Light with an energy flux of 20 W/cm² falls on a non-reflecting surface at normal incidence. If the surface  has an area of 30 cm² , then the total momentum delivered (for complete absorption) during 30 min is

Given, energy flux,  $\varphi =20 \mathrm{W}/{\mathrm{cm}}^2$

Area, A = 30 cm²
Time, t = 30 min = 30 x 60 s
Now, total energy falling on the surface in time t,
$u=\varphi At=20\times 30\times (30\times 60)\mathrm{J}$

Momentum of the incident light $P=\frac{u}{c}$

$=\frac{20\times 30\times (30\times 60)}{3\times {10}^8}$

$=36\times {10}^{-4} \mathrm{kg}-\mathrm{m}/\mathrm{s}$

Momentum of the reflected light = 0

$\therefore$  Momentum delivered to the surface = 36 x 10^-4 - 0

= 36 X 10^-4 kg-m/s