Light intensity formula using inverse square law

[deleted-134275]

Dear Science Buddies,

I was captivated by the experiment to measure light intensity by building the Joly photometer, and decided that I had to do this project. I am certain that I understand the experiment, yet would like to challenge myself and be original by changing it slightly. The formula used to calculate the light intensity of different wattages, includes distance. However, if I were to measure the light intensity of compact fluorescent and incandescent bulbs how would I do it?

The same formula would not apply, and was wondering if there was some other alternative way to calculate the light intensity, because in order to calculate it's efficiency I have to first determine the light intensity.

Any help would be appreciated, preferably if one could do so before Monday. I thank Science Buddies for the wonderful opportunity to post questions online. Would anyone happen to know also, how long it will be before I get a response?

Sincerely Thanking You,
Taniesha, fellow grade 7 student

[deleted-71882]

Hello Computer369,

I'm not sure I understand your question because I don't see exactly why you think the experiment has to be changed.

I assume you are following the experiment outlined at https://www.sciencebuddies.org/science- ... background.

That experiment gives you the relative intensity of two sources by measuring only the distances. Bulbs are almost always labeled by how much power they consume. The only factor you have to compute from the result is the efficiency. In other words,

I = P x E

I is the intensity measured by the photometer, P is the power consumed as labeled, and E is the bulb efficiency.

Please elaborate on your question if this is not a sufficient answer.

Good luck, WW

[deleted-134275]

Dear Science Buddies,

At first, I thought I had understood the inverse square law and the formula to calculate light intensity, but now I am not so sure. What I do understand is that if the distance1 of intensity1 is 4cm, and the distannce2 of intensity2 is 2cm, then the intensity of distance2 will be a quarter of that of intensity1.

I have tried researching this topic and have found numerous links and topics, and am really not sure which one to use or which one is correct. The first one states that we must first find the surface area of a square (4(pi)r*2) and then divide it by the watts contained. However this formula gives you an answer of watts per square and since it is already divided by the watts, won't give you the efficiency.

The second link says that in order to calculate the light intensity of the second bulb, one must divide its distance squared by that of the distance squared of the standard bulb's intensity (other bulb) and then multiply it by the standard bulb's intensity, which was assigned a comparative number of one. However, at the end of the page, it says that this is not the real intensity and that you may replace the digit 1 with its actual intensity, which I am not sure how to figure out.

I am puzzled at which one to use. A thought occurred to me that maybe I should incorporate both, but then the formula wouldn't work because the first link has already been divided by the watts. Should I just calculate the surface area of a sphere, then multiply that by the quotient of distance1 squared and distance2 squared?

Also, in the science buddies page it states that we need three testing bulbs of different wattages, but only mentions two in the actual experiment?

Thanks for your help and consideration,
Computer369

[deleted-134275]

Dear Science Buddies,

At first, I thought I had understood the inverse square law and the formula to calculate light intensity, but now I am not so sure. What I do understand is that if the distance1 of intensity1 is 4cm, and the distannce2 of intensity2 is 2cm, then the intensity of distance2 will be a quarter of that of intensity1.

I have tried researching this topic and have found numerous links and topics, and am really not sure which one to use or which one is correct. The first one states that we must first find the surface area of a square (4(pi)r*2) and then divide it by the watts contained. However this formula gives you an answer of watts per square and since it is already divided by the watts, won't give you the efficiency.

The second link says that in order to calculate the light intensity of the second bulb, one must divide its distance squared by that of the distance squared of the standard bulb's intensity (other bulb) and then multiply it by the standard bulb's intensity, which was assigned a comparative number of one. However, at the end of the page, it says that this is not the real intensity and that you may replace the digit 1 with its actual intensity, which I am not sure how to figure out.

I am puzzled at which one to use. A thought occurred to me that maybe I should incorporate both, but then the formula wouldn't work because the first link has already been divided by the watts. Should I just calculate the surface area of a sphere, then multiply that by the quotient of distance1 squared and distance2 squared?

Also, in the science buddies page it states that we need three testing bulbs of different wattages, but only mentions two in the actual experiment?

Thanks for your help and consideration,
Computer369

[deleted-71882]

Computer369,

The inverse square law states how the intensity of a light source changes with the distance from the source. It says that if you double the distance from the source, the intensity drops by four. Just to be clear about an important detail, this law applies only when the observations are made at distances that are large compared to the size of the source.

The Joly photometer tells you about only the relative intensity of two sources. It cannot tell you about the absolute intensity because it simply compares one source to another. To measure the absolute intensity of a source, you would have to measure the intensity over a sphere surrounding the source or make a measurement at one angle and somehow know how the intensity varies over the sphere.

You can use the relative intensity measurements to infer relative efficiency. For example if you have two sources that each consume the same electrical power, but you measure that one of them has twice the intensity of the other, then you know that the more intense one is twice as efficient as the other.

If one of your sources is a "standard bulb," then by comparing the intensity of it to a second source, you can infer then absolute intensity of the second source. Its absolute intensity is the intensity of the standard bulb times the ratio of the intensities you measured.

You can't "figure out" the intensity of a standard bulb unless you make absolute intensity measurements. A standard bulb intensity is usually specified by the bulb's manufacturer. Commercial bulbs usually have a specified "lumens per watt," but this number is not precise: the intensity changes significantly with the angle of emission, and other factors make it only an approximation.

The Project Ideas on Science Buddies are not complete recipes for a science project. They are guides to how to do something. In the Science Buddies project description of the Joly photometer, the procedure describes how to compare two light sources. With three sources, A, B, and C, you can use this procedure to measure the relative intensity of A versus B, B versus C, and A versus C. It is your job to turn the methods given in the project description into a complete science project.

Good luck, WW

[deleted-134275]

Dear WW,

Thank you so very much. I quite understand it now and will apply the knowledge given to conduct my experiment. I wasn't expecting a reply so quickly and was very buoyful when you had. It all makes sense now. I will still use that experiment yet will adapt it and will instead compare incandascent bulbs to compact flourescent bulbs.

I really do not know what I would've done without your help - thank you very much.

Computer369