For this use, you want the object-space size of one pixel.

Using magnification with a thin lens, the subject detail corresponding to one pixel is approximately:

subject detail = pixel size × (subject distance − focal length) / focal length

With your numbers at about 100 mm working distance:

5.6 µm × (100 mm − 16 mm) / 16 mm ≈ 29 µm

So one pixel corresponds to about 0.029 mm, roughly 0.03 mm on the surface.

Across your 90–110 mm range, it’s about:

• 90 mm: 5.6 µm × 74/16 ≈ 26 µm
• 110 mm: 5.6 µm × 94/16 ≈ 33 µm

So the theoretical sampling resolution is roughly 0.026–0.033 mm per pixel.

Important: that is a best-case geometric estimate, not guaranteed measurement accuracy. Real resolving power can be worse because of lens sharpness, focus, contrast, lighting, noise, and the surface-analysis method. If you need to reliably resolve 0.01 mm features, this setup is likely insufficient at 100 mm distance; you’d need higher magnification (for example, a shorter working distance or different optics).