Question

A parallel beam of light is incident normally on a plane surface absorbing 40% of the light and reflecting the rest. If the incident beamm carries 60 W of power, the force exerted by it on the surface is
A. `3.2xx10^(-8)N`
B. `3.2xx10^(-7)N`
C. `5.12xx10^(-7)N`
D. `5.12xx10^(-8)N`

Answer 1

Correct Answer - B
Momentum of iincident light per second
`p_(1)=(E)/(c)=(60)/(3xx10^(8))=2xx10^(-7)`
momentum of reflected light per second
`p_(2)=(60)/(100)xx(E)/(c)`
`=(60)/(100)xx2xx10^(-7)=1.2xx10^(-7)`
`therefore`Force on the surface=change in momentum per second
`=p_(2)-(-p_(1))=p_(2)+p_(1)`
`=(2+1.2)xx10^(-7)=3.2xx10^(-7)N`