I am conducting an experiment in which the power meter reading of $410\,nm$ narrow bandpass stimulus is noted to be 30 $\frac{\mu W}{cm^2}$ at a distance of 1 inch away from the light source.

I wish to convert this to $\frac{\text{photon}}{ cm^{2} s}.$

Can anyone tell me how to do this?

  • $\begingroup$ can you be more specifig? you need to say how big is your radiometer? what solid angle does it cover 1 inch away from the source. is the light source radiating a diverging beam? or is the beam collimated, then the distance does not play a role. $\endgroup$ – elcojon Jan 29 '13 at 17:28
  • $\begingroup$ the conversion from Power in [Watt] to photons per second is: Power/Photonenergy = [Photons/sec]. $\endgroup$ – elcojon Jan 29 '13 at 17:29
  • $\begingroup$ Read the datasheet for the instrument. There is no substitute. There are detector performance factors which we can not advise you about but which the manufacturer has presumably made some basic measurements of. $\endgroup$ – dmckee Jan 29 '13 at 17:32
  • $\begingroup$ @elcojon Times quantum efficiency folded with possible non-linearities in the response of either the device (say near threshold) or the electronics backing it. $\endgroup$ – dmckee Jan 29 '13 at 17:33
  • $\begingroup$ @dmckee: you are right, I was assuming he was reading of the calibrated flux intensity. that's not the general case. $\endgroup$ – elcojon Jan 29 '13 at 17:34

Assuming a calibrated device (for this wavelength) it's a back of the hand calculation. Just pre-compute the photon energy $E_{Photon}$

Photon energy $E_{Photon} = h \cdot \frac{c}{\lambda} \approx 4.84\cdot 10^{-19}\,J \approx 3.02\,eV$

Irradiance $[L] = \frac{W}{m^2}$, which is the value on the power meter (pay attention to conversion from $cm^2$ to $m^2$).

Photon number per second $[R] = \frac{photons}{cm^2 \cdot s}$

You get this number of photon rate (photons per second) $$R = \frac{L}{E_{Photon} \cdot \eta(\lambda)} = \frac{L\cdot \lambda}{h\cdot c \cdot \eta(\lambda)}$$

$h$ beeing Planck constant. $c$ speed of light.

$\eta(\lambda)$ beeing quantum efficiency of your detector. Power meters have an internal look-up table for $\eta$. Assuming calibration $\eta=1$ yields $R\approx 6.20\cdot 10^{13}\frac{1}{cm^2 \cdot s} = R\approx 6.20\cdot 10^{17}\frac{1}{m^2 \cdot s}$. Compare this result to the order of Planck's constant.

Neglecting non-linearities and calibration errors, mentioned in comments. Short burts are averaged by your power meter. I assume your unspecified source perpendicular hits your detector and losses on front-surface are taken into account.


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