# Inverse Square Law: 1mW laser range?

#### Saibottomus

Hello PF! I've got a strange question for you physics boffins; assuming for a moment that lasers obeyed the inverse square law, what range would a typical 1mW red laser have in the atmosphere?

Related Other Physics Topics News on Phys.org

#### cepheid

Staff Emeritus
Gold Member
Welcome to PF,

Only things that emit isotropically (i.e. the same in all directions) obey the inverse square law. Also, it's unclear what you mean by "what range" would the laser have? Do you mean how far away could you be from the source of the laser and still detect emission from it? You'd have to specify what the sensitivity limit of your detector was, and it would also depend on what the background noise level was.

#### Matterwave

Gold Member
To maybe clarify on cepheid's point a little more, one of the main properties of a laser (as we are all familiar with them) is that it DOESN'T emit light in all directions equally, but emits its light in a specific direction, and spreading out very little.

#### Antiphon

Laser emissions obey the inverse square law. It's just that the effective origin of the source is located very far behind the beam aperture of the laser.

The ideal plane wave or cylindrical wave is a mathematical entity. Once the observer is far enough from any source, the inverse square law applies.

#### Saibottomus

Sorry my question was poorly worded. To clarify a few things and reword the question:
Assume the brightness of the laser pointer (1mW red laser equivalent at the source), falls off with the square of the distance. At approximately what distance would the laser not be detectable if shone directly into the observer's eye.

(Sorry, I know it's a bit of an obscure question).

#### sophiecentaur

Gold Member
Welcome to PF,

Only things that emit isotropically (i.e. the same in all directions) obey the inverse square law. Also, it's unclear what you mean by "what range" would the laser have? Do you mean how far away could you be from the source of the laser and still detect emission from it? You'd have to specify what the sensitivity limit of your detector was, and it would also depend on what the background noise level was.
Actually, it not the "isotropic" bit that counts. It's whether or not they behave as a point source in the same location that counts. A directional radio antenna with a phase centre at its physical centre or even a lamp in a black sphere with a hole in it will produce a flux that follows the inverse square law - in their main beam.
A Laser has a phase centre that could be looked at as a virtual image, formed by the multiple reflections between the mirrors - i.e. a long way behind the Laser itself. Once focussed at a point, that point acts as a phase centre from which the emerging light will follow the ISL. You will always get some spreading (due to diffraction) and this will give you inverse square illumination - eventually, as the approximation gets closer and closer.

#### mfb

Mentor
At approximately what distance would the laser not be detectable if shone directly into the observer's eye.
Depends on the laser design and the eye.

In this distance, the laser diameter is not resolvable for the eye, so basically all radiation comes from a single direction. For the visibility limit I can find more numbers than I want to, but something like ~1000 photons/s seems realistic if the observer is in perfect darkness. As comparison, stars of apparent magnitude 6 give ~2000 photons/s.
The eye can see photons which arrive in a disk of ~3mm diameter.

1mW of ~700nm corresponds to ~4*10^15 photons/s.
Without absorption, we can distribute this over 2*10^12 times the area of the eye, this is a disk of ~4km diameter. As the ratio "distance/spot diameter" is something like ~1cm/(10m) or better, this would correspond to a laser pointer distance of ~4000km, not possible with a direct line of sight on earth. Without absorption and perfect adaption to darkness, you could see the laser pointer everywhere on earth within the line of sight.

What about absorption? Some graphs indicate that our atmosphere absorbs ~20% of red light in vertical direction, which roughly corresponds to 10km of air at sea level. This gives ~0.8^10 = 10% transmission after 100km and 1% after 1000km. We can combine this with the spread of the laser pointer and get a visibility range of >500km.

Keep in mind that this requires an eye which has enough time to adapt in perfect darkness, something you will not get outside of buildings.

With daylight, things are different - but at least you can see that you can ignore absorption there.

#### Saibottomus

~1000 photons/s seems realistic
[...]
1mW of ~700nm corresponds to ~4*10^15 photons/s.
[...]
>500km.
Thanks for your reply mfb. So if (hypothetically) the "brightness" (somehow) fell off with the square of the distance of the observer from the source, what would that reduce the range to?

#### mfb

Mentor
So if (hypothetically) the "brightness" (somehow) fell off with the square of the distance of the observer from the source
It does so for large distances, and I already took this into account with the assumption that the spread is ~1cm over 10m distance.

With unphysical (!) perfect parallel waves, you would not get any relevant limit at all and even a laser pointer on the moon could be dangerous (if it happens to hit your eye).

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving