A unpolarized beam of light having flux 10⁻² W falls normally on a rotating polarizer of cross-sectional area 3 10⁻⁴ m ^2 . If the energy of light passing through the polarizer per revolution is 10⁻² J, then the polarizer is rotating with the angular frequency of

Step-by-step solution

The incident power (radiant flux) of the unpolarized light is P₀ = 10⁻² W. When unpolarized light passes through a polarizer, the transmitted intensity is half of the incident intensity. Therefore, the transmitted power is P = P₀ 2 = 10⁻² 2 W. The energy transmitted per revolution is given by E = P T , where T is the time period of one revolution. Substituting the given values: 10⁻² = 10⁻² 2 T T = 2 s The angular frequency of the rotating polarizer is = 2 T . = 2 2 = rad/s. Answer: rad /s