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A beam of plane polarised light falls no...

A beam of plane polarised light falls normally on a polariser (cross sectional area `3 xx 10^(-4)m^(2)`) which rotates about the axis of the ray with an angular velocity of `31.4 rad//s`. Find the energy of light passing through polariser per revolution and the intensity of emergent beam, if flux of energy of the incident ray is `10^(-3) W`.

Text Solution

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If `I_(0)` is the intensity of plane polarised light incident on the polariser, then intensity of emerging light is given by
`I=I_(0)cos^(2) theta`
The average value of I over one revolution can be calculated as
`I_(av)=1/(2pi) int_(0)^(2pi)I d theta`
`=1/(2pi)int_(0)^(2pi)I_(0)cos^(2)theta d theta`
`=(I_(0))/2`
Intensity is given by
`I_(0)=("Power")/("area")`
`=(10^(-3))/(3xx10^(-4))=10/3W//m^(2)`
`:.I_(av)=(I_(0))/2=5/3W//m^(2)`
The energy of light passing through the polariser per revolution
`E=I_(av)xxAxxT=I_(av)xxAxx(2pi)/(omega)`
`=5/3xx(3xx10^(-4))xx(2pi)/31.4`
`=10^(-4)J`
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