A unpolarized beam of light having flux 10^ -2 W falls normally on...

The incident power (radiant flux) of the unpolarized light is $P_0 = 10^{-2}$ W.

When unpolarized light passes through a polarizer, the transmitted intensity is half of the incident intensity. Therefore, the transmitted power is $P = \dfrac{P_0}{2} = \dfrac{10^{-2}}{2}$ W.

The energy transmitted per revolution is given by $E = P \times T$, where $T$ is the time period of one revolution.

Substituting the given values:

$10^{-2} = \dfrac{10^{-2}}{2} \times T$

$T = 2$ s

The angular frequency of the rotating polarizer is $\omega = \dfrac{2\pi}{T}$.

$\omega = \dfrac{2\pi}{2} = \pi$ rad/s.

Answer: $\pi$ rad /s