Question 178 of 191

A beam of plane polarized light of large cross-sectional area and uniform intensity of $3.3W{m}^{–2}$ falls normally on a polarizer (cross-sectional area $3\times {10}^{–4}{m}^{2}$), which rotates about its axis with an angular speed of $31.4rad{s}^{-1}$. The energy of light passing through the polarizer per revolution, is close to:

A$1.0\times {10}^{-5}J$B$1.0\times {10}^{-4}J$C$1.5\times {10}^{-4}J$D$5.0\times {10}^{-4}J$

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Answer: (B) $1.0\times {10}^{-4}J$

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
${I}_{avg}={I}_{0}\dfrac{1}{T}{\int }_{0}^{T}{\cos }^{2}\left(\omega t\right)dt$
$\Rightarrow {I}_{avg}={I}_{0}\dfrac{1}{T}{\int }_{0}^{T}\left[\dfrac{1+\cos \left(2\omega t\right)}{2}\right]dt$
$\Rightarrow {I}_{avg}={I}_{0}\dfrac{1}{2T}\left[{\int }_{0}^{T}dt+{\int }_{0}^{T}\cos \left(2\omega t\right)dt\right]$
$\Rightarrow {I}_{avg}={I}_{0}\dfrac{1}{2T}\left[T+0\right]$
$\Rightarrow {I}_{avg}=\dfrac{{I}_{0}}{2}...(2)$
Using equation (1) and (2) we get energy per revolution is given by
$E=\dfrac{{I}_{0}A}{2}T=\dfrac{{I}_{0}A}{2}\dfrac{2\pi }{\omega }$
$E=\dfrac{3.3\times 3\times {10}^{-4}}{2}\dfrac{3.14}{31.4}$
$E≃1\times {10}^{-4}J$