More theory: pinhole optimisation

steveo_mcg

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I've decided the best way to try and use my glass plates is to try two things at once, I quite fancy a shot at pinhole shooting and I've nothing else that'll take these quarter plates so... How does one calculate the optimum focal length and aperture? Assuming I want it quite small for a semblance of sharpness and want decent coverage of the plate. I'd quite like it wider than normal if I'm being picky.
 
Very handy.

Helps to refine my question. How does one ensure that the image circle is large enough to cover the film?

edit: ok i think i've got this, to cover the plate I'd need a focal length of 136mm this looks to be a "normal" focal length for this size film, would I be able to adjust this and go wider?
 
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I think that this as wide as you can go to cover a glass quarter plate

Focal Length 76mm
Pinhole Diameter 0.37mm
f Stop f 207
Film Dimension 136mm
Angle Of View 83.8 Deg
Coverage 145.92mm

I has a play here to work it out http://www.mrpinhole.com/wiz.php
 
Could you explain why? I'm guessing it's to do with the coverage and the diagonal of the plate.
 
Having gone through the calculator again I think that these are the correct figures

Focal Length 71mm
Pinhole Diameter 0.36mm
f Stop f 200
Film Dimension 136mm
Angle Of View 87.7 Deg
Coverage 136.32mm

Anything less than a focal length of 71mm results in an image circle that does not cover the film area.
 
I understand, thanks.

Now to find a box with sufficent dimentions, or do I make one. I've not done any wood work in aaaggees.
 
As far as I'm aware, there isn't an optimum focal length. Given that a pinhole doesn't refract light, there is no focal length as commonly understood - the distance from lens to image plane at which an object at infinity is brought into focus. Any pinhole/film plane distance will result in an equally sharp image of an object at infinty (with one priviso that I'll come to shortly). The difference that the distance of pinhole to image makes is the size of the image, or putting it another way, the angle of view.

The proviso is that for every different distance, there is an optimum size of pinhole. What is that optimum size? Unfortunately, if you search the literature, you'll find at least three different formulae to use. If you're sufficiently interested, I'll post them once I've worked out how to handle mathematical symbols in a post.
 
As far as I'm aware, there isn't an optimum focal length. Given that a pinhole doesn't refract light, there is no focal length as commonly understood - the distance from lens to image plane at which an object at infinity is brought into focus. Any pinhole/film plane distance will result in an equally sharp image of an object at infinty (with one priviso that I'll come to shortly). The difference that the distance of pinhole to image makes is the size of the image, or putting it another way, the angle of view.

The proviso is that for every different distance, there is an optimum size of pinhole. What is that optimum size? Unfortunately, if you search the literature, you'll find at least three different formulae to use. If you're sufficiently interested, I'll post them once I've worked out how to handle mathematical symbols in a post.

One such formula is set out in the pinhole designer program, the link to which I gave in my earlier post. I also don't know how to write the formula in a post.:)
Edit: realised I can copy and past the formula :)
Another formula is:

d=sqrt(2f x λ)

where d is pinhole diameter, f is focal length (distance from pinhole to image plane) and λ is the wavelength of light.
 
Thanks for the suggestion. Copy and paste it is...

D = √(3.6vλ) where D is the diameter, v is the pinhole to image distance and λ is the wavelength of the light used. Other values are suggested in the literature, both higher and lower. For example, D = 2√(2vλ) or D = √(2vλ). Sidney Ray in Applied Photographic Optics gives a simpler to use empirical formula, D = k√v where k is a constant of around 0.03-0.04 and D and v are expressed in mm.
 
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