Question

Light with an intensity of 1 kW/m2 falls normally on a surface with an area of 1 cm2 and is completely absorbed. The force of the radiation on the surface is

Answer 1

The force of the radiation on the surface is 3.33 X 10⁻¹⁰ N

Step-by-step explanation:

Given;

intensity of light, I = 1 kw/m²

area of the surface, A = 1 cm² = 1 x 10⁻⁴ m²

Power of the incident light, P = I x A

Power of the incident light, P = (1 kw/m²) x (1 x 10⁻⁴ m²)

Power of the incident light, P = 1 x 10⁻⁴ kW = 0.1 W

Power of the incident light is given by;

P = Fv

where;

F is the force of the radiation on the surface

v is the speed of light = 3 x 10⁸ m/s

F = P/ v

F = (0.1) / (3 x 10⁸)

F = 3.33 X 10⁻¹⁰ N

Therefore, the force of the radiation on the surface is 3.33 X 10⁻¹⁰ N