JEE Main 2026 — 4 April, Evening Shift — Question 16

Without third polarizer: $I_2 = I_0/2 \cos^2(90^\circ-30^\circ?)$ Actually first at 30°, second at 90°, angle difference = 60°. So $I_{\text{without}} = \frac{I_0}{2}\cos^2 60^\circ = \frac{I_0}{2}\cdot\frac{1}{4} = \frac{I_0}{8}$. With third at 60° between them: first (30°) to third (60°) angle 30°, third to second (90°) angle 30°. $I_{\text{with}} = \frac{I_0}{2}\cos^2 30^\circ \cdot \cos^2 30^\circ = \frac{I_0}{2}\cdot\frac{3}{4}\cdot\frac{3}{4} = \frac{9I_0}{32}$. Ratio = $(9/32)/(1/8)=9/4$.