MinnesotaStateUniversity
Newbie level 5
I am in the process of building a night vision device. IIT's are quite expensive, so I was hoping to build an IR illuminator. However, instead of using IR throw LED's, I was hoping to use a single IR laser diode. The IR beam would then be sent through some type of lens (like a biconcave), & would be dispersed into a larger area. I've done a bit of research, but I'm still unsure on how to accurately calculate watts being diverted.
Below, is a "theory" on how to "accurately represent" watts being diverted. I believe this was linked from lucidscience off hackaday. I'm a bit skeptical & was hoping to clear some things up. Anyways, here it is:
"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 lens & 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 = Pi * (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 (Pi * (.25")^2) / (Pi 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?"
^Math's correct, but isn't there a flaw behind the logic. From my understanding, when you use use a lens to disperse a laser, it is not the size of the lens, but rather the index (material), lenticular power (diopters), and the focal distance that gives you the dispersion equivalent. Therefore, in order to to calculate the dispersion factor, one would use a downstream vergence equation. However, this leads me to another question. It will not calculate watts being diverted. It calculates the dioptric equivalent. Instead of using Diopter, could I substitute with watts? So lets say I have 1W hitting a diverging lens, couldn't I determine that 16mW is at a point "x"mm behind the lens?
One last thing. Pupil aperture. A pupil is X distance from your retina. When a laser of 1W hits your retina, that's when one would go blind. I understand if I am going to disperse a beam, this aperture will come into play. This will block the already previously determined X factor depending on how fast the lens disperses the laser (dioptric power of the lens).
The character who wrote the theory used a diode out of a cd burner. I don't have the specs off hand, but I believe IR diodes out of burners rate around 250mW @ 900nm. Most high powered (150mW+) IR diodes have a fairly significant fast axiz, so the ouput will square. It looks like "area of a cirlce" is even more irrelevant.
One quick note: Assume I were to use a diode exceeding 150mW (multimode). How difficult would it be to utilize a diffusing filter, & eliminate all hot spots present throughout the nature of the beam profile? I know this is far from practical, but one way to get very even power distribution across the beam profile while at the same time circularizing the beam is to couple the output into a short piece of single-mode fiber and then use a lens to expand the fiber output. It took the guy 3 months to set up. Of course, he was dealing with a class 4a 1W laser, but is there an easier way, perhaps w/ a diode ranging from 150-250mW?
Below, is a "theory" on how to "accurately represent" watts being diverted. I believe this was linked from lucidscience off hackaday. I'm a bit skeptical & was hoping to clear some things up. Anyways, here it is:
"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 lens & 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 = Pi * (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 (Pi * (.25")^2) / (Pi 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?"
^Math's correct, but isn't there a flaw behind the logic. From my understanding, when you use use a lens to disperse a laser, it is not the size of the lens, but rather the index (material), lenticular power (diopters), and the focal distance that gives you the dispersion equivalent. Therefore, in order to to calculate the dispersion factor, one would use a downstream vergence equation. However, this leads me to another question. It will not calculate watts being diverted. It calculates the dioptric equivalent. Instead of using Diopter, could I substitute with watts? So lets say I have 1W hitting a diverging lens, couldn't I determine that 16mW is at a point "x"mm behind the lens?
One last thing. Pupil aperture. A pupil is X distance from your retina. When a laser of 1W hits your retina, that's when one would go blind. I understand if I am going to disperse a beam, this aperture will come into play. This will block the already previously determined X factor depending on how fast the lens disperses the laser (dioptric power of the lens).
The character who wrote the theory used a diode out of a cd burner. I don't have the specs off hand, but I believe IR diodes out of burners rate around 250mW @ 900nm. Most high powered (150mW+) IR diodes have a fairly significant fast axiz, so the ouput will square. It looks like "area of a cirlce" is even more irrelevant.
One quick note: Assume I were to use a diode exceeding 150mW (multimode). How difficult would it be to utilize a diffusing filter, & eliminate all hot spots present throughout the nature of the beam profile? I know this is far from practical, but one way to get very even power distribution across the beam profile while at the same time circularizing the beam is to couple the output into a short piece of single-mode fiber and then use a lens to expand the fiber output. It took the guy 3 months to set up. Of course, he was dealing with a class 4a 1W laser, but is there an easier way, perhaps w/ a diode ranging from 150-250mW?
