 BetterPhoto Member |
aperture what is the difference in the exposure of a film between f5.6 anf f11
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Justin G. |
what test or application are you filling out? A: 2 stops.
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Pete H |
Are we assuming the same shutter speed? LOL
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Michael H. Cothran |
Assuming the shutter speed does remain the same, there are two answers. One was given above - 2 stops. The second answer is f11 allows in 1/4 the amount of light as f5.6, or conversely, f 5.6 allows 4 times the amount of light as f 11. Now, let's go a little further. Perhaps this will help you and others understand the origins of the "numbers" used in the aperture systems on lenses. There are really two sets of apertures at play on your lens. Those deriving from a balanced lens of f1, and those deriving from the pi ratio of f1, which is f1.4 Changing from one f-stop to another in the same set of apertures results in either a 4x increase in light or a 1/4x decrease in light, depending on which direction you are going. If you intermingle the two sets, as is done on all camera lenses, the difference from an f-stop in set one to the next f-stop in set two is a 2x increase, or 1/2x decrease in light, again depending on which direction you are heading.
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Justin G. |
the pi ratio of f1? could you ellaborate? I never knew that, I just can't think of how it looks. (and I know this reply sounds like a "polite way" of telling you you're wrong) but it's not I promise, I just like math and it's interesting how it and the logarithms play in photography. and i've seen your other explanations in different threads, very informative! thanks! justin.
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Pete H |
Yes.. and let's not forget Newtonian physics which states that all energy (e) including sub particles (-e) will integrate so long as we do not disregard primary derivitives. Further..as the state of entropy approaches 0.0, we place Dx/dt + sub 1-c^2 to approximate (v)-m^-2(sin-1) Therefore; as c= or approaches (1-(dx)+1) we differentiate the two sub particles including those of atomic wts of 10^-12/pi-2 arc radians...again assuming mV-md^-2(Gm+)-Gm(2arccos)
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anonymous |
Boys..... Boys .... Boys.... enough with all that mathematical cr*p now. LOL Michael - just learnt that in Tafe last week. Remembered it for about 2 hours, then poof out of the head, I don't have the spare grey matter to use remembering it unfortunately.
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Alan N. Marcus |
The mystery of the f number set. Few understand the math. The camera lens gathers light and projects an image on the film plane. The intensity of the projected image is based in part on the lens’s surface area. Since most iris diaphragms are circular, consider the following math problem. The working diameter of a lens is measured and found to be 10mm. Its area is 78.5 sq. mm. Your hearts desire is to enlarge the lens to increase the light energy at the film plane by a factor of 2. What must the revised lens diameter be to achieve twice the light energy.? Geometric solution! Multiply measure diameter by the square root of 2 which by the way is 1.4142. OK to round to 1.4 for convenience. Multiplying, the answer is about 14. Now find the surface area of a 14mm lens, the answer is about 157 sq. mm. which has twice the surface area. Construct any circle; multiply its diameter by 1.4. Construct a new circle with this adjusted diameter and your new circle is 2x larger. Conversely, dividing by 1.4 constructs a circle with half the area. The f number set is each number multiplied by 1.4142 and then rounded.
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