F stop ratios

[Ryan K]

Forgive the elementary nature of the question but can anyone give me an explanation as to why F stop ratios are expressed in multiples of 1.4 (2.8, 4, 5.6 etc) when each increase in stop halves the amount of light passing through the aperture? I'm afraid the mathematics here have short circuited my brain. ;)

 

Many thanks indeed.

[David Mullen ASC]

F-stops are calculated by dividing the focal length by the effective aperture size (I guess the diameter is used), so as the hole gets bigger the number gets smaller. As to why we used "2.8" etc. I suppose it depends on what number you start with that determines the rest of the increments.

 

I'm guessing that if you start with f/1.0, then you end up with f/1.4, f/2.0, f/2.8, etc. when each aperture is doubled in size each step. Multiplying the diameter by 2X gives you 4X the amount of light since mathematically if you doubled the diameter while keeping the focal length the same, you go from f/4 to f/2, for example, which is 2-stops.

[Chris Keth]

F-stops are calculated by dividing the focal length by the effective aperture size (I guess the diameter is used), so as the hole gets bigger the number gets smaller.  As to why we used "2.8" etc.  I suppose it depends on what number you start with that determines the rest of the increments. 

 

I'm guessing that if you start with f/1.0, then you end up with f/1.4, f/2.0, f/2.8, etc. when each aperture is doubled in size each step. Multiplying the diameter by 2X gives you 4X the amount of light since mathematically if you doubled the diameter while keeping the focal length the same, you go from f/4 to f/2, for example, which is 2-stops.

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I believe the series is derived from the area of the circle inside the aperture. The numbers are formulated how David said, but I think those particular numbers are used because each larger number is representative of half the area for light to pass through the lens.

 

 

For example a 55mm lens at F2.0 would have a diaphragm diameter of 27.5mm. This yields an area inside the diaphragm of 593.957mm^2.

 

The same lens at F2.8 would have a diaphragm diameter of 19.64mm. That yields an area of 303.039mm^2.

 

That gives a ratio of larger area to smaller of ~1.96 (where theoretically we should get exactly 2.00), and the margin of error could easily be due to my hasty rounding-off of numbers.

[John Sprung]

I'm guessing that if you start with f/1.0, then you end up with f/1.4, f/2.0, f/2.8, etc. when each aperture is doubled in size each step. Multiplying the diameter by 2X gives you 4X the amount of light

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That's right. Doubling the diameter gives you four times the area (half of pi times the square of the radius). Four times the area lets thru four times the light. To double the area and light, you need to multiply the radius by a number that will give you two when you square it. That's the square root of two, which is 1.41426..... For our purposes, 1.4 is close enough.

 

 

 

-- J.S.

[Dominic Case]

why F stop ratios are expressed in multiples of 1.4 (2.8, 4, 5.6 etc) when each increase in stop halves the amount of light

Quite simply, the f stop tells you how much light passes through the lens and therefore how bright the image is. THe two factors are the focal length of the lens, which determines the image size (a longer lens forms a larger image, therefore less bright, or less intensity of light) and the size of the aperture (or diaphragm) (a larger aperture lets in more light). The Fstop is calculated as the diameter divided by the focal length, as it turns out that this results in a constant exposure for all lenses.

 

Double the diameter of the aperture, you get four times the area of the lens that can collect light, therefore four times the amount of light (two stops). Similarly, halve the focal length, you get a half-sized image, or a quarter the area of image: once again, four times as bright.

 

So to get steps that increase in just one stop intervals, you need a factor of the square root of 2 - strictly 1.41428....., but 1.4 does it pretty well.