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Chemistry of Photography. The light sensitive emulsion The latent image Developing the image Fixing the image. The emulsion. AgNO 3 + KBr = AgBr + KNO 3 in gelatin AgBr precipitates (WHY??) and remain in the gelatin to form minute grains.

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chemistry of photography

Chemistry of Photography

The light sensitive emulsion

The latent image

Developing the image

Fixing the image

the emulsion
The emulsion
  • AgNO3 + KBr = AgBr + KNO3 in gelatin
  • AgBr precipitates (WHY??) and remain in the gelatin to form minute grains.
  • AgBr is light sensitive, forming a latent image that can be “developed”
  • But how?
The sensitivity of the grains are proportional to their sizes. If all the grains were the same size, there would be no shades of grey at all! Typical densities of grains are about 5 x 108 grains per cm2. If you consider a grain to be equivalent to a “pixel”, you see that photographic film (taken by itself) it quite a bit more capable of “resolving” detail than our current digital cameras.
the latent image
The latent image
  • For many years, it was thought that 2AgBr + light = Ag2Br + Br (the “sub-haloid” hypothesis…). But there was never evidence of a chemical change. Less than 5 silver atoms are involved at any site!!
  • X-ray spectroscopy finally showed that silver is liberated

Br - + light  Br + e –

The electron then migrates to a shallow “trap” (called a sensitivity site).

Ag + + e -  Ag

Species produced include: Ag2+, Ag2o, Ag3+, Ag3o, Ag4+, Ag4o

Why doesn’t it go the other way? i.e. why is it stable?

the mechanism of development
The Mechanism of Development
  • The photographic process depends upon the fact that the reaction:

Ag + + e  Ag

(i.e. the reduction of silver ion to metallic silver by a developing solution), proceeds much more easily for an exposed silver halide grain than for an unexposed grain. The “gain” can be ~109.


Development- Continued…

The reduction potential of the developer must be such that it will develop those exposed silver halide grains, but not large enough to develop them all. (A “fogging” developer…)

What actually happens?

C6H4(OH)2 + Na2SO3 + 2AgBr +NaOH C6H3(OH)2SO3Na +2NaBr+H2O +2Ag

Hydroquinone sodium sulphite silver bromide sodium hydroxide hydroquinone sulphonate sodium bromide water SILVER!

| |

stabilizer ya gotta do something for the bromine! (plus it adjusts the pH)

Chemical velocity: ΔT = 1o C  Δvchem = 10%.

hurter and driffield d vs log e
Hurter and Driffield…D vs. log E
  • Reflectivity ≡ amount of light reflected / amount of light incident = R
  • Transmissivity ≡ 1 – R = T
  • T = (1 – amount reflected) / amount incident
  • = (amount incident – amount reflected) / amount incident
  • = amount transmitted / amount incident
  • Opacity ≡ 1/ T
  • Density ≡ log Opacity = D
  • Exposure = amount of energy incident on each square centimeter of film
  • Question: Why do we use logarithms?
fixing the image
Fixing the image

The biggest problem after the invention of photography in the 1830’s was the lack of permanency. You have to get rid of that remaining bromide, or eventually the photograph will go black. There are no true solvents of AgBr. When sugar is dissolved in water, and then evaporated, the sugar is recovered. This never happens with AgBr. The residue left behind is always a transformed salt. So what we need to do is make sure the transformed salt is soluble, so it can be washed away.

AgBr + Na2S2O3 = AgNaS2O3 + NaBr (only slightly soluble)

But if we have a more liberal solution of sodium thiosulphate:

2AgBr + 3 Na2S2O3 = Ag2Na4(S2O3)3 + 2 NaBr (bingo!)

Does anything else work? KCN. We won’t go there….