A rectangular glass slab ABCD, of refractive | vedclass

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(A) For the ray to emerge from the surface $CD,$ it must not undergo total internal reflection at the surface $AD.$ The limiting condition is that the

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 from the surface $CD,$ it must not undergo total internal reflection at the surface $AD.$ The limiting condition is that the angle of incidence at surface $AD$ $(r_2)$ must be equal to the critical angle $C.$Applying Snell's law at surface $AB$:$n_2 \sin \alpha_{\max} = n_1 \sin r_1 \Rightarrow \sin \alpha_{\max} = \frac{n_1}{n_2} \sin r_1$  --- $(i)$In the triangle formed inside the slab,$r_1 + r_2 = 90^{\circ}.$ For the ray to just emerge from $CD,$ $r_2$ must be the critical angle $C,$ where $\sin C = \frac{n_2}{n_1}.$Thus,$r_1 = 90^{\circ} - C = 90^{\circ} - \sin^{-1}\left(\frac{n_2}{n_1}ight).$Substituting $r_1$ into equation $(i)$:$\alpha_{\max} = \sin^{-1}\left[\frac{n_1}{n_2} \sin\left(90^{\circ} - \sin^{-1}\frac{n_2}{n_1}ight)ight]$Since $\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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