Wave Optics

JEE Main level, single correct, medium

167 of 191

Question 167 of 191

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

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Answer: (D) 45∘45^{\circ}

Let I0{I}_{0} be intensity of unpolarised light incident on first polaroid.
I1={I}_{1}= Intensity of light transmitted from 1st{1}^{\text{st}} polaroid
=I02=\dfrac{{I}_{0}}{2}
If θ\theta be the angle between 1st{1}^{\text{st}} and 2nd{2}^{\text{nd}} polaroid and ϕ\phi be the angle between 2nd{2}^{\text{nd}} and 3rd{3}^{\text{rd}} polaroid, then
θ+ϕ=90∘...(1)\theta +\phi =90^{\circ}...\left(1\right)
If I2={I}_{2}= Intensity from 2nd{2}^{\text{nd}} polaroid,
I2=I1cos2θ=I02cos2θ\begin{matrix}{I}_{2} & = & {I}_{1}co{s}^{2}\theta \\ & = & \dfrac{{I}_{0}}{2}co{s}^{2}\theta \end{matrix}
If I3={I}_{3}= Intensity from 3rd{3}^{\text{rd}} polaroid
I3=I2cos2ϕ=I1cos⁡2θcos⁡2ϕ=I02cos2θcos2(90∘−θ)=I02cos2θsin2θ=I02[2sin⁡θcos⁡θ2]2=I08sin⁡22θ\begin{matrix}{I}_{3} & = & {I}_{2}co{s}^{2}\phi \\ & = & {I}_{1}{\cos }^{2}\theta {\cos }^{2}\phi \\ & = & \dfrac{{I}_{0}}{2}co{s}^{2}\theta co{s}^{2}\left(90^{\circ}-\theta \right) \\ & = & \dfrac{{I}_{0}}{2}co{s}^{2}\theta si{n}^{2}\theta \\ & = & \dfrac{{I}_{0}}{2}{\left[\dfrac{2\sin \theta \cos \theta }{2}\right]}^{2} \\ & = & \dfrac{{I}_{0}}{8}{\sin }^{2}2\theta \end{matrix}
Now, I3{I}_{3} will be maximum when
sin⁡2θ=1\sin 2\theta =1
⇒2θ=90∘\Rightarrow 2\theta =90^{\circ}
⇒θ=45∘\Rightarrow \theta =45^{\circ}

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