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

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 $$1.57\times10^{-4}\,\mathrm W$$.

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

So I calculated the irradiance dividing the $$1.57\times 10^{-4}\,\mathrm W$$ 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.

• This will depend on the state of the eye lens (its optical power), spectrum of the LED, age of the person being irradiated. And the irradiance will not be the uniform across the surface of retina. So currently this question lacks detail. – Ruslan Aug 21 '19 at 10:55
• Hi @Ruslan. I am assuming a 1.66 mm eye focal length. I am assuming a monochromatic 460 nm blue LED of 6 lumen. I converted into radiometric units with the eye curve sensitivity, and taking into account the angle percentage that reach the eye, I achieved the 0.006W value – az_ Aug 21 '19 at 11:13
• Monochromatic LED? That doesn't sound right. Either it's a laser, or it's not monochromatic. Typical 460 nm LEDs have FWHM ~20 nm. And you still haven't specified what kind of irradiance you're interested in – average over some area, maximum, or something else. All this info should be edited into the question so that potential answerers could see it without need to read comments. – Ruslan Aug 21 '19 at 11:21
• @Ruslan. I added more relevant information to the problem. Do you have some suggestion? – az_ Aug 21 '19 at 15:47

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 $$4.326\times10^{-19} \text{J}$$ which gives about $$n=1.16\times10^{17}$$ photons $$\text{cm}^{-2}\text{s} ^{-1}$$. Now take the log of that value, $$\log_{10}\left(n\right)\approx 17$$.