JEE Main 2024 — 27 January, Shift 2 — Question 45

Let $\mathrm{I}_{0}$ be intensity of unpolarised light incident on first polaroid.

$\mathrm{I}_{1}=$ Intensity of light transmitted from $1^{\text {st }}$

polaroid $=\frac{\mathrm{I}_{0}}{2}$

$\theta$ be the angle between $1^{\text {st }}$ and $2^{\text {nd }}$

polaroid $\phi$ be the angle between $2^{\text {nd }}$ and $3^{\text {rd }}$

polaroid $\theta+\phi=90^{\circ}$ (as $1^{\text {st }}$ and $3^{\text {rd }}$

polaroid are crossed) $\phi=90^{\circ}-\theta$ $\mathrm{I}_{2}=$

Intensity from $2^{\text {nd }}$ polaroid $I_{2}=I_{1} \cos ^{2} \theta=\frac{I_{0}}{2} \cos ^{2} \theta$ $\mathrm{I}_{3}=$ Intensity from $3^{\text {rd }}$ polaroid $\mathrm{I}_{3}=\mathrm{I}_{2} \cos ^{2} \phi$ $\mathrm{I}_{3}=\mathrm{I}_{1} \cos ^{2} \theta \cos ^{2} \phi$ $\mathrm{I}_{3}=\frac{\mathrm{I}_{0}}{2} \cos ^{2} \theta \cos ^{2} \phi$ $\phi=90-\theta$ $I_{3}=\frac{I_{0}}{2} \cos ^{2} \theta \sin ^{2} \theta$ $\mathrm{I}_{3}=\frac{\mathrm{I}_{0}}{2}\left[\frac{2 \sin \theta \cos \theta}{2}\right]^{2}$ $\mathrm{I}_{3}=\frac{\mathrm{I}_{0}}{8} \sin ^{2} 2 \theta$

$\mathrm{I}_{3}$ will be maximum when $\sin 2 \theta=1$

$2 \theta=90^{\circ}$

$\theta=45^{\circ}$