Itisconfusing. So I'll try to see if I can unmuddy the waters, at the risk of making it worse..

Image circle is a result of the construction of the lens, and the focal length of the lens. The lens does not care what film format it is used on, so we put the film sizes aside for later.

Lenses of the same construction have the sameimage angle, which when projected on the focal plane gives a circle with a diameter depending on the focal length of the lens.

For the mathematically minded, tan(Theta/2)=D/(2xF) , where Theta is the image angle, D the diameter of the image circle, and F the focal length.

Coverage also increases as the aperture is stopped down, since the unavoidable "weaknesses" towards the edge of the image circle become less important - using a smaller area of the glass means that a lot of aberrations are drastically reduced.

And next illumination:

The physical construction of the lens - barrel, filter mount, et cetera - mean that at full aperture, some of the light that would otherwise illuminate the edges of the image circle are blocked on the way. This isvignetting, not to be confused with light fall-off.

Light fall-off is caused by geometry only, and even without a lens you can see it.

Because when seen from the side the (round) aperture becomes a narrow oval, the area becomes smaller with the cosine of the angle. Because the light exiting the aperture travels farther and spreads over a larger area, it also is reduced by the square of the cosine of the angle (between the lens axis and the line the light travels from the aperture). And at the moment I can't remember where the fourth cosine drop-off is from, but I'm sure I will remember shortly after I hit "submit"...

A perfect pinhole would show fall-off proportional to cos^4(Phi), where Phi is the angle mentioned above - and Theta/2 is the limit of coverage and thus Phi.

Wide-angle lenses of "modern" design - the ones with huge front elements - use a thick strong negative outer element to "tilt" the image of the aperture so that the effect ofthat particularlight loss becomes less.

In short, all reasonably well constructed lenses have fall-off somewhere between cos^3(Phi) and cos^4(Phi) - that's just basic optics, really.

Note that the angle is very prominent in all the formulas here: That's why a "normal" lens doesn't show prominent light loss, whereas wide angle lenses do. It's simply that Phi becomes smaller with a longer focal length on the same film format!

Now let's look at otherwise similar lenses - let's say a 90mm lens for 617 format. The diagonal of the format is 179mm, so I'll round it off to 180mm since I like to be able to do calculations like these in my head.

The necessary image diameter is 180mm, which gives a radius of 90mm. D/2F equals 1 in this case, which gives us a Phi (or Theta/2 ) of 45 degrees, so we need a lens with at least 90 degrees angle of coverage.

Cos^4(45) = 0.25

Cos^3(45) = 0.35

A simple lens, like indeed an old Angulon 90mm, will lose two full stops in the corners of a 617 frame. A lens of more modern construction (like a Super Angulon), willapproach1.5 stops loss, but no real lenses are quite as good as that. The real light loss will be somewhere between 1.5 and 2 stops forall90mm WA lenses used on 617 format!

Lens manufacturers give curves showing the light fall-off, usually with the cos^4(Phi) curve drawn in to show that their lenses do better than this. But you have to remember that the curves show 0-100 diameter, where 100 is the manufacturer's intended diameter of coverage (or the angle of coverage, depending on who made the diagram). Reading and comparing curves from different manufacturers is not always straightforward - you may end up hand-drawing the curves to your own scale to make sense of it.

To summarise - all lenses have light fall-off, and depending on the type of lens it will always be between cos^3 and cos^4. The only reason this only becomes apparent on wide angle lenses is that only there does the angle get large enough to see the difference. At 30 degrees off axis, cos^4 gives 0.56 and cos^3 gives 0.65 - a relatively much smaller difference. About a stop in either case. Even closer to the axis, say at 10 degrees, we get 0.94 and 0.955 - a difference you wouldn't notice.