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

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$\sin^{- 1} \frac{μ_{2}}{μ_{1}}$

$\sin^{- 1} \frac{μ_{1}}{μ_{2}}$

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

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

answer is A.

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Detailed Solution

Ray comes out from CD, when it get reflected from AD
$\begin{array}{l} \frac{n_{1}}{n_{2}} = \frac{\sin α_{max}}{\sin r_{1}} \\ \Rightarrow α_{max} = \sin^{- 1} [\frac{μ_{1}}{μ_{2}} \sin r_{1}] \end{array}$
Also r₁ = 90^o - $θ$ = 90^o - C
$\begin{array}{l} or r_{1} = 90^{0} - \sin^{- 1} (\frac{1}{_{2} μ_{1}}) or r_{1} = 90^{0} - \sin^{- 1} (\frac{μ_{2}}{μ_{1}}) \\ ∴ α_{max} = \sin^{- 1} [\frac{μ_{1}}{μ_{2}} \sin \{90^{0} - \sin^{- 1} \frac{μ_{2}}{μ_{1}}\}] = \sin^{- 1} [\frac{μ_{1}}{μ_{2}} \cos \{\sin^{- 1} (\frac{μ_{2}}{μ_{1}})\}] \end{array}$