Tell me more ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.
up vote 0 down vote favorite

Possible Duplicate:
Photometer: measured Irradiance L converted to photon rate

I am conducting a experiment where stimulus output of $470\ nm$ is measured by a radiometer at $30\ \mu W\ cm^{-2}$. The stimulus is $1$ inch from the detector.

Any suggestion on how I might go about converting $\mu W\ cm^{-2}$ to log photon $cm^{-2} s^{-1}$?

share|improve this question
1  
Hi user20373. Welcome to Physics.SE. Here, we use an unique TeX markup called MathJax. The markup is very much helpful in understanding equations, etc. Please have a look here for an introductory, or atleast have a look at our FAQ for an overview. For now, I'll help revising your post. – Ϛѓăʑɏ βµԂԃϔ Jan 30 at 1:06
1  
Bruce, I'm merging your two accounts. For future reference you should not repost questions. If you feel that you have not received responsive answer you can edit to clarify (which will bump the question on the active queue)--if necessary use the comments (on the original, please) to understand how you question has been misunderstood before you edit. – dmckee♦ Jan 30 at 5:25

marked as duplicate by dmckee♦ Jan 30 at 5:24

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1 Answer

up vote 0 down vote

First you need to calculate the energy per photon.

The photons from your stimulus are of wavelength 470nm.

Therefore, use the equation $E = hf$, where $h$ is plank's constant and $f$ is frequency and the equation $c = \lambda f$, where $\lambda$ represents wavelength, to arrive at the equation:

$E = \frac{hc}{\lambda}$ which will tell you the amount of energy per photon. Convert this answer to micro-joules (by dividing by 1 000 000).

We know that a 30 $\mu W$ is the same as 30 $\mu J/s$. So if we divide this by the energy obtained from the previous equation (assuming we have converted to micro joules), then we will arrive at the number of photons per second per square cm.

Finally, we want to take the log of this answer to give the units you require.

share|improve this answer
Which is only right if you assume a perfectly linear device with perfect QE. See the comments and answer on the earlier instance. – dmckee♦ Jan 30 at 5:25
Thanks for the response! So, if I wanted to know the microwatts cm^-2 value of a stimulus for example of 14.2 log photon cm^-2 sec^-1, would I take the antilog of 14.2 and mutiply this value by the energy level for the wavelength examined? – Bruce Gaynes Jan 30 at 22:39
Yes that's right, assuming you have a perfectly linear device with perfect QE. Theoretically though, yes. – Chris Jan 30 at 23:37

Not the answer you're looking for? Browse other questions tagged or ask your own question.