# Cylindrical Light Intensity

1. Jun 16, 2010

### jakeman7676

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 intensity at 6in away over 18in^2.

All I could find in a quick search online is intensity from a point source instead of my cylindrical source. I'm not sure how this changes everything.

Thanks for any help,
Jake

2. Jun 16, 2010

### Andy Resnick

Without knowing any details of the source, assume it radiates all it's power uniformly into a cylindrical shell. 2.6W gets spread over a cylindrical surface with radius 6" and length 18". That will give you W/in^2.

There are some subtleties, since you are asking about the irradiance onto a flat surface, not a curved surface. And the illumination pattern of your cylindrical source is not exactly what I modeled above. But it's a start.

3. Jun 16, 2010

### jakeman7676

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 shell and then over my object then I think I've solved the problem. I get .0038 W/in^2. Spread that over my 18in^2 object and I have 2.13(10)-4 W/in^2. For my application this value is not quite enough....

How will the intensity over my object change if I were to put a reflector behind the light to direct all light towards the object.

Just fyi, I'm trying to expose an object to UV light via fluorescent UV lamps. The lamp I have picked out has a total UV out put of 2.6W. I'm trying to compare length of exposure under my lamp to length of exposure in sunlight but first I need a way to find the intensity at a given distance.

4. Jun 16, 2010

### Andy Resnick

I calculated a total of 6.9 *10^-2 W onto your 18 in^2 object.

5. Jun 16, 2010

### jakeman7676

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.

6. Jun 16, 2010

### Andy Resnick

The best approach, given the lack of detailed information, is to simply use geometry and conservation of energy. For example- does the reflector take 1/2 of the light and re-direct it forward? Then the incidance is approximately doubled.

Now, you mentioned comparison with sunlight. At high noon, the incidance on the earth is about 1 kW/ m^2. That's *total* power, so you need to know what fraction of sunlight is in the UV (it's about 0.5%)

Again, absent accurate information about the radiant intensity of the bulb (and reflector geometry), you can't do much more than this.

7. Jun 17, 2010

### jakeman7676

Ok, I'm having alot 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.

8. Jun 17, 2010

### Andy Resnick

You don't have a line source.

But, a good source for this is Wolfe's "Introduction to Radiometry".

9. Jun 17, 2010

### jakeman7676

Can it not at least be approximated as one? I'm only looking for ball park answers for this.

10. Jun 18, 2010

### jakeman7676

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 the bulb to get the collective intensity at the center. If my x axis is set where the origin is in the middle of my object and pointing along the object, each point on the bulb would have an increasingly smaller distance to the center of my object as you approched the center of the bulb which would have a distance of 6 inches. My integral would be something like:

Integral from r1 to r2 of Power/Area dr

Integral from r1 to r2 of 2.6W/ 4 pi r^2 dr

where r^2 = (36 + (9-x)^2), this is the changing hypotenuse of a triangle 6in tall and (9-x)in wide.

then Integral from 0 to 9 of 2.6W/ 4 pi (36 + (9-x)^2) dx

This integral is only one half of the bulb so multiply by 2 and I get .0677W/in^2. This number is nothing like what we got for the cylinder method which either means one of my approachs are wrong or I can't integrate. Which method is more representative of my situation? I realize that since my source isn't an ideal line source the intensity will differ along my object but I think the center will be a good enough approximation.

Let me know if this doesn't make any sense!

My best attempt at a picture:
18in
===*============== bulb
......|.\......|
......|...\r...|
....6|....\...|6in
......|.....\..|
......|(9-x)\.|
......|.____\|_____ object
12in

11. Jun 18, 2010

### Andy Resnick

It's not clear to me why you are making the problem more complicated than it needs to be. Just use conservation of energy- which is what we initially did.

12. Jun 18, 2010

### jakeman7676

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 (.5mW/cm^2). This is just really surprising to me and doesn't seem right. It may be that it is right and I just grossly underestimated these lamps but this number doesn't even come close to agreeing with the number we got at 6in away either. Any idea whats going on here?

13. Jun 18, 2010

### Andy Resnick

yeah- your lamp is not the sun.

Seriously, it's an arc lamp (a low pressure arc, but it's a discharge lamp), so most of the radiance is in discrete spectral bands.

I can sit outside all day long and not get sunburned- can you hold the lamp over your arm and say the same thing?

14. Jun 18, 2010

### jakeman7676

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.

Last edited: Jun 18, 2010
15. Jun 18, 2010

### Andy Resnick

The source of the inaccuracy is our imprecise knowledge of the radiance of the bulb. Knowing the radiance, you can compute any other radiometic (or photometric) quantity.

The radiance, plus geometry, tells you how much energy is incident onto a surface.

16. Jun 21, 2010

### jakeman7676

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.

17. Jun 21, 2010

### rcgldr

The math would be similar to this:

http://hyperphysics.phy-astr.gsu.edu/%E2%80%8Chbase/electric/elelin.html#c1 [Broken]

Last edited by a moderator: May 4, 2017