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^(-4)` Joule
B. `10^(-3)` Joule
C. `10^(-2)` Joule
D. `10^(-1)` Joule

Answer

A

Explanation

Using malus law,
`I = I_(0) cos^(2) theta`
`:. I_(av) = (1)/(2pi) int_(0)^(2pi) I d theta = (1)/(2pi) int_(0)^(2pi)I_(0)cos^(2) theta d theta`
`I_(av') = (I_(0))/(2)`
where `I_(0) = (P)/(A) = (10^(-3))/(3xx 10^(-4)) = (10)/(3) ("watt")/(m^(2))`
`:. T = (2pi)/(omega) = (2xx 3.14)/(31.4) = (1)/(5)sec`
Energy passes per revolution
`= I_(av) xx "area" xx T`
`=(5)/(3) xx 3 xx 10^(-4) xx (1)/(5) = 10^(-4)J`