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Oct 5, 2016 at 22:57 comment added Nathan Reed @PeteUK it comes about because you're projecting a light source that's a distance $r$ away onto the unit sphere. (The solid angle is equivalent to area on a unit sphere.) So, distances get divided by $r$. If that doesn't help, I can draw a diagram.
Oct 5, 2016 at 21:22 comment added PeteUK I'd really like to absorb this answer but I'm getting stuck on the bit about approximating the solid angle as an area of circle of radius $r_\text{light}/r$ using similar triangles. I don't get the denominator, and I can't see how similar triangles come into play. Any help appreciated.
May 11, 2016 at 18:04 history edited Nathan Reed CC BY-SA 3.0
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May 11, 2016 at 17:49 history answered Nathan Reed CC BY-SA 3.0