On Sep 19, 2011, at 8:02 PM, David Parsons wrote: > On Mon, Sep 19, 2011 at 10:50 PM, Larry Colen <[email protected]> wrote: >>> >> Diffusion or diffraction? >> It was my understanding that diffraction happens at any opaque edge within a >> range that is determined by the wavelength. Therefore the larger the >> aperture the smaller the percentage of it that will be subject to the >> effects of diffraction. >> >> f/64 on a 320 mm lens would be an aperture of 5mm, which would be the same >> as f/10 on a 50mm lens or f/8 on a 40mm. >> >> I suppose that it's possible that the angle of incidence of the light beams >> hitting the aperture could affect the amount of diffraction, and therefore >> it's not a linear function that diffraction becomes a problem with apertures >> of 2mm or smaller in diameter. >> >> It's also been roughly 30 years since I took a course in either physics, or >> field equations, so I could be completely off base. > > Aperture values are ratios, not absolute values. f/64 on LF is going > to be a different size than f/64 on 35mm. > > On a 320mm lens, f/64 has an aperture diameter of 5mm. On a 50mm > lens, f/64 would have an aperture diameter of .78mm. (This doesn't > account for the fact that the aperture blades aren't necessarily at > the nodal point in a lens, and you are talking about a simple lens, > reality is a bit messier.) > > Your numbers are completely wonky.
320mm/64 = 5mm, 50mm/10 = 5mm, 40mm/8 = 5mm How is that wonky? -- Larry Colen [email protected] sent from i4est -- PDML Pentax-Discuss Mail List [email protected] http://pdml.net/mailman/listinfo/pdml_pdml.net to UNSUBSCRIBE from the PDML, please visit the link directly above and follow the directions.
