Calculate Retinal irradiance (W/cm² to log photons/cm²/s)

Question

I have been trying to calculate what retinal irradiance value I get with a 1 blue LED system.

Since the manufacturer didn´t give the spectral distribution information, I will approximate LED as a monochromatic one (using the 460 nm peak).

From the datasheet, the LED intensity range goes from 6 lumen to 30 lumen.
For the 6 lumen case, I used scotopic eye sensitivity curve (i want to apply the stimulus in a dark room) to convert the lumen value to radiant power (W). I reached a value of $0.006,mathrm W$.

I know from the ray tracing software that only 3.7º of the LED (total from the center) reach the pupil. So, from the spacial LED curve, the $0.006,mathrm W$ become approximately $1.57times10^{-4},mathrm W$.

The illuminated retinal area is $0.0031,mathrm{cm}^2$.

So I calculated the irradiance dividing the $1.57times 10^{-4},mathrm W$ by the area of $0.0031,mathrm{cm}^2$, having a value of $0.05,mathrm{W}/mathrm{cm}^2$
However, I need this value in $log mathrm{photons}/mathrm{cm}^2/mathrm s$.

Can you help me with my problem?
Thanks.

boyfarrell · Answer

Irradiance has units $text{J}text {cm}^{-2}text{s}^{-1}$ and you want to know photons $text{cm}^{-2}text{s}^{-1}$. Under the monochromatic assumption this is really easy, simply divide your irradiance value by the energy per photon,

$$
E = frac{hc}{lambda}
$$

The wavelength of 460nm has energy of about 2.7eV or $4.326times10^{-19} text{J}$ which gives about $n=1.16times10^{17}$ photons $text{cm}^{-2}text{s} ^{-1} $. Now take the log of that value, $log_{10}left(nright)approx 17$.

Calculate Retinal irradiance (W/cm² to log photons/cm²/s)

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