A beam of plane polarised light of large cross-sectional area and uniform intensity of $3.3 W m^{- 2}$ falls normally on a polariser (cross sectional area $3 \times 10^{- 4} m^{2}$ ) which rotates about its axis with an angular speed of 31.4 rad/s. The energy of light passing through the polariser per revolution, is close to :

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$1.0 \times 10^{- 5} J$

$1.0 \times 10^{- 4} J$

$1.5 \times 10^{- 4} J$

$5.0 \times 10^{- 4} J$

answer is B.

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Detailed Solution

Malus law for polarised light $\Rightarrow I = I_{0} \cos^{2} θ$
Here $I_{0} = 3.3 W / m^{2}$
$θ = ω t$ where $ω = 10 π r a d / s$
Energy $E = \int I A d t$

$\begin{array}{l} = I_{0} A \int \cos^{2} ω tdt = \frac{I_{0} A}{ω} \int_{0}^{2 π} \cos^{2} ω t (ω t) \\ = \frac{I_{0} A}{ω} π = \frac{(3.3) (3 \times 10^{- 4})}{10} = 0.99 \times 10^{- 4} J \end{array}$