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Old 10-04-2015, 04:49 AM #1
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Default Let's talk about divergence (mrad)

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.



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Old 10-04-2015, 05:18 AM #2
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Default Re: Let's talk about divergence (mrad)

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.
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Old 10-04-2015, 05:33 AM #3
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Default Re: Let's talk about divergence (mrad)

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%.
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Old 10-04-2015, 05:38 AM #4
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Default Re: Let's talk about divergence (mrad)

^^ Number cruncher
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Old 10-04-2015, 05:27 PM #5
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Default Re: Let's talk about divergence (mrad)

Quote:
Originally Posted by Alaskan View Post
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%.
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Old 10-04-2015, 07:03 PM #6
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Default Re: Let's talk about divergence (mrad)

...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.

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Old 10-04-2015, 07:41 PM #7
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Default Re: Let's talk about divergence (mrad)

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 mē 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.
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Old 10-04-2015, 07:59 PM #8
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Default Re: Let's talk about divergence (mrad)

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.
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Old 10-05-2015, 12:17 AM #9
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Default Re: Let's talk about divergence (mrad)

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

https://what-if.xkcd.com/13/
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Old 10-05-2015, 12:42 AM #10
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Default Re: Let's talk about divergence (mrad)

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.
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Old 10-05-2015, 01:53 AM #11
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Default Re: Let's talk about divergence (mrad)

Quote:
Originally Posted by CDBEAM777 View Post
MMMmmmm ???? My friend Zippy just could not resist some infantile humor !!!

Hahahahaha....See pic !!
Still only 4mm.

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Old 10-05-2015, 02:05 AM #12
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Default Re: Let's talk about divergence (mrad)

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Still only 4mm.

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Old 10-05-2015, 02:35 AM #13
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Default Re: Let's talk about divergence (mrad)

Safest to go with " Deer in the Headlights " :/
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Old 10-05-2015, 06:29 AM #14
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Default Re: Let's talk about divergence (mrad)

Quote:
Originally Posted by Benm View Post
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)ē=41550 milesē
41550 milesē = 1.08*10^11 mē

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

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.
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Last edited by Cyparagon; 10-05-2015 at 06:34 AM.
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