A) 10?4 Joule
B) 10?3 Joule
C) 10?2 Joule
D) 10?1 Joule
Correct Answer: A
Solution :
Using Matus law, \[I={{I}_{0}}{{\cos }^{2}}\theta \] As here polariser is rotating i.e. all the values of q are possible. \[{{I}_{av}}=\frac{1}{2\pi }\int_{0}^{\,2\pi }{I\,d\theta }=\frac{1}{2\pi }\int_{\,0}^{\,2\pi }{{{I}_{0}}{{\cos }^{2}}\theta \,d\theta }\] On integration we get \[{{I}_{av}}=\frac{{{I}_{0}}}{2}\] where \[{{I}_{0}}=\frac{\text{Energy}}{\text{Area }\times \text{ Time}}=\frac{p}{A}=\frac{{{10}^{-3}}}{3\times {{10}^{-4}}}=\frac{10}{3}\frac{Watt}{{{m}^{2}}}\] \ \[{{I}_{av}}=\frac{1}{2}\times \frac{10}{3}=\frac{5}{3}\,Watt\] and Time period \[T=\frac{2\pi }{\omega }=\frac{2\times 3.14}{31.4}=\frac{1}{5}sec\] \ Energy of light passing through the polariser per revolution \[={{I}_{av}}\times \text{Area}\times T\]\[=\frac{5}{3}\times 3\times {{10}^{-4}}\times \frac{1}{5}\]\[={{10}^{-4}}\,J.\]
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