Re: Pinhole Calculations

Hello photoian at earthlink.net:

Trying to answer all your questions, after Richard answers would be redundant.
I will though, answer a question Richard did not addressed:

----- Original Message -----
From: "{USER_FIRSTNAME} {USER_LASTNAME}" <photoian@earthlink.net>
>
> 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
Received on Mon Mar 11 15:07:27 2002