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

Answer:

E = 149.92 N/C

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 :

$I=\dfrac{P}{4\pi r^2}\\\\I=\dfrac{60}{4\pi \times (0.4)^2}\\\\I=29.84\ W/m^2$

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

$I=\dfrac{E_o^2c \epsilon_o}{2}$

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

$E_o=\sqrt{\dfrac{2I}{c\epsilon_o}} \\\\E_o=\sqrt{\dfrac{2\times 29.84}{3\times 10^8\times 8.85\times 10^{-12}}}\\E_o=149.92\ N/C$

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