JEE Main 2026 — Wave Optics Question with Solution

Intensity of light after passing through the first polarizer is:

$I_1=\frac{I_o}{2}$

Let the optically active solution rotate the plane of polarization by an angle $\theta$. Since the angle between the polarizer and the analyser is $0^{\circ }$, the angle between the plane of polarization of the light emerging from the solution and the transmission axis of the analyser is also $\theta$.

According to Malus's law, the intensity of light emerging from the analyser is:

$I_2=I_1{\cos }^2\theta =\frac{I_o}{2}{\cos }^2\theta$

Given that the final intensity is $\frac{3}{8}{I}_o$, we have:

$\frac{I_o}{2}{\cos }^2\theta =\frac{3}{8}{I}_o$

${\cos }^2\theta =\frac{3}{4}$

$\cos \theta =\frac{\sqrt{3}}{2}$

$\theta =30^{\circ }$

Answer: $30$