Trevor
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- Jul 17, 2009
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I remember reading here that you shouldn't look at a dot from a laser more powerful than 30mW. I was interested in seeing why this is, so I did some conservative math, and it didn't seem to work out. Here's my thought process:
Say we shine a laser pointer of 200mW at a perfect 90° angle at some sort of (flat) miracle matte surface that disperses the light evenly in all directions. The area of reflection would be a hemisphere whose radius is the observer's distance from the dot. The formula would be (4πr²)/2, or just 2πr².
Say we are observing at night from a distance of 2m. The area of the surface of the hemisphere whose surface touches the observer's eye(s) would be 25.132m². To make this number more manageable, let's turn it into cm². This yields 251,327cm². The intensity of the dispersed laser light would be 200/251327 mW/cm², or 0.000795mW/cm².
Even if the retina is as large as 1cm², the amount of light reaching the retina seems like it would be much less. It would take an error of nearly 1,000,000% before the power goes above the generally accepted "safe"(ish) 5mW.
My question to you is - what have I missed? Is a whole lot more light reflected back to the person holding the pointer, since no surface is perfect like the one I described? Have I missed an important physics concept in my crude "back-of-the-envelope" calculations? Is my error approaching 1,000,000%?
I figured I'd ask, since an entire forum of laser enthusiasts is very rarely wrong...
Trevor
Say we shine a laser pointer of 200mW at a perfect 90° angle at some sort of (flat) miracle matte surface that disperses the light evenly in all directions. The area of reflection would be a hemisphere whose radius is the observer's distance from the dot. The formula would be (4πr²)/2, or just 2πr².
Say we are observing at night from a distance of 2m. The area of the surface of the hemisphere whose surface touches the observer's eye(s) would be 25.132m². To make this number more manageable, let's turn it into cm². This yields 251,327cm². The intensity of the dispersed laser light would be 200/251327 mW/cm², or 0.000795mW/cm².
Even if the retina is as large as 1cm², the amount of light reaching the retina seems like it would be much less. It would take an error of nearly 1,000,000% before the power goes above the generally accepted "safe"(ish) 5mW.
My question to you is - what have I missed? Is a whole lot more light reflected back to the person holding the pointer, since no surface is perfect like the one I described? Have I missed an important physics concept in my crude "back-of-the-envelope" calculations? Is my error approaching 1,000,000%?
I figured I'd ask, since an entire forum of laser enthusiasts is very rarely wrong...
Trevor
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