# Measuring the intensity of light

1. Sep 13, 2009

### N0b0dy

Hi, I'm trying to find out how changing the current fed through to a filament (e.g. sodium) of a lamp would change the light intensity. To determine the relationship, I will first need to find out how to measure the intensity of light. I've thought of one possible way to do it, but the measurement would not give me a direct expression of the intensity of light. Here it is if anyone wants to give me some potential feedback as to whether this would work:

Shine the light onto a cathode made of a metal that the light can emit electrons from. Electrons will then be emitted and will be attracted towards an anode connected to a galvanometer. The current measured will be related to the intensity of the light, where a higher intensity obviously yields a higher current.

Is there any other practical (and relatively simple) way to measure the intensity of light that will result in measurements that are direct expressions of the intensity?

Cheers.

2. Sep 13, 2009

### Born2bwire

A CCD or a photodiode would do the job I should think.

3. Sep 13, 2009

### N0b0dy

So is my description of a potential setup okay? I gave it some further thought and I believe that if I can obtain a value for current, I should able to observe the number of photons and hence the intensity because I=q/t. Hence if I divide the total charge per second by the charge of an electron I should be able to get the number of electrons per second and therefore the number of photons per second. The number of electrons per second is directly proportional to the number of photons per second, correct? So my measure of intensity will be number of photons per second?

4. Sep 14, 2009

### ZapperZ

Staff Emeritus
The only issue here is that there are no photocathodes that have 100% quantum efficiency. If you're using a metal as your photocathode, then that's even worse, since metals, at best, have a QE of the order of $10^{-2}$%. So there will be plenty of photons that didn't produce a single electrons.

Photodiodes, photomultipliers, photodetectors, etc. tend to be more efficient than your current set up.

Zz.

5. Sep 14, 2009