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Why does the LASER beam quality factor (M^2) use squared value?

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?

One Answer

From the RP Photonics Encylopedia website, Beam Parameter Product,

$$BPP = omega_0theta$$ Or $$omega_0 = BPP/theta$$

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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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