Re: RESOLUTION AND PINHOLE

----- Original Message -----
From: "Philippe Monnoyer" <monnoyer@imec.be>

> Image point radius = 0.61 * light_wavelength * focal_length /
> pinhole_radius
> Indeed, according to that equation, the bigger the pinhole, the smaller
> the Image point radius and the sharpest the image.

Mea culpa! I admit.

What I called "Image point radius" is actually "Airy disk radius", which is the
radius if the 1st order diffraction ring (I believe that's what is called). The
key word here is "diffraction", in that respect, is understandable that the
bigger the pinhole radius, the smaller the Airy disk's one, in other words, the
smaller the diffraction effect. And the smaller the pinhole the greater the
Airy disk and the diffraction effect, which is what is happening when you use a
170µm, not for its "optimum" FL of 21mm or so, but for a FL of 100mm.

By the way, I called it "Image point radius", so later I could equate that to
the image point produced by a large pinhole, which for will not be affected by
diffraction (for our purposes) and that will project image points equal to the
size of the large pinhole itself.

The so called optimum pinhole happens when the curve:

Airy disk radius = 0.61 * Lambda * FL / pinhole_radius

intersects the straight line:

Image point radius = pinhole_radius

Therefore:

pinhole_radius = 0.61 * Lambda * FL / pinhole_radius

or

Pinhole_radius = SQRT (0.61 * Lambda * FL / pinhole_radius)

or

when using Lambda = 0.000555mm

Pinhole_diameter = 0.03679 SQRT( FL )

Sorry for the confusion,

Guillermo
Received on Mon Mar 18 09:00:11 2002