Is there a way to relate the $1/e^2$ width of a Gaussian to the standard deviation $\sigma$?

Question

I am trying to create a 3D Gaussian fit so that the fit has the same 1/e^2 diameter as the original data.

The fit parameters need to specify a standard deviation $\sigma$. Is there a way to relate $\sigma$ to the $1/e^2$ diameter?

Thanks

Answer 1

I think I found a way to do it, please let me know if this is correct.

The intensity of a Gaussian beam (laser beam in this case) is modeled by:

$I=I_0e^{-2r^2/w^2}$ where $w$ is the beam RADIUS.

I found on wikipedia that $FWHM=2\sqrt{2\ln(2)}$ where $FWHM$ is the width at half the max intensity value. So let $r=FWHM/2$

$0.5=e^{-2(FWHM/2)^2/w^2}$

$-2\ln(1/2)=FWHM^2/w^2$

$w=FWHM*\sqrt{2\ln(2)}$

Plug in FWHM

$w=\sigma * 2\sqrt{2\ln(2)}/\sqrt{2\ln(2)}=2\sigma$

So the beam diameter $2w=4\sigma=D$