Hi Philippe,
There's a excellent article by a physicist named Matt Young that might be
helpful to you:
http://www.???????/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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