Grix
0
- Joined
- Dec 9, 2008
- Messages
- 2,190
- Points
- 63
You are right, the eye (pupil) is 7mm wide at night and 4mm wide during the day.
So a 100mw laser with a divergence of 1.5mrads =
800 ft = 243 Meters
divergence of 1.5mm per Meter * 243 Meters = Diameter of:
1.5mm + 1.5*243 = 367 mm diameter of the dot at 800 ft.
Area of dot in Meters = Pie * .367 ^2 = .423 square meters. I am not sure about your math, I think that this is the correct formula/answer for the area at 800ft.
The area of a fully dilated pupil is Pie * .007 M ^2 = .000154 M^2
so the percentage of 100mw entering each individual eye is .000154 / .423 = .036 PERCENT
.00036 * 100mw = .036 mw
That means that .036 mw is entering each eye at 800ft, through the cockpit, at a moving target. A worse case scenario of .05 seconds exposure is being severely generous and I would think overestimates the ability of a handheld laser to be kept on target at that distance. Its a lot worse then what Grix said (at .0007 mw,) but that sounds an awful lot like a night light to me.
I would consider this mathmatical proof that there is little to no danger from a theoretical $45 laser at the stated distances. It would most likely show up as a flashlight (.423 m^2 area) on the outside of the aircraft. Get a little closer like take off and landing and you have a totally different story. I think thats where the true teeth of the law belongs. Only someone with malicious intent (terrorist) would point a laser at a pilot focusing on the runway.
Whops, somehow I thought the beam diameter would be 3,6m after 800 feet, not 0,36m. Thanks for correcting me.
But yeah, the article is two years old, and two years ago a 100mW laser did not cost £30. It was probably a 5-30 mW laser.