Question

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

Answer 1

The radiation pressure of the light is 3.33 x 10⁻⁶ Pa.

Step-by-step explanation:

Given;

intensity of light, I = 1 kW/m²

The radiation pressure of light is given as;

`Radiation \ Pressure = (Flux \ density)/(Speed \ of \ light)`

I kW = 1000 J/s

The energy flux density = 1000 J/m².s

The speed of light = 3 x 10⁸ m/s

Thus, the radiation pressure of the light is calculated as;

`Radiation \ pressure = (1000)/(3*10^(8)) \\\\Radiation \ pressure =3.33*10^(-6) \ Pa`

Therefore, the radiation pressure of the light is 3.33 x 10⁻⁶ Pa.