1.

When unpolarized light incidents on polarizer 3.14 rad/s angular speed and area 3 xx 10^(-4) m^(2) with energy 3 * 10^(-3) J, then find energy of polarized light in each rotation

Answer»

`47.1xx10^(-4)J`
`27.1xx10^(-4)J`
`37.1xx10^(-4)J`
`17.1xx10^(-4)J`

Solution :`omega=3.14 RAD"s^(-1)`
`:.(2pi)/(T)=3.14`
`:.T=(2xx3.14)/(3.14)`
`:. T=2s`
Average intensity
`lt I gt =(I_(0)+0)/(2)=(I_(0))/(2)`
`=(3xx10^(-3))/(3xx10^(-4)xx1xx2)`
`lt I gt -(10)/(2)J`
Energy polarized in each rotation
`E= lt I gt xx "AREA" xx` Angular SPEED
`=(10)/(2)xx3xx10^(-4)xx3.14=47.1xx10^(-4)J`


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