When a polaroid sheet is rotated between two crossed polaroids then...

Let ${I}_{0}$ be intensity of unpolarised light incident on first polaroid.
${I}_{1} =$ Intensity of light transmitted from ${1}^{\text{st}}$ polaroid
$= \dfrac{{I}_{0}}{2}$
If $\theta$ be the angle between ${1}^{\text{st}}$ and ${2}^{\text{nd}}$ polaroid and $ϕ$ be the angle between ${2}^{\text{nd}}$ and ${3}^{\text{rd}}$ polaroid, then

$\theta + ϕ = 90 ° . . . \left(1\right)$

If ${I}_{2} =$ Intensity from ${2}^{\text{nd}}$ polaroid,

$\begin{cases}{I}_{2} & = & {I}_{1}co{s}^{2}\theta \\ & = & \dfrac{{I}_{0}}{2}co{s}^{2}\theta\end{cases}$

If ${I}_{3} =$ Intensity from ${3}^{\text{rd}}$ polaroid

$\begin{cases}{I}_{3} & = & {I}_{2}co{s}^{2}ϕ \\ & = & {I}_{1}\cos^{2}\theta\cos^{2}ϕ \\ & = & \dfrac{{I}_{0}}{2}co{s}^{2}\thetaco{s}^{2}\left(90 ° - \theta\right) \\ & = & \dfrac{{I}_{0}}{2}co{s}^{2}\thetasi{n}^{2}\theta \\ & = & \dfrac{{I}_{0}}{2}{\left[\dfrac{2sin\theta cos\theta}{2}\right]}^{2} \\ & = & \dfrac{{I}_{0}}{8}\sin^{2}2\theta\end{cases}$

Now, ${I}_{3}$ will be maximum when

$\sin 2\theta = 1$

$⇒ 2\theta = 90 °$

$⇒ \theta = 45 °$