50W/m^2 energy density of sunlight is normally incident on the surface of a solar panel. Some part of incident energy (25%) is reflected from the surface and the rest is absorbed. The force exerted on 1m^2 surface area will be close to?

Correct option is D. 20 × 10⁻⁸ N
Force on the surface (25% reflected, 75% absorbed):
Incident power = 50 W/m²
Reflected power = 0.25 * 50 W/m² = 12.5 W/m²
Absorbed power = 0.75 * 50 W/m² = 37.5 W/m²
Radiation pressure due to reflection = (2 * Reflected power) / c
Radiation pressure due to absorption = (Absorbed power) / c
Total radiation pressure = (2 * Reflected power + Absorbed power) / c
where c is the speed of light (3 × 10⁸ m/s)
Total radiation pressure = (2 * 12.5 W/m² + 37.5 W/m²) / (3 × 10⁸ m/s) = (25 + 37.5) W/m² / (3 × 10⁸ m/s) = 62.5 W/m² / (3 × 10⁸ m/s)
Force = Pressure × Area = (62.5 W/m² / (3 × 10⁸ m/s)) * 1 m² = 62.5 / (3 × 10⁸) N
Force ≈ 20.83 × 10⁻⁸ N ≈ 20 × 10⁻⁸ N