The intended output of many lasers in laser scanning is Gaussian. At distance $z$ from the waist, the radius of a Gaussian beam is calculated as $$w(z) = w_0 \sqrt{1+(z/z_R)^2},$$ where $w_0$ is the waist radius, and $z_R = \pi w_0^2/\lambda$ is the Rayleigh range, depending on the waist $w_0$ and the wavelength $\lambda$. When distance $z$ is considerably larger than $z_R$, the radius $w$ grows approximately linearly, $$w(z) \approx \theta z,$$ where $$\theta = \frac{\lambda}{\pi w_0}$$ is the divergence angle.

I am modeling a laser beam from a laser scanning device as a Gaussian beam. I am not sure how to decide the waist radius $w_0$. For example, it is given that laser beam footprint at exit is $5$ mm and the divergence is reported to be 0.5 mrad. Wavelength is 1500 nm.

Then we may calculate $$w_0 = \frac{\lambda}{ \pi \theta} = \frac{1500 nm}{\pi \times 0.5 \times 10^{-3}} = 0.0009549... m$$ and $$z_R = \pi w_0^2/\lambda = \frac{\pi \times (0.0009549... m)^2}{1500 nm}= 1.909... m.$$

But then the radius is equal to the output radius $5mm/2 = 2.5 mm$ at distance $z \approx 4.6209 m$, which is insane. The device is surely not 5 meters long. So have I understood the parameters of Gaussian beam incorrectly, or is there some optical tricks happening inside the laser device? Or is it so that the Gaussian model is accurate only for the Gaussian shape, not for the radius calculation?

  • $\begingroup$ When you say "the device is surely not 5 meters long", what "device" are you talking about? This is the first you've mentioned a "device" so its hard to understand why you brought it up at this point. $\endgroup$
    – The Photon
    Jan 5 at 21:53
  • $\begingroup$ @ThePhoton Sorry, I edited my question now. I am modeling laser beams in the context of laser scanning. $\endgroup$
    – mathslover
    Jan 5 at 22:12

2 Answers 2


It is usually assumed that the waist of the laser beam is at the exit pupil. However, that may not be true in your case. The way you phrase it (which I assume is how the laser specs are stated) does not necessary imply that the 5 mm at the exit means that the waist diameter is 5 mm. It could be that the device contains some lenses producing a converging beam at the exit. That would make sense in the context of laser scanning.

What you can do is to make some measurements to get a rough estimate of the beam size at different distances. A plot of these beam sizes as a function of distance should give you an idea of the location of the waist.

  • $\begingroup$ What does it mean for the beam to be converging at the exit? Especially, what does it mean for the beam parameters, e.g. $\theta$, $w_0$ and $z_R$? $\endgroup$
    – mathslover
    Jan 6 at 12:10
  • $\begingroup$ I would imagine that the beam is divergent at exit, because $\theta>0$. Convention is that usually negative divergence for converging beam, and positive for divergent. So, for @mathslover, in principle the waist is a virtual waist inside your device. It's not "optical tricks", its just lenses or mirrors. Like flippiefanus said, if you can measure your beam at a few locations after your source: if it only gets bigger, virtual waist inside the device, if it gets smaller, then your waist will be 4.6m after the exit pupil. I think its the case where the beam will only get bigger. $\endgroup$ Jan 6 at 17:03

A Gaussian beam does not have a sharp edge. It is brightest at the center and fades away as you get farther from the beam axis. So the intensity isn't $0$ outside the beam radius.

This makes it hard to describe the beam diameter. The way it is done is to pick the diameter where the intensity has dropped by a factor is $1/e^2$.

You need a bigger aperture than the beam diameter to avoid cutting off the outer portion of the beam. If you do that, you have passed the beam through a large pinhole. A pinhole causes diffraction. It isn't as bad with a large pinhole, but it is enough that the beam has a larger divergence angle than it should.

To avoid this, the aperture should be large enough that the bean intensity at the edge is at most $1$ % of the central intensity. That works out to be a aperture $1.5$ times larger than the beam diameter.

If the aperture is $5$ mm, the beam diameter is at most $3.33$ mm, and the beam radius $1.66$ mm. A beam with this radius would have a smaller divergence angle than advertised.

The beam could be smaller. A beam radius of $0.95$ mm does match the divergence angle.

Or the beam could be something other than a perfect Gaussian beam. The presence of higher modes would make a beam with a $1.66$ mm radius have a larger than ideal divergence angle.

Near the beam waist, rays follow hyperbolic paths. Far from the beam waist, the hyperbolic beam approximates a cone. The divergence angle is the angle of the vertex of the cone. See the RP Photonics Encyclopedia article Gaussian Beams for more. It has a beam calculator.

The beam waist is often at or near the exit aperture, but it doesn't have to be.


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