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

Answer:

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

Explanation:

Given;

intensity of light, I = 1kw/m² = 1000 W/m²

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

Since the light is completely absorbed, the force of the radiation is given by;

F = P/c

where;

c is the speed of light = 3 x 10⁸ m/s

But P = IA

F = IA /c

F = (1000 X 1 X 10⁻⁴) / 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

Answer 2

The force of radiation will be "3.33 × 10⁻¹⁰ N"

Intensity and Force

According to the question,

Intensity of force, I = 1 kW/m² or,

                               = 1000 W/m²

Area of surface, A = 1 cm² or,

                              = 1 × 10⁻⁴ m²  

Speed of light, c = 3 × 10³ m/s

As we know the relation,

→ F = $\frac{P}{c}$

or,

  P = IA

or,

  F = $\frac{IA}{c}$

By substituting the values, we get

     = $\frac{1000\times 1\times 10^{-4}}{3\times 10^3}$

     = 3.33 × 10⁻¹⁰ N

Thus the response above is correct.

Find out more information about intensity here:

https://brainly.com/question/1444040