#### photostranger

##### Member

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John- you are likely right, I have spent more time in the digital darkroom than out shooting. I am a digital media major that is about to switch to photography.

However, I could have sworn that 1 stop increase doubles the amount of light on the film. This is evident when you stop down and require a twice as long exposure.

If each stop down DOES cut the light in half, then each stop up DOUBLES the amount of light, right?

If this is correct, then:

f5.6=+0% differential (1:1)

f4.5=+50% differential (1.5:1)

f4.0=+100% differential (2:1, twice as much total light)

f2.8=+200% differential (4:1)

Ratios are expressed as "light on film plane:light available"

If you take a value of 50 and increase it by 25% (more light) then the result is 62.5 ((50x0.25)+50) or calculated from ratio and simplified (50x1.25). If "62.5" expresses the luminance on the focussing screen this would not be 1/2 stop, a value of 75 would be (50x1.5) and a value of 100 (50x2) would be one stop as 1 stop expresses twice as much light entering the film plane. This is assuming my ratios and factors are correct.

As a value of 50 would already be present (light available), the additional differentiated(sp) value would be added, resulting in the values 62.5 (12.5+50), 75 (25+50) and 100 (50+50).

In the case of a 25% or 1/4 stop increase (based on that 100%=2:1), would result in a +12.5 differential, as, 12.5 is 1/4 of 50 (50x0.25=12.5). A 25% increase could be expressed as 1.25:1.

So unless i'm TOTALLY off, which is entirely possible, 20% would be slightly less than 1/4 stop?

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