Laser pointers manufacturers claim that some pointers have range of several kilometers.

Okay, they use a powerful laser, but that powerful laser usually has power less than one watt. Okay, the laser beam is very focused.

But what about atmosphere particles like dust and vapor? Why don't these particles diffuse the beam and make it lose energy?

How does it happen that a less than one watt laser can have several kilometers range in atmosphere?

  • $\begingroup$ I don't think that the test is carried in the atmosphere. I suspect it is carried in a clean and dry room (possibly, under vacuum) and use mirrors. This is much like the tests for batteries which are carried under ideal conditions (temperature, humidity etc.) $\endgroup$ – Yotam Aug 4 '11 at 9:10
  • 3
    $\begingroup$ This is more about luminance than simply power. Your question could be rephrased to "why can I see a candle from several kilometers away in the night". The radial distribution of the light "eats" a lot more than clear air absorbs but still enough light reaches your eye to be made out (take a look at the stars, their power is enormous but your candle is brighter: it is not about power alone). Enter absorbers like fog/clouds and you won't see the candle/star/laser/sun ... or mountain/moon/landscape anymore. $\endgroup$ – Leonidas Aug 4 '11 at 13:27
  • $\begingroup$ Sure the intervening atmosphere saps energy of the beam and spreads it out, but what do you mean by "range"? The distance at which its holder can see its spot? or the distance at which someone at the other end can see it? $\endgroup$ – Mike Dunlavey Jul 3 '12 at 18:49
  • $\begingroup$ @Mike Dunlavey: Since it's a pointer I guess it's the distance at which the holder can see the spot. $\endgroup$ – sharptooth Jul 4 '12 at 6:18

A laser pointer's energy and lights are concentrated in a very small light cone to reach a quite high intensity. So its labelled Wattage is much smaller than a usual bulb while its light is very strong at one point. And the labled power on laser pointers is the output power. The typical input power of a 50mW pointer is about 0.5 W. This make the difference looks bigger.

Actually common laser pointers used by teachers and lecturers are less than 5mW. The farthest distance to see their lights is about dozens of meters. But the laser pointers used by professional and amateur astronomers are much more powerful. You can see the light path caused by Tyndall effect directly as below image(source, though this impressive light path is not generated by a handheld pointer, the scence is similar.)

At Mount Hopkins in Arizona, a bundle of five lasers is shot into the atmosphere to improve the imaging of the 6.3-meter MMT telescope

The energy reduction is mainly caused by Rayleigh scattering, which is relatively small compared with laser's intensity. So the beam can easily reach several miles away with the power higher than 50 mW.

Update: Georg insist the previous image can not be used to discribe laser pointers. So I add another real pointer picture here. But I have no feeling about distance with this one.

enter image description here

  • $\begingroup$ ""The Wattage does not refer to the the amount of light. It only indicates the amount of electrical energy consumed by the device."" This is rubbish! Your idea applies for light bulbs or a electric heater or the motor of Your lawn mover. For lasers the "wattage" is the power of the light coming out! -1 $\endgroup$ – Georg Aug 6 '11 at 9:24
  • $\begingroup$ Yep, I made a mistake here. The labled power on laser pointers is output power. The deviation between input power and output value is conversion efficiency. The typical input power of a 50mW pointer is about 0.5 W. But This is not the point. $\endgroup$ – gerry Aug 6 '11 at 12:40
  • $\begingroup$ Laser pointers are those handheld things used to point. The thing in Your picture is a artificial star to control adaptive optics, which has nothing to do with a pointer! Range of pointers is limited by chep lenses, thats all. $\endgroup$ – Georg Aug 6 '11 at 13:02
  • $\begingroup$ Those laser pointers (there are 5 actually) are used to play as artificial stars. They are more powerful (5W each) than handheld device, but ARE still pointers and have a range of several kilometers. Why do you think they are different ? $\endgroup$ – gerry Aug 6 '11 at 13:24
  • $\begingroup$ Because they are not used to point! Read "laser pointer" in wikipedia! $\endgroup$ – Georg Aug 6 '11 at 13:26

What about the elephant in the room, called coherence?

Laser light and ordinary light differ in the amount of coherence of the beam. Incoherent beams lose intensity as 1/r**2, where r is the distance from the source. Coherent beams in vacuum disperse slowly according to optical equations. In the atmosphere there will be absorption and scattering, as discussed in other answers.

Have a look at http://en.wikipedia.org/wiki/Lunar_Laser_Ranging_experiment to see laser light reflected from the moon, as far as distance travelled goes.

  • $\begingroup$ As far as I know that's not correct, only beams that illuminate the whole unit sphere ($4\pi$ steradian) lose 'intensity' with $1/r^2$. The difference is in the divergence of the beams, the more convergent the beams are, the less they lose 'intensity' ('intensity' is a bit tricky here because there are various units that can be used). $\endgroup$ – Tim Aug 8 '11 at 7:30
  • $\begingroup$ @Tim en.wikipedia.org/wiki/Inverse-square_law . You are mistaken, if a beam is incoherent it follows the law. $\endgroup$ – anna v Aug 8 '11 at 9:41
  • $\begingroup$ You're right, I had some things mixed up. But I fail to see why this does not apply to coherent light. Even laser beams diverge (slightly) and thus decay with $1/r^2$. $\endgroup$ – Tim Aug 8 '11 at 10:18
  • $\begingroup$ Laser beams diverge according to the plan of the optics,as in the link I gave in my answer, plus a bit of scattering and absorption will reduce the intensity a bit more. Note the first formula for intensity has a negative exponential for the fall along the axis. $\endgroup$ – anna v Aug 8 '11 at 13:06
  • $\begingroup$ @annav Thanks for the flag, but I haven't found any traces of voting abuse, and this is not mods role to judge the meritoric aspects of users' voting decisions. $\endgroup$ – user68 Aug 8 '11 at 19:26

It is most likely the same reason as why the sky is blue. More specifically, Rayleigh scattering is proportional to the fourth inverse power of the wavelength, so if you choose the wavelength of the laser to be large (i.e. towards the red end of the visible spectrum), then you can suppress Rayleigh scattering significantly. In that case the range of the beam can be very large.


Upshot: the range of a laser is greater because the light is concentrated in a very narrow beam.

Without going into much detail of the power of the laser, I think the problem here has to do with the definition of 'intensity'. Wikipedia lists several different units for intensity, the one that most closely matches 'brightness' as perceived by the eye is probably radiance.

The units of radiance are

$$ L = W·sr^{−1}·m^{−2} $$

or watt per steradian per square metre, a 'steradian' being a two-dimensional angle on the sky.

Now given a fixed power for a certain light source, the difference between ordinary lighting and a laser are two-fold:

  1. The surface ($m^{2}$) of a laser is much smaller (probably a factor of 10)
  2. The divergence ($sr$) of the laser beam is also much smaller (factor ~100 or more).

Both these factors mean that the radiance for the same power ($W$) a laser beam is at least 1000 times as 'bright' as a regular light source. Since the brightness is higher, the beam will be visible up to a longer distance.

It is important to note that this result is regardless of the absorbance or scattering that occurs, which in principle happens for both laser and non-laser beams.

  • $\begingroup$ Aha, what about that in vacuum of space? The question is here about "reach" in the atmosphere (what ever that means) $\endgroup$ – Georg Aug 8 '11 at 8:04
  • $\begingroup$ In a vacuum (which space is not), light travels more or less uninterruptedly. But because of non-zero divergence of any beam, the 'brightness' will slowly go to zero. $\endgroup$ – Tim Aug 8 '11 at 8:08
  • 1
    $\begingroup$ Please people, learn about the difference of lasers to other sources. It was a revolution at the time it appeared. en.wikipedia.org/wiki/Laser $\endgroup$ – anna v Aug 8 '11 at 9:44
  • $\begingroup$ Anna I'm not sure why you think these answers are wrong. Beyond the rayleigh range (on the order of 10 meters for a pointer) a gaussian laser beam still falls in intensity as 1/r^2, the only difference is the smaller divergence angle. $\endgroup$ – user2963 Oct 19 '11 at 20:56

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.