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QJEEAdv23Ph2542: JEE Advanced 2023 Physics - Class 12 JEEAdv23Ph2542 Solution: A plane polarized blue light ray is incident on a prism such…

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_{\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?

JEE Advanced 2023 Paper 1 Online Physics - Geometrical Optics Question 7 English

Multiple Choice Options

  • A The angle of refraction for blue light in air at the exit plane of the prism is 60∘60^{\circ} .
  • B The blue light is polarized in the plane of incidence.
  • C The angle of the prism is 45∘45^{\circ} .
  • D The refractive index of the material of the prism for red light is 2\sqrt{2} .

Step-by-Step Verified Solution

When light strikes a surface at Brewster's angle, it becomes polarized perpendicular to the plane of incidence.

Therefore, there is no reflection of blue light observed.

This implies that the light must be polarized within the plane of incidence.

JEE Advanced 2023 Paper 1 Online Physics - Geometrical Optics Question 7 English Explanation

tan⁡iamp;=3iamp;=60∘δamp;=i+e−A60amp;=60+e−A⇒Aamp;=e\begin{aligned} \tan i & =\sqrt{3} \\\\ i & =60^{\circ} \\\\ \delta & =i+e-A \\\\ 60 & =60+e-A \\\\ \Rightarrow A & =e\end{aligned}

sin⁡isin⁡r1amp;=3⇒sin⁡60sin⁡r1amp;=3⇒r1=30∘\begin{aligned} \frac{\sin i}{\sin r_1} & =\sqrt{3} \\\\ \Rightarrow \frac{\sin 60}{\sin r_1} & =\sqrt{3} \Rightarrow r_1=30^{\circ}\end{aligned}

r2=A−r1rsin⁡(A−r1)=sin⁡e⇒3sin⁡(A−30)=sin⁡Ar3[sin⁡Acos⁡30∘−cos⁡Asin⁡30]=sin⁡A⇒32(3−cot⁡A)=1⇒cot⁡A=3−23=13⇒cot⁡A=13⇒A=60∘\begin{array}{rlrl}r_2 =A-r_1 \\\\ r \sin \left(A-r_1\right) =\sin e \\\\ \Rightarrow \sqrt{3} \sin (A-30) =\sin A \\\\ r_3\left[\sin A \cos 30^{\circ}-\cos A\right. \sin 30]=\sin A \\\\ \Rightarrow \frac{\sqrt{3}}{2}(\sqrt{3}-\cot A) =1 \\\\ \Rightarrow \cot A =\sqrt{3}-\frac{2}{\sqrt{3}}=\frac{1}{\sqrt{3}} \\\\ \Rightarrow \cot A =\frac{1}{\sqrt{3}} \\\\ \Rightarrow A =60^{\circ}\end{array}

So, e=60∘\quad e=60^{\circ} (for blue)

For red,

μamp;=sin⁡(δm+A2)sin⁡A2amp;=sin⁡(30+602)sin⁡602=sin⁡45sin⁡30=1/21/2μred amp;=2\begin{aligned}\mu & =\frac{\sin \left(\frac{\delta_m+A}{2}\right)}{\sin \frac{A}{2}} \\\\& =\frac{\sin \left(\frac{30+60}{2}\right)}{\sin \frac{60}{2}}=\frac{\sin 45}{\sin 30}=\frac{1 / \sqrt{2}}{1 / 2} \\\\\mu_{\text {red }} & =\sqrt{2}\end{aligned}

(A), (C), (D) are correct.

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