A beam of plane polarized light falls normally on a polarizer of cross sectional area `3xx10^-4m^2`. Flux of energy of incident ray in `10^-3W`. 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^-7`Joule
B. `10^-3`Joule
C. `10^-2` Joule
D. `10^-1` Joule

Answer

A

Explanation

Using Matus law, `I=I_0cos^2theta`
As here polariser is rotating, i.e., all the values of `theta` are possible.
`I_(av)=(1)/(2pi)int_0^(2pi)Idtheta=(1)/(2pi)int_0^(2pi)I_0cos^2thetadtheta`
On integration we get `I_(av)=I_0/2`
where `I_0=(En ergy)/(AreaxxTime)=p/A=(10^-3)/(3xx10^-4)=(10wat t)/(3m^2)`
`:. I_(av)=1/2xx10/3xx5/3wat t`
and Time period `T=(2pi)/(omega)=(2xx3.14)/(31.4)=1/5 sec`