If I have a green laser (500 nm wavelength) with initial beam diameter 3 mm, with proper lenses I can easily focus it to 1 mm or 0.5 mm for small distances like 10 cm.

My question is can I focus it from 3mm to 1 mm on long distances like 1 km or 100 m? From common sense, I feel that it is not possible from diffraction limits point of view, but I could not find exact explanation and equation how to calculate it. There are equations how to calculate maximum possible resolution for microscope/telescope, but not the minimum size of focal spot depending on the focal length/wave length of the lens.

For example, in the picture below if X is a few centimetres it looks possible, but if X is 100 meters or 1 km is it possible or not theoretically and practically?

enter image description here


Let's consider a simple case when I have a laser beam, which usually has Gaussian energy distribution, but for simplicity let's assume I cut sides of this Gaussian beam with the diaphragm and it almost flat (not really Gaussian). Anyway, 80% of energy is in the middle of Gaussian beam and if I cut the sides I will lose only 20%.

Or I can ask how to calculate what is the minimal Gaussian beam waist can be achieved by lens depending on focal length of the lens, beam wavelength and initial beam waist size. If I read this article, I could not find the answer as well.


For lasers there is the only way to calculate it properly is Ray Transfer matrices for Gaussian beams as written here. It depends not only on initial beam size, but on initial beam curvature (how fast it diverges in space).

I asked one professor specialized in lasers and the answer was: In reality, focusing the laser beam to small size can be achieved for several meters with some optics (lenses etc), but long distance is practically not possible.


Let's be more practical. I bought a green laser pointer 5 milliwatts from Aliexpress with 0.5 $\mu m$ (500 nm) wavelength. In a normal condition in a 1 km distance, the spot size of this laser is 10 cm. In 1 meter the spot size is 3mm.

Can I put any lens (one or many) and make the spot size 3mm in 1 km distance?

If yes, which lens should I put (what is the lens focal distance)?

I do not know anything else about this laser. If I need to do any extra measurements to do the calculations, then which measurements?

  • 1
    $\begingroup$ Laser beam cross section is characterized by a Gaussian intensity and behaves somewhat different than ray optics. Google laser waist for additional information. $\endgroup$
    – npojo
    Commented Apr 19, 2018 at 8:22
  • $\begingroup$ I know about Gaussian beams. I will update the question to make it more clear. $\endgroup$
    – Zlelik
    Commented Apr 19, 2018 at 11:28
  • 1
    $\begingroup$ There is a lot of overthinking going on here. Just invert the beam path and consider this as a telescope resolution problem. You want to resolve a 3mm spot in 1km distance. What is the necessary angular resolution? 0.6 seconds of arc, I believe. What is the necessary telescope aperture to achieve that angular resolution at that distance. There is your answer and it sounds like that you will need a fairly large optics to achieve this (200mm). $\endgroup$ Commented Sep 28, 2022 at 6:33

3 Answers 3


A laser beam is described as a gaussian beam. A gaussian beam is described by a spot size that changes as you move along the length of the beam. This is obvious when a laser beam is being focused. The beam is big near the lens but gets smaller as you approach the focus. The equation describing the size of the spot is:

\begin{align} w(z) =& w_0\sqrt{1+\left(\frac{z}{z_R}\right)^2}\\ z_r =& \frac{\pi w_0^2}{\lambda} \end{align}

Here $z$ is distance along the propagation direction away from the focus, $w_0$ is the waist where the beam is smallest, $\lambda$ is the wavelength of light and $z_R$ is what is known as the Rayleigh range. Note that if the waist is small then the Rayleigh range is small and vice versa.

For $z\ll z_R$ we have that $w(z) \approx w_0$. That is, within the Rayleigh range the size of the beam is approximately constant. Consider, for example, a beam with wavelenght $\lambda \approx 1 \text{ $\mu$m}$ with waist $w_0 = 1 \text{ mm}$. The Rayleigh range would then be $z_R \approx 3 \text{ m}$. This means that for 3 meters the beam would keep a size of roughly 1 mm. However, at larger distances the beam would begin to diverge.

Looking at the equation above we see that for $z\gg z_R$ we have

$$ w(z) \approx \frac{w_0}{z_R} z $$

That is the waist grows linearly as the beam propagates. This is what is meant by divergence of a beam. The divergence angle is given by

$$ \theta = \arctan\left(\frac{\omega_0}{z_R}\right) \approx \frac{\omega_0}{z_R} = \frac{\lambda}{\pi w_0} $$

Where the small angle approximation holds for $\lambda \ll w_0$ which is typically the case. If $\omega \approx \lambda$ then the paraxial description of a Gaussian beam begins to break down. Notice that a beam with a smaller waist diverges more rapidly.

With these equations you see that if you know $w_0$ and $\lambda$ you can calculate the waist $w(z)$ everywhere. Alternatively, if you know $w(z)$ at some point $z$ outside the Rayleigh range as well as $\theta$ you can calculate both $w_0$ as well as the location of the waist.

One final note. I've hinted at this above but I want to say it clearly. You can see there is no such thing as a truly collimated beam. For large enough distances beams always diverge. The relevant length scale is $z_R$. If we are only concerned with the properties of a beam over short distances ($z< z_R$) then it is possible to think about the concept of a collimated beam, but we should be careful in case we get confused.

edit: To address the question, with the equation for $\theta$ as a function of $w_0$ you should be able to calculate the size of the beam at the lens for your 1 mm waist as a function of distance between the focus and the lens you are using to focus down the beam. As @flippiefanus points out, the farther away your lens is the larger the beam is going to need to be. You quickly reach technical impracticality.

  • $\begingroup$ Let's be practical. I bought a red laser pointer 5 milliwatt with 0.623 $\mu m$ wavelength. In a normal condition in 1 km distance, the spot size of this laser is 10 cm. In 1 meter the spot size is 3mm. Can I put any lens (one or many) and make the spot size 3mm in 1 km distance? If yes, Which lens should I put? $\endgroup$
    – Zlelik
    Commented Jul 20, 2020 at 16:11
  • $\begingroup$ It's helpful to work backwards from your desired spot. Using the formulas I give above if you have a 1.5 mm beam waist then the Rayleigh range is about 11.8 m which means that 1km away the beam is going to have a size of about 13 cm. The question is if you can focus this 13 cm beam down to match the beam coming out of your laser (working backwards). $\endgroup$
    – Jagerber48
    Commented Jul 20, 2020 at 17:37
  • $\begingroup$ To think about exact lenses and distances I would need more time than I have. One way to do it would be to put a 1 km focal length lens, this would collimate the 11.8m beam. You could then use a lens to focus it down to use a lens with a short positive focal length to rapidly focus this beam and then a lens with a short negative focal length to collimate this rapidly converging beam to the 1.5 mm collimated beam coming out of your laser. $\endgroup$
    – Jagerber48
    Commented Jul 20, 2020 at 17:39
  • $\begingroup$ Try playing around with this app a little bit: gaussianbeam.sourceforge.net. $\endgroup$
    – Jagerber48
    Commented Jul 20, 2020 at 17:40
  • $\begingroup$ So the optics scheme would be: laser -> diverging lens -> collimating lens to collimate beam at 13 cm size -> long focal length lens to focus 1 km away. My guess is that this system will be really sensitive to alignment making it technically difficult if not impossible to construct but I'm not sure. $\endgroup$
    – Jagerber48
    Commented Jul 20, 2020 at 17:41

If you make the input beam size big enough you can, but you would probably have to make the beam so large it would be impractical.

  • $\begingroup$ Do you mean that if I have a beam 10 km diameter, then I might be able to focus it to 1 mm in 1 km distance? $\endgroup$
    – Zlelik
    Commented Apr 19, 2018 at 11:27
  • $\begingroup$ Yes but you might need a lens with a 100 m diameter for instance. Have a look at diffraction limit equations and what the variables are. $\endgroup$
    – MJC
    Commented Apr 19, 2018 at 11:32
  • $\begingroup$ Thanks, but I check diffraction limits and they do not tell about focal spot size. They tell only about the resolution of microscope/telescope. Anyway, I asked for exact equation how to calculate it, I cannot accept your answer as correct one. $\endgroup$
    – Zlelik
    Commented Apr 19, 2018 at 11:37
  • $\begingroup$ They do tell you about the focal spot size, you have obviously either not found the relevant equations or not understood what they mean. If you need help understanding the equations let me know and i'll try to explain them as they are not very complicated. $\endgroup$
    – MJC
    Commented Apr 19, 2018 at 14:52
  • $\begingroup$ Send me link please. I could find anything in wikipedia article about Difraction limit for example. $\endgroup$
    – Zlelik
    Commented Apr 23, 2018 at 11:26

One can derive an expression for the location of the waist (focus) of the Gaussian beam behind a lens, as a function the beam waist location in front of the lens. The maximum distance for the location of the waist behind the lens (sometimes refered to as beam throw) is given by the focal length plus the Rayleigh range $$z_{\rm out} = f + z_{R,{\rm out}} = f + \frac{\pi w_{\rm out}^2}{\lambda} , $$ where $w_{\rm out}$ is the beam waist radius behind the lens. This condition is obtained when the input waist is also located at the focal length plus the input Rayleigh range from the lens. If the beam behind the lens has a large size, then the Rayleigh range would also be large. The size of the output Gaussian beam behind the lens under maximum beam throw conditions, is given by $$ w_{\rm out} = \frac{f\lambda}{\sqrt{2} \pi w_{\rm in}} , $$ where $w_{\rm in}$ input beam waist radius in front of the lens, and the output Rayleigh range under these conditions, is given by $$ z_{R,{\rm out}} = \frac{f^2}{2 z_{R,{\rm in}}} , $$ where $z_{R,{\rm in}}$ is the input Rayleigh range in front of the lens.

  • $\begingroup$ Thank you very much. I got very interesting results. If I have green light 500 nm, and use 10 cm focus lens, with 5 mm initial beam waist, then I get only 5 wavelength (2250 nm) beam waist, which looks too small. I think I need to use Ray transfer matrices for Gaussian beams (en.wikipedia.org/wiki/…) $\endgroup$
    – Zlelik
    Commented Apr 20, 2018 at 7:29
  • $\begingroup$ No, you should not need to use ray tracing. Why do you think the focal point it is too small? $\endgroup$ Commented May 6, 2018 at 10:43
  • $\begingroup$ It is not just a geometry optics ray tracing, it is an adopted ray transfer matrices for Gaussian beams. What is "input beam waist radius" in this formulas? Is it 3mm on my original picture? Output beam size should depend on these 4 parameters: focal distance of the lens, input beam diameter, input beam curvature (how fast it is expanding in space). without these 4 parameter any equation will not give proper results. For example, if there are 2 beams with the same beam diameter, but different input beam curvature, then I believe result will be different. $\endgroup$
    – Zlelik
    Commented May 7, 2018 at 20:13
  • 1
    $\begingroup$ A Gaussian beam does not have any curvature in its waist. Usually, when one specifies the radius of a Gaussian beam, it is the radius at its waist. So if you specify the position of the waist, you don't need to worry about the curvature. $\endgroup$ Commented May 8, 2018 at 5:42

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