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As the focal length decreases, the depth of field increases as well. Why is this? I'm not so much interested in a physics lesson as I am interested in a simple, down-to-earth explanation.

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    \$\begingroup\$ Several similar questions have been closed as duplicates of photo.stackexchange.com/questions/37, but that question a) is more about what than why and b) lacks a good answer. \$\endgroup\$
    – mattdm
    Oct 2, 2011 at 12:46
  • \$\begingroup\$ There are some related technical/why answers here: photo.stackexchange.com/questions/9624 \$\endgroup\$
    – mattdm
    Oct 2, 2011 at 12:46
  • \$\begingroup\$ I think you mean the depth of field _de_creases. When something's very close to the lens it's very easy to throw out of focus – depth of field is shallow. When it's far away it's more difficult – depth of field is deep. \$\endgroup\$ Oct 2, 2011 at 12:48
  • \$\begingroup\$ No, I meant the focal length. A lens at 24mm has more depth of field at 200mm. \$\endgroup\$
    – Daniel T.
    Oct 2, 2011 at 13:10
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    \$\begingroup\$ It's essential to state whether you're changing focal length in isolation, or whether you are also changing the subject distance to compensate for the different angle of view (keeping the subject approximately the same size in the final image) as the answer is different in each case. \$\endgroup\$
    – Matt Grum
    Oct 2, 2011 at 14:21

6 Answers 6

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Pretty sure I answered this one before but I cannot find it.

  • As focal-length gets longer, the angle of view gets smaller.
  • With a smaller angle of view, rays forming the image are closer to being parallel.
  • With less variation of angle between rays, light has to travel more before being sufficiently out of focus.

This is a little oversimplified but I hope it is easy to visualize at least.

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    \$\begingroup\$ That's a very good intuition into what's going on (which I think is what the questioner was after), and about as succinct as I've ever seen it put! \$\endgroup\$
    – Matt Grum
    Oct 2, 2011 at 14:23
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    \$\begingroup\$ Your answer seems to suggest DOF increases as focal length increases. Shouldn't it be the other way around? \$\endgroup\$ Aug 20, 2018 at 21:22
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    \$\begingroup\$ The last point is confusing and seems to cause misunderstanding the facts! if the light rays are harder to be out of focus (which is wrong), then this means the DOF is greater!! \$\endgroup\$ Jul 2, 2019 at 4:35
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    \$\begingroup\$ This reasoning leads to the opposite of what happens in reality, unless I'm missing something. According to this reasoning, longer focal length leads to increased depth of field. But in reality, longer focal length leads to decreased depth of field! \$\endgroup\$ Jul 16, 2019 at 20:03
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For this discussion, the aperture should be considered the same, since the variance we're discussing is focal length.

So, a telephoto lens focussed on the same subject from the same distance as a wide angle lens will have shallower depth of a field as a result of higher magnification, but the angle of view between the two images is vastly different as a result. A telephoto lens and a wide angle lens focussed on the same subject with the same angle of view, will have the same depth of field (there is variation but it is negligable).

The difference here? angle of view. So, this is about your distance to the subject, really, not specifically the focal length. Vary the distance to accomodate the differences in focal length and the depth of field remains virtually the same. What does change, though is the foreground and background ratios for it. Wider angles tend to have more of the background in focus and telephotos tend have more of the foreground. The result of this behaviour can create an illusion of shallower depth as the telephoto would magnify the background blur. That's one reason that landscape photographers don't stand way back with a telephoto (haze and other factors would also play a role, probably more significantly).

You can test out my information at various sites that offer a DoF calculator such as DOFMaster for example. For example: for a distance of 10m (@ f/8 ) then 10mm DoF = Infinite and 100mm DoF = 3.08m. Now, move the 100mm lens to 100m (10 times further away) and the 100mm DoF is now equal to infinite. The angle of view of the 100mm lens is now the same as the 10mm lens.

In summary, wide angle lenses do not have more depth of field than telephoto lenses and this is demonstrated by both showing the same DoF for the same angle of view.

You can get some more detailed (and not math oriented) explanations at Cambridge in Colour and Luminous Landscape. The second link has sample images too, kind of handy for visually seeing it.

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  • \$\begingroup\$ "In summary, wide angle lenses do not have more depth of field than telephoto lenses and this is demonstrated by both showing the same DoF for the same angle of view." If that is true, how do you explain that smart phones and point-and-shoot cameras (with a small sensor and therefore short focal length) have much larger depth of field than full frame or medium format cameras (with large sensors and therefore long focal length for the same angle of view)? \$\endgroup\$ Jul 16, 2019 at 20:06
  • \$\begingroup\$ @RoelSchroeven A smaller sensor means we're cropping out more of the picture. So, take phone's f2.8 4mm lens, put it on a full frame camera. The subject lives inside the tiny area in the center of the frame. We're 10cm away. If we keep the subject size the same, but switch to a 80mm lens, we need to step 20x! as far away. So we step 200cm away using 80mm lens. In both cases you can imagine if the subject was even further the DOF would be very, very large. The phone is merely cropping out a tiny area in the super wide angle 4mm lens. \$\endgroup\$
    – xpirle
    Oct 27, 2023 at 0:24
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Depth of field is affected only by actual aperture size, but actual aperture size is not f stop. When we say "aperture" we actually mean "aperture ratio" or "f-stop", not aperture size.

This "aperture ratio" is what is required to calculate image brightness, but actual aperture size is need to calculate depth of field.

For any given f-stop value, the longer the focal length, the larger the actual aperture size in mm.

F stop is the ratio of aperture to focal length and is calculated by f-stop = focal-length / aperture.

To get actual aperture size from an f-stop ... aperture-size = (1 / f-stop) * focal-length

So for a 50mm f1.4 lens ..actual aperture size = 1(1.4 * 50) = 35mm aperture size.

The aperture size being the size of the hole the light passes through. To build a 100mm f1.4 lens, a 70mm aperture is required, which makes a really large diameter lens.

So the bigger the actual aperture, the smaller the depth of field, and for any given f-stop value, the longer the focal length, the larger the actual aperture opening being used.

F-stop was invented to make the calculation of exposure brightness easier, but complicates actually calculating depth of field. But before automatic cameras, calculating the desired f-stop and shutter speed was workable, but would have been a real pain if working with actual aperture size!

Note: As some others answers have discussed, as the distance to a subject increases, then the light from that subject will be more parallel. This means the further away a subject is the greater the depth of field. This will counter the effect of the longer lens with same f stop having a smaller depth of field. Consider the 50mm and and 100mm f1.4 lenses. The 100mm has a larger aperture size in mm, but if you have to move 2x further away to take the photo, the increased distance will counter increased actual aperture size and the depth of field will be similar as using the 50mm lens at a closer distance.

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  • \$\begingroup\$ There are lots of technical discussions about photography where different situations are compared, and conclusions are drawn from there. But often the discussions fail to mention exactly what changes and what doesn't change between the different situations, and that leads to lots of misunderstandings. You do state what changes and what doesn't, leading to the only correct result. +1. \$\endgroup\$ Jul 16, 2019 at 20:12
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Why do longer lenses have shallower dof ... in short, because they require physically bigger apertures to maintain the same f-stop number. (remember, the f stop value "f" = focal length / aperture.

Let's start by thinking about a true pinhole camera. It has no lens so no focal length and requires a really small pinhole to make a decent focussed image. if the pinhole is too big then nothing will be in focus. (ie seriously shallow dof !)

Now, if we put a lens in front of our pinhole box, we need to open up the aperture a bit so as to let enough light through - without diffracting our image. (remember we have to keep the image focused and the wavelengths of light are set by the laws of physics).

So, as the lens gets longer (while projecting onto the same sensor) it gets proportionately narrower in terms of its length relative to the size of its back end. (remember same size sensor) - this makes the lens darker. So to make it comparable to the light capturing ability of the shorter lens (ie same f=stop value) the aperture must be increased (to let more light through to the sensor) in proportion to the change in length.

As this progresses, the physical size of the aperture (in mm) is increasing in relation to the size of the sensor. So (remember the oversized pinhole) it gets much harder to keep things in focus. Hence longer fl lenses with wide apertures are complex, usually large in size and often much more expensive.

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  • \$\begingroup\$ this is not the real reasons, as changing focal length with a same aperture will also have less DOF (because of lower angle of view) \$\endgroup\$ Jul 2, 2019 at 4:06
  • \$\begingroup\$ @S.Serpooshan: that's because the aperture measure used in photography, the f-stop, is actually the ratio of focal length to effective aperture diameter. That's because that number is much more useful for exposure calculations. But what matters for depth-of-field is the effective aperture. Changing focal length therefore changes the effective aperture; if you adjust for that by change the f-stop so that the effective aperture stays constant, you'll see that depth of field stays constant too. \$\endgroup\$ Jul 16, 2019 at 20:17
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A pretty simple yet good explanation is as following:

  • When the focal length increased, we actually do a zoom-in and so the field of view (the area which fit in the frame) will be smaller.

  • This will cause that a less area behind the subject projected in the camera sensor.

  • Since the camera sensor size is the same, this means that those far out-of-focus objects from the background will be stretched more to fill the sensor area. In other words, those far blurry objects in the background (that are not in focus range in either case of focal length) will be blurred more as they magnified/stretched more.

enter image description here

Note that to have same image size of an object in the frame when we double the focal length, we should also double the distance to subject. Although, this does not matter directly here but this is only required for better comparison. Anyway, the background will be more blur with higher f.

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  • \$\begingroup\$ That reasoning seems to work for objects behind the focal plane, but does it still work and give the correct conclusion if you apply it to objects in front of the focal plane? \$\endgroup\$ Jul 17, 2019 at 10:26
  • \$\begingroup\$ @RoelSchroeven Yes, if you look at the figure above, for lower focal length we have a wider angle of view (even behind or in front of focal plane), and more objects can fit in each distance, so it looks less blurry. \$\endgroup\$ Dec 11, 2019 at 13:22
  • \$\begingroup\$ @S.Serphooshan: That does not seem right. The assumption in the figure is that the size of the subject stays the same since the subject-camera distance was adapted. That means that foreground objects in the long-focal-length case appear smaller than in the short-focal-length case, as opposed to the background objects. According to your theory that should mean they are less blurred, but in reality the opposite happens (I think; I'm not 100% certain about what happens when the distance is changed to keep the subject the same size). \$\endgroup\$ Dec 11, 2019 at 13:39
  • \$\begingroup\$ @RoelSchroeven I think your explanation is not right. although we have to go back to get same image size from the subject, but it doesn't mean that other objects in front of the subject will be smaller, because with higher focal-length, we have higher zoom and narrower field of view! that narrow view angle means less objects will fit in the frame which means more stretch and so more blurry. It can be seen from the highlighted (pink) color in above figure. \$\endgroup\$ Dec 16, 2019 at 11:28
  • \$\begingroup\$ The diagrams are misleading, because the photos suggest that the subject (the camera in the pictures) stays the same size, while the amount of flowers in the diagrams suggest the subject size does not stay the same. Try drawing the diagram for the case where the subject size stays the same; you'll see what happens. Apart from that, your theory assumes that blurriness is somehow a property of the real-world objects, which gets enlarged by the lens. That's not what happens in a real camera. \$\endgroup\$ Dec 16, 2019 at 12:36
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This is a great question! I have been at this stuff for more than 65 years and I have yet to read what I think is a respectable answered. To this end I challenge my peers to post a good explanation.

But wait, I think I have had enlightenment – anyway let me give it a go.

The lens projects an image of the outside world onto the surface of film or digital sensor. If you closely examine this image, you will find that it consists of countless circles, each varying in intensity and color. When we observe or capture this image, it will appear to be uniform and tack sharp only if these circles are too small to be perceived. We are talking about circles of confusion. So named, because under the microscope they are seen as ill-defined and they overlap. Nevertheless, when viewed from a suitable distance, we recognize that they merge to form a good-looking image.

When we think about the size of these circles, sooner or later, it dawns that the working diameter of the iris diaphragm (the aperture) will set the stage as to how big these circles are when projected on the surface at the focal plane of our camera.

Now we know that if we set our camera to f/11 or f/16 or f/22 we are shrinking down the working diameter of the camera’s aperture. In doing so, we gain depth-of-field because the outcome is smaller circles of confusion. Now the f-number and the focal length are intertwined. The f-number is derived by dividing the focal length by the working diameter of the lens. Suppose you mount a 50mm and set the f-number to f/16. The working aperture diameter is 50 ÷ 16 = 3.125mm. Such a lash-up delivers respectable depth-of-field, because the circles of confusion at the image plane will be tiny, provided the camera is focused accurately.

Now switch to a 28mm wide-angle. If the shutter speed and ISO are held constant, the same aperture setting of f/16 does this deed. However what has happened to the working diameter of the iris diaphragm to achieve f-16? The revised working diameter becomes 28 ÷ 16 = 1.75mm.

It’s that simple --- shorter focal length at the same f-number yields smaller working aperture, and this results in a smaller circle of confusion -- thus the span of depth of field expands.

Everything has plusses and minuses. If the working diameter becomes super small, the result will be infinite depth-of-field. The minus is: The twin demons of diffraction and interference step in and the image degrades.

Factorial – Maximum sharpness will occur when the camera lens is approximately stopped down two f-stops from maximum (wide-open).

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