Physics Q42 – JEE-MAIN 2026

An unpolarized light of certain intensity passes through a combination of two polarizers whose transmission axes are at

30^{\circ}

and

90^{\circ}

, respectively, with respect to the horizontal axis. A third polarizer with its transmission axis at

60^{\circ}

with the horizontal axis is placed between the two existing polarizers. The ratio of the output intensities with and without the third polarizer is _______.

Clue 1 of 3: Polarization of Light

Understand how unpolarized light's intensity changes after passing through a polarizer and how polarized light's intensity changes when passing through another polarizer.

🧠 Full Solution Path

Step 1: Calculate output intensity without the third polarizer✦ Active

Let the initial intensity of unpolarized light be $I_{0}$ . When unpolarized light passes through the first polarizer (P1) at $30^{\circ}$ , its intensity becomes $I_{1} = I_{0} / 2$ . This light is now polarized at $30^{\circ}$ . The second polarizer (P2) is at $90^{\circ}$ . The angle between P1 and P2 is $θ_{1} = 90^{\circ} - 30^{\circ} = 60^{\circ}$ . Using Malus's Law, the intensity after P2 is:

I_{o u t, 1} = I_{1} \cos^{2} θ_{1} = \frac{I_{0}}{2} \cos^{2} (60^{\circ}) = \frac{I_{0}}{2} {(\frac{1}{2})}^{2} = \frac{I_{0}}{8}

Step 2: Calculate output intensity with the third polarizer○ Expand

With the third polarizer (P3) at $60^{\circ}$ placed between P1 and P2: After P1, intensity is $I_{1} = I_{0} / 2$ , polarized at $30^{\circ}$ . The angle between P1 ( $30^{\circ}$ ) and P3 ( $60^{\circ}$ ) is $θ_{13} = 60^{\circ} - 30^{\circ} = 30^{\circ}$ . The intensity after P3 is:

I_{2} = I_{1} \cos^{2} θ_{13} = \frac{I_{0}}{2} \cos^{2} (30^{\circ}) = \frac{I_{0}}{2} {(\frac{\sqrt{3}}{2})}^{2} = \frac{I_{0}}{2} (\frac{3}{4}) = \frac{3 I_{0}}{8}

This light is now polarized at $60^{\circ}$ . The angle between P3 ( $60^{\circ}$ ) and P2 ( $90^{\circ}$ ) is $θ_{32} = 90^{\circ} - 60^{\circ} = 30^{\circ}$ . The final intensity after P2 is:

I_{o u t, 2} = I_{2} \cos^{2} θ_{32} = \frac{3 I_{0}}{8} \cos^{2} (30^{\circ}) = \frac{3 I_{0}}{8} {(\frac{\sqrt{3}}{2})}^{2} = \frac{3 I_{0}}{8} (\frac{3}{4}) = \frac{9 I_{0}}{32}

Step 3: Determine the ratio of output intensities○ Expand

The ratio of the output intensities with and without the third polarizer is $I_{o u t, 2} / I_{o u t, 1}$ :

Ratio = \frac{I_{o u t, 2}}{I_{o u t, 1}} = \frac{9 I_{0} / 32}{I_{0} / 8} = \frac{9}{32} \times \frac{8}{1} = \frac{9}{4}

💡 Teacher's Secret Hint

Ensure to correctly identify the angles between successive polarizers for Malus's Law.

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Physics Question 42 – JEE-MAIN 2026

Clue 1 of 3: Polarization of Light

🧠 Full Solution Path