[Physics] Relation between intensity of light and amplitude of electric field

electromagnetismintensityvisible-light

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.

Best Answer

As light is an electromagnetic wave, it is a combination of both electric field and magnetic field. So intensity of light is basically the power transmitted through electric and magnetic field divided by the cross section area of that light beam.

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: $$\epsilon_0 E^2_{\rm RMS}c$$ $$\frac{1}{2}\epsilon_0 E^2 c$$

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