Diffraction Limit Question

[EdRosch]

Hi,

I've been reading the various articles and comments posted about the subject and I have a question that I hope someone more versed in physics than I can answer.

In all the discussion the diffraction limit seems to be a function of f/stop without regard to the focal length.  On the other hand, my understanding was that diffraction in general was related to the relationship between the size of the interfering object and the wavelength, that is the ratio of the absolute sizes.  Thus, say if you're looking several  lenses at f/10 (to make the math easy)  a one meter lens would have a working aperture of 100mm and a 100mm lens would have a working aperture of 10mm, a 10mm superwide, one of only 1mm!   It would seem that, assuming that the working wavelengths of light are the same,  the amount of diffraction through a 1mm hole and its effect upon the sensor would be quite different than coming through a 100mm one.

Now I did look at the airy disk math and this is apparently not the case, and the fact is that for a given camera the loss of sharpness at f/10 would be identical for both the 10mm and 1,000 mm lens.   I just don't understand why.

Thanks

[ejmartin]

The angular spread of the diffracted beam is approximately the ratio of the wavelength to the diameter of the aperture.  This spreading beam of light makes a spot on the focal plane whose size is the angular spread times the distance from the iris to the focal plane (ie the further it travels, the proportionally more it spreads).  Thus the size of the spot is roughly

diffraction spot diameter ≈ wavelength * (iris-focal plane distance/iris diameter)

The quantity in parenthesis is approximately (depending on the lens design, it could differ some) the f-ratio.  Thus the size of the diffraction spot is proportional to the f-ratio.  There will be an overall numerical factor that accounts for details of the lens design, but it is typically not too far from one.

EDIT:  Corrected the formula (had the f-ratio upside down).

[BJL]

The angular spread of the diffracted beam is approximately the ratio of the wavelength to the diameter of the aperture.

Thanks. Note that in photographing the same composition with different formats (different focal length, different size image recorded by the sensor) it is angular resolution of the subject that counts, and so effective aperture diameter is the best measure of the effect of diffraction on image resolution.

[bjanes]

The angular spread of the diffracted beam is approximately the ratio of the wavelength to the diameter of the aperture.  This spreading beam of light makes a spot on the focal plane whose size is the angular spread times the distance from the iris to the focal plane (ie the further it travels, the proportionally more it spreads).  Thus the size of the spot is roughly

diffraction spot diameter ≈ wavelength * (iris diameter/iris-focal plane distance)

The quantity in parenthesis is approximately (depending on the lens design, it could differ some) the f-ratio.  Thus the size of the diffraction spot is proportional to the f-ratio.  There will be an overall numerical factor that accounts for details of the lens design, but it is typically not too far from one.

The Cambridge in Colour site gives an intuitive explanation of Emil's math. The site is well worth visiting.

"Since the physical size of the lens aperture is larger for telephoto lenses (f/22 is a larger aperture at 200 mm than at 50 mm), why doesn't the size of the airy disk vary with focal length?  This is because the distance to the focal plane also increases with focal length, and so the airy disk diverges more over this greater distance.  As a result, the two effects of physical aperture size and focal length cancel out."

Bill

[Daniel Browning]

NOOOO! Consecutive responses from ejmartin, BJL, and bjanes!? The concentration of expertise is reaching dangerous levels. Please disperse immediately to other web sites or the informative value of this thread may reach critical mass and go supernova.

I will try to water it down with the usual inaccurate rules of thumb, common misconceptions, and old wive's tales.  Just give me a minute to come up with some inappropriate ones.

EDIT: Sorry, I just couldn't come up with anything that was incorrect or misleading, so I'm going try for on-topic and informative instead.

On the other hand, my understanding was that diffraction in general was related to the relationship between the size of the interfering object and the wavelength, that is the ratio of the absolute sizes.  Thus, say if you're looking several  lenses at f/10 (to make the math easy)  a one meter lens would have a working aperture of 100mm and a 100mm lens would have a working aperture of 10mm, a 10mm superwide, one of only 1mm!   It would seem that, assuming that the working wavelengths of light are the same,  the amount of diffraction through a 1mm hole and its effect upon the sensor would be quite different than coming through a 100mm one.

Now I did look at the airy disk math and this is apparently not the case, and the fact is that for a given camera the loss of sharpness at f/10 would be identical for both the 10mm and 1,000 mm lens.   I just don't understand why.

I'm going to repeat what BJL said in a different way. The amount of diffraction "per subject detail" *does* scale with iris diameter. To demonstrate:

In the 1000mm f/10, the subject (say a bird) is very large and nearly fills the frame. Diffraction softens just the minutest feather details.

In the 100mm f/10, the bird is now much smaller, only occupying the tiny center portion of the frame. Diffraction blur is still the same number of pixels, but since the subject is now covering a smaller number of pixels, it is more greatly affected by the diffraction.

In the 10mm f/10, the bird is so small in the frame that nearly all the detail is obscured by diffraction.

Some photographers are not aware of this, so you will hear things like "I crop digitally instead of use a 2X TC because the two-stop increase in diffraction causes too much detail loss." But diffraction is actually the same.

Hope that helps.

[bjanes]

I'm going to repeat what BJL said in a different way. The amount of diffraction "per subject detail" *does* scale with iris diameter. To demonstrate:

In the 1000mm f/10, the subject (say a bird) is very large and nearly fills the frame. Diffraction softens just the minutest feather details.

In the 100mm f/10, the bird is now much smaller, only occupying the tiny center portion of the frame. Diffraction blur is still the same number of pixels, but since the subject is now covering a smaller number of pixels, it is more greatly affected by the diffraction.

In the 10mm f/10, the bird is so small in the frame that nearly all the detail is obscured by diffraction.

Hope that helps.

Daniel,

Your analogy does make sense. Astronomers measure the resolving power of their telescopes by objective diameter and not the f/number because they are interested in resolving small angular separations in the object plane, not in the image plane. The formula for Angular Resolution does not involve the focal length, and takes the form sin θ = 1.220 λ / D, where θ is the angular resolution λ is the wave length of the light, and D the diameter of the lens aperture.

Spatial resolution does involve the f/number as shown in the link.

Bill