A plane polarized blue light ray is incident on a prism such that there is no reflection from the surface of the prism. The angle of deviation of the emergent ray is $δ = 60^{0}$ (see Figure-1).The angle of minimum deviation for red light from the same prism is $δ_{\min} = 30^{0}$ (see Figure-2).The refractive index of the prism material for blue light is $\sqrt{3}$ . Which of the following statement(s) is(are) correct?

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The blue light is polarized in the plane of incidence.

The refractive index of the material of the prism for red light is $\sqrt{2}$

The angle of refraction for blue light in air at the exit plane of the prism is $60^{0}$

The angle of the prism is $60^{0}$

answer is A, B, C, D.

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Detailed Solution

$μ = \tan i_{p} i_{p} = \tan^{- 1} \sqrt{3} = 60^{0} δ = 60^{0} δ = i + e - A \Rightarrow 60^{0} = 60^{0} + e - A \Rightarrow e = A ....... (i)$

At incident surface, $\sin 60^{0} = \sqrt{3} \sin r_{1}$
$\Rightarrow r_{1} = 30^{0}$
$∴ r_{1} + r_{2} = A$
$\Rightarrow r_{2} = A - 30^{0}$
At emergent surface,
$\sqrt{3} \sin (A - 30^{0}) = \sin A$
$\frac{3}{2} \sin A - \frac{\sqrt{3}}{2} \cos A = \sin A$ $\Rightarrow \tan A = \sqrt{3}$
$A = 60^{0} \Rightarrow e = 60^{0}$
For Figure 2
For minimum deviation
$\frac{\sin (\frac{A + δ_{m}}{2})}{\sin \frac{A}{2}} = μ_{R} \Rightarrow \frac{\sin 45^{0}}{\sin 30^{0}} = μ_{R} (o r) μ_{R} = \sqrt{2}$