1
$\begingroup$

I read that longer laser cavities generally produce more collimated beams. Is this true? And, if so, then why? The closest source of information that I could find is this Wikipedia article on collimated beam:

Laser diodes emit less-collimated light due to their short cavity, and therefore higher collimation requires a collimating lens.

However, this obviously does not explain why this is the case.

$\endgroup$
4
  • $\begingroup$ Would be interested in a derivative answer too. Hand waving argument here but I am unsure: a long cavity is like an angular filter. Moreover, for light to bounce between two mirrors far apart the solid angle is smaller and therefore the number of optical modes which can pass between the mirrors are fewer. This will give more a more collimate beam because fewer modes contribute to lasing. $\endgroup$
    – boyfarrell
    Mar 30, 2021 at 18:25
  • $\begingroup$ Laser diodes emit uncollimated light because their aperture is very small. It's just diffractive divergence first of all. See Why does a laser beam diverge?. $\endgroup$
    – Ruslan
    Apr 1, 2021 at 15:00
  • $\begingroup$ @Ruslan I already read that, but I don't think it answers this question. My understanding is that laser diodes often come equipped with a collimating lens (aspheric lenses en.wikipedia.org/wiki/Aspheric_lens, I think) for exactly this reason; but, again, this doesn't answer the question at hand. $\endgroup$ Apr 1, 2021 at 15:01
  • $\begingroup$ Sincerely, I think that the problem is the definition of "collimation". Collimation is a broad term we use to say that the beam is the least diverging possible at a certain location, for a certain beam waist. It's like "focusing to infinity", but not really, because a laser will always diffract. It also depends a lot on your cavity design, you might even make a cavity 20m long, still the beam can come out "uncollimated". Depends on many many things, and I think that if you want a design with lower output divergence, then yes, making it longer helps, but the design ultimately dictates how much $\endgroup$ Apr 1, 2021 at 19:15

1 Answer 1

1
+50
$\begingroup$

The Wikipedia article is being a little misleading here. The beam from a laser diverges by an amount which depends on its width not length. The divergence is due to diffraction, and the angular spread of the beam is given approximately by $$ \Delta \theta = \frac{\lambda}{w} $$ where $\lambda$ is the wavelength and $w$ the width. But if a laser has a long cavity then clearly the angular divergence of the beam could not be very large or it would spread outside the laser cavity. In other words, a long cavity implies, in practice, that the laser mode is wide enough for not much spreading to be happening as it propagates inside the cavity. This implies that the beam waist cannot be all that small, so in practice the longer cavity will be associated with a less diverging beam. However there is no basic physical requirement that says it has to be this way. One could design a long cavity with a very diverging mode; the concentric design is like this.

Of course if you let the beam spread a bit and then put a suitable lens, then after the lens you will get a nice collimated beam no matter how divergent it was coming out of the laser cavity itself.

$\endgroup$
7
  • $\begingroup$ Thanks for the answer. Is the divergence of a laser beam due to diffraction, or is it due to the Heisenberg uncertainty principle? $\endgroup$ Apr 2, 2021 at 2:47
  • $\begingroup$ This physics.stackexchange.com/a/79469/141502 seems to say that beam divergence is due to the HUP. And then, to make matters more confusing, this physics.stackexchange.com/a/114300/141502 answer seems to suggest that the HUP and diffraction are not the same: "strictly speking, the physics of diffraction cannot be explained as the HUP (i.e. as arising from the canonical commutation relationships) because there is no position observable $\hat{X}$ for the photon, so you can't think of $\Delta\,x\,\Delta p$." $\endgroup$ Apr 2, 2021 at 3:04
  • $\begingroup$ So, although I often see it said that diffraction causes beam divergence, I'm confused as to whether that's actually true, or whether it is actually the HUP. $\endgroup$ Apr 2, 2021 at 3:05
  • $\begingroup$ @ThePointer Diffraction is a statement about waves which applies to all waves. HUP brings in a connection between wavevector $\bf k$ and momentum $\bf p$, namely ${\bf p} = \hbar {\bf k}$. That relationship cannot be derived from diffraction so it is different from diffraction. But both agree that $\Delta x \Delta k \ge 1/2$. $\endgroup$ Apr 2, 2021 at 14:39
  • 1
    $\begingroup$ @ThePointer No, not separate. Quantum theory relates $\bf k$ to $\bf p$ and then the nature of diffraction implies or results in the HUP. It is what follows from diffraction, or more generally, from Fourier analysis of any function. The specifically quantum theory contribution here is to relate momentum to wave-like properties in the first place. $\endgroup$ Apr 2, 2021 at 16:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.