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A beam of unpolarised light having flux \mathrm{10^{-3}} watt falls normally on a polarizer of cross sectional area \mathrm{3 \times 10^{-4} \mathrm{~m}^2}. The polarizer rotates with an angular frequency of \mathrm{31.4 \, \mathrm{rad} \mathrm{s}^{-1}}. The energy of light passing through the polarizer per revolution will be:

Option: 1

\mathrm{10^{-4} \text { joule }}


Option: 2

\mathrm{10^{-3} \text { joule }}


Option: 3

\mathrm{10^{-2} \text { joule }}


Option: 4

\mathrm{10^{-1} \text { joule }}


Answers (1)

best_answer

Here, \mathrm{\omega=31.4\, \mathrm{rads}^{-1}}
\mathrm{\therefore \quad}  Time period of revolution, \mathrm{\mathrm{T}=\frac{2 \pi}{\omega}=\frac{2 \times 3.14}{31.4}=0.2 \mathrm{~s}}

Energy transmitted/revolution \mathrm{=(\mathrm{IA}) \mathrm{T}=\left(\frac{\mathrm{I}_0}{2} \mathrm{~A}\right) \mathrm{T}=\frac{\phi_0 \mathrm{~T}}{2}=\frac{10^{-3} \times 0.2}{2}=10^{-4}\, \, joule.}

Posted by

Irshad Anwar

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