Photodiode current estimate in an Infrared Proximity sensor

[vst98]

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

E_d = L_s * π * sin²(θ_1/2)

(Field guide to Radiometry p.23, Irradiance from a Lambertian disk)

I think L_s can be written as

L_s = M/π = R*E_s/π

where R is the reflectance of the Lambertian surface (gray card) and E_s is the irradiance of the disk.
Since all of the radiated power of the VSMY2850 is contained within its half angle

E_s= 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(θ_1/2)=a/d , I have

E_d ~ 1/d² *cos²(θ_1/2)

So for θ_1/2=10°
E_d ~ 1/d² * 0.970

and for say θ_1/2=3°
E_d ~ 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.

 

[vst98]

By using a narrow angle LED (or an additional lens on the LED), the LED beam spot would start to overspill the target at greater distances then when using wider angle LED.
This is where I can see an advantage if I choose a narrow angle LED.
But if the whole LED beam spot is on the target, difference between 3° and 10° half angle LED's are not much, right ?

d is the distance between the sensor plane and the target and the distance under consideration are from about ~0.5m to maximum I can get.
(forgot to write this in my first post)