JEE Advanced 2023 — Paper 1 — PHY-I solution

JEE Advanced2023Paper 1PHY-I

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 $δ = 6 0^{\circ}$ (see Figure-1). The angle of minimum deviation for red light from the same prism is $δ_{min} = 3 0^{\circ}$ (see Figure-2). The refractive index of the prism material for blue light is $3$ . Which of the following statement(s) is(are) correct?

A
The blue light is polarized in the plane of incidence.
B
The angle of the prism is $4 5^{\circ}$ .
C
The refractive index of the material of the prism for red light is $2$ .
D
The angle of refraction for blue light in air at the exit plane of the prism is $6 0^{\circ}$ .

Solution

Polarization condition. Zero reflection at the entry face means the blue light is incident at Brewster's angle and is polarized in the plane of incidence (p-polarized). Hence option (A) is correct.

Angle of incidence for blue light. Brewster's law gives $tan i_{B} = μ_{B} = 3$ , so $i_{B} = 6 0^{\circ}$ .

Refraction at entry face. Using Snell's law: $sin 6 0^{\circ} = 3 sin r_{1} \Rightarrow sin r_{1} = \frac{1}{2} \Rightarrow r_{1} = 3 0^{\circ}$ .

Geometry of the prism. With prism angle $A$ , $r_{1} + r_{2} = A$ and the total deviation $δ = (i + e) - A = 6 0^{\circ}$ . So $e = A$ . Snell's law at exit: $3 sin r_{2} = sin e$ , i.e. $3 sin (A - 3 0^{\circ}) = sin A$ . Expanding: $3 (sin A cos 3 0^{\circ} - cos A sin 3 0^{\circ}) = sin A$ , giving $\frac{3}{2} sin A - \frac{3}{2} cos A = sin A$ , so $tan A = 3$ and $A = 6 0^{\circ}$ .

Therefore the exit angle in air is $e = 6 0^{\circ}$ , confirming option (D). Option (B) (which claims $A = 4 5^{\circ}$ ) is incorrect.

Refractive index for red light. At minimum deviation with $A = 6 0^{\circ}$ and $δ_{min} = 3 0^{\circ}$ : $μ_{R} = \frac{sin ( \frac{A + δ _{min}}{2} )}{sin ( A /2 )} = \frac{sin 4 5 ^{\circ}}{sin 3 0 ^{\circ}} = 2$ . Option (C) holds.

Correct options: (A), (C), (D).

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