let's consider a simple lens without diaphragm.
For a lens like this, the book Science for Curious Photographer states that:
The important fact is that the illumination of the image is proportional to the brightness B of the object and the square of sin θ' [that is f/2r = F Number squared]. Here the diameter 2r refers to the effective aperture of the lens. Th is is not the diameter of the front lens element or even the iris diaphragm. The effective aperture or entrance pupil is the image of the aperture stop as seen from the front of the lens.
Well, there is nothing intuitive for me. Let's start from the basics.
The luminous subject has a luminance (you may express it as Watt or Lumen per square meter, or per steradians, as you prefer). This luminance is represented by its emitted rays, which the lens make converge to produce another luminance value on the sensor plane.
Not all the rays enter the lens. I'd say that a lens limits the possible input rays by its diameter. I'd say that 2r = Lens diameter. If the lens were infinite, all the input luminance would be directed towards the sensor.
The limited rays acceptance circle is just due to the lens diameter. I'd say that the focal length is considered because, if you increase it, some portion of image goes outside the sensor. So, it is not a problem of the lens. It is just the sensor that is too small. But, if that's true, shouldn't it be embodied into luminance definition? Luminance is Watt or Lumen per square meter. Big sensor -> more exposure because luminance is multiplied by a bigger area.
Now let's introduce the diaphragm. Why should I consider for D the diameter of the entrance pupil? Why should the luminance be attenuated by the diameter of what I see in front of the lens? What does it have to do with the accepted power of the lens?
Every guide says the "brightness attenuation scales as (f/D)^2",... but I've not found an intuitive explanation for my points.