A rectangular glass slab $A B C D$ of refractive index $n_{1}$ is immersed in water of refractive index $n_{2} (n_{1} > n_{2})$ . A ray of light is incident at the surface $A B$ of the slab. The maximum value of the angle of incidence $α_{\max},$ such that the ray comes out only from the other surface $C D$ , is given by :

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$\sin^{- 1} [\frac{n_{1}}{n_{2}} \cos (\sin^{- 1} (\frac{n_{2}}{n_{1}}))]$

$\sin^{- 1} [n_{1} \cos (\sin^{- 1} (\frac{1}{n_{2}}))]$

$\sin^{- 1} (\frac{n_{1}}{n_{2}})$

$\sin^{- 1} (\frac{n_{2}}{n_{1}})$

answer is A.

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Detailed Solution

From figure $r_{1} + r_{2} = 90^{0}$
${(r_{1})}_{\max} = 90^{0} - {(r_{2})}_{\min}$
$= 90^{0} - θ_{C} f o r T I R a t A D$
$\sin θ_{C} = \frac{n_{2}}{n_{1}}$
Snell’s law at $A, B$
$\frac{n_{1}}{n_{2}} = \frac{\sin α_{\max}}{\sin {(r_{1})}_{\max}}$
$= \frac{\sin α_{\max}}{\sin (90^{0} - θ_{C})} = \frac{\sin α_{\max}}{\cos θ_{C}}$
$\sin α_{\max} = (\frac{n_{1}}{n_{2}}) (\cos θ_{C})$
$\frac{n_{1}}{n_{2}} . \cos (\sin^{- 1} \frac{n_{2}}{n_{1}})$
$α_{\max} = \sin^{- 1} [\frac{n_{1}}{n_{2}} \cos \sin^{- 1} (\frac{n_{2}}{n_{1}})]$
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