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Question

(D) Let $C$ be the critical angle for the interface between the cube $(\mu_2)$ and the liquid $(\mu_1)$.For total internal reflection $(TIR)$ at point

Vedclass · Accepted Answer

$	heta < \sin^{-1} \sqrt{\frac{\mu_2^2}{\mu_1^2} - 1}$

Vedclass · Answer

(D) Let $C$ be the critical angle for the interface between the cube $(\mu_2)$ and the liquid $(\mu_1)$.For total internal reflection $(TIR)$ at point $E$,the angle of incidence $i'$ at face $BC$ must be greater than the critical angle $C$.Thus,$i' > C$.From the geometry,the angle of refraction $r$ at face $AB$ is related to $i'$ by $r = 90^\circ - i'$.Since $i' > C$,we have $90^\circ - r > C$,or $r Applying Snell's Law at face $AB$: $\mu_1 \sin 	heta = \mu_2 \sin r$.Since $\sin$ is an increasing function,$\sin r Therefore,$\mu_1 \sin 	heta We know $\sin C = \frac{\mu_1}{\mu_2}$,so $\cos C = \sqrt{1 - \sin^2 C} = \sqrt{1 - \frac{\mu_1^2}{\mu_2^2}} = \frac{\sqrt{\mu_2^2 - \mu_1^2}}{\mu_2}$.Substituting this into the inequality: $\mu_1 \sin 	heta $\sin 	heta Thus,$	heta < \sin^{-1} \sqrt{\frac{\mu_2^2}{\mu_1^2} - 1}$.

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