NAwlins Contrarian

The same amount of noise, while capturing less signal (due to less exposure), means lower signal-to-noise ratio, which means more visible appearance of noise. My sense is that what we perceive is not so much noise as signal-to-noise ratio. The thing we sense is somewhat indirect. This is a far from perfect analogy, but it's somewhat like: we don't actually feel hot or cold. What we feel is heat transfer, which is proportional to relative hot or cold but also affected by other factors. That's why the air inside a 450 degree F oven is more tolerable to our bare skin than 150 degree F water in a pot. In contact with skin, water transfers heat much more efficiently than air, so there's more / faster transfer of heat to our skin from 150 degree F water than from 450 degree F air. Similarly, our eyes (and ears) don't perceive noise so much as they perceive signal-to-noise ratio. Or at least, that's my impression / understanding.

Talking about inkjet photo papers, there are two ways to look at this, and a basic issue underlying the discussion. The basic issue is that although paper white is the same regardless of the printer / inks, maximum black can vary to a substantial / visible degree depending on the printer / inks. Now the two ways to look at it: (1) The easy thing to do is to compare the ICC profiles for the printer + inks + driver settings + paper combinations of interest. Various tools can do this, and if the ICC profiles are the older-type v. 2 profiles instead of the newer-type v. 4 profiles, then the free ICC View website will calculate paper white (and gamut volume) and let you optically inspect maximum black (bot darkest neutral and darkest color). You might find e.g. that a particular combination has, on the L* scale from 0 to 100, a range of 90, e.g., L* min = 6 and L* max = 96. With my little Epson, the papers with the largest range (Hahnemühle Photo Rag Pearl, Mitsubishi Pictorico Pro Hi-Gloss White Film, Epson Legacy Platine, and Epson Ultra Premium Glossy) have a range of 93, although some are on the whole darker (range 3 to 96) or brighter (range 5 to 98). At the other end of the performance envelope, the the paper with the smallest range (Epson Ultra Premium Matte) has a range of 77 (19 to 97). (2) The more traditional thing to do is measure optical density. I'll use the examples above to compare. If as stated Red River Polar Matte has a density range in some unspecified printer of 0.05 to 1.65, then right, that's a difference of 1.60 and a dynamic range of 10^1.60 = 39.8:1. To convert instead to f-stops or EV, 1.60 / log(2) = 1.60 / 0.301 = 5.3 f-stops. Likewise, if Red River UltraPro Gloss has a density range (again) in some unspecified printer of 0.07 to 1.98, then right, that's a difference of 1.91 and a dynamic range of 10^1.91 = 81.3:1:1. Again, to convert instead to f-stops or EV, 1.91 / log(2) = 1.91 / 0.301 = 6.3 f-stops. So with this unspecified printer, Red River Polar Matte has printable range of about 5.3 f-stops, and Red River UltraPro Gloss improves that by about (just a hair more than) 1 stop to about 6.3 f-stops. FWIW, in most cases non-matte and matte papers have relatively similar paper whites, but non-matte papers can usually show substantially darker blacks. And if there's a way to convert between L* and optical density, I don't know what it is--but I'd love to know.

I'm also interested in the basic issue because I have an old Epson 3200 (about three generations older than your V550), and I'd like to be able to scan 4x4 cm frames from 127 film. That said, two points: (1) Film height is an issue. AFAIK in film mode, the scanner focuses to the height of film in am one of he Epson film carriers. So if you put the film directly on the scanner glass, the scan may be out of focus. One option might be to put a thin piece of glass between the film and the scanner glass. Another option might be if the scanner software has a mode intended for scanning regular transparencies, like were used decades ago on overhead projectors, and which (being much larger than anything other than the largest photographic film) presumably would sit right on the glass. (2) The suggestion, "I used sheet of white sheet of paper on top of the negative, because I think you want the light to reflect off the background and back through the negative," is wrong, and fundamentally misunderstands how film scanning on a flatbed is supposed to work. Your V550, my 3200, and others among these sorts of scanners do not illuminate the film the same way they illuminate documents. You remove a cover in the top, which lets a second light source in the transparency unit backlight the film. The lack of a transparency unit is why most flatbed scanners do such a terrible job scanning film.

I'm not sure I understand the question, but I'll take a crack at it. If the photographer reduces exposure, then that causes reduced signal. The photons coming through the lens and striking the sensor are the signal. The photographer can reduce the signal by shortening the exposure time (i.e. increasing the shutter speed) and/or reducing the size of the aperture (i.e. stopping down).

[Coming in late, but interested] Underexposure doesn't produce noise, or at least, not precisely-speaking; but it does reduce signal, which lowers signal-to-noise ratio, which appears (above some threshold of visibility, and especially insofar as postprocessing equalizes overall image brightness) as relatively more noise. Agree? Disagree? I think the similar-but-different test that might show very similar noise would be both at f/8 and 1/125 s but one at ISO 200 and the other at ISO 1600, and then equalize brightness in postprocessing. And I stress "might" because dual-gain sensors and non-linear gain ("ISO") settings can substantially change that behavior / result. Agree? Disagree?