> Yes, sin(theta) = atan(r/f). This relation was given because it is more > useful to the original poster. should read: Yes, arctan(r/f) is the angle of the ray from the pinhole to the element away from the normal 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: "Bob Blakely" <Bob@Blakely.com> To: "List for Photo/Imaging Educators - Professionals - Students" <photoforum@listserver.isc.rit.edu> Sent: Saturday, August 23, 2003 6:40 AM Subject: Re: Minimizing pinhole image falloff > 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 > > |> > |> > > |> > |> > > |> > > > |> > > > |> > > > |> > > |> > > |> > > > > > > > > >
