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What should the f# of a lens aperture
be that transmits 100, 1000 or any factor in between more light in relation to any
starting f#?
Andrew Davidhazy School of Photo Arts and Sciences Rochester Institute of Technology So let's start with f/16 and we want to select an aperture that transmits 10 time more than f/16 1. So, we find the logarithm of 16 which is 1.2 2. Next we find the logarithm of 10 which is 1.0 3. Then we divide the log of 10 by 2 and we get .5 4. Then we subtract .5 from 1.2 and we get .7 5. We find the inverse logarithm of .7 and ... and we come up with 5 or f/5!! which is 3 and 1/3 stops more aperture than f/16 and f/5 delivers 10x more light to image than f/16 (this is also 3 1/3 stops larger aperture) - perfect! so 16 11 is 2x more light than 16 8 4x 5.6 8x 5 10x <- it works!! 4 16x So now how about selecting an off or odd factor ... like 300? log 16 = 1.2 the log of 300 is 2.47 we divide that by 2 ... 2.47 / 2 = 1.235 we subtract that as follows: 1.2 - 1.235 = -.035 and the antilog of this is: .92 so 16 11 is 2x more light than 16 8 4x 5.6 8x 4 16x 2.8 32x 2 64x 1.4 128x 1 256x .92 300x <- it works!! .7 500x OK - so now let's start with another f# - how about f/45 and we want the aperture that will deliver 800 times more light than f/45? 1. the log of 45 is 1.65 2. the log of 800 is 2.90 3. we divide 2.90 by 2 and we get 2.90 / 2 = 1.45 4. we subtract that from 1.65 and we get 1.65 - 1.45 = .20 5. and the antilog of .20 is ... 1.58 45 32 2x 22 4x 16 8x 11 16x 8 32x 5.6 64x 2 128x 1.4 256x 1 500x 1.58 800x <- it is here!! it works again!! .7 1000x Procedure independently developed on January 20, 2011 between 4 pm and 8 pm - surely this has been formulated by others in the past but for me: a first! |