Pinholes and image circle/vignetting

[Zacariaz]

From a purely theoretical perspective, I've been trying to figure out a thing or two in regards to pinhole photography, and I more or less have what I need, that is except when it comes to the actual coverage, image circle/diameter/area or whatever you which to call it.

I wanted to figure out the widest feasible focal length for use with a full frame sensor, and quickly found and online calculator, which told me that a focal length somewhere around 24mm and an aperture of 200µ would be the limit. Of course being who I am, I studied the subject a bit further, and quickly found that people don't exactly agree on this.

I'm fairly intelligent and have been thinking long and hard about what a mathematical model might look like, but apparently I'm not that smart, nor have I been able to find any help online, not any comprehensible help anyway. It is always a matter of opinion, which I care little about in this case. I simply what to know where the "edge" on the projection is, and by "edge" of course I mean and arbitrary limit to avoid noticeable vignetting, and how it is calculated.

I thought some of you might be able to shed some light on the subject.

That's basically it, though please remember that I am very much aware that in the end, experimentation is the answer.

Thanks in advance and best regards.

 

[Tom Axford]

From a purely theoretical perspective, I've been trying to figure out a thing or two in regards to pinhole photography, and I more or less have what I need, that is except when it comes to the actual coverage, image circle/diameter/area or whatever you which to call it.

I wanted to figure out the widest feasible focal length for use with a full frame sensor, and quickly found and online calculator, which told me that a focal length somewhere around 24mm and an aperture of 200µ would be the limit. Of course being who I am, I studied the subject a bit further, and quickly found that people don't exactly agree on this.

I'm fairly intelligent and have been thinking long and hard about what a mathematical model might look like, but apparently I'm not that smart, nor have I been able to find any help online, not any comprehensible help anyway. It is always a matter of opinion, which I care little about in this case. I simply what to know where the "edge" on the projection is, and by "edge" of course I mean and arbitrary limit to avoid noticeable vignetting, and how it is calculated.

I thought some of you might be able to shed some light on the subject.

A pinhole doesn't have an image circle in the same way that most lenses have. Light can pass through the pinhole at almost any angle and not be obstructed as it is in a typical compound lens (by the edges of other lens elements). So a pinhole has no well-defined image circle, but the light intensity on a flat sensor will fall off steadily as the angle between the light ray and the optical axis increases.

How fast it falls off in practice will depend on the characteristics of the sensor (e.g. any directional variation in sensitivity) and also how good the pinhole is (e.g. the thickness of the plate through which the pinhole is drilled, relative to the diameter of the pinhole). Of course, diffraction effects will further complicate the calculation.

So, defining an image circle for the pinhole will depend on how much light falloff you are prepared to accept between the centre of the image and the edge of the "image circle".

 

[Zacariaz]

A pinhole doesn't have an image circle in the same way that most lenses have. Light can pass through the pinhole at almost any angle and not be obstructed as it is in a typical compound lens (by the edges of other lens elements). So a pinhole has no well-defined image circle, but the light intensity on a flat sensor will fall off steadily as the angle between the light ray and the optical axis increases.

How fast it falls off in practice will depend on the characteristics of the sensor (e.g. any directional variation in sensitivity) and also how good the pinhole is (e.g. the thickness of the plate through which the pinhole is drilled, relative to the diameter of the pinhole). Of course, diffraction effects will further complicate the calculation.

So, defining an image circle for the pinhole will depend on how much light falloff you are prepared to accept between the centre of the image and the edge of the "image circle".

Well that's pretty much what I would have thought, but because since the popular tool at mrpinhole.com provide an image circle diameter, I assumed otherwise.

Anyway, what I hear you say is that the falloff, as you call it, is basically linear, and thus no arbitrary limit can be set as it really depends upon exposure time and so on and so forth.

In a way it seems to make sense, but then again not.

Suppose I'll have trouble sleeping tonight, but thanks for your answer anyway.

 

[Tom Axford]

A pinhole doesn't have an image circle in the same way that most lenses have. Light can pass through the pinhole at almost any angle and not be obstructed as it is in a typical compound lens (by the edges of other lens elements). So a pinhole has no well-defined image circle, but the light intensity on a flat sensor will fall off steadily as the angle between the light ray and the optical axis increases.

How fast it falls off in practice will depend on the characteristics of the sensor (e.g. any directional variation in sensitivity) and also how good the pinhole is (e.g. the thickness of the plate through which the pinhole is drilled, relative to the diameter of the pinhole). Of course, diffraction effects will further complicate the calculation.

So, defining an image circle for the pinhole will depend on how much light falloff you are prepared to accept between the centre of the image and the edge of the "image circle".

Well that's pretty much what I would have thought, but because since the popular tool at mrpinhole.com provide an image circle diameter, I assumed otherwise.
Anyway, what I hear you say is that the falloff, as you call it, is basically linear, and thus no arbitrary limit can be set as it really depends upon exposure time and so on and so forth.

No, it is certainly not linear. I seem to recall that for film it is something like the square of the cosine of the angle of the ray to the optic axis, but as I said it also depends on how good the pinhole is and any directional variations in the sensitivity of the sensor. So it falls off much faster the further you move away from the centre of the image. Which is why some people talk about an image circle.

As you say, it ultimately comes down to experiment unless you have a great deal of detailed data about the pinhole and the sensor characteristics (plus the mathematical skills to work it out).

 

[Mark S Abeln]

As you say, it ultimately comes down to experiment unless you have a great deal of detailed data about the pinhole and the sensor characteristics (plus the mathematical skills to work it out).

Factors which aren't mentioned in the formulae are the characteristics of the sensor, including the micro lenses, and the thickness and refractive index of the glass cover plate, which will have a varying effect, depending on the angle which the light rays strike the sensor. As Tom says, good results will depend on experiment.

Also, some newer pinhole theories say that subject distance has an effect on optimal pinhole width.

 

[D Cox]

I think you would get the most even lighting on a hemispherical sensor with the pinhole at its centre.

When I made a pinhole camera to use with 10x8 inch paper, I used a paint can so that the paper was curved at least in one direction. This gave more even exposure than having the paper flat.

+++++++++++

You could make an image by putting half a ping-pong ball over the pinhole, to see how uneven the result is. This image could then be converted to a negative and used as a mask to correct exposure on other pictures.

 

[Doctor J]

A pinhole doesn't have an image circle in the same way that most lenses have. Light can pass through the pinhole at almost any angle and not be obstructed as it is in a typical compound lens (by the edges of other lens elements). So a pinhole has no well-defined image circle, but the light intensity on a flat sensor will fall off steadily as the angle between the light ray and the optical axis increases.

How fast it falls off in practice will depend on the characteristics of the sensor (e.g. any directional variation in sensitivity) and also how good the pinhole is (e.g. the thickness of the plate through which the pinhole is drilled, relative to the diameter of the pinhole). Of course, diffraction effects will further complicate the calculation.

So, defining an image circle for the pinhole will depend on how much light falloff you are prepared to accept between the centre of the image and the edge of the "image circle".

Well that's pretty much what I would have thought, but because since the popular tool at mrpinhole.com provide an image circle diameter, I assumed otherwise.
Anyway, what I hear you say is that the falloff, as you call it, is basically linear, and thus no arbitrary limit can be set as it really depends upon exposure time and so on and so forth.

No, it is certainly not linear. I seem to recall that for film it is something like the square of the cosine of the angle of the ray to the optic axis, but as I said it also depends on how good the pinhole is and any directional variations in the sensitivity of the sensor. So it falls off much faster the further you move away from the centre of the image. Which is why some people talk about an image circle.

As you say, it ultimately comes down to experiment unless you have a great deal of detailed data about the pinhole and the sensor characteristics (plus the mathematical skills to work it out).

Here is a very simple model that gives the light fall-off as approximately proportional to the cosine of the angle squared, as you say:



J

 

[Tom Axford]

Here is a very simple model that gives the light fall-off as approximately proportional to the cosine of the angle squared, as you say:



J

Thank you for that elegant argument. Unfortunately there are some problems with it.

We do not have a situation that can reasonably be described as a point source located at the pinhole (unless you produce an argument to explain why it is reasonable).

Instead, I think a better model would be to consider a point source at infinity, which will produce a beam of parallel light rays passing through the pinhole. If the pinhole is a perfectly circular hole in an infinitesimally thin sheet, then a cross section of the beam after passing through the pinhole will be an ellipse with max diameter d (the diameter of the pinhole) and min diameter d cos theta. The total light flux in the beam will be proportional to cos theta. This, of course, ignores diffraction effects.

This beam will produce a circular image of diameter d when it hits the sensor plane (no matter how far away it is, again ignoring diffraction). But the total light flux in that circular image is proportional to cos theta.

I'm not sure where the extra factor of cos theta comes from (to give cos squared theta as I originally suggested) or whether my memory is wrong in thinking that the formula is cos squared theta.

 

[Doctor J]

Like I said - "a very simple model"! There are plenty of more accurate models around if you want to do some reading, but they go a bit beyond answering a simple dpreview question about the nature of light fall-off from a pin hole.

J

 

[Zacariaz]

Also, some newer pinhole theories say that subject distance has an effect on optimal pinhole width.

 

[Zacariaz]

Thanks a bunch. I'm not really good with Greek letters, but I'n sure I'll figure it out.

 

[Zacariaz]

I think you would get the most even lighting on a hemispherical sensor with the pinhole at its centre.

When I made a pinhole camera to use with 10x8 inch paper, I used a paint can so that the paper was curved at least in one direction. This gave more even exposure than having the paper flat.

+++++++++++

You could make an image by putting half a ping-pong ball over the pinhole, to see how uneven the result is. This image could then be converted to a negative and used as a mask to correct exposure on other pictures.

I do intent to go digital, and thus a curved plane isn't really an option, sadly.

Also, what I wan't to do in the end goes much much further than simply poking a hole i a tin can.

Using half a ping pong ball for testing, is actually a rather nice idea. Was already thinking about how to handle that scenario, but I guess I'm node thinking.

 

[Zacariaz]

Here is a very simple model that gives the light fall-off as approximately proportional to the cosine of the angle squared, as you say:



J

One thing though, if I'm reading you correctly, the formula can be written as such:

light intensity * cos⁴(theta) / D²

But isn't light intensity what we're trying to calculate here? Where's the equal sign? Or is the light intensity perhaps some arbitrary constant?

I'm sorry, but while I am rather good at math, my education is somewhat lagging

 

[Gandolphi]

According to my old Ilford manual of photography, "the optimum pinhole diameter is about 1/25th of the square root of the distance from the pinhole to the screen. With a150mm camera this gives a pinhole of 0.5mm.

 

[Doctor J]

Here is a very simple model that gives the light fall-off as approximately proportional to the cosine of the angle squared, as you say:



J

One thing though, if I'm reading you correctly, the formula can be written as such:

light intensity * cos⁴(theta) / D²

But isn't light intensity what we're trying to calculate here? Where's the equal sign? Or is the light intensity perhaps some arbitrary constant?

I'm sorry, but while I am rather good at math, my education is somewhat lagging

Sorry - that funny symbol means "proportional to". In words, light intensity is proportional to one divided by the distance R squared. Substituting the value for R = D / cos theta, we get that light intensity is proportional to the cosine squared of the angle theta. So relative to the centre of the image, the light intensity falls of roughly as cosine squared of theta.

This model assumes a point source, so the reason it isn't a very accurate model is that the pinhole isn't a point source, and the size of the pinhole has an effect, but it gives a sort of "rough order of magnitude" estimate of what is going on. There is a nice discussion about pinhole cameras and resolution in the geometrical optics and diffraction limits here.

J

 

[alanr0]

Here is a very simple model that gives the light fall-off as approximately proportional to the cosine of the angle squared, as you say:



J

One thing though, if I'm reading you correctly, the formula can be written as such:

light intensity * cos⁴(theta) / D²

But isn't light intensity what we're trying to calculate here? Where's the equal sign? Or is the light intensity perhaps some arbitrary constant?

I'm sorry, but while I am rather good at math, my education is somewhat lagging

Sorry - that funny symbol means "proportional to". In words, light intensity is proportional to one divided by the distance R squared. Substituting the value for R = D / cos theta, we get that light intensity is proportional to the cosine squared of the angle theta. So relative to the centre of the image, the light intensity falls of roughly as cosine squared of theta.

This model assumes a point source, so the reason it isn't a very accurate model is that the pinhole isn't a point source, and the size of the pinhole has an effect, but it gives a sort of "rough order of magnitude" estimate of what is going on. There is a nice discussion about pinhole cameras and resolution in the geometrical optics and diffraction limits here.

J

In addition to the inverse square law factors, there are additional cos(theta) factors associated with the inclination of the incident light to the pinhole aperture, and the angle of incidence on the sensor plane. The net dependence is on the fourth power of the cosine, as indicated by Zacariaz.

http://nvlpubs.nist.gov/nistpubs/jres/39/jresv39n3p213_A1b.pdf

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--
Alan Robinson

 

[Zacariaz]

Think I get it know, simply read it wrong.

Thanks

 

[Doctor J]

Think I get it know, simply read it wrong.

Thanks

The simplest correction for a small but finite pinhole gives an extra factor of cos squared, so as someone said, cos theta to the fourth power is probably a better model to use. Beyond that the formulas get a lot more complicated, but probably don't add much correction for a typical pinhole of f/100 of whatever you end up with.

J

 

[Zacariaz]

Think I get it know, simply read it wrong.

Thanks

The simplest correction for a small but finite pinhole gives an extra factor of cos squared, so as someone said, cos theta to the fourth power is probably a better model to use. Beyond that the formulas get a lot more complicated, but probably don't add much correction for a typical pinhole of f/100 of whatever you end up with.

J

I know there are other factors at play, but one has to take one step at the time. I am currently trying to figure out what a good limit would be, that is how bad vignetting actually is when it's noticeable.

After that there's of course the optimal pinhole size, which people can't seem to agree on either. There's the wavelength of light to use for the calculations. Usually 550nm is what people use, but I'm thinking it's better to think worst was scenario and use 700nm instead, thus avoiding any diffraction at all. These also the matter of subject distance, which no one seems to care about, and so on and so forth...

All in all I don't really know what I'm doing, but I find it quite interesting.

 

[Zacariaz]

According to my old Ilford manual of photography, "the optimum pinhole diameter is about 1/25th of the square root of the distance from the pinhole to the screen. With a150mm camera this gives a pinhole of 0.5mm.

I suppose that may be true for that particular scenario, but as a general rule I don't think so. If anything, it's more like 1/100, but of course that's not the entire story either. The larger the pinhole, the blurrier the image, but more light is let in, and of course if the pinhole is too small you get diffraction, which also makes for a blurry image. And of course, the wider the lens, the more light is let in, but the vignetting is also more severe.

All in all it really a big compromise, and the trick is to find the right balance.

Bu hey, what do I know, I'm just a theoretician.