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Exposure Controls, Part 2

The Standard f-

At the end of Part 1, we obtained this relationship:

f-

This section explains how the standard f-

Aperture diameter and f-

Remember the factor-

What must you do to the diameter of a circle in order to double the area? Multiply by 2? No. That would make the area 4 times greater because the area varies as the square of the diameter.

To double the area of a circle, multiply its diameter by the square root of 2, which is 1.414, rounded off.

Going in the other direction, what must happen to the diameter in order to cut the area in half? And the answer is, . . .

To cut the area of a circle in half, divide the diameter by the square root of 2, or equivalently, multiply by 0.707.

What’s the point? It is desirable to find a series of f-

Suppose we have an unusual lens, one whose diameter (maximum aperture) is the same as its focal length. We’ll take the diameter and focal length to be 100 mm just to be specific. This represents a sizable chunk of glass because 100 mm is just about 4”.

At maximum aperture, what will be the f-

f-

= 1 It’s an f/1 lens.

Let’s express the aperture diameter in terms of the f-

aperture diameter = focal length / f-

To check this out, let’s apply it to our rather unusual lens:

aperture diameter = 100 mm / 1

= 100 mm We knew this.

Suppose we reduce the aperture diameter enough to cut the area, and light transmission, in half. From the explanation above, we should divide the diameter by the square root of two, which is 1.414:

New aperture diameter = 100 mm / 1.414

= 70.7 mm

Calculate the f-

f-

= 1.414

What this tells us is that the f-

Let’s do another click, so to speak. What should be the new aperture diameter that will reduce the exposure by one half from what it is with the 70.7 mm aperture?

To find the new diameter, divide the present one of 70.7 mm by 1.414, and you will
get 50 mm. To find the corresponding f-

f-

= 2

Thus, the third in our series of standard f-

Do another one. Divide 50 mm by 1.414 to get the aperture that will transmit half
as much light as the 50 mm aperture. The result is 35.36 mm. Then calculate the
corresponding f-

f-

= 100 mm / 35.36 mm

= 2.828

This is f/2.8, with a little rounding off, and is the fourth in the series of standard
f-

Take a look at the series so far: 1, 1.4, 2, 2.8

What’s the relationship between these numbers? If you have one, such as f/2, how do you get the next one?

Answer: To get the next one in the series, multiply by 1.414.

The f-

Here’s the complete series of standard f-

1, 1.4, 2, 2.8, 4, 5.6, 8, 11, 16, 22, 32, 45, 64, 90

And what is their significance? In going from one to the one next to it, the exposure is either cut in half, if you go to a larger number, or doubled, if you go to a smaller number.

There is another relationship that allows you to calculate the standard f-

1, 2, 4, 8, 16, 32, 64, 128, 256, . . .

If you take the square root of each of these numbers, you get a standard f-

What’s an f-