Thank you. We didn't do Bessel functions except in fm radio theory. The theory that I don't remember involved an approximation. When I tried to do the integration myself just now I got the integral of the sin of a sin which I think is the differential form of the Bessel function. Very interesting. I'll keep quiet now..... 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: 23 August 2003 15:07 |> To: List for Photo/Imaging Educators - Professionals - Students |> Subject: Re: Minimizing pinhole image falloff - Correction |> |> |> > 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 |> > > |> > |> |> > > |> > |> |> > > |> > |> > > |> > |> > > |> > |> > > |> |> > > |> |> > > |> |> > > |> > > |> > > |> > |> > |> > |> |>
