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

Correct Answer - `30^(@) ; 45^(@)`
Let `theta =` angle between transmission axis of `P_(1) and P_(2)`
`phi =` angle between transmission axis of `P_(2) and P_(3)`.
`:. theta + phi = 90^(@)`
or `phi = (90^(@) - theta)` … (i)
Here, `I_(0) = 32 Wm^(-2)`
`:. I_(1) = (1)/(2)I_(0) = 16 Wm^(-2)`
`I_(2) = I_(1) cos^(2) theta and I_(3) = I_(2) cos^(2)phi`
`:. I_(3) = I_(1) cos^(2) theta cos^(2) phi`
`= I_(1)cos^(2) theta cos^(2)(90^(@) - theta)`
`= I_(1) cos^(2)theta sin^(2)theta = 16 cos^(2)theta sin^(2)theta`
`3 = 4 (sin 2 theta)^(2)`
`sin 2 theta = sqrt((3)/(4)) = (sqrt(3))/(2) = sin 60^(@), :. theta = 30^(@)`
`I_(3)` will be maximum when
`sin 2 theta = max. = 1 = sin 90^(@) :. theta = 45^(@)`