Question

A light source radiates 60.0 W of single-wavelength sinusoidal light uniformly in all directions. Find the amplitude of the electric field of this light at a distance of 0.400 m from the bulb

Answer 1

E = 149.92 N/C

Step-by-step explanation:

Given that,

The power emitted by a sinusoidal light is 60 W

Intensity is equal to the power per unit area. It can be given by :

`I=(P)/(4\pi r^2)\\\\I=(60)/(4\pi * (0.4)^2)\\\\I=29.84\ W/m^2`

The relation between the amplitude of the electric field and the intensity is given by :

`I=(E_o^2c \epsilon_o)/(2)`

Here,
`E_o` is the amplitude of the electric field

`E_o=\sqrt{(2I)/(c\epsilon_o)} \\\\E_o=\sqrt{(2* 29.84)/(3* 10^8* 8.85* 10^(-12))}\\E_o=149.92\ N/C`

So, the amplitude of the electric field is 149.92 N/C.