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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:

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

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

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    $\begingroup$ 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. $\endgroup$
    – Carl Witthoft
    Commented Feb 16, 2016 at 13:24
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    $\begingroup$ 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 $\endgroup$
    – Ronan Tarik Drevon
    Commented Feb 16, 2016 at 15:30
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    $\begingroup$ Check any college textbook on wave optics. $\endgroup$
    – Ján Lalinský
    Commented Feb 16, 2016 at 21:49
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    $\begingroup$ @CarlWitthoft yeah everybody knows that, i was interested in its mathematical derivation. $\endgroup$
    – Marcus
    Commented Feb 17, 2016 at 18:37

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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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    $\begingroup$ Posting images of derivations or any mathematical content in general is discouraged since the site supports mathjax, which both looks much better and makes the content searchable. $\endgroup$
    – Kyle Oman
    Commented Mar 4, 2019 at 8:07
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    $\begingroup$ I am new to this site. So i don't know much about how to write with mathjax. I will try next time $\endgroup$
    – lowkeyy
    Commented Mar 4, 2019 at 8:09
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    $\begingroup$ if you know how to markup math in latex, it's basically the same. Otherwise, you can read the post I linked and/or find some posts with math content and click 'edit' to see how it was done. Not terribly complicated for simple equations. $\endgroup$
    – Kyle Oman
    Commented Mar 4, 2019 at 8:11
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    $\begingroup$ And there's an example for you with your own post. $\endgroup$
    – Kyle Oman
    Commented Mar 4, 2019 at 8:21
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    $\begingroup$ A more detailed derivation/explanation can be found here: phys.libretexts.org/Bookshelves/University_Physics/… The one here is a bit confusing, as it does not introduce $E_{RMS}$ and the formula actually differs from the one in the question, which I think is incorrect, as $E_{RMS}$ is usually defined as $E_{RMS}=E_{0}/\sqrt{2}$, no? $\endgroup$
    – KIAaze
    Commented Jul 14, 2023 at 9:51

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