JEE Main 2026 — Wave Optics Question with Solution

Let the intensity of the incident unpolarized light be $I_0$.

After passing through the first polarizer, the intensity becomes $I_1=\frac{I_0}{2}$.

Without the third polarizer, the light passes directly from the first polarizer (at $30^{\circ }$) to the second polarizer (at $90^{\circ }$). The angle between their transmission axes is $90^{\circ }-30^{\circ }=60^{\circ }$.

Using Malus's law, the output intensity without the third polarizer is:
$I_{without}=I_1{\cos }^2(60^{\circ })=\frac{I_0}{2}\times {\left(\frac{1}{2}\right)}^2=\frac{I_0}{8}$

When the third polarizer (at $60^{\circ }$) is placed between them, the light first passes from the first polarizer to the third polarizer. The angle between their axes is $60^{\circ }-30^{\circ }=30^{\circ }$.

The intensity after the third polarizer is:
$I_3=I_1{\cos }^2(30^{\circ })=\frac{I_0}{2}\times {\left(\frac{\sqrt{3}}{2}\right)}^2=\frac{3I_0}{8}$

The light then passes from the third polarizer to the second polarizer. The angle between their axes is $90^{\circ }-60^{\circ }=30^{\circ }$.

The final output intensity with the third polarizer is:
$I_{with}=I_3{\cos }^2(30^{\circ })=\frac{3I_0}{8}\times {\left(\frac{\sqrt{3}}{2}\right)}^2=\frac{9I_0}{32}$

The ratio of the output intensities with and without the third polarizer is:
$\text{Ratio}=\frac{I_{with}}{I_{without}}=\frac{\frac{9I_0}{32}}{\frac{I_0}{8}}=\frac{9}{4}$

Answer: $9/4$