Physics Q40 – JEE-MAIN 2025

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 particular angle between

P_{2}

and

P_{3}

. The transmitted intensity of the light passing the through all three polarisers is maximum. The angle between the polarisers

P_{2}

and

P_{3}

is :

Clue 1 of 3: Malus's Law

The intensity of plane-polarized light after passing through an analyzer is proportional to the square of the cosine of the angle between the transmission axes of the polarizer and the analyzer.

🧠 Full Solution Path

Step 1: Initial Setup and Malus's Law Application✦ Active

Initially, polarizers $P_{1}$ and $P_{2}$ are crossed, meaning the angle between their transmission axes is $π / 2$ . Let the transmission axis of $P_{1}$ be at $0^{\circ}$ . Then $P_{2}$ is at $90^{\circ}$ . When unpolarized light of intensity $I_{0}$ passes through $P_{1}$ , its intensity becomes $I_{1} = I_{0} / 2$ . Let $P_{3}$ be inserted between $P_{1}$ and $P_{2}$ such that its transmission axis makes an angle $θ$ with $P_{1}$ . The intensity after $P_{3}$ is $I_{3} = I_{1} \cos^{2} θ = (I_{0} / 2) \cos^{2} θ$ . The angle between $P_{3}$ and $P_{2}$ is $(π / 2 - θ)$ . The final intensity $I_{F}$ after passing through $P_{2}$ is given by Malus's Law:

I_{F} = I_{3} \cos^{2} (π / 2 - θ) = \frac{I_{0}}{2} \cos^{2} θ \sin^{2} θ

Step 2: Expressing Intensity in terms of a single trigonometric function○ Expand

Using the identity $\sin θ \cos θ = \frac{1}{2} \sin (2 θ)$ , the final intensity can be rewritten as:

I_{F} = \frac{I_{0}}{2} {(\frac{1}{2} \sin (2 θ))}^{2} = \frac{I_{0}}{8} \sin^{2} (2 θ)

Step 3: Maximizing the Intensity and finding the required angle○ Expand

For the transmitted intensity $I_{F}$ to be maximum, $\sin^{2} (2 θ)$ must be maximum. The maximum value of $\sin^{2} (2 θ)$ is 1. This occurs when $2 θ = π / 2$ , which implies $θ = π / 4$ . The question asks for the angle between polarisers $P_{2}$ and $P_{3}$ . This angle is $(π / 2 - θ)$ .

Angle between P_{2} and P_{3} = \frac{π}{2} - θ = \frac{π}{2} - \frac{π}{4} = \frac{π}{4}

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Physics Question 40 – JEE-MAIN 2025

Clue 1 of 3: Malus's Law

🧠 Full Solution Path