Light enters an isosceles right triangular pr | vedclass

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Question

(C) For total internal reflection $(TIR)$ to occur at the face $BC$,the angle of incidence $i$ must be greater than or equal to the critical angle $C$

Vedclass · Accepted Answer

$1.42$

Vedclass · Answer

(C) For total internal reflection $(TIR)$ to occur at the face $BC$,the angle of incidence $i$ must be greater than or equal to the critical angle $C$.From the geometry of the isosceles right-angled prism,the light ray enters face $AB$ normally and strikes face $BC$ at an angle of incidence $i = 45^{\circ}$.For $TIR$ to occur,the condition is $i \geq C$,where $C$ is the critical angle.Therefore,$45^{\circ} \geq C$.Taking the sine on both sides,$\sin 45^{\circ} \geq \sin C$.We know that $\sin C = \frac{1}{\mu}$,where $\mu$ is the refractive index of the prism.So,$\sin 45^{\circ} \geq \frac{1}{\mu}$.$\frac{1}{\sqrt{2}} \geq \frac{1}{\mu}$.$\mu \geq \sqrt{2}$.Since $\sqrt{2} \approx 1.414$,the minimum value of the refractive index $\mu$ is approximately $1.42$.

Light enters an isosceles right triangular prism at normal incidence through face $AB$ and undergoes total internal reflection at face $BC$ as shown below. The minimum value of the refractive index of the prism is close to

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