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(C) For total internal reflection $(TIR)$ to occur at the second surface for all $	heta \leq 60^{\circ}$,the condition must be satisfied at the minim

Vedclass · Accepted Answer

$1.50$

Vedclass · Answer

(C) For total internal reflection $(TIR)$ to occur at the second surface for all $	heta \leq 60^{\circ}$,the condition must be satisfied at the minimum angle of incidence within the prism. Let $r_1$ be the angle of refraction at the first surface and $r_2$ be the angle of incidence at the second surface. From the geometry of the prism,$r_1 + r_2 = A = 75^{\circ}$. For $TIR$ at the second surface,the angle of incidence $r_2$ must be greater than or equal to the critical angle $C$,where $\sin C = \frac{n}{n_0}$. As $	heta$ decreases,$r_1$ decreases,and consequently $r_2 = 75^{\circ} - r_1$ increases. The condition for $TIR$ is most easily satisfied at the maximum value of $r_2$,which occurs at the minimum value of $r_1$. However,the problem states $TIR$ occurs for $	heta \leq 60^{\circ}$. The boundary condition is at $	heta = 60^{\circ}$. At $	heta = 60^{\circ}$,by Snell's law at the first surface: $1 \cdot \sin 60^{\circ} = \sqrt{3} \cdot \sin r_1 \Rightarrow \frac{\sqrt{3}}{2} = \sqrt{3} \sin r_1 \Rightarrow r_1 = 30^{\circ}$. Then,$r_2 = 75^{\circ} - 30^{\circ} = 45^{\circ}$. For $TIR$ at this surface,$r_2 \geq C$,so $45^{\circ} \geq C$,which means $\sin 45^{\circ} \geq \sin C = \frac{n}{\sqrt{3}}$. $\frac{1}{\sqrt{2}} \geq \frac{n}{\sqrt{3}} \Rightarrow n \leq \sqrt{\frac{3}{2}}$. The limiting value is $n = \sqrt{1.5}$,so $n^2 = 1.5$.

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