Over the years I have seen a lot written about Depth of Field and recently I’ve seen it explained wrongly but with great passion. So I thought I would post the basic formulae, show how they are derived and explain how they work in practical cases.
So first: a definition. When part of an image is out of focus that’s not an on/off state. There’s massively blurred, a little blurred, and such a small amount of blur it still looks sharply focused: if we magnify the slightly parts we can see that they are not sharply focused. Depth of field is a measure of how far either side of the point of focus, appears to be properly focused (even though it is very slightly out of focus).
When a lens focuses and image, the lens to subject distance, D, and lens to image distance, d, are related to the focal length with the equation
1/D + 1/d = 1/f
We can rearrange this to derive the distance to the subject (D) in terms of focal length (f) and image distance (d)
D = df/(d-f)
Since d is always further than f, we can write the difference as Δ and replace d with f+Δ. Putting that in the previous equation makes it
D = (f2+Δf)/Δ which re-arranges to
D = (f2/Δ)+f
This can be rearranged as the Newtonian form of the equation
D-f = f2/Δ , therefore
Δ(D-f) = f2 and since Δ = (d-f)
(d-f)(D-f) = f2
We can work out our a focus scale for a lens using D = (f2/Δ)+f . Assume we have a 60mm lens, and it moves in or out 1/3mm for each 30 degrees of turn ;
When the lens is 60mm from the image we can mark ∞ at the 12 O’clock position: Δ = 0 and D= ∞,
if we turn the ∞ mark to the 1 O’clock position (30 degrees) Δ = 1/3 and D= 3600/(1/3) = 10800 + 60 = 10.86 M, so we can write 10.9 in at the new 12 O’clock position
turn the ∞ mark to the 2 O’clock position (60 degrees) Δ = 2/3 and D= 3600/(2/3) = 5400 + 60 = 5.46 M , so we can write 5.5 at the latest 12 O’clock position
turn the ∞ mark to the 3 O’clock position (90 degrees) Δ = 1 and D= 3600 + 60 = 3.66 M, so this time we write 3.7 at 12 O’clock
turn the ∞ mark to the 4 O’clock position (120 degrees) Δ = 4/3 and D= 3600/(4/3) = 2700 + 60 = 2.76 M, so 2.8 goes on the scale at 12 O’clock
turn the ∞ mark to the 5 O’clock position (150 degrees) Δ = 5/3 and D= 3600/(5/3) = 2160 + 60 = 2.22 M
turn the ∞ mark to the 6 O’clock position (180 degrees) Δ = 2 and D= 3600/2 = 1800 + 60 = 1.86 M so we can 2.2 and 1.9 to the scale to finish the job
And so on. For simplicity of calculation we often consider the extra 60mm insignificant and D ≈(f2/Δ) is usually close enough. It’s also worth noting that the roles of D as subject distance and d as image distance can be swapped – the whole arrangement is symmetrical.
In the diagram above, the blue lines show the lens is focused at a distance D, which has a lens to image distance (d) of (f+Δ) and D=f2/Δ +f , something further away than D will not come to a sharp focus at the image plane, but some distance in front of it (something nearer than D will come to a focus behind the image plane). Focused rays of light form a cone: if the point of the code is not on the image plane, the rays form a disc which is called “a circle of confusion”. The red lines in the diagram illustrate the case for something at infinity and show how smaller a aperture width (bigger f/ number) leads to a smaller circle of confusion.
The only factors which determine the size of the circle that is formed are focal length, aperture, and the distance between the lens and the image (i.e. the distance at which the lens is focused) Two set-ups using the same focal length, and same aperture, focused at the same distance will produce the same size circle regardless of the size of the recording medium which captures the image, the size of the image circle produced by the lens or any other factor.
A point at infinity will form an image at a distance f behind the lens (that’s the definition of focal length) and so we know it forms a image Δ in front of the film/sensor in the setup in the diagram.
The red lines form two similar triangles between the lens and the image. The “base” of the large one is w (the aperture width) and its "height" is f.
We normally write aperture as a ratio between width and focal length, e.g. f/2 means the aperture’s width is half the focal length.
So f = aw (where a is the f/ number) , so we can say this triangle has a base of w and a height of w*a
The base of smaller triangle is the circle of confusion from the mis-focused point at infinity.
This circle’s diameter is normally written as c, so using similar triangles the height of the smaller triangle must be its base * a, so:
Δ = c * a
As the lens moves further away from the image, the circle for the point at infinity gets bigger: a small enough circle looks like a point and but there comes a size where we can see it is a circle.
If we know that size we, can calculate the value of Δ as c*a and since we know that D = (f2/Δ) + f, we can define the subject distance when a point at infinity starts to look out of focus as
(f2/ca) + f . This distance is known as the hyperfocal distance (H) strictly, H = (f2/ca ) + f, but it usually accurate enough to write H ≈ f2/ca ;
It later we’ll use a rearrangement of this: since Δ = ca, this simplified form of the equation can be turned into Δ≈f2/H
We can see that we get the same size circle if the image plane is c*a in front of where the image would focus as well as c*a behind it, so we can say
(1) for an subject at distance is D, the lens to image distance is approximately (f2/D)+f (more accurately it is (f2/(D-f))+f ) and
(2) the zone of apparent sharp focus runs from anything which would be in focus at (f2/D)+f -ca to anything which would be in focus at (f2/D)+f + ca
This formula is accurate enough for most purposes: but it would be more accurate to say the range runs ((f2/(D-f))+f)*(f+ca)/f to ((f2/(D-f))+f)*(f-ca)/f because this accounts for Δ getting slightly bigger as d increases for nearer and nearer subjects. The error is biggest at short distances with wide apertures.
A 35mm frame with c=0.03 and an aperture of f/32, gives c*a ≈ 1. If we focus a 50mm lens at 1m (f2/D)+f = 52.5,
So the simple form of the formula would say the in-focus zone ran form 51.5-53.5mm. The long form is 51.45- 53.55.
So instead of the dof extending from 1.716M to 0.764m it actually goes from 1.774 to 0.754m.
Since we only measure distance to 1 or two significant figures, aperture to 1 or 2 significant figures (and f/22 is really f/23) and focal length to the nearest whole mm (and stated focal length can be inaccurate by 1 or 2 mm) the simple formula gives the the point where most people kind-of feel that the image isn’t really properly focused to enough accuracy.
It’s also worth noting that the if we have a focus scale like the one out lined above the same distance either side of a focus will give the same Δ, so we can calculate Δ for each aperture mark, and put depth of field scale marks on a lens.
∞ 11. 5.5 3.7 2.8 2.2 1.9M
^ | ^
If we want to work out D.o.F numbers (e.g. to make our own tables) , we know that the lens to image distance for the far point (df ) is (f2/Df)+f and for the near point, (dn) it is (f2/Dn)+f
therefore, f2/Df + f = f2/D + f – Δ (or + Δ for the near point)
we can remove +f from each side and get f2/Df = f2/D – Δ ;
since Δ = f2/H, we can rewrite this it as f2/Df = f2/D – f2/H ;
the f2 terms cancel out so we get 1/Df = 1/D – 1/H , for the far point and for the near point 1/Dn = 1/D + 1/H ;
We can rewrite these as 1/Df =(H-D)/(H*D), for the far point and for the near point 1/Dn =(H+D)/(H*D) so
Df = HD/(H-D), for the far point and for the near point Dn =HD/(H+D)
These produce an interesting series
|Focus Distance (D)||Near Point(Dn)||Far Point(Df)|
In other words, if the focus distance is H/x the near point is H/(x+1) and the far point is H/(x-1).
[These formulae: Dn = H/((H/D)+1) , Df = H/((H/D)-1) can be re-arranged to Dn = H/((H+d)/D) , Df = H/((H-D)/D) and then to Dn = HD/(H+d), Df = HD/(H-D) – the original formulae ]
This can useful for doing a quick mental d.o.f calculation. A 50mm lens @ f/8 on full frame has a hyperfocal distance of roughly 10m (502/(.03*8) +50 = 10.46M). If I focus at 1M (roughly H/10) the near point is H/11 = 0.90909M and the far point is 1.1111M so I have roughly 9CM in front and 11 CM behind
Earlier I said “As the lens moves further away from the image, the circle gets bigger: a small enough circle looks like a point and but there comes a size where it starts looking like a circle. If we know that size…”
How much of the image the circle occupies, determines whether it is judged to be still in focus, or a long way out of focus. So value for c must be proportional to the size of the image, after any cropping has been done.
By convention 35mm film used c=0.03mm and APS-C crop sensor cameras use c=0.02mm. Changing image (sensor) size changes allowable circle size c, and so changes Δ , and so the depth of field scale on a lens designed for one size of of image needs to be adjusted if used on a camera where the image is different size (on an APS-C camera reading the scale for 1 stop wider aperture than actually set will give roughly the right reading).
Size of the circle formed does not depend on image size but allowable circle size does and hyperfocal distance and apparent depth of field change when c changes
Changing sensor size (keeping same position with the same lens and accepting a change of framing).
If we use two different cameras – i.e. use a different circle size – at the same spot, focused on the same place and we use the same focal length and same aperture on both, then the one with the smaller image has less depth of field. It doesn’t matter how we get to the smaller image, whether it is by cropping a big one or starting with a smaller film/sensor size.
We get less D.o.F because c has become smaller, so f2/ca – the hyperfocal distance has moved further away. When you look at f2/ca, a smaller value of c needs a larger value of a to compensate.
Changing sensor size and focal length (getting the same framing from same position)
If we use two cameras – with different circle size – and use different focal lengths to give the same angle of view, but keep the same aperture then the larger image will have less depth of field because the f and c have gone up by the same factor , but f is squared in the equation. A larger value of a is needed to compensate for f being squared.
So: a 50mm @ f/8 on full frame has the approximate field of view and depth of field of a 35mm @ f/5.6 on APS-C . If that’s we want, the full frame camera needs to use a slower shutter speed or higher ISO to compensate, which have their own side effects.
If we want the depth that comes from the 35mm @ f/32 on APS-C , the 50 might not stop down to f/44 to give the same depth on Full Frame.
But if we use the 50 @ f/1.4 to isolate a subject from the background on full frame the 35 probably doesn’t open up to f/1
Changing focal length and camera position
People often think of perspective as a function of the angle of view of the lens. Strictly that isn’t correct : perspective is a function of the ratios of subject to camera distances. If you have two items the same size with one a meter behind the other and you stand a meter from the nearer one, the far one is 2M away, and will appear 1/2 the size. If you stand 10 meters from the first (and therefore 11 meters from the second), the far object will appear 10/11ths of the size. It doesn’t matter what else is in the frame. But: if you fit a wider lens the natural response is to move closer to the subject : it is that change of viewpoint which causes that the change of perspective. Changing focal length and keeping position constant means the perspective is constant, and the framing changes. Changing focal length and keeping framing constant means a change of position and with it a change of perspective.
If you have two lenses for the same camera and a choice between standing close with a wide angle lens or further away with a telephoto (and accepting the change of perspective for the same framing) we can work out the distances.
Let’s say with the short lens, H is 10 and you stand 5 meters away.
The near point is (10 * 5) / (10+5) = 3.33 : 1.67 meters in front
The far point is (10 * 5) / (10-5) = 10 : 5 meters behind = 6.67 in total
If we double the focal length and stand twice as far away the hyperfocal distance increases 4 fold (if the circle size and aperture don’t change), so we get a d.o.f zone like this
(40*10) / (40+10) = 8 : 2 meters in front
(40*10) / (40-10) = 13.33: 3.33 meters behind =5.33 in total.
Notice the background is more out of focus with the long lens, but there is actually MORE in focus in front of the subject. The wider lens includes more "stuff" in the background and it is sharper – which is why long lenses are thought of as better at isolating a subject from the background.
Changing camera position and sensor size.
If you only have one lens and your choice is to move further away and crop the image (or use a smaller sensor) or come close and use a bigger image what we can calculate that too: keeping the full image / close position as the first case from the previous example we would keep the near point 1.6667 meters in front and a far point 5 meters behind = 6.67 in total
If we use half the sensor width, we halve c and double H, if we double the distances we have doubled every term in the equation.
(20*10) / (20+10) = 6.6667 : 3.33 meters (in front)
(20*10) / (20-10) = 20 : 10 meters (behind) – 13.33Meters in total, so you get twice as much in the zone either side of the image but dropping back and cropping.