JEE Main · 2020 · 07 Jan 2020 · Shift 1 · Wave Optics

When unpolarised light passes the first polariser its intensity drops to a fixed value $I_1$; the analyser then transmits $I=I_1{\cos }^2\theta$, where $\theta$ is measured from the position of maximum transmission. The output is $10\%$ of $I_1$, so ${\cos }^2\theta =0.1$, giving $\theta \approx 71.6^{\circ }$. Extinction occurs when the analyser is crossed at $90^{\circ }$ to that maximum, so it must be turned through a further $90^{\circ }-71.6^{\circ }\approx 18.4^{\circ }$.