QJEEM24Ph7425: JEE Main 2024 Physics - Class 12 JEEM24Ph7425 Solution: When a polaroid sheet is rotated between two crossed polaroids…

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 $1^{\mathrm{st}}$ 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

$\mathrm{I}_2=\mathrm{I}_1 \cos ^2 \theta=\frac{\mathrm{I}_0}{2} \cos ^2 \theta$

$\mathrm{I}_3=$ Intensity from $3^{\text {rd }}$ polaroid

$\mathrm{I}_3=\mathrm{I}_2 \cos ^2 \phi$

$\mathrm{I}_3=\mathrm{I}_1 \cos ^2 \theta \cos ^2 \phi$

$\mathrm{I}_3=\frac{\mathrm{I}_0}{2} \cos ^2 \theta \cos ^2 \phi$

$\phi=90-\theta$

$\begin{aligned}& I_3=\frac{I_0}{2} \cos ^2 \theta \sin ^2 \theta \\& I_3=\frac{I_0}{2}\left[\frac{2 \sin \theta \cos \theta}{2}\right]^2 \\& I_3=\frac{I_0}{8} \sin ^2 2 \theta\end{aligned}$

$\mathrm{I}_3$ will be maximum when $\sin 2 \theta=1$

$\begin{aligned}& 2 \theta=90^{\circ} \\& \theta=45^{\circ}\end{aligned}$