2011.08.25 19:20, Dainiushas rašė:
> o kiek yra jau pavojinga?

išvadas darykit patys:

A 100 W light bulb puts out about 5 to 7 W of visible light and another 
35 to 40 W in the near-IR which is also relevant since it passes through 
glass, water, and the anterior structures of the eye can be focused on 
the retina. The rest is mid to far-IR and heat with a small amount of UV 
tossed in. All of this radiation is more or less uniformly distributed 
in every direction. However, at any reasonable distance from the light 
bulb, the power density (e.g., W/mm2) entering the eye is much lower 
than for a collimated laser beam of even very low power. And, it takes 
significant effort to produce any sort of truly collimated beam from 
such a non-point source such as is present with even the filament of a 
clear light bulb. For a frosted light bulb, insert another factor of a 
thousand or so. :) Without collimation, even the portion of that 
additional 35 to 40 W of near-IR that enters the eye isn't going to 
cause damage. However, for a helium-neon laser, the collimation is such 
that the entire beam (total power output of the laser) will still be 
small enough to enter the eye even at a distance of several meters.

For example, at 10 cm from a 100 W bulb (which would be a very 
uncomfortable place to be just due to the heat), the power density of 
the visible light (assuming 5 total watts) would be only about 0.05 
mW/mm2. At 1 m, it would be only 0.0005 mW/mm2 or 500 mW/m2. Based on 
this back-of-the-envelope calculation, a 5 mW laser beam spread out to a 
circular spot of 0.1 m diameter (i.e., 1 mR divergence at a distance of 
100 m - without external optics) will appear brighter than the 100 W 
light bulb at 1 m! And, close to the laser itself, that beam may be only 
1 *mm* in diameter and thus 10,000 times more intense! (And note that 
the other invisible radiation that passes through to the back of the eye 
is still not nearly as dangerous as the beam from the 1 mW laser because 
it isn't focused to a tiny spot by the lens.)

Intensity of a 1 mW Laser versus the Sun

Here is a comparison between the maximum intensity on the retina of the 
Sun and the beam from a 1 mW HeNe laser. (Adapted from one of Simon 
Waldman's optics lectures.)

Standard Sun:
     Maximum intensity of sunlight at ground level (directly overhead, 
no smog, etc.) = 1 kW/m2 or 1 mW/mm2.
     Assuming pupil diameter is 2 mm (i.e., radius of 1 mm), the area is 
approximately 3 mm2. So, the power of the sunlight through the pupil = 3 mW.
     Focal length of eye's lens = approximately 22 mm. Angular size of 
Sun from Earth = 0.5 degree = 9 mR. Thus, diameter of image formed = 22 
mm x 9 mR = 0.2 mm and the area of image = 0.03 mm2.
     The intensity of the Sun on the retina (Power/Area) = 3 mW/0.03 mm2 
= 100 mW/mm2.

Typical 1 mW HeNe laser (or laser pointer):
     Power (P) = 1 mW, wavelength (l) = 633 nm, radius of beam (w) = 1 
mm, focal length of eye (f) = 22 mm. So, the diameter of spot = (2 x f x 
l)/(w x pi) = 9 x 10-3 mm and the area of spot = 6 x 10-5 mm2.
     The power density of the HeNe laser on the retina is 1 mW/(6 x 10-5 
mm2) = 16,667 mW/mm2 = 16.667 watts/mm2.

http://www.repairfaq.org/sam/lasersaf.htm#safyor0

-- 
ejs