Question

A light bulb emits light that travels uniformly in all directions. Detailed measurements show that at a distance of 56 m from the bulb, the amplitude of the electric field is 3.78 V/m. What is

Answer 1

Complete question:

A light bulb emits light that travels uniformly in all directions. Detailed measurements show that at a distance of 56 m from the bulb, the amplitude of the electric field is 3.78 V/m. What is the average intensity of the light?

Answer:

The average intensity of the light is 0.02 W/m²

Explanation:

Given;

Amplitude of the electric field, E₀ = 3.78 V/m

The average intensity of the light is calculated as follows;

$I_{avg} = \frac{c\epsilon_0 E_0^2}{2}$

where;

$I_{avg}$ is the average intensity of the light

c is speed of light = 3 x 10⁸ m/s

$I_{avg} = \frac{(3\times 10^8)(8.85 \times 10^{-12}) (3.78)^2}{2} \\\\I_{avg} = 0.01897 \ W/m^2\\\\I_{avg} = 0.02 \ W/m^2$

Therefore, the average intensity of the light is 0.02 W/m²