Discuss the intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids?

Explanation

Situation given in the statement is depicted in the following figure.

In above figure, `P_(1), P_(2), P_(3)` are the three polarisers placed successively with their pass axes in the vertical planes. Since `P_(1)` and `P_(3)` are crossed, angle between their optic axes (or pass axes, as per the textbook) `A_(1)` and `A_(3)` will be `(pi)/(2)` rad. Now in this situation, if angle between `A_(1)` and `A_(2)` is `theta` then angle between `A_(2)` and `A_(3)` will be `((pi)/(2)-theta)`.
Here if intensity of light transmitted by `P_(1)` is `I_(1) = I_(0)` then intensity of light transmitted by `P_(2)` is given by Malus law,
`I_(2)=I_(1)cos^(2)theta`
`=I_(0)cos^(2)theta" "("":.I_(1)=I_(0))" "....(1)`
Similarly intensity of light transmitted by `P_(3)` will be :
`I_(3)=I_(2)cos^(2)((pi)/(2)-theta)`
`=(I_(0)cos^(2)theta)cos^(2)((pi)/(2)-theta)` [From equation (1) ]
`=I_(0)cos^(2)thetasin^(2)theta{"":.cos((pi)/(2)-theta)=sintheta}`
`:.I_(3)=I_(0)((1+cos(2theta))/(2))((1-cos(2theta))/(2))`
`:.I_(3)=I_(0)((1-cos^(2)(2theta))/(4))`
`:.I_(3)=((I_(0))/(4))sin^(2)(2theta)" "......(2)`
In above figure, when `P_(2)` is rotated in vertical plane (keeping `P_(1)` and `P_(3)` crossed and stationary), `theta` will change and hence `I_(3)` will change as per equation (2)
Note : From equation (2),
`(dI_(3))/(d""theta)=((I_(0))/(4))(d)/(d""theta){sin^(2)(2theta)}`
`:.(dI_(3))/(d""theta)=((I_(0))/(4))2sin(2theta)cos(2theta)(2)`
`:.(dI_(3))/(d""theta)=I_(0)sin(2theta)cos(2theta)`
For `I_(3)`= maximum,
`(dI_(3))/(d""theta)=0`
`:.sin(2theta)cos(2theta)=0`
`impliessin(2theta)=0" OR "cos(2theta)=0`

Equivalently, from equation (2), for `I_(3)=max`,
`sin^(2)(2theta)=max`.
`impliessin(2theta)=max`
`impliessin(2theta)=1`
`implies2theta=90^(@)`
`impliestheta=45^(@)`