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

Let ${I}_{0}$I0​ be intensity of unpolarised light incident on first polaroid.

${I}_{1}=$I1​= Intensity of light transmitted from ${1}^{\text{st }}$1st  polaroid

$$=\frac{{I}_{0}}{2}$$=2I0​​

If $\theta$θ be the angle between ${1}^{\text{st }}$1st  and ${2}^{\text{nd }}$2nd  polaroid and $\phi$ϕ be the angle between ${2}^{\text{nd }}$2nd  and ${3}^{\text{rd }}$3rd  polaroid, then

$$\theta +\phi =90^{\circ}...(1)$$θ+ϕ=90∘...(1)

If ${I}_{2}=$I2​= Intensity from ${2}^{\text{nd }}$2nd  polaroid,

$$\begin{matrix}{I}_{2} & = & {I}_{1}co{s}^{2}\theta \\ & = & \frac{{I}_{0}}{2}co{s}^{2}\theta \end{matrix}$$I2​​==​I1​cos2θ2I0​​cos2θ​

If ${I}_{3}=$I3​= Intensity from ${3}^{\text{rd }}$3rd  polaroid

$$\begin{matrix}{I}_{3} & = & {I}_{2}co{s}^{2}\phi \\ & = & {I}_{1}{\mathrm{cos}}^{2}\theta {\mathrm{cos}}^{2}\phi \\ & = & \frac{{I}_{0}}{2}co{s}^{2}\theta co{s}^{2}(90^{\circ}-\theta ) \\ & = & \frac{{I}_{0}}{2}co{s}^{2}\theta si{n}^{2}\theta \\ & = & \frac{{I}_{0}}{2}{[\frac{2\mathrm{sin}\theta \mathrm{cos}\theta }{2}]}^{2} \\ & = & \frac{{I}_{0}}{8}{\mathrm{sin}}^{2}2\theta \end{matrix}$$I3​​======​I2​cos2ϕI1​cos2θcos2ϕ2I0​​cos2θcos2(90∘−θ)2I0​​cos2θsin2θ2I0​​[22sinθcosθ​]28I0​​sin22θ​

Now, ${I}_{3}$I3​ will be maximum when

$$\mathrm{sin}2\theta =1$$sin2θ=1

$$\Rightarrow 2\theta =90^{\circ}$$⇒2θ=90∘

$$\Rightarrow \theta =45^{\circ}$$⇒θ=45∘