A beam of plane polarized light of large cross-sectional area and...

Given,

Intensity of plane polarized light of large cross-sectional area is ${I}_{0} = 3 . 3W{m}^{- 2}$

Area of cross-section of polarizer is $A = 3 \times {10}^{- 4}{m}^{2}$

Angular speed of rotation is $\omega = 31 . 4rad{s}^{- 1}$

The intensity of light is the given by the definition,

$I = \dfrac{E}{AT} . . . ( 1 )$

Intensity of light passing through the polarizer as polarizer rotates is given by

${I}_{avg} = {I}_{0}\dfrac{1}{T}{∫}_{0}^{T}\cos^{2}\left(\omegat\right) d t$

$⇒ {I}_{avg} = {I}_{0}\dfrac{1}{T}{∫}_{0}^{T}\left[\dfrac{1 + \cos \left(2\omegat\right)}{2}\right] d t$

$⇒ {I}_{avg} = {I}_{0}\dfrac{1}{2T}\left[{∫}_{0}^{T} d t + {∫}_{0}^{T}\cos \left(2\omegat\right) d t\right]$

$⇒ {I}_{avg} = {I}_{0}\dfrac{1}{2T}\left[T + 0\right]$

$⇒ {I}_{avg} = \dfrac{{I}_{0}}{2} . . . ( 2 )$

Using equation (1) and (2) we get energy per revolution is given by

$E = \dfrac{{I}_{0}A}{2}T = \dfrac{{I}_{0}A}{2}\dfrac{2\pi}{\omega}$

$E = \dfrac{3 .3 \times 3 \times {10}^{- 4}}{2}\dfrac{3 . 14}{31 . 4}$

$E ≃ 1 \times {10}^{- 4}J$