10. Wave optics — Physics STD 12 Science — Question

$I_1=$ Intensity of light transmitted from $1^{\text {st }}$ polaroid

$=\frac{I_0}{2}$

$\theta$ 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

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

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

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

$\phi=90-\theta$

$I_3=\frac{I_0}{2} \cos ^2 \theta \sin ^2 \theta$

$I_3=\frac{I_0}{2}\left[\frac{2 \sin \theta \cos \theta}{2}\right]^2$

$I_3=\frac{I_0}{8} \sin ^2 2 \theta$

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

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

$ \theta=45^{\circ}$