu \sin \theta = 1 \times \sin 90^\circ$ $\Rightarrow \mu \sin \theta = \text{constant} \quad \cdots (1)$ $\text{By geometry, } r + \theta = 90^\circ$ $\Rightarrow \theta = 90^\circ - r$ $\Rightarrow \mu \sin (90^\circ - r) = 1$ $\Rightarrow \mu \cos r = 1$ $\Rightarrow \cos r = \frac{1}{\mu} \quad \cdots (2)$ $\text{By applying Snell's law at } Q, \text{ we get;} 1 \sin 45^\circ = \mu \sin r$ $\Rightarrow \mu \sin r = \frac{1}{\sqrt{2}}$ $\Rightarrow \mu \sqrt{1 - \cos^2 r} = \frac{1}{\sqrt{2}}$ $\Rightarrow \mu \sqrt{1 - \frac{1}{\mu^2}} = \frac{1}{\sqrt{2}} \quad \left[ \cos r = \frac{1}{\mu} \right]$ $\Rightarrow \mu^2 \left( 1 - \frac{1}{\mu^2} \right) = \frac{1}{2}$ $\Rightarrow \mu^2 - 1 = \frac{1}{2} \Rightarrow \mu^2 = \frac{3}{2}$ $\Rightarrow \mu = \sqrt{\frac{3}{2}}$ $\text{Hence, option (3) is the correct answer.}$