It would be difficult to fit Gaussian peaks to this because the peak contributions are not well-defined. Calculating the FWHM of the single peak however would be relatively straightforward. [pkmax,ixp] = max(ybc); ixv = find(diff(sign(ybc-(pkmax/2))); fork...
Gaussian laser beams don’t have a simple cut-and-dry beam width. If you’re not familiar with the math behind a “Gaussian” you might wonder why this would be. Here’s why:
A thermal ensemble of atoms has a Gaussian distribution of velocities in the light propagation direction, given by [49](6)g(v)=1Uπexp(−v2U2), where v is the component of velocity of the atom in the light propagation direction and U is the root mean square (rms) speed of the ...