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(C) Let the refractive index of the prism be $\mu$. Applying Snell's law at the first surface:$1 	imes \sin 45^{\circ} = \mu 	imes \sin(90^{\circ} -

Vedclass · Accepted Answer

$\sqrt{\frac{3}{2}}$

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(C) Let the refractive index of the prism be $\mu$. Applying Snell's law at the first surface:$1 	imes \sin 45^{\circ} = \mu 	imes \sin(90^{\circ} - 	heta)$$\frac{1}{\sqrt{2}} = \mu \cos 	heta \Rightarrow \cos 	heta = \frac{1}{\sqrt{2}\mu}$For total internal reflection at the second surface,the angle of incidence $	heta$ must be greater than or equal to the critical angle $C$:$	heta \geq C \Rightarrow \sin 	heta \geq \sin C = \frac{1}{\mu}$Using the identity $\sin 	heta = \sqrt{1 - \cos^2 	heta}$:$\sqrt{1 - \left(\frac{1}{\sqrt{2}\mu}ight)^2} \geq \frac{1}{\mu}$$\sqrt{1 - \frac{1}{2\mu^2}} \geq \frac{1}{\mu}$Squaring both sides:$1 - \frac{1}{2\mu^2} \geq \frac{1}{\mu^2}$$1 \geq \frac{1}{\mu^2} + \frac{1}{2\mu^2} = \frac{3}{2\mu^2}$$\mu^2 \geq \frac{3}{2} \Rightarrow \mu \geq \sqrt{\frac{3}{2}}$Thus,the minimum refractive index is $\sqrt{\frac{3}{2}}$.

For the given incident ray as shown in the figure,what is the minimum refractive index of the prism required for total internal reflection of this ray?

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