ABC is a right-angled prism kept in air. A ray (1) is incident on the face AB along the normal. Refractive index of the material of prism is the minimum value that will be required so that ray (1) undergoes total internal reflection at the face AC. Another ray (2) is incident on the face AB such that it emerges from face AC along the normal to AC. A third ray (3) falls on the face BC and emerges from face AC such that its angle of emergence is the same as that of incidence. Assuming light (1), (2) and (3) have the same wavelength, then:

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Deviation in degree suffered by ray (2) is 60⁰

Deviation in degree suffered by ray (3) is 120⁰

Refractive index of the material of prism is 2

Angle of incidence in degree of ray (2) is 90⁰

answer is A, B, C, D.

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Detailed Solution

$(A) μ = \frac{1}{\sin C}$
For T.I.R of ray (1)
30^o > C
$\Rightarrow \sin 30^{\circ} > \sin C \Rightarrow \frac{1}{2} > \frac{1}{μ} \Rightarrow μ > 2 ∴ μ (min) \overset{}{>} 2$
(B) Snell’s law for Ray (2), 1sini = 2sin30^o
$\Rightarrow \sin i = 1 ∴ i = 90^{\circ}$
(C) Deviation of Ray (2) = i - r = 90^o - 30^o = 60^o
(D) For Ray (3), $i = e \Rightarrow δ = 2 i - ∠ C = 2 i - 60^{\circ}$
$r_{1} = r_{2} = \frac{A}{2} = 30^{\circ} \Rightarrow i = 90^{\circ} ∴ δ = 180^{\circ} - 60^{\circ} = 120^{\circ}$