Average force exerted on a non-reflecting surface at normal incidence is $$2.4 \times 10^{-4} \text{ N}$$. If $$360 \text{ W/cm}^2$$ is the light energy flux during span of 1 hour 30 minutes, Then the area of the surface is:

Find the area of a non-reflecting surface given the radiation force and intensity.

For a perfectly absorbing (non-reflecting) surface, the radiation pressure is:

$$P_{\text{rad}} = \frac{I}{c}$$

where $$I$$ is the intensity (power per unit area) and $$c = 3 \times 10^8$$ m/s is the speed of light.

The force on the surface is: $$F = P_{\text{rad}} \times A = \frac{I \times A}{c}$$.

$$I = 360$$ W/cm$$^2 = 360 \times 10^4$$ W/m$$^2 = 3.6 \times 10^6$$ W/m$$^2$$.

$$F = \frac{I \times A}{c} \implies A = \frac{Fc}{I}$$

$$A = \frac{2.4 \times 10^{-4} \times 3 \times 10^8}{3.6 \times 10^6} = \frac{7.2 \times 10^4}{3.6 \times 10^6} = 0.02 \text{ m}^2$$

The correct answer is Option (4): 0.02 m$$^2$$.