Physics Asked by Minhyeok Kim on February 10, 2021
Why does the LASER beam quality factor (M^2) use squared value? Why not using just M?
i.e., Why define beam quality factor as
$M^2 = frac{BPP_{Real}}{BPP_{fundamental}}$
not
$M = frac{BPP_{Real}}{BPP_{fundamental}}$ ?
Is there any particular reason to use squared M?
From the RP Photonics Encylopedia website, Beam Parameter Product,
$$BPP = omega_0theta$$ Or $$omega_0 = BPP/theta$$
For a perfect Gaussian beam (the perfect beam for many lasers), $omega_0 = lambda/(pitheta)$. So
$$BPP_{perfect} = lambda/pi$$
Originally, BPP was a measure of how well a beam could be focused. For example, lens aberrations or thermal effects could produce a larger that perfect focal spot, and therefore a larger than perfect BPP.
Let us define M as
$$M = frac{BPP}{BPP_{perfect}}$$
So $M = 1$ for a perfect Gaussian beam.
For many applications where a small focal spot is important, the beam intensity is the important parameter. E.G. A cutting laser needs high intensity focal spot.
$I = Power/area$
For a fixed beam power,
$$I/I_{perfect} = biggl(frac{omega_{0,perfect}}{omega_0}biggr)^2 = biggl(frac{BPP_{perfect}}{BPP}biggr)^2 = 1/M^2$$
So $M^2$ is the measure we want. A small $M^2$ means a high intensity.
Once $M^2$ became accepted as the measure of beam quality, it was used where intensity is not the primary goal. For example, communication lasers need a small divergence angle. Electronics manufacturers need a laser to uniformly illuminate a chip.
A perfect Gaussian beam has spherical wave fronts and a Gaussian intensity profile. If it did not, the beam would not converge to a minimal spot. Likewise, it would not have a minimal divergence angle nor most uniform intensity cross section. So $M^2$ is reasonable to use as a measure of beam quality there too. See Beam Quality for more.
Answered by mmesser314 on February 10, 2021
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