I am trying to estimate a current which I can get from the photodiode in a simple proximity sensor application having a IR LED emitter and a photo-diode detector. The LED and the photodiode are mounted in the same plane, on a PCB board, and separated about 10mm. If I have a flat Lambertian surface (like a gray card) parallel to the PCB and a LED emitter like VSMY2850 where all of the emitted power is inside the half angle, then irradiance at the photodiode (detector) can be found from Ed = Ls * π * sin2(θ1/2) (Field guide to Radiometry p.23, Irradiance from a Lambertian disk) I think Ls can be written as Ls = M/π = R*Es/π where R is the reflectance of the Lambertian surface (gray card) and Es is the irradiance of the disk. Since all of the radiated power of the VSMY2850 is contained within its half angle Es= P/A = P/(a2*π) a - radius of Lambertian disk P - radiant power emitted on the disk from the LED (~ 55mW for 100mA) but since tan(θ1/2)=a/d , I have Ed ~ 1/d2 *cos2(θ1/2) So for θ1/2=10° Ed ~ 1/d2 * 0.970 and for say θ1/2=3° Ed ~ 1/d2 * 0.997 This made me a little confused, considering the same power, if I used LED with 3° (typical for highly focused LED) or If I placed a lens (like a PMMA Plano convex) on the VSMY2850 to get more focused beam from it, not much would change in terms of the current I would get from the photodiode. I'm not sure if this is correct.