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03-16-2012, 01:42 AM #1
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BrittanyGulden
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Some Of You May Remember Me. I am back W/ more information! DIY Night Vision

Some of you may recall the theory that I provided in my previous DIY night vision posts (below):

*Assume I have a laser diode optical output of 1w, & a pupil size of .25" in my eye (@ Night). If I shined that 1W Laser directly into my eye, I will be blind instantly. However, what If I put that 1w through a lense & disperse it out into a circular area. I should only need to have a large enough circle so that the fraction that would go into my eyes is a safe amount, correct? Area of a Circle = Pie * (R)^2, So I can just figure the fraction of the pupil's area to the larger circle's area. I only need an "Eye-Safe Level," which I'll call 1mW. So I have (Diode Power) x (Pie * (.25")^2) / (Pie x r^2) which equals 1mW. Hence, 1000 x (1/16) / (r^2) = 1. Radius = about 8" or 16" in diameter. This means that If I have 1 watt of light, going through a lense , & spreading out into a 16" circle of light, and that shines directly into my Eye, my eye will see 1 milliwatt correct?

^ I have some more info on this "theory." Special thanks to Gene Clark from New Ulm for helpin' me out I am half way through writting my response to Gene. I'm curious on input you guys may have. Anyways, when I emailed Gene Clark this "theory," this is the response I was given:

*Tyler,

I received your question from Ann. WOW. Tough question and I don't know if I can give you the right answer, but I do want to clarify something's that may help.

1. When use use a lens to disperse your laser it is not the size of the lens but the index(material), lenticular power(diopters), and the focal distance(in meters) that gives you the dispersion equivalent. Have you ever burnt ants with a magnifying glass or started a fire with one? As light enters the magnifying glass it converges light to a point (focal point). This is how the eyeball works assuming the eye is emmotropic. If you hold the magnifier to far or to close is does not get hot enough because the light is being dispersed differently before and after its focal point. This is true for a large magnifier or small magnifier you just have to hold each magnifier at different focal points in order to burn the ants.

Thus to calculate the dispersion factor you could use a down stream vergence equation. The problem is that it will not calculate the watts as it is diverted. It calculates the dioptric equivalent. But this may still work. Instead of using Diopter you would maybe able to substitute with watts. So lets say you have 1W hitting a diverging lens (I would recommend biconcave lens (which I have many)) you might be able to determine that 16mW is at a point "x"mm behind the lens.

2. Now the hard part. Your pupil is an aperture of 0.25" (strongly recommend you use metric in your equations). Your pupil is x distance from your retina. When your laser of 1W hits your retina that's when you go blind. Thus if you are going to disperse the laser this aperture will come into play. This will block the already previously determine X factor depending on how fast the lens disperses the laser (dioptric power of the lens). This concept is difficult to explain with words it really is best explained with a ray diagram.

Good luck.

^Sorry for the lengthy response. I am finishing up my reply, but curious to see what your guys input is as none of this "new info" was brought up in my previous threads.

Thanks

Last edited by BrittanyGulden; 03-16-2012 at 01:43 AM.

03-16-2012, 03:43 AM #2
 Class 3R Laser Join Date: Dec 2011 Location: South Cackalackey, USA Posts: 2,065 Rep Power: 140
tsteele93
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Rep Power: 140
Re: Some Of You May Remember Me. I am back W/ more information! DIY Night Vision

Quote:
 Originally Posted by BrittanyGulden Some of you may recall the theory that I provided in my previous DIY night vision posts (below): *Assume I have a laser diode optical output of 1w, & a pupil size of .25" in my eye (@ Night). If I shined that 1W Laser directly into my eye, I will be blind instantly. However, what If I put that 1w through a lense & disperse it out into a circular area. I should only need to have a large enough circle so that the fraction that would go into my eyes is a safe amount, correct? Area of a Circle = Pie * (R)^2, So I can just figure the fraction of the pupil's area to the larger circle's area. I only need an "Eye-Safe Level," which I'll call 1mW. So I have (Diode Power) x (Pie * (.25")^2) / (Pie x r^2) which equals 1mW. Hence, 1000 x (1/16) / (r^2) = 1. Radius = about 8" or 16" in diameter. This means that If I have 1 watt of light, going through a lense , & spreading out into a 16" circle of light, and that shines directly into my Eye, my eye will see 1 milliwatt correct?
Um, ok. But you go first!

Quote:
 ^ I have some more info on this "theory." Special thanks to Gene Clark from New Ulm for helpin' me out I am half way through writting my response to Gene. I'm curious on input you guys may have. Anyways, when I emailed Gene Clark this "theory," this is the response I was given:
Who is Gene Clark? Is he the lead singer for KISS?

Quote:
 2. Now the hard part. Your pupil is an aperture of 0.25" (strongly recommend you use metric in your equations). Your pupil is x distance from your retina. When your laser of 1W hits your retina that's when you go blind. Thus if you are going to disperse the laser this aperture will come into play. This will block the already previously determine X factor depending on how fast the lens disperses the laser (dioptric power of the lens). This concept is difficult to explain with words it really is best explained with a ray diagram.
Aim high on that diameter. At night your pupil may be larger than 1/4 inch. Give yourself some room for error.

Good luck.
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