Tai cia jei plemas butu 1 mm diametro. O svieciant i objekta uz keliu simtu metru-kilometro plemas jau centimetrais matuojamas, plius spindulio silpimas einant per ne idealiai skaidria terpe ... Jei tas plemas butu centimetro, tai galia/ploto vienetui kristu 100 kartu ir is 17 W/cm2 jau liktu 170 mW/cm2. Kiek tas diametras butu uz kilometro, pasakyti sunku, bet perzvelgus video, panasu, kad jis tikrai didesnis, nei centimetro skersmens. 

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"ejs" <ejs@no.where> wrote in message news:j36dg6$sbu$1@trimpas.omnitel.net...
> 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