Re: Angle of ligh

Since you brought this up, there are two factors influencing the intensity
of light at the film plane, the distance from the pinhole and the angle off
axis. As you move off axis of a flat film plane, the distance from the
pinhole to the film grows, and the apparent shape of the pinhole changes
from round to narrower and narrower. The so called fourth power of the
cosine law governs. The intensity at any point on a flat film plane equals
the intensity at the axis point times the cosine, to the fourth power, of
the angle off axis. When you curve the film around the pinhole you
counteract half of it because the film is always the same distance, and the
only darkening you get at the edges is due to the change in the apparent
shape of the pinhole.
 ----- Original Message -----
From: "ragowaring" <ragowaring@btinternet.com>
To: <pinhole-discussion@pinhole.com>
Sent: Tuesday, February 12, 2002 2:33 PM
Subject: Re: [pinhole-discussion] Angle of ligh

> Dear Joao
>
> I'm no mathematician but I think you will find that the parts of the film
> nearest the pinhole will receive a greater amount of light for a given
area
> than parts of the film further away
>
> This is because of the inverse square law, which states simply that the
> radiation falling on a surface from a point source will decrease inversely
> proportinally with distance by 1/xsquared where x is the distance.
>
> This means that for every doubling of the distance from the source, the
> amount of radiation reaching a given area is quartered (that is because
the
> same radiation has to cover four times the area covered at half the
> distance).
>
> Imagine the area covered by a cone (of radiation if you like) - it is
> actually easier to imagine this as a four sided pyramid, so I shall
continue
> with this visualisation. The square at the base of the pyramid is 1
square
> unit. The point of the pyramid is the source. Radiation will reach the
base
> at a given rate, say one unit of radiation per second.
>
> If you double the height of the pyramid, which is equivalent to doubling
the
> distance from the source of radiation, you will find that the base of the
> pyramid is now four times the area of the first pyramid - four squares of
> one unit each or one big square four times the area.
>
> Now come the fun part. The radiation reaching this larger square in a
given
> time is the same as that reaching the 1 unit square at half the distance.
> That is to say, one unit of radiation per second. But this time that one
> unit has to cover four times the area as the radiation spreads out.
>
> Therefore each square unit at double the distance recieves a quarter of
the
> radiation per second. Therefore a doubling in the distance from a point
> source of radiation results in one quarter of the radiation falling on one
> unit area!
>
> This explains why on wide angle pinhole photographs, the sides of the
> negative come out less dense - because they are further away from the
> pinhole and therefore less light reaches them per given time. It is this
> per given time that is all important when calculating exposures with focal
> lengths etc.
>
> Now, when the film is parallel to the plane of the pinhole, i.e. at the
back
> of the camera, normally the inverse square law has a small effect,
> particularly if the angle of acceptance or vision is small.
>
> However, if you put the film on the camera side walls, the effect becomes
> very significant indeed. The parts nearer the pinhole will need a
> considerably shorter exposure that those further away.
>
> This however, can be compensated for if the side wall of the camera are
> short, that is to say, the camera has a short focal length.
>
>
> Enough of theory, the thing is to EXPERIMEMT!
>
> It is so much easier with pictures
>
> By the way, the above explanation is an approximation because in real life
> the base of the pyramid would be curved and not flat, but it is close
> enought to get the picture - sorry no pun intended
>
> Alexis
>
>
>
>
>
>
>
>
>
> on 12/2/02 5:40 pm, Joao Ribeiro at jribeiro@greco.com.br wrote:
>
> > Thanks Bill and Guillermo for your answers.
> >
> > But ...
> >
> >> Geometrically/mathematically speaking, the angle changes when the
> >> pinhole diameter changes, the change is so small tho, that in practice
you
> >> can dismiss it. Since you want to calculate the "cone angle",
otherwise
> >> known as "angle of view", here is a formula I just derived that takes
the
> >> pinhole diameter into consideration:
> >>
> >> Cone angle = ArcTan [ (D+P) / (2 * B) ]
> >>
> >> Where"
> >> D = Diagonal of your film format
> >> P = Pinhole diameter
> >> B = Bellows extension (or focal length)
> >>
> >> As you can see, the effect of adding P to D is very small, i.e., for
8x10,
> >> "D" would be equal to about 325mm if you add to that a "P" of 0.5mm,
you get
> >> 325.5mm, again, not a big change. The same happens if you change the
> >> pinhole diameter.
> >
> > I'm not sure this is the answer to my question. If I could send a
drawing
> > attached to the list it would be easier, but this is what I want:
> >
> > I imagine a light entering the box/camera and forming a cone. This cone
will
> > be
> > independent of film size. I believe it will vary with pinhole diameter
but
> > maybe
> > not in a meaningful way.
> > Let's say I made a very long focal distance box, no matter the film
size, and
> > at
> > the end of the box, parallel to the pinhole plane usually is put the
film. But
> > I'll be using just a fraction of the image formed by this cone, the rest
will
> > be
> > absorbed by the black walls of the camera box.
> > Now, what I intend to do is to place 2 sheets of film not at the end of
the
> > camera, but at the walls that are perpendicular to the pinhole plane. I
> > imagine
> > that the light absorbed by these walls will also form images, certainly
> > distorted (I believe someone in this list have already done that). How
far
> > from
> > the pinhole plane the film should be to be completely covered by light?
My
> > imaginations tells me that if it is too close I'll have a "V" shape
image.
> > Is this formula you sent me able to give me that info? I'm asking
because it
> > takes into account the film diagonal and I believe this cone is
independent of
> > the film diagonal, I'm not sure.
> >
> > Sorry for this long post
> >
> > Joao
> >
> >
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>
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Received on Tue Feb 12 16:54:28 2002