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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`. |
Answer» 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`. |
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