Let's talk about divergence (mrad)

[Tre]

I don't know how to say all that I want to but I will try, please correct me if I am wrong. I did not understand the measurement mrad before so I think this will be good for people to understand what it means
1 mrad (milliradian) = 0.057296°
Assuming the starting diameter is negligible and the far field is near the aperture A one mrad beam will have a diameter of 1 meter at 1 km, 2mrad = 2m @ 1km and so on. Typically lower divergence is desired except in special uses such as signalling devices Rescue Laser Flare Magnum.
Anything else to add?
Just for fun- if you pointed a 1 mrad laser at the moon the 'dot' would be as big as the distance between Bakersfield and San Francisco, California or about 1̶/̶6̶ (1/10) the diameter of the moon.

 

[Laser Chick]

I always look at it in the smaller scale so that beam divergence measures how much the laser beam expands per meter. For example, a laser with a beam divergence of 1.0mRad will have a beam that expands 1.0mm per meter in distance.

 

[Laser Chick]

^^ Number cruncher

 

[CDBEAM777]

At closest, the moon is 225,000 miles away, at furthest about 252,000 miles. When at 225,000 miles a 1 mRad 4mm wide beam leaving a laser pointer theoretically produces a 225 mile wide spot (of course atmosphere would interfere some and the spot would never be visible from a pointer) at 252,000 miles a 252 mile wide spot. The diameter of the moon is roughly 2150 miles wide. A 1 mRad laser pointed at the moon at perigee would be about 9.5% its diameter, at apogee 8.5%.

MMMmmmm ???? My friend Zippy just could not resist some infantile humor !!!

Hahahahaha....See pic !!

 

[Alaskan]

...at its closest approach, the spot size of a 1 mRad laser pointer that far out would be about 1.6 million miles and would take a little under two and a half hours to get there.

 

[Tre]

Yeah I know that the dot would not be visible, If you really want to do some math you could probably calculate the average number of photons striking the moon every second per m² and then figure out the chance of a single photon reflecting back into your eye in a certain amount of time at various power levels and excluding the effects of the atmosphere of course... you know what nevermind that.
Laser Pointer

 

[Alaskan]

Actually, I was wondering about that myself, but I'm not going to dig that deep with my basic math abilities! Even with a calculator in hand.

 

[Cyparagon]

XKCD did an interesting write-up on this subject a while back.

https://what-if.xkcd.com/13/

 

[Benm]

It depends a bit on how you calculate divergence. At long distance 1 mrad is either a 'spot' with a diameter or radius of 1m per km distance. This is fairly important since there is a factor of 4 in illumnination per m2 between them.

Obviously you would not be able to see the reflection of a portable laser on the moon, not even when its a new moon and there is no sunlight reflecting off it that would overwhelm the bit of light from the laser.

Another thing to consider is: could you see a laser fired from earth if you were standing on the moon. I think this might be possible if the laser was operated from a dark spot (remote island in the middle of the great ocean that has few other light sources do drown it out).

At a much shorter distance this has been tested: A 1 watt blue laser is visible from the ISS. While only 400 km or so away, it was clearly the brightest light source coming from a small town and easily visible to the naked eye.

Sadly we have no eyes in space beyond the ISS as the moment, but an attempt could be done using space telescopes. Obviously Hubble could be used for this, but there is also the no longer funded but still operational canadian MOST satellite that orbits at 800 km or so.

Neither was built to take images of earth, but they could be directed to do so if desired.

The hard part of such an experiment would probably to point the laser the right direction - MOST is especially tiny and can probably not be seen from the ground even using telescopes, so you're aiming at coordinates, not something that gives a visible reflection when you manage to hit it.

 

[Pi R Squared]

MMMmmmm ???? My friend Zippy just could not resist some infantile humor !!!

Hahahahaha....See pic !!

Still only 4mm.:crackup:

Alan

 

[CDBEAM777]

Still only 4mm.:crackup:

Alan

WOW....Zippy thanx you !! Being a Man of Science himself !!! Albiet he is really such a Doltz !!!

The rest of LPF is either too shocked to answer....thinks is just tooooo stupid to reply to....or...just " Dear in the Headlights " response !!:crackup::crackup::crackup::crackup::crackup:

 

[Laser Chick]

Safest to go with " Deer in the Headlights " :/

 

[Cyparagon]

Another thing to consider is: could you see a laser fired from earth if you were standing on the moon. I think this might be possible if the laser was operated from a dark spot

CHALLENGE ACCEPTED. Physics time!

Given 1mRad, the spot size is roughly 230 miles.
230 mile diameter is:
pi(230/2)²=41550 miles²
41550 miles² = 1.08*10^11 m²

We'll assume maximum possible luminous efficacy of 683 lm/w.
683lm/1.08*10^11 m² = 6.3* 10 ^ -9 lux per watt (lux is the same as cd/m²)

According to the studies done on absolute threshold, you need at least 10^-6 lux even under perfect conditions for the human eye to detect the light source.

6.3* 10 ^ -9 lux per watt * X watts = 10^-6 lux,
X = 160 watts.



Conclusion: ONLY if the entire face of the earth you see was in total darkness, and ONLY if the surface of the moon you were standing on was in complete darkness (does this ever even happen? Not even during a lunar eclipse I don't think), and assuming no atmospheric interference, the man on earth would need to be pointing a 160 watt green laser right at your face, and you MIGHT see it.