Why camera settings go 2, 2.8, 4, 5.6
Tonight's idea
A camera lens has a round opening that can be made wider or narrower, called the aperture. Its size is given as an f-number \(N\): the focal length \(f\) of the lens (a fixed length, such as 50 mm) divided by the diameter \(d\) of the opening. So \(d = f/N\), and the setting is written f/4. A bigger f-number means a smaller hole.
The light that gets through depends on the area of the opening, and area goes with the diameter squared.
Light let in is proportional to \(d^2\), so to \(\dfrac{1}{N^2}\).
Photographers want each step, called a stop, to halve the light. Halving the diameter would cut the area to a quarter. To halve the area, divide the diameter by \(\sqrt2\), which means multiplying \(N\) by \(\sqrt2\).
Start at 1 and keep multiplying by \(\sqrt2\): 1.41, 2, 2.83, 4, 5.66, 8, 11.3, 16. Rounded, that is the scale on a lens: 1.4, 2, 2.8, 4, 5.6, 8, 11, 16. Every second number doubles, because \(\sqrt2 \times \sqrt2 = 2\).
Watch one: a 50 mm lens
At f/2 the opening is \(50 \div 2 = 25\) mm across. Its area is \(\pi \times 12.5^2 \approx 491\) mm².
One stop on is f/2.8, which is really \(2\sqrt2\). The diameter is \(25 \div \sqrt2 \approx 17.7\) mm and the area is \(\pi \times 8.84^2 \approx 245\) mm²: half.
At f/4 the diameter is 12.5 mm and the area about 123 mm²: a quarter. Check with the rule: \(N\) doubled from 2 to 4, and \(1/2^2 = \tfrac14\).
Your turn
- (a) Work out \((\sqrt2)^n\) for \(n = 1\) to 8, to 2 decimal places. Which number on the lens scale is not what correct rounding would give? (b) Find the next two stops after f/16.
- A lens has focal length 100 mm. (a) How wide is the opening at f/4 and at f/8? (b) How many times as much light does f/4 let in? How many stops apart are they?
- (a) How many times as much light does f/2 let in as f/16? How many stops is that? (b) The total light in a photo is the light let in each second times the time the shutter stays open. A photo is right at f/4 for \(\tfrac{1}{500}\) s. What time gives the same total at f/8?
- A 50 mm lens and a 200 mm lens are both set to f/4. (a) Find each opening's diameter. (b) How many times the area is the larger opening? (c) The 200 mm lens makes everything in the picture 4 times as wide and 4 times as tall. Why are the two pictures equally bright?
Stretch: thirds of a stop
Many cameras click in thirds of a stop: three equal multiplying steps take \(N\) from one full stop to the next. What does each click multiply \(N\) by? Find the two settings between f/4 and f/5.6.
Answers
- (a) 1.41, 2, 2.83, 4, 5.66, 8, 11.31, 16. To two figures 5.66 rounds to 5.7, but lenses are marked 5.6. The others round as marked. (b) \(16\sqrt2 \approx 22.6\), marked f/22, then f/32.
- (a) \(100/4 = 25\) mm and \(100/8 = 12.5\) mm. (b) The diameter is twice as big, so the area and the light are \(2^2 = 4\) times as much. Two stops, since each stop is a factor of 2.
- (a) \(N\) is 8 times bigger, so the light ratio is \(8^2 = 64 = 2^6\): six stops. Count them: 2, 2.8, 4, 5.6, 8, 11, 16. (b) f/8 lets in a quarter of the light each second, so the shutter must stay open 4 times as long: \(\tfrac{4}{500} = \tfrac{1}{125}\) s.
- (a) \(50/4 = 12.5\) mm and \(200/4 = 50\) mm. (b) The diameter is 4 times as big, so the area is 16 times as big. (c) The long lens collects 16 times the light but spreads it over \(4 \times 4 = 16\) times the area. The light on each bit of the picture is the same. That is why photographers quote \(N\) and not the diameter: equal f-numbers give equal brightness on any lens.
Stretch. Call the multiplier \(m\). Three clicks make one stop, so \(m^3 = \sqrt2 \approx 1.414\), and the cube root gives \(m \approx 1.122\). (Since \(m^6 = 2\), it is the sixth root of 2.) From 4: \(4 \times 1.122 \approx 4.49\) and \(4.49 \times 1.122 \approx 5.04\), then 5.66. Cameras show f/4.5 and f/5.0.