Integrating sphere as a uniform backlight

Hi all,

As mentioned in the first post in this forum thread, a finite planar light source (such as an LED panel) will always result in an inhomogeneous light distribution on film. A planar light panel a give a homogeneous light distribution on film, provided it is infinitely large compared to the film, which is not practical for table-top solutions. It took me a while to get this done, but I just calculated the brightness inhomogeneity on a film that results from using a planar light source that is not infinitely large, like so:

Screenshot 2024-06-18 at 16.55.49

The details of this calculation are provided elsewhere:

Typical results for my custom trichromatic light source (A custom-built trichromatic light for DSLR film scanning - arnogodeke) with 68 x 68 mm^2 homogeneous and diffused light, at 20 mm in front of a 24x36 mm^2 negative are like so:

This should be comparable to what most of us are doing: 13.1% light fall-off in the extreme corners, and significant fall-off around the edges of the frame. This will lead to all kinds of commonly observed issues, such as brightness variations towards the edges of the frame in B&W scanning, as well as color shifts in color scanning.

The tool obviously allows me to move the light around: When it is placed closer to the frame things improve, but not by a lot. When it is placed at 205 mm away from the negative, the fall-off can be as low as 1% in the last mm^2 in the corner of the frame, like so:

Hence, for all of you out there that are using LED panels for your scans, please be aware. This is quite significant, and the calculations seem to match closely what I indeed see when I scan a blank frame.

Solutions are to create a brightness mask in your editor, create a hardware mask that counters the light fall-off, tune the LED panel to provide more brightness towards the edges (if at all possible), or move the panel further away where things will improve (at the cost of overall brightness for scanning obviously).

I had to create some code in MS Excel to solve the quite nasty integrals that are needed to calculate this, and might share that file if there is sufficient interest.

Cheers,

Arno