A ray falls on a prism ABC (AB = BC) and trav | vedclass

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(B) From the figure,the prism is a right-angled isosceles triangle with $\angle B = 90^\circ$ and $AB = BC$,which implies $\angle A = \angle C = 45^\c

Vedclass · Accepted Answer

$\sqrt{2}$

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(B) From the figure,the prism is a right-angled isosceles triangle with $\angle B = 90^\circ$ and $AB = BC$,which implies $\angle A = \angle C = 45^\circ$.The ray strikes the face $AC$ at an angle of incidence such that it undergoes total internal reflection at the face $BC$. For total internal reflection to occur,the angle of incidence $i$ at the face $BC$ must be greater than or equal to the critical angle $	heta_C$.From the geometry,the angle of incidence at face $BC$ is $45^\circ$. Thus,for total internal reflection,we must have $i \ge 	heta_C$,which means $45^\circ \ge 	heta_C$.Since $\sin 	heta_C = \frac{1}{\mu}$,we have $\sin 45^\circ \ge \frac{1}{\mu}$.$\frac{1}{\sqrt{2}} \ge \frac{1}{\mu} \implies \mu \ge \sqrt{2}$.Therefore,the minimum refractive index is $\mu = \sqrt{2}$.

$A$ ray falls on a prism $ABC$ $(AB = BC)$ and travels as shown in the figure. The minimum refractive index of the prism material should be

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