Welcome to Laser Pointer Forums - discuss green laser pointers, blue laser pointers, and all types of lasers

Buy Site Supporter Role (remove some ads) | LPF Donations

Links below open in new window

FrozenGate by Avery

What OD for what mw?

Joined
Nov 12, 2008
Messages
89
Points
0
Is there some sort of chart showing what OD rating you need in goggles to protect from the mw of a laser? So if you have N amount of mw, what Od do you need?

Thanks
 





Since you don't want to research it (as it's all over the safety section), the actual calculation is:

OD = log[sub]10[/sub](E[sub]i[/sub]/E[sub]t[/sub])

Where E[sub]i[/sub] is the incident beam energy (generally in W/cm[sup]2[/sup]) and
where E[sub]t[/sub] is the transmitted beam energy (generally in W/cm[sup]2[/sup])

Because you are dividing E[sub]i[/sub] by E[sub]t[/sub] you have a ratio and thus can use any power rating as long as the incident and the transmitted power are the same units.

Here is an example:

Laser Maximum Output: 200mW
Maximum Transmitted Power: 5mW (keep the exposure at or below Class IIIa levels)

OD = log[sub]10[/sub](200mW/5mW)

OD = log[sub]10[/sub](40)

OD = 1.602059991

Thus if I have a 200mW laser and I want to allow no more than 5mW to pass through the lens I will need protective eyewear rated at least OD 1.60205991 at the laser wavelength.
 
I alway heard divide by 10. Example: 100mW through OD2 would be 1mW. 300mW throughe OD3 would be 0.3mW
 
FrothyChimp said:
Since you don't want to research it (as it's all over the safety section)

Heh, you enabler Frothy! ;)

Seriously though, why don't people research more before asking questions?

I take pride in being able to find my own answers, without having to pose public questions. Posing a public question is kinda like admitting defeat - I lost my personal battle of finding the answers for myself. ;)
 
randomlugia said:
I alway heard divide by 10. Example: 100mW through OD2 would be 1mW. 300mW throughe OD3 would be 0.3mW

Not exactly, at some OD, that does work. But for other values it would not work. Use the equation listed above.
 
You can also calculate how much power will be transmitted through the filter using:

P[sub]transmitted[/sub] =  P[sub]incident[/sub] * 10[sup]-OD[/sup]

Ex:

Your laser outputs 100mW, and you have a filter with OD = 2.5 for that wavelength.  The power of the filter's transmitted light will be:

0.100W * 10[sup]-2.5[/sup] = 0.00032W = 0.32mW

So it isn't quite just divide-by-10.
 





Back
Top