The rectangular glass slab ABCD of refractive index $n_{1}$ is immersed in water of refractive index $n_{2} (n_{1} > n_{2})$ . A ray of light is incident in the surface AB of the slab as shown maximum value of the angle of incident $α_{\max}$ such that the ray comes out only from the other surface CD is given be

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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

$n_{1} sinC = n_{2} \sin 90 ° sinC = n_{2} / n_{1} C = \sin^{- 1} (n_{2} / n_{1}) \to (1) Refraction at AB : n_{2} sinα = n_{1} sinγ sinr = \frac{n_{2}}{n_{1}} sinα \to (2) r + i = 90 ° r = 90 ° - i Substitute ‘ r ’ in eq (2) \sin (90 ° - i) = \frac{n_{2}}{n_{1}} sinα cosi = \frac{n_{2}}{n_{1}} sinα cosC = \frac{n_{2}}{n_{1}} sinα [∵ i = c at AD] \cos [\sin^{- 1} \frac{n_{2}}{n_{1}}] = \frac{n_{2}}{n_{1}} sinα \frac{n_{1}}{n_{2}} \cos [\sin^{- 1} (\frac{n_{2}}{n_{1}})] = sinα α = \sin^{- 1} [\frac{n_{1}}{n_{2}} \cos [\sin^{- 1} (\frac{n_{2}}{n_{1}})]]$