I'm a working cinematographer from Australia. One of the first properties of light 'law's' we are taught early in our career is the inverse square law. So an object double the distance receives a quarter of the energy.

This however is from a point source, or - "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%"

My question is how would one calculate light fall-off from a source substantially larger than the subject.

For example I recently rigged 50 Par64 Cans (A PAR Can is an 1kw globe with a PAR reflector). I rigged them in a row spanning 12 metres with a layer of Light Grid directly in front (a diffusion material) this source was 14 metres in width and 2 metres in height. How could one accurate calculate light fall-off from said source.

My current approximation is using the inverse square law from the original source while taking into consideration the -0.8 Stop of light loss from the diffusion material (almost halving the total output).

Thanks Gabriel


If your diffuser is good enough that the light from any portion of it close to uniform from any direction, and the light from all portions is equal in intensity, then the total illumination on your point is proportional to the solid angle from the point that intercepts the light. At large distances (small angles) this will become equal to inverse square.

At shorter distances, the falloff is slower. Very close to the source, the intensity becomes nearly uniform with distance.

The actual solid angle formula is not one that I tend to use much, but there are some formulas and references on MathSE

  • $\begingroup$ Sir, do excuse the ignorance. So say if I calculated my solid angle from diffuser to subject. If I had my total output of fixtures before hitting the diffusion in Foot-candles or LUX how would one integrate that into an equation. $\endgroup$ – Devereux Gabriel Jan 14 at 14:18
  • $\begingroup$ Presumably, that would just be a linear multiple. I'm just suggesting the distance relationship. If you measured it at one point, you can then calculate it at another point. To go directly from the source intensity would require more details about the diffuser and such. $\endgroup$ – BowlOfRed Jan 14 at 17:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.