Unpolarised light of intensity `32 Wm^(-2)` passes through three polarisers such that the transmission axis of the last polariser is crossed with first. If the intensity of the emerging light is `3Wm^(-2)`, the angle between the axes of the first two polarisers is

Explanation

If `theta` is the angle between the transmission 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)" "".........."(i)`
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)" ""............"(ii)`
Now as angle between transmission axes of polaroids `P_1" and "P_(2)` is `theta P_(2)" and "P_(3)` is `phi`, light transmitted through `P_3` will be `I_(3)= (I_0)/(2) cos^(2) theta* cos^(2) 90-theta= (I_0)/(8)sin^(2) 2theta`
According to given proble, `I_(3)= 3 W"/"m^(2)`
So, `4(sin 2 theta)^(2)= 3`
`i.e., sin 2theta= (sqrt(3)"/"2)" or "2theta= 60^(@) i.e., theta= 30^(@)`.