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I always see diagrams of how a camera obscura works where the projected image neatly stops before or at the edges of the wall opposite the pinhole.

But when I look at pictures and videos of rooms converted into a camera obscura (usually as a fun experiment), the image often continues projecting onto the roof, side walls and floor. For example, there is a great picture here demonstrating this:

Walk-In Camera Obscura - Empty Kitchen

Even though we're primarily interested in the wall directly opposite the pinhole (since in a camera, that's usually where the film or sensor is) doesn't it also mean those diagrams are not entirely accurate then? Should they look less like this:

illustration showing the camera obscura image on only the opposite wall, with 1 foot of margin on the top and bottom

And more like this?:

diagram showing the camera obscura image projected a little onto the floor and ceiling

Or even like this? For I have heard in various places that the angle of view can even be as much as 180°:

diagram showing the camera obscura image covering most of the floor and ceiling

(Apologies for my crude diagrams but I think they get the point across)

(I originally posted this question in the Mathematics Stack Exchange, but decided it might be better here instead. I hope this is the right place for it)

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  • $\begingroup$ To stop it, no. To facilitate an image being projected onto the side walls would require some combination of lenses and mirrors and why might that be difficult? $\endgroup$
    – Robbie Goodwin
    Apr 3 at 22:32

4 Answers 4

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By placing a tube in front of the pinhole you should be able confine the solid angle that gets imaged into the room, such that only the wall is illuminated.

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    $\begingroup$ BTW, even if the pinhole has zero thickness (tube length in this answer's terms), the image suffers from heavy vignetting due to projected area of the pinhole being smaller when looked at from the sides compared to looking along the normal. $\endgroup$
    – Ruslan
    Mar 30 at 18:20
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    $\begingroup$ @Ruslan and for mathematically thin apertures the obliquity factor helps at least a little bit as well. $\endgroup$
    – uhoh
    Apr 2 at 6:04
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It's not just a problem with pinholes: lenses can accept light from outside the intended camera field of view. That can, through scattering and reflection, find its way to the image sensor, contaminating the image. In this collage, the four rectangular objects on the TESS spacecraft are not the lenses for the photometric cameras, but elaborate baffles that (mostly) prevent unwanted light entering the lenses. The lenses themselves are hidden behind the baffles.

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    $\begingroup$ A cut-away view of a TESS camera here does indeed seem to show circular lenses space.stackexchange.com/q/30742/12102 And I guess that these were circular too; Wide angle camera of Lunar Reconnaissance Orbiter, were rectangular lenses used? Even optics for scanning photocopy machines and Deep UV scanning projection photolithography lenses are circular. $\endgroup$
    – uhoh
    Apr 2 at 6:22
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    $\begingroup$ @uhoh The TESS lenses are circular. The camera bodies are circular, too. And, while the focal planes are rectangular, the electronics to support them are on stacks of circular circuit boards, inside the cameras. The baffles are rectangular to match the rectangular focal planes. $\endgroup$
    – John Doty
    Apr 2 at 12:58
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You are right, there would be pictures at at all 4 walls. But usually you are not in the box but look from outside mostly the back wall, and you put a half transparent paper on it, so you see only the picture on the transparent side. the rest has to be dark. so no other light comes in.

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If the pinhole were in infinitely thin material and there were no other things blocking the light, the angle of view of the projection is 180 degrees--i.e., all the inside surfaces of the rectangular solid. But a real pinhole is a little tunnel of non zero length and width, which lets light in from a field of view less than 180 degrees; similar geometry applies on the other side of the pinhole. And you can deliberately limit the angles with exterior or interior masking.

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  • $\begingroup$ Thank you for your reply (and to everyone else who replied of course) When you say "all the inside surfaces of the rectangular solid" does that include the inner surface/wall which contains the actual pinhole inside it? When I first wrote this post, I had assumed it would be impossible for there to be any projection on this side (even theoretically) due to the rectilinear propagation of light, but now I'm not so sure. $\endgroup$
    – CinnabarToffee
    Apr 3 at 12:31

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