Re: IR and pinhole

[Wayde Allen <allen@xxxxxxxxxxxxxxxx>]

There are two competing issues at work with the pinhole.  Ignoring
diffraction, the smaller the pinhole the better the resolution.  You can
think of the pinhole as a light source with a brightness related to what
you can see through the hole.  Pick a location on the film plane and draw
a line through the pinhole.  The line will show you the light ray that
will expose that spot on the film.  The smaller the hole, the smaller the
spot.  That is until diffraction enters the picture.

The image radius of the pinhole for small sizes (diffraction) is the
radius of the Airy disk, or (0.61 * wavelength * f)/s

   where:  f is the focal length
           s is the pinhole radius 

For large holes diffraction becomes negligible, and the image size of the
pinhole is essentially the size of the pinhole itself.  What you do is
look for the point at which these two equations are equal.  Above this
point the size of the pinhole spot is getting bigger since the hole is
bigger, and below this point the size of the pinhole spot is getting
bigger due to diffraction.  The equation is: 

   (0.61 * wavelength * f)/s = s

or

   f = (1.6 * s^2) / wavelength

If you approximate this as f = s^2 / wavelength you get the basis for the
equation I posted earlier.  PLEASE NOTE THAT WAVELENGTH IS PART OF THE
EQUATION!  In other words, you can find the optimum hole size for an IR
pinhole.  Yes this optimum resolution point varies with the focal length
(lens to pinhole distance) as well.

I'm getting most of this stuff from a paper written by one of my collegues
here at NIST.  The paper is:

   Young, Matt, The Pinhole Camera - Imaging without Lenses or Mirrors,
   The Physics Teacher, December 1989

I can probably get you a copy if you'd like. 

- - Wayde
  (allen@xxxxxxxxxxxxxxxx)

 


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