6
\$\begingroup\$

I was wondering whether film grain is depending on the incident amount of light.

There is the Schwarzschild-effect (link to Wikipedia), which will make a difference in very low light, that will result in a more uneven distribution of brightness values.

That aside, are there any processes that will result in more or less agglomeration or splitting of the photosensitive grains on the film? If there are any of these processes (let's asume in the low illumination areas), is there anything wrong in assuming the print also finds the grainstructure more pronounced in the reversed area of the histogram (in this Gedankenexperiment the highlights)?

As an extra: If one does prints from negatives, are there any different processes concerning grain involved? I would assume that paper grain is negligible in size and impact compared to the one of the film, is that correct?

Improve this question
\$\endgroup\$

2 Answers 2

Reset to default
3
\$\begingroup\$

As you know, photo film is a suspension of salts of silver halogens (Swedish for salt maker) in gelatin. These silver salts are quite tiny. When exposed to light, the bond that holds the silver salt crystal together is weakened. The developer is a reducing agent that has an affinity for oxygen and it is able to reduce crystals with weakened bonds. We are taking a timed reaction. Given sufficient time the developer will reduce all however, in the time allotted it reduces essentially exposed crystal.

The reduction process liberates the halogen component and it is dissolved away into the water of the developer. The silver component of the crystal is thus liberated as a tuff of metallic silver. It is this opaque tuff of metallic sliver that makes the image you see. The tuff of metallic silver is not the grain structure you see. Grain is a clumping the many tuffs of metallic silver. Thus the degree of granularity is proportional to the density of the tuffs of metallic silver in any unit area.

Now the ISO of a particular film will be the average sensitivity of silver halogen crystals. Some will be totally insensitive to light thus they will not be reduced by the developer. Some will have weakened bonds (developable) from the start. These will be reduced forming an overall background density comprised of metallic silver. We call this a fog level or base fog level.

The bottom line is the granularity you see is based on the count of metallic silver clumps. Thus granularity is a variable based on the density of the image in any one unit area.

Grain as seen in the final print is not uniform as to granularity. Grain is more obvious in mundane areas like expanses of sky. In other words, the subject matter has much to do with our perception of granularity in a picture. The type of illuminant in the enlarger is a major contributor to our perception of granularity. An enlarger equipped with a condenser light source beams through the image on the emulsion. The directness of the light mitigates emulsion turbidity. This action adds to our perception of sharpness and granularity. A diffuser type enlarger light source scatters the illuminant. This method mitigates to some degree the granularity of the negative. Granularity in the print is seen in proportion to the density of the silver that forms the image.

Improve this answer
\$\endgroup\$
4
  • 1
    \$\begingroup\$ Perhaps you could connect the dots between the various brightness of a specific area to the density of the tuffs of metallic silver in that same area. You've got a good start on the background theory, but you haven't really applied that to the specific particulars of the question as it has been asked. \$\endgroup\$
    – Michael C
    Commented Dec 10, 2015 at 0:10
  • 1
    \$\begingroup\$ I don’t know how to connect the dots any better. The densities seen on the negative are proportional to scene brightness. Thus granularity is more noticeable in the dense areas of the negative. Reduced silver in the image translates to a finer grain. This is accomplished using low speed films and fine grain developers. These contain silver solvents that whittle away some of the silver. Some tone the image a warm olive, this adds printing density to the abridged silver image. \$\endgroup\$
    – Alan Marcus
    Commented Dec 10, 2015 at 3:19
  • 1
    \$\begingroup\$ Could you edit your answer to reflect that information in your last comment? Especially with regard to which areas (bright or dark in a print) have a higher count of silver clumps, and how film speed might affect granularity? Your answers seem to reflect an expectation that the reader already knows as much or more about the subject as you do and can thus draw inferred conclusions from your information. You present a lot of general knowledge but don't really tell anyone with less knowledge of the subject how that applies to the specific question being asked. \$\endgroup\$
    – Michael C
    Commented Dec 10, 2015 at 4:35
  • \$\begingroup\$ I have been coming back to this question now and reading the answer again, I still can't completely connect all the implications. I still miss the connection of blacks in a scene equal dense areas of the negative (had to look it up again). You say less light equals less weakened bonds equals less reduction of the silver by the developer, hence less metallic silver that can be washed away. Metallic silver stays in developement, the rest gets washed away. Is that so far correct? It is not clear to me why the metallic silver tuffs now build up colloids/clumps in denser areas than less dense ones. \$\endgroup\$
    – kamuro
    Commented Apr 12, 2016 at 9:57
0
\$\begingroup\$

Larger ISO has a larger graininess, but the grain is most visible in areas with little exposure. With more exposure, the individual grains become less conspicuous.

Improve this answer
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.