<<[snip]... "Slide film maps a density range of 0:2.7 to an intensity range
of
1:500, but negative film maps a smaller density range of 0:2.4 to a larger
intensity range of 1:4000."
Trnaslation for the digitally uninitiated, please, and comment on the
implications.>>
I quoted this because I don't know too much but I will attempt to render my
understanding of this.
Density range is measured logarithmically meaning that you get a doubling
every 0.3 in density increase. The intensity range goes by doubling the
measured amount of light in the scene (lamberts or foot-candles or whatever)
in a doubling progression of 1, 2, 4, 8, 16, etc.,, with each number in the
sequence denoting one f stop more of light. So, 2.7 density range means slide
film can have 8 f stops of difference in light as measured on the film. But
the relationship between film density and the exposure range of the film is
not 1 to 1 but is compressed somewhat. So mapping a density range of 0:2.7 to
an intensity range of 1:500 means you can cram 10 stops worth of information
into 8 stops worth on the transparency. Likewise with negative film you can
cram a wider range of information in your scene onto the negative. In other
words, 13 f stops worth of information can fit into 7 stops on the negative.
"Latitude" by itself means what f stop range a particular film can handle.
For the negative, it has a wide latitude because it can accommodate 13 stops
of info. Likewise slide film has less latitude.
"Exposure latitude" is an error factor. If the scene you are taking has
all the essential information with a 6 stop range, then you have an extra 7
stops to play with on your negative. As a point in fact, you can still get a
very printable negative if you overexpose by as much as 4 to 5 stops. I think
Popular Photo pointed this out about a year ago. This is the theory behind
Kodak Max film. It is actually a 800 ISO film but you can overexpose it at
100 ISO an still get good results.
Warren
-hopefully I am half-way correct and not too confusing
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