According to my calculations, 2.8,4,5.7,8,11.3,16, and 22.6 should be the standard f-stops that half or double area of the diaphragm. How did we get the standard f-stops we have, and did we come to expect they half and double light on sensors?
#1. "RE: Standard f-Stops" In response to Reply # 0
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Successive f/stop values do indeed represent a halving (or doubling) of the area of the lens' diaphragm opening, and therefore a corresponding difference in the amount of light transmitted. That's assuming that the lens is correctly adjusted, of course
#2. "RE: Standard f-Stops" In response to Reply # 1
I figured out the error in my calculations. I was using radius in places where I should have been using diameter. Thus,1.4, 2.0, 2.8, 4.0, 5.7, 8.0, 11.3, 16.0, and 22.6 are what I get for the exact doubling or halving of light. Now, I am trying to figure out where f-1.8 comes from and how lens makers get away with the cost difference between 1.8 and 1.4. It seems just as easy to make diaphram elements that go open up to either.
#7. "RE: Standard f-Stops" In response to Reply # 6
Another factor to consider is the equation which includes the focal length of the lens: lens focal length/diameter of the aperture = f-stop. The f-stop number is a convenient way of describing the exposure settings rather than noting the area of the aperture. And, a given f-stop produces the same exposure regardless of the lens focal length (for a given shutter speed). For example, 1/500th of a second at f/5.6 on a 50mm lens gives the same exposure as those settings on a 300mm lens.
This becomes obvious if you think about an exposure meter - it only gives shutter speed and aperture regardless of the lens or camera type.