Calculating Light Fall-Off From Large Source

Question

"the inverse square rule is often still a useful approximation; when the size of the light source is less than one-fifth of the distance to the subject, the calculation error is less than 1%"

When calculating light-off from a source substantially larger than normal (6m x 1m, 10m x 5m) with the subject being 1m - 10m away from said source how would one approximate calculating format in a practical sense. Assuming all portions of the diffuser used is uniform I've read "then the total illumination on your point is proportional to the solid angle from the point that intercepts the light".

If, for example, I had a source 6 metres in width & 1.5 metres in depth. With my subject being approximately 10 metres away with total output being 6600 FC, 1' away from the source or 71041.81 LUX, 30.5cm away from the source. How would one put that into a practical equation.

Thanks Gabriel

Answer 1

Unfortunately, the answer to your direct question (how would one put this into a practical equation) is: one wouldn't.

After reading this answer, if you want to relax the definition of 'practical', and get some help with the math involved, comment as such and I can delete this answer. No guarantees that I'll actually help with that math, though. ;)

You will want to research 'Far Field' vs 'Near Field'. Wikipedia has a reasonable start to this concept, though focused on general EM applications and not optics.

Optical far field (as I have understood it) is typically 10x the source size (not 5x as your quote says). Once you know your subject is in the far field relative to your source(s), you can treat a source as an infinitesimally small point source, and the calculations become much simpler.

A 6m max source size at 10m away means that you are definitely not in the far field (distance is only ~1.5x the source size), and you can no longer treat the source as a point. If the source is not illuminating uniformly from every emission point - which it almost always is not - then the lux generated from the 'close' side of the source will be different from the 'far' side of the source to your subject. (The pictures from the Wikipedia link illustrate this effect, even if not created for optics.)

Even if you are to assume your source is truly uniform (very good diffuser), as the commenter @CarlWitthoft has said - the math is still not 'practical'. You would need an intensity vs angle plot for your source: you need to know what the luminous intensity values are for a range of angles (since each component of the source will have a different angle to the subject).

If you want to assume your diffuser is fully Lambertian (as 'perfect' as possible), then perhaps this calculation could be done. But this then becomes an exercise in math and concepts, rather than a 'practical' equation, and we are back to defining what you mean by 'practical'.