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(B) Let the prism angle be $A = 90^{\circ}$ and refractive index be $\mu = \sqrt{2}$.For total internal reflection $(TIR)$ at the second face, the ang

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$90$

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(B) Let the prism angle be $A = 90^{\circ}$ and refractive index be $\mu = \sqrt{2}$.For total internal reflection $(TIR)$ at the second face, the angle of incidence at the second face $r_2$ must be greater than or equal to the critical angle $C$.The critical angle $C$ is given by $\sin C = \frac{1}{\mu} = \frac{1}{\sqrt{2}}$, which gives $C = 45^{\circ}$.Thus, for $TIR$, $r_2 > 45^{\circ}$.From the geometry of the prism, $r_1 + r_2 = A = 90^{\circ}$.Substituting $r_2 > 45^{\circ}$, we get $r_1 Using Snell's law at the first face: $\sin i = \mu \sin r_1$.Since $r_1 Therefore, $\sin i However, for the light to enter the prism and satisfy the condition, we look for the boundary condition where $r_2 = 45^{\circ}$, which implies $r_1 = 45^{\circ}$.Then $\sin i = \sqrt{2} \sin 45^{\circ} = \sqrt{2} 	imes \frac{1}{\sqrt{2}} = 1$, so $i = 90^{\circ}$.Since the question asks for the condition for $TIR$, any angle of incidence $i  45^{\circ}$ which leads to $r_2 < 45^{\circ}$ (no $TIR$). Thus, the boundary is at $i = 90^{\circ}$.

The angle of a prism made of a material of refractive index $\sqrt{2}$ is $90^{\circ}$. The angle of incidence for a light ray on the first face of the prism such that the light ray suffers total internal reflection at the second face is (in $^{\circ}$)

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