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A ray of light is incident at an angle $i$ on the boundary of media I and II as shown in Fig. It travels from medium I to medium IV of reflective indices μ₁, μ₂, μ₃and μ₄ respectively. If μ₁ > μ₂ > μ₃> μ₄, the value of $i$ for which the ray just misses entering medium IV is

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$\sin^{- 1} (\frac{μ_{4}}{μ_{1}})$

$\sin^{- 1} (\frac{μ_{4}}{μ_{3}})$

$\sin^{- 1} (\frac{μ_{2} μ_{3}}{μ_{1}})$

$\sin^{- 1} (\frac{μ_{4}}{μ_{3}} \times \frac{μ_{2}}{μ_{1}})$

answer is A.

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Detailed Solution

The ray will not enter medium IV if the angle of refraction r₃ = 90°.

Applying Snell's law at the interfaces of the media, we have

$μ_{1} \sin i = μ_{2} \sin r_{1} = μ_{3} \sin r_{2} = μ_{4} \sin 90 ° \Rightarrow μ_{1} \sin i = μ_{4} \Rightarrow i = \sin^{- 1} (\frac{μ_{4}}{μ_{1}})$

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