Optics Question 186
Question: A beam of plane polarized light falls normally on a polarizer of cross sectional area $ 3\times {{10}^{-4}}m^{2} $ . Flux of energy of incident ray in 10?3 W. The polarizer rotates with an angular frequency of 31.4 rad/sec. The energy of light passing through the polarizer per revolution will be
Options:
A) 10?4 Joule
B) 10?3 Joule
C) 10?2 Joule
D) 10?1 Joule
Show Answer
Answer:
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 }{Id\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. $
