- Original Message -
From: HypoBob hypo...@pacbell.net
Guillermo,
In a posting on 13 December you stated that an 8x10 pinhole image rivaling
the resolution of a lens image (i.e., 5 lp/mm) could be obtained with a
focal length of 120 mm or less and the optimum pinhole.
Since one could place any size negative material in such an arrangement,
aren't you in effect saying that a 120 mm focal length is the maximum for
obtaining 5 lp/mm resolution?
Yes, but those 5 lp/mm are on the film which is more often than not and
intermediate step to the final result, the latter being, more often that
not, a positive print. So although the resolution you may have on your
35mm format made with the optimum size pinhole 50mm body cap is better than
5 lp/mm when you enlarge it to 8x10 you will get a print with just about 1/8
the resolution on film. If you didn't have to enlarge and view the print at
the normal distance of comfortable vision, then, the 5 lp/mm are enough.
Now, if you make a 16x20 pinhole camera and view that print at a distance
equal to twice the normal distance of comfortable vision, you will then
only need 2.5 lp/mm on the print, in which case 480mm focal length would be
the maximum.
Is there also a minimum focal length for this resolution?
Not mathematically, but practically there is. For a given format, there is
a minimum limit of the focal length that would acceptably cover the film
format, as well as there are practical limits forced by how small a camera
we can make, a manufacturer of laser drilled pinholes once sent me a 0.001
pinhole good for an optimal focal length of under 0.5mm!! I couldn't
possible make a camera with such focal length.
Do you have some equations you can share with us that give the
relationship between resolution, focal distance, and pinhole size? I'm
guessing
that this all may be based on the usual equations with a properly chosen
value for the circle of confusion.
Resolution = 1 / (1.22 * wavelength * fstop)
if you use wavelength = 0.00055mm, it becomes:
Resolution = 1490 / fstop
Assuming you know the resolution and want to know what fstop would give you
that resolution:
fstop = 1490 / resolution
but, fstop is also equal to:
fstop = Focal length / pinhole diameter
and:
pinhole diameter = 0.03679 * SQRT(Focal length) {units in mm}
therefore:
fstop = Focal length / (0.03679 * SQRT(Focal length)
or also written as:
fstop = 27.1813 * SQRT(Focal length)
therefore:
Focal length = ( fstop / 27.1813 )^2
How did I get 120mm as the maximum focal length I talked in my post?:
I started with the need of 5 lp/mm, therefore:
fstop = 1490 / resolution
fstop = 1490 / 5
fstop = 298
Focal length = ( fstop / 27.1813 )^2
Focal length = ( 298 / 27.1813 )^2
Focal length = 120mm (aprox)
In any event, it seems that the conditions will hold only 'near' the
center of the image because the corners and edges are beyond the optimum
focal
distance.
That is true, but you could have a semicircular film plane with the pinhole
in the center of the semicircle, in which case the distance pinhole film
would be constant (in one axis, at least). Make the film plane half an
sphere and you will get same distance everywhere on the film plane :-)
Also, I'm thinking that you may be assuming a certain viewing distance for
the observer since smaller images tend to be held closer to
the eye than do larger ones. That is, the 8x10 size may be necessary to
get the viewer to hold the 5 lp/mm print about 20 inches from his eyes
rather than the 12 inches at which he may hold a smaller 5 lp/mm print.
That is correct.
Thanks for getting me off onto this tangent. I always enjoy your posts,
Thanks, I enjoy posting them, it keeps my brain from getting rustier.
BTW, corrections to the above are welcome.
Guillermo
