Reply #30: Hey it's sunday morning after the night before :) But here's a quote:

[FULL_METAL_HAT]

But you would have to have a sufficient power density at that large a distance. A good laser will give you a divergence angle of about a milliradian (probably can do better, but that's typical).
Some basic calculations:
tan(theta) = y / x -> y = x tan(theta)
So the distance multiplied by the tangent of the angle will give us an indication (ballpark) of the beam's size at 5 miles' distance: (5 miles)(5280 ft/mi)(12 in/ft)(tan 0.001) = 316 inches
The power density of a 10-Watt laser at that distance would be (10 W) / (pi * (316 inches) ^ 2 / 4) = 127 microwatts / square inch.
Pretty darn small. Not going to hurt you. The sun dumps about a kilowatt per square meter on earth, so that works out to 657 milliwatts / square inch (4000 times higher), if I'm doing the math correctly. I'm not figuring numerical aperture and all those fun things that you have to take into account for careful calculations, but it should be a decent estimate. And of course, that's a $100,000 laser.
http://64.233.161.104/search?q=cache:MDF018JEu-IJ:www.sarahphrase.com/+laser+watt+blind+pilot&hl=en

If the inverse-square law does not apply to lasers, I would love to stand corrected! :)

All the best GAspnes in the new year!
{B^)
FMH