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ArcticMyst Security by Avery

Math for Laser Incident? (And a warning to others for eye safety)

Joined
Oct 19, 2010
Messages
4
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So, I'm an idiot and managed to do something stupid, but read on before you facepalm. I planned on using my 1W 445 for a short outdoor sky/distant object pointing session. The laser did not have a battery in it at first, so I went back inside to grab one. I'm very experienced with this pointer (I've owned it for 4+ years now without incident), know my way around eye safety, and even had goggles on hand out of habit... In my pocket.

Went about changing the battery outside, and somehow had managed to click my rear button without feeling it whilst I was changing (Yes, the only switch on my pointer. Perhaps many will frown upon this.). I was not fully unaware of the possibility of incident, so the pointer was pointed away from me, downward. Looked away as I put the battery in, but had the device in my peripheral vision in order to thread the rear cap back on. Of course, the device lights up when I make contact, and I manage to catch a glimpse, in my far peripheral vision of a diffuse reflection (goggles weren't on at the time) off of what ended up being my knee or some part of my lower leg (The dot was by no means focused to such a distance, but it certainly wasn't a large defocused splotch either). I immediately withdrew contact and was so turned off by the incident that I packed all my laser stuff away and scurried over to these forums for advice, and grabbed a calculator. I don't believe any damage has been done (The spot artifact left in my eye faded in less than 10 seconds, and I'm reading/typing this using only the eye in question). However, I am curious about how to correctly model a situation such as this.

The fairly conservative method I chose for a quick calculation was a point source of full power (~1 Watt) emitting in a hemispherical fashion. Diffuse reflection was caught at approximately 3 feet, and assumed full pupil dilation of 9mm. So I came out with ((1[Watt])/(2*pi*(0.9144[m])^2))=1.9E-7[W/mm^2]. Multiplied by the area of my entire pupil I get (pi/4)*(9[mm])^2=0.012[mW]. I would think this crude model would certainly over-approximate the power incident on my pupil, however I haven't done anything to account for lens focusing on the retina, yet. Any rule of thumb suggestions (I've seen them around before, I think) to add a "fudge factor" to this?

Anyway, this may as well serve as a warning to everyone else that accidents do happen, even when you're very familiar with your equipment. In my case, I should have went about putting in the battery inside with my goggles on, and the laser completely defocused.
 
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Joined
Mar 17, 2015
Messages
824
Points
63
I'm glad to read that you have no seeing issues at present. But my main question is how have you only posted 4 times in 5 years?
 




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