Radiance is a conservative magnitude. W=dQ/dt => L=dW/dwdS

Q energy, t time, W power-flux, L radiance, w solid angle, S surface

If w_e is emitter solid angle(SA), w_r is reception SA, S_e emitter surface and S_r receptor surface. For a flux W passing though S_e:

{dw_e = dS_r/r², dw_r=dS_e/r² }

L_e = dW/(dw_e dSe) = r²dW/(dS_r dS_e) = r²dW/(dS_r r²dw_r) = sW/(dS_rdw_r) = L_r

For that the radiance (bright) of a source is the same that its image. You can put apertures and reimaging systems that the radiance is preserved.

The bright of the image of a star seem through a telescope is the same that the bright of the star (the star is in the infinity and the optical system is symmetric)

If you search about Etendu o Throughput, for example:

http://www.horiba.com/scientific/products/optics-tutorial/throughput-etendue/

https://en.wikipedia.org/wiki/Etendue

https://www.spie.org/publications/tt48_27_throughput?SSO=1

You can see some samples of maths and geometry what that implies.

I hope what that clarify your curiosity.

Thank you for the explanation and the links.

My question is if we can still apply it when one of the image distance goes to infinity, as a limit. And how can I prove that?

If the image is formed in the infinity, the squared distance in the solid angle on the image side goes to zero. How does that affect the definition L = radiant flux/(area*solid angle) ?

I think my question comes more from the optics side.

Normally we use the expression h'/h to define the magnification, in which h is the object height, and h' the image correspondent. And we have the relation h's=hs', in which s and s' are the distances from object to lens and from image to lens, respectively.

Does the fact that s' goes to infinity change anything or how do we deal with it mathematically?