A collimated beam of light of flux density `3kWm^(-2)` is incident normally on `100mm^(2)` completely absorbing screen. If P is the pressure exerted on the screen and `trianglep` is the momentum tranferred to the screen during a 1000 s interval, then

Options

A. `P=10^(-3)Nm^(-2)`
B. `P=10^(-4)Nm^(-2)`
C. `trianglep=10^(-4)kgms^(-1)`
D. `trianglep=10^(-5)kgms^(-1)`

Answer

B, D

Explanation

Since `P=(I)/(c )=10^(4)Nm^(-2)`
`P=(E)/(A)=(1)/(A)(trianglep)/(trianglet)`
`trianglep=Patrianglet=10^(-5)kgms^(-1)`