does distance to subject affect vignetting (independent of focus/zoom/apertuer)

[spacediver]

Suppose you are imaging a perfectly flat light emitting surface, and you set the camera a certain distance from this surface and take an image. Now suppose you move the camera further away from this surface, but don't change anything (same ISO, aperture, zoom, focus) and take another image.

Will the "vignetting function" be identical? Or would it only be identical with such a distance change if the surface in question is perfectly Lambertian?

 

[480sparky]

The vignetting caused by the lens will be exactly the same.

What will change is how much the vignetting impacts the flat surface. Move away far enough, and you won't see any vignetting.

I have no idea what you mean by 'vignetting function'.

 

[spacediver]

By function, I just meant the mathematical description of this change in luminance across the image (the radial drop off in brightness). If the vignetting is identical in both cases, why will you see no vignetting if you're far away enough? Or did you just mean that because if you're far away enough, the surface won't occupy the full frame anymore?

 

[unpopular]

I agree, I am confused by the latter technical points, but yes, any variation to image circle diameter is dependent on focus and zoom with vignette in the typical photographic sense, being created by variation in brightness over the image field, and is independent entirely of subject distance alone.

Now if the source itself has a vignette quality, then moving closer might blur the source sufficiently to allow it's luminance to average to some degree.

 

[spacediver]

excellent, this is good news, thanks for the responses.

 

[480sparky]

By function, I just meant the mathematical description of this change in luminance across the image (the radial drop off in brightness). If the vignetting is identical in both cases, why will you see no vignetting if you're far away enough?

If you're assuming the flat surface is of infinite size, then moving an infinite distance will make no difference.

I was assuming the flat surface to be of a finite size. Move back far enough, and the surface takes up less of the image area.

 

[spacediver]

thanks, yea I thought about it some more and figured that's what you meant (see my edit).