5
$\begingroup$

I am 60 years old. I'm not a student and this is not homework (despite somebody trying to close it as a "homework question".)

I'm interested in building a long-focal-length pinhole camera (by "long" I mean 1 to 10 meters). I'm aware that will involve very long exposures and am planning to used a Peltier-cooled CCD for an imager (except maybe for solar viewing).

But I'd like to know if it's worth bothering - how much resolution can I expect, given a pinhole size and focal length (hole-to-sensor distance)? Will diffraction make this entirely pointless?

How would I go about calculating the expected resolution - clearly (to me) there's a contribution from the hole size (resolution will drop with a bigger hole), but diffraction will increase as the hole gets smaller.

Surely there must be some well-known formula for calculating this - what is it?

Improve this question
$\endgroup$
6
  • 1
    $\begingroup$ A decade ago I found a source that went through this analysis in the context of solar observations. The result was that if you use a standard paper hole punch in a piece of paper or tinfoil, you can get a hand- sized image of the sun about ten meters away with enough resolution to track sunspots from day to day. If I am able to find the link I'll add it later (unless someone beats me to it). $\endgroup$
    – rob ♦
    Commented Jul 21, 2022 at 17:20
  • $\begingroup$ I'd guess the limit is due to diffraction by the pinhole, which forms an Airy disk. Then the angular resolution will be given by $\sin\theta = 1.22\lambda/d$, where $\lambda$ is the wavelength of the light and $d$ is the pinhole diameter. $\endgroup$
    – John Rennie
    Commented Jul 21, 2022 at 17:25
  • $\begingroup$ @JohnRennie Clearly not all pinhole camera configurations are diffraction limited. Consider a 1 inch "pinhole" positioned 1 inch from the film/sensor. (focal length 1 inch). The blurring in that case is obviously dominated by the size of the hole, not diffraction. $\endgroup$
    – nerdfever.com
    Commented Jul 21, 2022 at 17:30
  • 2
    $\begingroup$ This PDF link is the document I remembered. There is a competition among the diffraction-limited resolution, the observing distance at which the pinhole is "small," and the brightness of the projected image. $\endgroup$
    – rob ♦
    Commented Jul 21, 2022 at 19:36
  • 1
    $\begingroup$ Whenever I see something about a pinhole camera resolution I think of this image from Optics by Hecht. $\endgroup$
    – Farcher
    Commented Jul 21, 2022 at 22:21

1 Answer 1

Reset to default
4
$\begingroup$

In general, both "wave effects" (i.e. diffraction effects) and "ray effects" (i.e. simply geometric effects) will be observed in a pinhole camera. The theoretical resolution is given by the sum of these two effects. For a pinhole camera of diameter $d$, depth $l'$, the resolution (or "spot width") $U$ is given by:

$$U_{tot} = U_{wave} + U_{ray} = \bigg(1.22l'\frac{\lambda}{d}\bigg) + d\frac{(l+l')}{l}$$

...for an object a lateral distance $l$ from the pinhole.

See: The Pinhole Camera and Image Resolution

(note a change in notation: this author uses $s$ and $s'$ to denote the camera depth and lateral distance to the object)

Improve this answer
$\endgroup$
1
  • 2
    $\begingroup$ Note that for the Sun, $l\gg l'$ and so the second term ist just $d$. You don't want either term to dominate, and so you want $1.22l'\lambda = d^2$ $\endgroup$
    – rfl
    Commented Aug 7, 2022 at 3:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.