Light with an energy flux of $18\, W/cm^2$, falls at normal incidence on a non-reflecting surface with area of $20\, cm^2$. If the light falls for 30 minutes on the surface, find the total momentum delivered to the surface.

$2.16 × 10^{-3}\, Kg\, ms^{-1}$

The correct answer is Option (4) → $2.16 × 10^{-3}\, Kg\, ms^{-1}$

Energy flux $= 18 \, \text{W/cm}^2 = 18 \times 10^4 \, \text{W/m}^2$

Area $= 20 \, \text{cm}^2 = 20 \times 10^{-4} \, \text{m}^2 = 2 \times 10^{-3} \, \text{m}^2$

Total power incident $P = (18 \times 10^4)(2 \times 10^{-3}) = 360 \, \text{W}$

Time $= 30 \, \text{min} = 1800 \, \text{s}$

Total energy incident $E = P \times t = 360 \times 1800 = 6.48 \times 10^5 \, \text{J}$

For absorption, momentum delivered $p = \frac{E}{c}$

$p = \frac{6.48 \times 10^5}{3 \times 10^8} = 2.16 \times 10^{-3} \, \text{kg·m/s}$

Final Answer: $2.16 \times 10^{-3} \, \text{kg·m/s}$