I like to use my 3W 445nm Sanwu outdoors at golf courses very late at night when no one is around. I aim at the tree line, far away from people or reflective objects several hundred feet away. I love to look at the beam outdoors without eye protection, but of course I never do this indoors or when I'm pointing at something unknown. The intensity of just the beam itself from my 3W blue got me thinking though...at what point does looking at the beam, not the dot, damage your eyes?
Ignoring fog or smoke, we can see the beam of a laser due to Rayleigh scattering off air molecules, and once scattered, the laser light becomes incoherent and subject to the standard inverse square law. That's why the beam becomes invisible when I'm standing orthogonal to it at about 50 feet away. Given that the dot of a green laser is visible in daytime at 1 mW, and that the beam doesn't become visible until at least 1000mW, I imagine that it would take quite a bit of power for the beam to actually damage your eyes.
Also, since the incoherent nature of light scattered off air molecules makes the distance from the beam and the angle of incidence extremely important, let's assume a typical scenario with the laser 1 foot (30 cm) from your eyes, and the angle of viewing at 0 degrees (looking down the beam). To solve this problem, I'm guessing that we'd need to use the density of air, the power density of the laser beam, collision theory for scattering off air molecules, and the inverse square law. I'm probably missing a few things and maybe stated something incorrectly. If anyone has equations for this already, or even a table with damage thresholds (for the beam, not that common hazard table we all know) that would save me the trouble of trying to calculate this for myself. Any ideas?
Ignoring fog or smoke, we can see the beam of a laser due to Rayleigh scattering off air molecules, and once scattered, the laser light becomes incoherent and subject to the standard inverse square law. That's why the beam becomes invisible when I'm standing orthogonal to it at about 50 feet away. Given that the dot of a green laser is visible in daytime at 1 mW, and that the beam doesn't become visible until at least 1000mW, I imagine that it would take quite a bit of power for the beam to actually damage your eyes.
Also, since the incoherent nature of light scattered off air molecules makes the distance from the beam and the angle of incidence extremely important, let's assume a typical scenario with the laser 1 foot (30 cm) from your eyes, and the angle of viewing at 0 degrees (looking down the beam). To solve this problem, I'm guessing that we'd need to use the density of air, the power density of the laser beam, collision theory for scattering off air molecules, and the inverse square law. I'm probably missing a few things and maybe stated something incorrectly. If anyone has equations for this already, or even a table with damage thresholds (for the beam, not that common hazard table we all know) that would save me the trouble of trying to calculate this for myself. Any ideas?