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Light from a denser medium 1 passes to a rarer medium 2. When the angle of incidence is $θ$ , the partially reflected and refracted rays are mutually perpendicular. The critical angle will be

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$\sin^{- 1} (\cot θ)$

$\sin^{- 1} (\tan θ)$

$\sin^{- 1} (\cos θ)$

$\sin^{- 1} (\sec θ)$

answer is B.

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Detailed Solution

$μ_{1} \sin θ = μ_{2} \times \sin (90^{\circ} - θ) \Rightarrow \frac{μ_{1}}{μ_{2}} = \tan θ$

$For θ_{e} \Rightarrow μ_{1} \times \sin θ_{C} = μ_{2} \times \sin (90^{\circ}) \sin θ_{c} = \frac{μ_{1}}{μ_{2}} = \tan θ \Rightarrow θ_{c} = \sin^{- 1} (\tan θ)$

$\Rightarrow θ_{c} = \sin^{- 1} (\tan θ)$

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