# Light focusing on long distances

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? UPDATE 1

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.

UPDATE2

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.

• Laser beam cross section is characterized by a Gaussian intensity and behaves somewhat different than ray optics. Google laser waist for additional information. – npojo Apr 19 '18 at 8:22
• I know about Gaussian beams. I will update the question to make it more clear. – Zlelik Apr 19 '18 at 11:28

## 2 Answers

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.

• 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? – Zlelik Apr 19 '18 at 11:27
• 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. – MJC Apr 19 '18 at 11:32
• 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. – Zlelik Apr 19 '18 at 11:37
• 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. – MJC Apr 19 '18 at 14:52
• Send me link please. I could find anything in wikipedia article about Difraction limit for example. – Zlelik Apr 23 '18 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.

• 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/…) – Zlelik Apr 20 '18 at 7:29
• No, you should not need to use ray tracing. Why do you think the focal point it is too small? – flippiefanus May 6 '18 at 10:43
• 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. – Zlelik May 7 '18 at 20:13
• 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. – flippiefanus May 8 '18 at 5:42