A perfectly reflecting mirror has an area of `1cm^(2)` Light enery is allowed to fall on it 1 h at the rate of `10 W cm^(2)`. The force that acts on the mirror is

Options

A. `3.35 xx 10^(-8) N`
B. `6.7 xx 10^(-8) N`
C. `1.34 xx 10^(-7)N`
D. `2.4 xx 10^(-4) N`

Answer

B

Explanation

Let, E = Energy falling on the surface per second `= 10 J`
Momentum of photons, `p = (h)/(lambda) = (h)/((c//v)) = (hv)/(c) = (E)/(c)`
On reflection, change in momentum per second `=` force
`= 2p = (2E)/(c) = (2 xx 10)/(3 xx 10^(8)) = 6.7 xx 10^(-8) N`