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A ray of light is incident normally on a right-angle prism of vertex angle 60° and refractive index 5/3, as shown, immersed in a liquid of refractive index 4/3 Then

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The angle of emergence from base, $θ_{2} is \sin^{- 1} (5 / 8)$

The angle of emergence from base, $θ_{2}$ is $\sin^{- 1} (5 / 4 \sqrt{3})$

T.I.R. will not take place at P if R.I. of liquid become $5 / 2 \sqrt{3}$

T.I.R. will not take place at P if R.I. of liquid becomes 5/6

answer is A, C.

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Detailed Solution

From geometry

$θ_{1} = 60^{\circ}$ such that

$\frac{5}{3} \sin 30^{\circ} = \frac{4}{3} \sin θ_{2} \Rightarrow θ_{2} = \sin^{- 1} (5 / 8)$

For $T.I.R θ_{1} > θ_{c}$ so, for no TIR

$\frac{5}{3} \sin 60^{\circ} = μ \sin 90^{\circ} \Rightarrow μ = \frac{5}{2 \sqrt{3}}$

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