Embarrasing Question which I should be Certain of

[photostranger]

On another post (RX v. RXII) my math was questioned about f-stops. I've been using manual exposure systems for a long time, you'd think I'd have no question what so ever, but I find myself second guessing myself. It's not 100% necissary, i get good exposures, but it deals with weather or not 1-stop can be expressed as 100% more light. Here is the original post, which i had over-complicated greatly:
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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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[Colin645]

Shawn, You are right. 1 stop is double or half the light E.g. 1/60 @f8 is double the exposure of 1/60 @ f11 (f9.5 = 1/2 stop or 50% more light;f11 = 1 stop or 100% more light.
Therefore a viewfinder brightness reduced by 20% is in fact 1/5th of a stop.
Colin

 

[photostranger]

Thanks, I was 99% certain i was correct, but there was that doubt that made me uneasy. If you could please post this at RXII>>RXII v. RX it would be greatly appreciated as it is cruitial for my advice to the original author (I argued that 1/5 stop is likely not going to be worth paying extra for an RXII over a demo or used RX)

thanks

 

[John_bird]

Hi Shawn Im afraid your not correct an increase cannot be equated to a loss when you open a lens 1 stop youn increase the light transmision by 100% on this point you are correct but when you stop down by 1 stop you reduce light transmision by 50% of the new value. Any value reduced by 100% is zero. As we are talking about losses(reduction) you will find I am correct.
Do hope this helps John