JEE Challenger
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PhyMedAdv 2023 • Paper 1

Polarization and Refraction of Blue and Red Light in Prism

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∘\delta = 60^\circ (see Figure-1). The angle of minimum deviation for red light from the same prism is δmin=30∘\delta_{\text{min}} = 30^\circ (see Figure-2). The refractive index of the prism material for blue light is 3\sqrt{3}. Which of the following statement(s) is(are) correct?

Question Diagram 1

Options

A

The blue light is polarized in the plane of incidence.

Correct
B

The angle of the prism is 45∘45^\circ.

C

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

Correct
D

The angle of refraction for blue light in air at the exit plane of the prism is 60∘60^\circ.

Correct

Topics & Concepts

Ray Optics and Optical InstrumentsPrismsPolarizationReflection and Refraction

Step-by-Step Solution

To find the correct statement(s), let us analyze each concept and condition step by step:

1. Brewster's Angle and Polarization (Option A)

When light is incident on an interface at Brewster's angle, the reflected ray is completely polarized perpendicular to the plane of incidence (s-polarized). If the incident light is plane-polarized in the plane of incidence (p-polarized), there is zero reflection at the interface, and the light is completely transmitted into the medium.

Given that there is no reflection at the first surface of the prism:

  • The blue light must be incident at Brewster's angle θB\theta_B.
  • The blue light must be polarized in the plane of incidence.

Therefore, Option (A) is correct.

2. Angle of Incidence and Prism Angle (Options B and D)

For the blue light:

  1. Brewster's Angle of Incidence: tan⁡i=μblue=3  ⟹  i=60∘\tan i = \mu_{\text{blue}} = \sqrt{3} \implies i = 60^\circ

  2. Angle of Refraction at the First Surface (r1r_1): Using Snell's law at the first surface: sin⁡i=μbluesin⁡r1\sin i = \mu_{\text{blue}} \sin r_1 sin⁡60∘=3sin⁡r1  ⟹  32=3sin⁡r1  ⟹  sin⁡r1=12  ⟹  r1=30∘\sin 60^\circ = \sqrt{3} \sin r_1 \implies \frac{\sqrt{3}}{2} = \sqrt{3} \sin r_1 \implies \sin r_1 = \frac{1}{2} \implies r_1 = 30^\circ

  3. Relation between Deviation, Prism Angle, and Emergence: The deviation δ\delta of a ray passing through a prism is given by: δ=i+e−A\delta = i + e - A Given δ=60∘\delta = 60^\circ and i=60∘i = 60^\circ: 60∘=60∘+e−A  ⟹  e=A60^\circ = 60^\circ + e - A \implies e = A

  4. Refraction at the Second Surface: The angle of incidence at the second surface inside the prism is: r2=A−r1=A−30∘r_2 = A - r_1 = A - 30^\circ Applying Snell's law at the exit surface: μbluesin⁡r2=1⋅sin⁡e\mu_{\text{blue}} \sin r_2 = 1 \cdot \sin e Substituting e=Ae = A and r2=A−30∘r_2 = A - 30^\circ: 3sin⁡(A−30∘)=sin⁡A\sqrt{3} \sin(A - 30^\circ) = \sin A 3(sin⁡Acos⁡30∘−cos⁡Asin⁡30∘)=sin⁡A\sqrt{3} \left(\sin A \cos 30^\circ - \cos A \sin 30^\circ\right) = \sin A 3(sin⁡A32−cos⁡A12)=sin⁡A\sqrt{3} \left(\sin A \frac{\sqrt{3}}{2} - \cos A \frac{1}{2}\right) = \sin A 32sin⁡A−32cos⁡A=sin⁡A\frac{3}{2} \sin A - \frac{\sqrt{3}}{2} \cos A = \sin A 12sin⁡A=32cos⁡A  ⟹  tan⁡A=3  ⟹  A=60∘\frac{1}{2} \sin A = \frac{\sqrt{3}}{2} \cos A \implies \tan A = \sqrt{3} \implies A = 60^\circ

  • The angle of the prism is A=60∘A = 60^\circ, which makes Option (B) incorrect.
  • The angle of refraction for blue light in air at the exit plane is the angle of emergence: e=A=60∘e = A = 60^\circ Therefore, Option (D) is correct.

3. Refractive Index for Red Light (Option C)

For red light, the angle of minimum deviation is given as δmin=30∘\delta_{\text{min}} = 30^\circ, and the prism angle is A=60∘A = 60^\circ.

Using the prism formula for minimum deviation: μred=sin⁡(A+δmin2)sin⁡(A2)\mu_{\text{red}} = \frac{\sin\left(\frac{A + \delta_{\text{min}}}{2}\right)}{\sin\left(\frac{A}{2}\right)} μred=sin⁡(60∘+30∘2)sin⁡(60∘2)=sin⁡45∘sin⁡30∘=1212=2\mu_{\text{red}} = \frac{\sin\left(\frac{60^\circ + 30^\circ}{2}\right)}{\sin\left(\frac{60^\circ}{2}\right)} = \frac{\sin 45^\circ}{\sin 30^\circ} = \frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}} = \sqrt{2}

Therefore, Option (C) is correct.

Conclusion:

The correct statements are (A), (C), and (D).

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