Re: RESOLUTION AND PINHOLE

Hi Philippe,

There's a excellent article by a physicist named Matt Young that might be
helpful to you:

         http://www.pinholevisions.org/resources/articles/Young/

- Gregg

At 02:18 PM 3/18/02 +0100, you wrote:
>Hello Guillermo and others,
>
>I have built a pinhole camera (half cylindrical) with a focal lenght of
>100 mm.
>The pinhole diameter is exactly 170 µm, laser-cut in a very thin
>metallic surface.
>I thought this is still too big to diffract visible light and the
>smaller the better.
>I made only one test 90 sec exposure on 400 ISO film (a band piece of 35
>mm HP5 from ILFORD)+ Stouffer step tablet.
>Exposure time is OK but image is blurred. Either my film moved during
>exposure or my pinhole is too small.
>According to the formula present in one of Guillermo last mails copied
>hereunder for reference, I should use a pinhole of 0.36 mm, indeed quite
>far from the 0.17 mm present diameter.
>The problem with Guillermo's formula is that I don't understand where
>this following line comes from:
>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.
>If one takes a gigantic pinhole that would give no image at all, the
>formula tells you that the Image point radius will be very small ??
>Where does this formula come from ?
>I really would like to understand this once for all. Would you have some
>recommended readings about the fundamental science of pinhole
>photography from the physics point of vue ?
>Is there really an optimal pinhole diameter for a given film-hole
>distance ? And if yes, why excatly ?
>Feel free to send deep explanations ! or just answers to these many
>questions.
>
>Yours,
>
>Philippe
>Photographic Chemist
>Ph.D.
>Belgium
>
>
>
> > The formula is pinhole(in) = square root FL x 0.0073 or pinhole(mm)= square
> > root FL x 0.03679.
> >
> > My question is; does this formula really give the sharpest image?
> >
> > The constant (.oo73 or .03679) is what determines the
> > answer. So, now the question is; How is the constant determined?
> > Does it give the "sharpest" image or is it just a trade off
> > between exposure time and pinhole size?
>
>The formula is a trade off between geometrical sharpness (the smaller
>the hole
>the "sharper" the image) and diffraction (the bigger the hole the less
>the image
>"sharpness is affected by diffraction).
>
>Science tells us the radius an image point imaged by a small aperture is
>given
>by the formula:
>
>Image point radius = 0.61 * light_wavelength * focal_length /
>pinhole_radius
>
>A purely geometrical analysis will show us that the image point radius
>would be
>the same size as the pinhole radius:
>
>Image point radius = pinhole radius
>
>So when the 2 formulas above are equal, we have a balanced "trade off"
>you talk
>about:
>
>pinhole radius = 0.61 * light_wavelength * focal_length / pinhole_radius
>
>therefore:
>
>pinhole_radius^2 = 0.61 * light_wavelength * focal_length
>
>or
>
>pinhole_radius = SQRT (0.61 * light_wavelength * focal_length)
>
>or
>
>Pinhole_diameter = 2 * SQRT (0.61 * light_wavelength * focal_length)
>
>If we select light with wavelength of 0.000555 mm, we would have:
>
>Pinhole_diameter = 2 * SQRT (0.61 * 0.000555 * focal_length)
>
>or
>
>Pinhole_diameter = 0.03679 * SQRT ( focal_length) millimeters
>
>The equivalent in inches is:
>
>Pinhole_diameter = 0.00073 * SQRT (0.61 * light_wavelength *
>focal_length)
>
>Guillermo
>
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Received on Mon Mar 18 08:31:58 2002