> If irradiance in above case is 2, what about the radiance than is
> reflected from that point to a certain direction? (That is equal no
> matter what the direction as it is lambertian).

No it's not equal, it's proportional to cos(x), that is what lambertian 
means.  It then "looks" the same brightness to a camera/eye/surface from any 
angle, because the area you see for each pixel/cone/rod/patch is 
proportional to 1/cos(x), so the cos factor cancels out leaving constant 
"brightness" from any angle.

In the case of irradiance of 2, the radiance is actually exactly cos(x), 
because if you integrate cos(x) over -pi/2 to pi/2, you will get 2, as you 
already found out.  So in fact your sky and ground are identically looking 
planes (assuming the sky is not reflective at all), which you proved 
already.

In 3D your radiance is now in units of power per solid angle per area, 
compared to power per angle per area.  Lambertian still means that the 
radiance is proportional to cos(x) (where x is the angle between the surface 
normal and radiated/reflected light), it's just you need to integrate over a 
hemi-sphere rather than a semi-circle.

For lambertian surfaces, in reality this just means that your units are 
different and you only integrate from 0 to pi/2.

So in 3D, if you have a lambertian surface with radiance function 1*cos(x) 
Watts/steradian/metre2, the total irradiance from a point will be 1 
Watt/metre2.

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