Question

A red LED emits light at $0.1Watt$ uniformly around it. The amplitude of the electric field of the light at a distance $1m$ from the diode is

Answer 1

Given data

1. The power of the LED is $P=0.1watt.$
2. The distance where the electric field has to be measured is $L=1m.$

Intensity of light

1. Intensity is the energy emitted per unit area in unit time. For a point light source the intensity is $I=\frac{P}{4\pi r^2}$, where, P is the power of the light and r is the distance where the intensity has to measure.
2. Intensity of a light source is defined by the form, $I=\frac{1}{2}{\epsilon }_o{E}^2\times c$, where, ${\epsilon }_o$is the permittivity of free space, $E$ is the electric field, and $c$ is the velocity of light in free space.

Finding the electric field

As we know, $\frac{1}{2}{\epsilon }_o{E}^2\times c=$$\frac{P}{4\pi r^2}$

So,

$$\begin{gathered} E^2=\frac{P\times 2}{4\pi {\epsilon }_o\times r^2\times c} \\ or{E}^2=\frac{0.1\times 2\times 9\times {10}^9}{1^2\times 3\times {10}^8}=6\left(\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}=Coulombcons\tan t=9\times {10}^9{C}^2/N/m^2\right) \\ orE=\sqrt{6}=2.45 \\ orE=2.45V/m. \end{gathered}$$

Therefore, the electric field of the light at a distance $1m$ from the diode is $2.45volt/m.$