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Deriving W/cm^2 of UV light exposure from lamp power rating

How can I derive UV light exposure as Watts per CM ^ 2 if I only know the power rating of the light source (i.e. flourescent lamp) and the distance of the object to the light source?

For example, If a flourescend UV light source that uses 9 Watts is placed 1 inches away from an EPROM chip, what would be the exposure in W/CM^2?

Many thanks,

ArcaArtem

ASKER

To clarify, power rating means the power that the light source uses. Here is a link to the actua power source:

http://www.lamptech.co.uk/Spec%20Sheets/Philips%20PL9.htm

ASKER CERTIFIED SOLUTION

rumi78

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ozo

It looks like that bulb about 9% efficient in converting input power into light energy,
with most if it in the visible and not uv.
Also, the bulb is longer than 1 inch so the area over which the light would be spread would be larger than a 1 inch sphere

ArcaArtem

ASKER

@rumi78: Cool, thanks for that. Actually, the lamp is nearly 15 inches long and 8 inches wide so the area is more close to a cylinder than a sphere or perhaps an ellipsoid, but yes, I get the idea. I believe calculating a cylinder's surface area with dimensions that is 1 inch greater than the lamp's dimensions would give me a more accurate exposure metric.

@ozo: Sorry my mistake, I should've been more careful with the link. The lamp I have is an UV light with exact specifications with the one on the link except that it's an UV lamp. I'm curios to how you deduced the 9% efficiency though...