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The critical angle for light wave going from a medium in which wavelength is $4000 A°$ to a medium in which its wavelength is $μ_{1} μ_{2} μ_{3}$ is

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30 °

45 °

60 °

\sin^{- 1} (2 / 3)

answer is D.

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Detailed Solution

Wave length in denser medium

μ_{d} = 4000 A°, μ_{r} = 6000 A°

critical angle c = ?

$\frac{μ_{d}}{μ_{h}} = \frac{1}{\sin c}$

$But μ \propto \frac{1}{λ}$

$\frac{λ_{l}}{λ_{d}} = \frac{1}{\sin c}$

$\frac{6000}{4000} = \frac{1}{\sin c}$

$\frac{3}{2} = \frac{1}{\sin c}$

$\sin c = \frac{2}{3}$

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

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