laser level safety googles

I just bought one of those laser levels and am very pleased with it. However I think it makes sense to get some laser safety goggles. The laser is "only" 1mW class II but still I think it makes sense. Anyone know somewhere (in uk) I can get such goggles? Thanks. btw this level is one that can rotate and project vert/horizontal lines. A useful piece of kit imho. 40 quid from Argos (no relation!)
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dave wrote:

It really, really isn't important to get goggles, honest.
I used to work with lasers (in a lab), and I can tell you categorically that 1mW red won't do your eyes any harm at all unless you stare into it for several seconds. In the same way a pencil won't do your eyes any harm unless you poke yourself in the eye with the sharp end.
A quick look into the beam won't do you any harm.
More to the point, at 1mW, any decent laser goggles will result in you not seeing the dot/line except for in very dark situations.
--
Grunff


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dave wrote:

Ive looked down one of those lasers. Its nothing. FAR worse to look at the Sun, or a welding torch. Looking at page 3 of the Sun has made some people go blind.
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Hold on a mo'. Strange you should compare the two because below is an extract from one of the laser FAQ's I've been looking at. This snippet copied without permission but in the hope the author will not object as is a safety issue(!).
==START 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.
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 /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 intensity of the HeNe laser on the retina is 1 mW/(6 x 10-5 mm2) = 16,667 mW/mm2.
So the 1 mW laser has the potential to produce an intensity on the retina 167 times that of direct sunlight! But there are many more factors to consider in determining the real risk of damage. In addition to those noted below, the actual focal point when looking at a laser at close range will not be at the retina so the spot size will most likely be much larger than the diffraction limit of the calculation. Even if the spot from the laser beam is smaller, natural eye movements or movement of the source (e.g., some moron waving a laser pointer) will result in it hitting any given point for a shorter time than the larger spot from the Sun (which usually doesn't move very quickly).
But, at least, perhaps you'll now have a bit more respect for that little HeNe laser or laser pointer!
OTHER:
The real problem behind this is that it is assumed that the power density is the significant factor in the thermal damage mechanism. The ability of the retina to dissipate heat is not dependent on the area covered, but the periphery (circumference) of the exposed area! The blood vessels are in the retina and not the sclera (the surface under the retina) - it is the blood flow that dissipates the heat and so can only act on the *edge* not the middle of the exposed area. In circumference terms, the ratio drops to 7 times. Furthermore because the larger spot is less efficient at dissipating heat, the effective power delivered by the laser beam is only about 2 times greater than that of the spot formed by the sun.
==END
As to the effect of page 3, I dunno how to calculate the effect of that :-)
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I want a refund! The one I bought from Lidl some months ago was a couple of quid dearer than that.
It does what I want of it, laying verticals for paper hanging for example, although I could probably get much better than the claimed +/- 0.5mm/m accuracy by paying 400 for one from my local builder's merchant.
Colin Bignell
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of
Thanks Colin. I KNEW I would not please everyone!
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