For example, this one. And there's lots of calculations like this, but this one is common enough in the optical world. It's got its own name, the Cosine to the Fourth Law. So let's walk through this. It uses all the math that we just worked out. But it tells you how similar to the point source radiating onto a plane. Now if you're using an optical system to image a Lambertian extended source. A source whose radiance does not depend on the angle that you look at it. Now what will you expect to fall on your film or your digital receiver over here? And the angle is it's even worse than cosine to the third. And so we could just walk through the math. So we'll describe our optical system as an exit pupil. Now hopefully you're comfortable with that now. The exit pupil is the image of the aperture stop. And the aperture stop controls the angles of the radiation getting off of our source. And in a well designed system, of course the film itself would be a window or the field stop. So we've described both of the two important apertures in our system, and when we're looking at the film we don't care about the aperture stop. We might not even be able to see it directly, there might be optics in the way. So we'll describe our whole system simply as radiating some light from an exit pupil, wherever that might be. We're some distance d from the exit pupil to the film. There's some projected distance s and we're interested in how power changes as we tilt up at an angle theta. So from the same geometry we just had, the projected distance goes up by 1 over cosine theta, check. The solid angle subtended by the exit pupil, we're looking back at the exit pupil from our camera. That's pretty simple at this point. And we understand that the exit pupil has some area. And then we divide by the distance to the exit pupil, s, just go along the radius of our sphere of course. But notice that the exit pupil itself seems foreshortened by the tilt. And so the apparent area of the cap, the normal that we would draw here Is smaller by a factor of cosine theta. So we initially see that the solid angle is falling off, the solid angle of the observed exit pupil including the projection here of its area, that's falling off like cosine cubed. Two from the projected, from the tilted distance going up, and one from the pupil itself, appearing smaller and smaller. Then we worry about, so now we've got a solid angle. Now we're asking how much power appears to fall on a unit of my image plane, maybe my pixel of my detector. And it turns out that's got another cosine theta in it because we also have a tilt here. So the pixels are capturing less of the total power and that's through one more factor of cosine theta. So when you put all that together you'd find that if you put a camera here and you're imaging a Lambertian source overall and its Lambertian because L itself does not depend on angle. That overall your film would find a fall off in intensity like a cosine to the fourth. So this is strong enough, it can look like that that we showed you before and it's not. So that's a pretty important thing to understand when you're designing a camera, for example it's supposed to image with a particular signal to noise, that you have for Lambertian objects a pretty significant falloff of power just because of the radiometry of the sources.