Question

Unpolarized light of intensity 32 `Wm^(-3)` passes through three polarizers such that the transmission axis of the last polarizer is crossed with the first. If the intensity of the emerging light is `3 Wm^(-2)`, what is the angle between the transmission axces of the first two polarizers ? At what angle will the transmitted intensity be maximum ?

Answer 1

If `theta` is the angle between the transission axes of first polaroid `P_(1)` and second `P_(2)` while `phi` between the transmission axes of second polaroid `P_(2)` and third `P_(3)`, then according to given problem. `theta+phi=90^(@) or phi= (90^(@)-theta)` ......(1)
Now if `I_(0)` is the intensity of unpolarized light incident on polaroid `P_(1)`, the intensity of light transmitted through it,
`I_(1)= (1)/(2)I_(0) =(1)/(2) (32) = 16 (W)/(m^(2))` ....(2)
Now as angle between transmission axes of polariods `P_(1)` and `P_(2)` is `theta` , in a accordance with Malus law, intensity of light transmitted through `P_(2)` will be
`I_(2)= I_(1) cos^(2)theta=16 cos^(2) theta` .......(3)
And as angle between transmission axes of `P_(2)`and `P_(3)` is `phi`, light transmitted through `P_(3)` will be
`I_(3)=I_(2)cos^(2)phi = 16cos^(2) theta cos^(2) phi` .....(4)
According to given problem, `I_(3)=3W//m^(2)`
So, `4(sin 2theta)^(2)=3 i.e., sin 2theta = (sqrt(3)//2)` or
`2theta=60^(@), i.e., theta = 30^(@)`.