JEE Main 2025 Apr 4 Shift 2, Physics Q15: Wave Optics

JEE Main2025Apr 4, Shift 2Physics

Two polarisers $P_{1}$ and $P_{2}$ are placed in such a way that the intensity of the transmitted light will be zero. A third polariser $P_{3}$ is inserted in between $P_{1}$ and $P_{2}$ , at the particular angle between $P_{2}$ and $P_{3}$ . The transmitted intensity of the light passing through all three polarisers is maximum. The angle between the polarisers $P_{2}$ and $P_{3}$ is :

A
$\frac{π}{4}$
B
$\frac{π}{6}$
C
$\frac{π}{8}$
D
$\frac{π}{3}$

Solution

Initially, the transmitted intensity is zero. Hence, the transmission axes of $P_{1}$ and $P_{2}$ are perpendicular.

Let the angle between $P_{1}$ and $P_{3}$ be $θ$ .

Then, the angle between $P_{2}$ and $P_{3}$ is $(9 0^{\circ} - θ)$ .

After passing through $P_{1}$ , the intensity becomes

$I_{1} = \frac{I _{0}}{2}$

After passing through $P_{3}$ , the intensity becomes

$I_{2} = I_{1} cos^{2} θ = \frac{I _{0}}{2} cos^{2} θ$

After passing through $P_{2}$ , the intensity becomes

$I = I_{2} cos^{2} (9 0^{\circ} - θ)$

$= \frac{I _{0}}{2} cos^{2} θ sin^{2} θ$

$= \frac{I _{0}}{8} sin^{2} 2 θ$

The intensity is maximum when $sin^{2} 2 θ = 1$ .

$\Rightarrow 2 θ = 9 0^{\circ}$

$\Rightarrow θ = 4 5^{\circ}$

$∴ ∠ (P_{2}, P_{3}) = 9 0^{\circ} - 4 5^{\circ} = 4 5^{\circ}$

$4 5^{\circ}$

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