dh003i

1-Sep-2010, 07:12

I've read many places that normal lenses for 35mm, optimized for say 1:20 front-forward, are optimized for 20:1 reversed. If a lens is diffraction limited at f/10 at 1:20, is it also diffraction limited when reversed at f/10 at 20:1?

e.g., if a symmetrical lens is diffraction limited at f/10 stated at 1:20, effective f-stop is f/10.5 and resolution is 1500/10.5 = 142.86 lp/mm. So, when reversed and used at f/10 stated at 20:1, the effective f-stop is f/210 (strictly speaking, a symmetrical lens doesn't have to be reversed for this example, but a asymmetrical one would need to be). Is the resolution then diffraction limited to 7.14 lp/mm? (note that this would be the resolution at 1:20 divided by 20).

More generalized, say resolution at f/10 stated at 1:20 is 71.43 lp/mm (50% of the diffraction limited). When reversed and used at f/10 stated at 20:1, would the resolution then be 3.5715 lp/mm (also 50% of the diffraction limit)?

e.g., for f = stated f/stop, m = magnification, R = resolution at m,

R = 1500 / f(1 + m)

Assuming r = resolution at 1/m

r = 1500/ f(1 + 1/m)

We can equate these two to eachother

1500 = R × f(1 + m)

1500 = r × f × (1 + 1/m)

R x f(1 + m) = r × f(1 + 1/m)

r = R × (1 + m) / (1 + 1/m)

r = R × (1 + m) / (m/m + 1/m)

r = R × (1 + m) / ([m + 1]/m)

r = R × (1 + m) × m / (1 + m)

r = R × m

Ergo, for any given stated f-stop, if the resolution at 1:10 is say 500 lp/mm, the resolution at 10:1 with the same stated f-stop would be 70 × 1/10 = 50 lp/mm. This makes sense based on what I've read; lenses optimized and very sharp at 1:10 make good macro lenses at 10:1.

Am I correct in my assumption here? Fundamentally, that a diffraction limited lens at stated f/10 at 1:20 is also diffraction limited at stated f/10 at 20:1 when reversed?...or that a lens limited to 50% the diffraction limit at 1:20 f/10 stated is limited to 50% the diffraction limit at 20:1 f/10 stated?

e.g., if a symmetrical lens is diffraction limited at f/10 stated at 1:20, effective f-stop is f/10.5 and resolution is 1500/10.5 = 142.86 lp/mm. So, when reversed and used at f/10 stated at 20:1, the effective f-stop is f/210 (strictly speaking, a symmetrical lens doesn't have to be reversed for this example, but a asymmetrical one would need to be). Is the resolution then diffraction limited to 7.14 lp/mm? (note that this would be the resolution at 1:20 divided by 20).

More generalized, say resolution at f/10 stated at 1:20 is 71.43 lp/mm (50% of the diffraction limited). When reversed and used at f/10 stated at 20:1, would the resolution then be 3.5715 lp/mm (also 50% of the diffraction limit)?

e.g., for f = stated f/stop, m = magnification, R = resolution at m,

R = 1500 / f(1 + m)

Assuming r = resolution at 1/m

r = 1500/ f(1 + 1/m)

We can equate these two to eachother

1500 = R × f(1 + m)

1500 = r × f × (1 + 1/m)

R x f(1 + m) = r × f(1 + 1/m)

r = R × (1 + m) / (1 + 1/m)

r = R × (1 + m) / (m/m + 1/m)

r = R × (1 + m) / ([m + 1]/m)

r = R × (1 + m) × m / (1 + m)

r = R × m

Ergo, for any given stated f-stop, if the resolution at 1:10 is say 500 lp/mm, the resolution at 10:1 with the same stated f-stop would be 70 × 1/10 = 50 lp/mm. This makes sense based on what I've read; lenses optimized and very sharp at 1:10 make good macro lenses at 10:1.

Am I correct in my assumption here? Fundamentally, that a diffraction limited lens at stated f/10 at 1:20 is also diffraction limited at stated f/10 at 20:1 when reversed?...or that a lens limited to 50% the diffraction limit at 1:20 f/10 stated is limited to 50% the diffraction limit at 20:1 f/10 stated?