# Relation between intensity of light and amplitude of electric field?

A question in my textbook involve finding the electric field amplitude at a point in space given the intensity of light. It uses the following equation to solve it: -

$$I=\frac{1}{2}\epsilon_{0}|E_{rms}|^2c$$

But where did this equation come from? I am unable to find an explanation for this anywhere.

• It came from observation and analysis. What is your question? The relationship between wave amplitude and power ( $I \propto A^2$ ) holds for pretty much any kind of wave, not just E-M. – Carl Witthoft Feb 16 '16 at 13:24
• It comes from the Poynting vector theorem which can be derived from the wave equation. Have a look on wikipedia and apply it for a plane TEM wave – Ronan Tarik Drevon Feb 16 '16 at 15:30
• Check any college textbook on wave optics. – Ján Lalinský Feb 16 '16 at 21:49
• @CarlWitthoft yeah everybody knows that, i was interested in its mathematical derivation. – Marcus Feb 17 '16 at 18:37

The energy density of the electric field is $$\frac{1}{2}\epsilon_0 E^2$$, and the energy density of the magnetic field is $$\frac{1}{2}\frac{B^2}{\mu_0}$$. The total energy density of an EM wave is then: $$\frac{1}{2}\left(\epsilon_0 E^2 + \frac{B^2}{\mu_0}\right)$$ The total energy transmitted per second per unit area is then: $$\frac{c}{2}\left(\epsilon_0 E^2 + \frac{B^2}{\mu_0}\right)\tag{1}$$ As we know $$|\vec{E}|=c|\vec{B}|$$ and $$c^2=\frac{1}{\epsilon_0 \mu_0}$$, so $$(1)$$ turns out to be: $$\epsilon_0 E^2_{\rm RMS}c$$ $$\frac{1}{2}\epsilon_0 E^2 c$$