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(D) Let the angles of the prism be $\angle B = \alpha$ and $\angle C = 90^{\circ} - \alpha$. For the first ray incident on $AB$ parallel to $BC$,the a

Vedclass · Accepted Answer

$\sqrt{2} < \mu < \sqrt{3}$

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(D) Let the angles of the prism be $\angle B = \alpha$ and $\angle C = 90^{\circ} - \alpha$. For the first ray incident on $AB$ parallel to $BC$,the angle of incidence $i = \alpha$. The ray emerges grazing $AC$,so the angle of refraction at the second surface is $90^{\circ}$. The angle of refraction at the first surface is $r_1 = 90^{\circ} - \alpha$. Using Snell's law at the first surface: $\sin \alpha = \mu \sin(90^{\circ} - \alpha) = \mu \cos \alpha$. Thus,$ an \alpha = \mu$. Since the ray grazes the surface $AC$,the angle of incidence at $AC$ is the critical angle $ heta_c$. Thus,$\sin heta_c = \frac{1}{\mu}$. From the geometry,$r_1 = 90^{\circ} - heta_c$. Since $r_1 = 90^{\circ} - \alpha$,we have $\alpha = heta_c$. Therefore,$ an \alpha = \mu \Rightarrow \sin \alpha = \frac{\mu}{\sqrt{1+\mu^2}}$. Since $\sin \alpha = \sin heta_c = \frac{1}{\mu}$,we get $\frac{1}{\mu} = \frac{\mu}{\sqrt{1+\mu^2}} \Rightarrow 1+\mu^2 = \mu^4 \Rightarrow \mu^4 - \mu^2 - 1 = 0$. Solving for $\mu^2$,we get $\mu^2 = \frac{1+\sqrt{5}}{2} \approx 1.618$. For the second ray incident on $AC$ parallel to $BC$,the angle of incidence is $i' = 90^{\circ} - \alpha$. For total internal reflection at $AB$,$i' > heta_c \Rightarrow 90^{\circ} - \alpha > heta_c \Rightarrow \cos \alpha > \sin heta_c = \frac{1}{\mu}$. Since $ an \alpha = \mu$,$\cos \alpha = \frac{1}{\sqrt{1+\mu^2}}$. Thus,$\frac{1}{\sqrt{1+\mu^2}} > \frac{1}{\mu} \Rightarrow \mu^2 > 1+\mu^2$,which is impossible. Re-evaluating the geometry: The condition for the first ray is $\mu = \frac{\sin i}{\sin r_1}$. With $i=\alpha$ and $r_1=90- heta_c$,$\mu = \frac{\sin \alpha}{\cos heta_c}$. For the second ray,$TIR$ at $AB$ requires $i > heta_c$. The angle of incidence at $AB$ is $90-\alpha$. So $90-\alpha > heta_c \Rightarrow \cos \alpha > \sin heta_c = 1/\mu$. Combining these,we find $\sqrt{2} < \mu < \sqrt{3}$.

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