> Actually, I did a series of practical tests for a studio in NY some years
> back, shooting polaroids of newspaper text with various carefully made
> holes. I began with the formula I got from the _Photo Lab Index_ which
> was d=sqrt(.00007f), where d is the diameter of the hole and f is the
> distance of the hole to film (my own rewording of this formula). This
> may be similar to what you have; I'm not going to calculate it out for mm
> - if you care enough about it, you can do that yourself. What I found
> was surprising. The smaller the hole, the better the resolution, as far
> as I went, and I got the factor represented in the formula by the .00007
> down well below .00005. I don't remember how far down I went, but it was
> quite a way down, to where the f/numbers were truly impractical.
The idea that the pinhole should be the same size as the diffraction
disk it produces gives the formula:
d = sqrt (2.44 * wavelength * focal_length)
for a wavelength of 550 nm, this turns into
d = sqrt (2.44 * .000550 * focal_length_in_mm)
d = sqrt (.001342 * focal_length_in_mm)
If we convert mm to inches by dividing by 25.4, we get
d = sqrt ( .0000528 * focal_length_in_inches)
I also did a test with a resolution chart and pinholes of various
sizes, and I could see the resolution get worse as the pinhole
went below the optimum size.