Photometer: measured Irradiance $L$ converted to photon rate 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? 
 A: 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 $E_{Photon} = h \cdot \frac{c}{\lambda} \approx 4.84\cdot 10^{-19}\,J \approx 3.02\,eV (\text{for }\lambda = 410\,nm)$
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 bursts are averaged by your power meter. I assume your unspecified source hits your detector perpendicularly  and losses on front-surface are taken into account.
