A rectangular glass slab ABCD,of refractive i | vedclass

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Question

(A) For the ray to emerge only from the surface $CD$,it must undergo total internal reflection at the surface $AD$.Applying Snell's law at the surface

Vedclass · Accepted Answer

$\sin^{-1} \left[ \frac{n_1}{n_2} \cos \left( \sin^{-1} \frac{n_2}{n_1} \right) \right]$

Vedclass · Answer

(A) For the ray to emerge only from the surface $CD$,it must undergo total internal reflection at the surface $AD$.Applying Snell's law at the surface $AB$:$n_2 \sin \alpha_{max} = n_1 \sin r_1 \implies \alpha_{max} = \sin^{-1} \left( \frac{n_1}{n_2} \sin r_1 ight) \dots (i)$At the surface $AD$,the angle of incidence is $r_2$. For total internal reflection to occur,$r_2$ must be at least the critical angle $C$,where $\sin C = \frac{n_2}{n_1}$.Since the surface $AD$ is perpendicular to $AB$,we have $r_1 + r_2 = 90^\circ$,so $r_1 = 90^\circ - r_2$.To ensure the ray reaches $CD$,we set $r_2 = C = \sin^{-1} \left( \frac{n_2}{n_1} ight)$.Thus,$r_1 = 90^\circ - \sin^{-1} \left( \frac{n_2}{n_1} ight)$.Substituting this into equation $(i)$:$\alpha_{max} = \sin^{-1} \left[ \frac{n_1}{n_2} \sin \left( 90^\circ - \sin^{-1} \frac{n_2}{n_1} ight) ight]$Using the identity $\sin(90^\circ - 	heta) = \cos 	heta$:$\alpha_{max} = \sin^{-1} \left[ \frac{n_1}{n_2} \cos \left( \sin^{-1} \frac{n_2}{n_1} ight) ight]$.

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