Without accurate information you can't get an accurate answer, makes sense.
Thanks again for all your help, this whole project has been quite a learning experience.
I was just really surprised at how much more intense the UV radiation from the lamp was compared to the sun but that makes sense.
I'm curious about why the conservation of energy method didn't provide at least a comparable number though.
I'm sorry I was completely over analyizing the problem.
I just found a much more detailed data sheet for my lamp straight from the manufacturer. It has the UVA irradiance at 20cm (8in) as 510 mW/cm^2. This number is more than 1000x bigger than the value your source had for the sun...
Would treating my bulb as a collection of point sources and then using inverse square law to find the intensity of each at a point, say the center of my object, be a better way of doing this? I would have the 1.5x12 object lined up length wise and centered under the bulb. Then integrate along...
Ok, I'm having a lot of trouble finding info on how intensity falls off from a line source. I need credible sources to back up all of my numbers and calculations. Do you know of any good sites or books that explain this in detail?
Thanks again.
Shoot, I divided by 18in^2. I got the same number as you now.
But now, how does the intensity on my object change with a reflector behind my light? With out a reflector only 4% (14.25* out of 360* are facing my object) of the light is hitting my object.
The intensity doesn't fall off with radius squared from a cylindrical source? I originally did the calculations like you stated but I thought it was wrong since I didn't take into account the radius squared. If the intensity can be reasonably approximated by spreading the 2.6W over a cylindrical...
I'm working on a project and don't have access to a physics book so I'm asking for help. I need to find the intensity of light at a certain distance over a given rectangular area from a cylindrical light source.
The source is 18in long, 1in in diameter, and outputs 2.6watts. I need the light...