Light with an energy flux` 20 W//cm^2` falls on a non-reflecting surface at normal incidence. If the surface has an area of `30 cm^2`. the total momentum delivered ( for complete absorption)during 30 minutes is

Options

A. `36 xx 10^(-5) kgm//s`
B. `36 xx 10^(-4)kg m//s`
C. `108 xx 10^(4) kg m//s`
D. `1.08 xx 10^(7) kg m//s`

Answer

B

Explanation

Energy flux, `phi = 20 W//cm^(2)`
Area `A = 30 cm^(2)`, time `t = 30 xx 60 sec`
U = Total energy falling in t sec = Energy flux `xx` Area `xx` time `= phi` At
`U = 20 xx 30 xx 30 xx 60 J`
Momentum of the incident light `= (u)/(c) = (20 xx 30 xx 30 xx 60)/(3 xx 10^(8)) = 36 xx 10^(-4) kg-ms^(-1)`
As no reflection from the surface and for complete absorption, momentum of reflected radiation is zero.
Momentum delivered to surface = Change in momentum
`= p_(f) - p_(i) = 0 - 36 xx 10^(-4)` kgm/s
`= -36 xx 10^(-4)` kgm/s
(-) sign shows the direction of momentum.