High Falls, Pigeon River

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The reason I ask is that f/18 on a 24mm lens = 1.5mm, and f/18 on a 180mm lens = 10mm. I thought diffraction is due to the small physical size of the aperture, rather than the f-ratio, yet I only ever see mention of the f-ratio in discussions of diffraction.

(The lens and camera in my case are an APS-C Nikon D300s and a Sigma 105mm f/2.8, which goes to f64.)

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Excellent question. It boils down to the nature of F-number, which is focalLength/physicalAperture, and the fact that longer focal lengths magnify more. Keep in mind that light projected through an aperture still has to travel from the aperture to the sensor. The greater the distance from aperture to sensor, the greater the magnification...including magnification of the airy disc. The difference between a 180mm lens and a 24mm lens is about 7.5x. To get the same amount of diffraction from a 180mm lens as you would from 24mm lens at f/18, the 180mm lens would need a physical aperture of about 11.25mm in diameter. Given that 180/18 = 10mm, the amount of diffraction present at the sensor is actually a little bit more than with the 24mm lens.

Regarding the Sigma 105/2.8 lens you mention. I believe that is a macro lens. When it comes to macro photography, things change a little bit. You tend to focus extremely close to your subjects with macro photography, so close that depth of field is incredibly small...sometimes millimeters thick. In such situations, it is often more desirable to deal with some diffraction softening as a trade-off for increasing depth of field. In other words, you trade perfect sharpness at the focal plane for additional sharpness beyond the focal plane.

Apertures of f/32 or even f/64 are sometimes necessary to even get a shot at all when involving extension tubes. Additionally, at macro scale, particularly with extension, the effective aperture is usually greater than the actual aperture, thereby requiring exposure compensation to get a proper exposure. A general rule of thumb is that you will need 2x the exposure to compensate at macro scale. This is true for 1:1 magnification, however if you add any extension, you will likely need more. The formula for computing effective aperture at macro scale is as follows:

Ne = N * (M + 1)

Where N is the selected f/#, M is the current magnification (i.e. 2x, 5x) and Ne is the effective aperture number. For the 105mm macro lens with enough extension tubes to produce 2x magnification, at an actual aperture of f/4, the effective aperture from an exposure and diffraction standpoint would be f/12. Most modern cameras will compensate for this automatically, given that they have built in metering. It is still useful to understand exactly how macro photography affects aperture, though...and the possible implications from a diffraction standpoint.

Generally, you will want to set an aperture that gives you the effective aperture (not the actual, or physical, aperture) you need to get the exposure and DOF you want, at an acceptable level of diffraction. For a 1:1 macro lens, you have to double the actual aperture to get the effective aperture. On the Nikon D300s, which has a 12.3mp APS-C sensor, the diffraction limit kicks in at around f/11, and becomes a visible problem by around f/22 or so. At f/32, diffraction will likely be a real problem. If you want to take a macro shot at f/22, you would need to set the actual aperture to f/16.

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hi jrista. i am currently taking macro pictures, and am trying to find the correct balance between dof and diffraction. having used the online dof calculator, the dof wide open at 1:1 is less than 1mm...am starting to use focus stacking to try and compensate, but first results came out unsharp, possibly due to diffraction –  rapscalli Sep 25 '11 at 10:44
hi again, so are we saying: as focal length increases aperture size for a given f stop, and also increases magnification of the airy disk, i can ignore focal length in determining diffraction limits? –  rapscalli Sep 25 '11 at 10:50
@rapscalli: Correct. The size of the Airy disc is about 1.2 * lambda * N, where lambda is the wavelength (around 550 nm for visible light) and N the aperture number, irrespective of focal length. For macro, you have to use the effective aperture number, including the so called “bellows factor”. –  Edgar Bonet Sep 25 '11 at 11:00
@rapscalli: Thats correct, focal length in isolation is not a factor in diffraction. As Edgar mentioned, at macro scale, diffraction is dependent upon the effective aperture, which can be different than physical aperture. In the normal case, the two are the same, however at 1:1 mag and beyond, the effective aperture number can be larger...sometimes much larger. I'll see if I can add some info on how to calculate that. –  jrista Sep 25 '11 at 16:03

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