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 $$\sin 2 \theta=1$$

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