Science Buddies: "Ask an Expert"

We found your great website when we were searching for a way to build a photometer. My project is comparing light intensities of two different types of light bulbs. I've completed my science fair testing and I've finished my data tables and graphs. I am wanting to use the inverse square law to compare the light intensities of incandescent light bulbs and compact fluorescent lamps (CFLs), but I am having trouble with the formula. I need to now how the formula goes from Light Intensity 1 / Distance 1 squared = Light Intensity 2 / Distance 2 squared to Light intensity 2 = Light intensity 1 * Distance 2 squared/Distance 1 squared. What happens to the "Light intensity 1" in the last formula when trying to solve for Light intensity 2? Do you throw it out or do you call it one? And, why?

I plugged in the data I got, which was the distance each light source was from the photometer. For one of my tests, the Incandescent light bulb was 85.5 cm from photometer. The CFL was 77.5 cm from photometer. After using the inverse square law formula, I got that the incandescent light bulb's light intensity was 1.25 times greater than the CFL's. Is this right? If a judge asks me what I did with the "light intensity 2" from the formula, I won't know how to answer the question.

I got the inverse square law formula(s) from your website.

Thank you for you help!

The way I interpret your project: the photometer needed to be 77.5cm from the CFL to get the same reading as when it was 85.5cm from the incandescent. Since you are comparing the relative intensity between the two bulbs, you can simply pick one bulb and call it your reference and say it has an intensity level of 1. You will see that the value and units you assign don’t matter. You can arbitrarily call it 28.372 Snapdragons if you want. But to keep things simple, let’s call it 1 “LightGuy”. (Now you’re famous!)

Let’s say we make the incandescent the reference intensity of 1 LightGuy. Then the equation says we compute the CFLs brightness as:

CFL intensity = 1 LightGuy * 77.5^2 / 85.5^2 = .82 LightGuys

So you can say that the CFL is 82% of the intensity of the incandescent.

And amazingly enough, it works the other way around, too, using the CFL with a reference intensity of 1 LightGuy. The equation then says:

incandescent intensity = 1 LightGuy * (85.5^2 / 77.5^2) = 1.22 LightGuys

That is, the incandescent has 22% more intensity than the CFL.

So, no, you can’t throw out Light intensity 1. But you can make it whatever you want since you are comparing one thing relative to another.

Hope that helps.

Mike

Wow! That really helps. Thank you so much! I finally get it!