When a polaroid sheet is rotated between two crossed polaroids then the transmitted light intensity will be maximum for a rotation of :

<p>Let $\mathrm{I}_0$ be intensity of unpolarised light incident on first polaroid.</p> <p>$\mathrm{I}_1=$ Intensity of light transmitted from $1^{\text {st }}$ polaroid $=\frac{\mathrm{I}_0}{2}$</p> <p>$\theta$ be the angle between $1^{\mathrm{st}}$ and $2^{\text {nd }}$ polaroid</p> <p>$\phi$ be the angle between $2^{\text {nd }}$ and $3^{\text {rd }}$ polaroid</p> <p>$\theta+\phi=90^{\circ}$ (as $1^{\text {st }}$ and $3^{\text {rd }}$ polaroid are crossed)</p> <p>$\phi=90^{\circ}-\theta$</p> <p>$\mathrm{I}_2=$ Intensity from $2^{\text {nd }}$ polaroid</p> <p>$\mathrm{I}_2=\mathrm{I}_1 \cos ^2 \theta=\frac{\mathrm{I}_0}{2} \cos ^2 \theta$</p> <p>$\mathrm{I}_3=$ Intensity from $3^{\text {rd }}$ polaroid</p> <p>$\mathrm{I}_3=\mathrm{I}_2 \cos ^2 \phi$</p> <p>$\mathrm{I}_3=\mathrm{I}_1 \cos ^2 \theta \cos ^2 \phi$</p> <p>$\mathrm{I}_3=\frac{\mathrm{I}_0}{2} \cos ^2 \theta \cos ^2 \phi$</p> <p>$\phi=90-\theta$</p> <p>$$\begin{aligned} & I_3=\frac{I_0}{2} \cos ^2 \theta \sin ^2 \theta \\ & I_3=\frac{I_0}{2}\left[\frac{2 \sin \theta \cos \theta}{2}\right]^2 \\ & I_3=\frac{I_0}{8} \sin ^2 2 \theta \end{aligned}$$</p> <p>$\mathrm{I}_3$ will be maximum when $\sin 2 \theta=1$</p> <p>$$\begin{aligned} & 2 \theta=90^{\circ} \\ & \theta=45^{\circ} \end{aligned}$$</p>