A unpolarized beam of light having flux 1 0 − 2 W falls normally on a rotating polarizer of cross-sectional area 3 × 1 0 − 4 m 2 . If the energy of light passing through the polarizer per revolution i

Question

A unpolarized beam of light having flux 1 0 − 2 W falls normally on a rotating polarizer of cross-sectional area 3 × 1 0 − 4 m 2 . If the energy of light passing through the polarizer per revolution is 1 0 − 2 J, then the polarizer is rotating with the angular frequency of

TestHub · Accepted Answer

The incident power (radiant flux) of the unpolarized light is $P_{0} = 1 0^{- 2}$ W.

When unpolarized light passes through a polarizer, the transmitted intensity is half of the incident intensity. Therefore, the transmitted power is $P = \frac{P _{0}}{2} = \frac{1 0 ^{- 2}}{2}$ 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:

$1 0^{- 2} = \frac{1 0 ^{- 2}}{2} \times T$

$T = 2$ s

The angular frequency of the rotating polarizer is $ω = \frac{2 π}{T}$ .

$ω = \frac{2 π}{2} = π$ rad/s.

Answer: $π$ rad /s

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