At an angle theta, the view of the aperture from the film plane is an ellipse whose semi-major axis, b, is: b = a * cos(theta) where a is the major axis and is the diameter of the aperture. Theta here is the angle of the ray off center. This has the same effect as reducing the area of the pinhole by cos(theta). The film in the example is flat and the ellipse of light strikes it at the same angle theta causing the elliptic "tube" of light from the pinhole to stretch back to a circle reducing intensity (light/area). Again, this is cos(theta). So, the total light fall off is: I = I0 * [cos(theta)]^2 Where I is the intensity of the light, and I0 is the center intensity The effect of the slant of the pinhole is, therefore, not neglected. This is the same as: I = I0 * cos^2[atan(r/f)] In my post I was actually I / I0, or relative intensity as stated in the post. Yes, sin(theta) = atan(r/f). This relation was given because it is more useful to the original poster. I've addressed diffraction in a separate post. There is another (relatively) minor effect. Defining this effect is left as an exercise to the reader. Regards, Bob... -------------------------------------------- "Do not suppose that abuses are eliminated by destroying the object which is abused. Men can go wrong with wine and women. Shall we then prohibit and abolish women?" -Martin Luther ----- Original Message ----- From: "Chris" <nimbo@ukonline.co.uk> To: "List for Photo/Imaging Educators - Professionals - Students" <photoforum@listserver.isc.rit.edu> Sent: Saturday, August 23, 2003 5:00 AM Subject: RE: Minimizing pinhole image falloff > I can't comment on this: but arctan(r/f) is the angle of the ray from the > pinhole to the element away from the normal. And the cosine of this angle is > the ratio of the perpendicular distance of the film from the pinhole over > the distance from the pinhole to the element. > > so this quantity is (f/d)^2 and the centre intensity is proportional to > A/f^2 thus making the element intensity proportional to > ((f/d)^2)(A/f^2)=A/d^2 > > This neglects the effect of the slant of the pinhole to the element > (Sin(Theta)) because the element sees a projection of the pinhole. > > There is also a diffraction effect due to the difference of the path of the > light across the pinhole. > > 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 Bob Blakely > |> Sent: 22 August 2003 22:15 > |> To: List for Photo/Imaging Educators - Professionals - Students > |> Subject: RE: Minimizing pinhole image falloff > |> > |> > |> The image intensity will fall off from center intensity approximately > |> according to the formula: > |> > |> I = cos^2[atan(r/f)] > |> > |> where: > |> I = intensity relative to center intensity. > |> r = distance in the film plane from center image. > |> f = distance from pinhole to center image. > |> > |> The fall off in stops is: > |> > |> s = log(I) / log(2) > |> = 3.322 * log(I) > |> > |> This assumes flat film plane with pinhole plane parallel to film plane. > |> Center image is defined as the point on the film plane where the normal > |> passes through center pinhole. > |> > |> > -----Original Message----- > |> > From: owner-photoforum@listserver.isc.rit.edu > |> > [mailto:owner-photoforum@listserver.isc.rit.edu]On Behalf Of Chris > |> > Sent: Friday, August 22, 2003 11:53 AM > |> > To: List for Photo/Imaging Educators - Professionals - Students > |> > Subject: RE: Minimizing pinhole image falloff > |> > > |> > > |> > 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 > |> > |> > |> > |> > |> > > |> > > |> > > |> > |> > |> > > >
