A beam of plane-polarized light falls on a polarizer which rotates about the axis of the ray with angular velocity `omega = 21rad//s`. Find the energy of light passing through the Polarizer per one revolution if the flux of energy of the incident ray is equal to `Phi_(0) = 4.0mW`.

Explanation

When the polarizer rotates with angualr velocity `omega` its instantaneous principle direction makes angle `omegat` from a reference direction which we choose to be along the direction of vobration of the plane polarized incident light. The transmitted flux at this instant is
`Phi_(0)cos^(2)omegat`
and the total energy passing through the polarizer per revolution is
`underset(0)overset(T)int Phi_(0)cos^(2)omegat dt, T = 2pi//omega`
`= Phi_(0)(pi)/(omega) = 0.6mJ`.