At a guess I would say the fall off was proportional to Sin(Theta) where theta is the angle of the ray away from the normal. There is no focal length for a pinhole. The brilliance of the image is proportional to the area of the pinhole. So the brightness at angle theta from the normal will be proportional to A.Sin(Theta)/d^2 where d is the distance of the element from the pinhole. Don't quote me I'm a beginner! Chris Web Page http://www.chrisweb.pwp.blueyonder.co.uk/ |> -----Original Message----- |> From: owner-photoforum@listserver.isc.rit.edu |> [mailto:owner-photoforum@listserver.isc.rit.edu]On Behalf Of Gregory |> Fraser |> Sent: 19 August 2003 16:19 |> To: List for Photo/Imaging Educators - Professionals - Students |> Subject: Minimizing pinhole image falloff |> |> |> I went to a web site that had a calculator for the image circle |> diameter of pinhole setups. I calculated that a focal length of |> 3 inches would give me an image circle that would cover 4x5 inch |> film. I forget the pinhole diameter. Then I remembered how |> drastic the falloff is at the edges of pinhole images so I |> thought perhaps by increasing the focal length, I would have |> more of the brighter central part of the image and that would |> reduce the effects of falloff. 'But wait,' I yelled, 'if this |> were the case wouldn't Guy have been able to find a hotel room |> long enough to prevent the falloff he experienced in Montreal? |> Certainly someone as intimate with pinholes as Guy would know |> about that.' |> |> So, does the light falloff of a pinhole camera image follow an |> inverse square rule? Will it always be an issue no matter how |> big your shoebox, cigar tube or Quaker Oats box is? |> |> Greg Fraser |> |>
