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A camera on a 5-foot-high tripod is placed in front of a 6-foot-high picture that is mounted 3 feet above the floor.

Aexpress angle theta as a function of the distance x from the camera to the wall.
Bthe photographer wants to use a particular lens, for which theta = 36 (pi/5 radians). How far should she place the camera from the wall to be sure the entire picture will show in the photograph?

Verified Answer :

AL2006

This question is a fascinating example.  I mean a fascinating example of
one that you want to jump right in and start cranking out answers, but as
soon as you do that, you run into lots of sneaky things that you'll need
in order to find answers but they're not given in the question.

a).  We think we kind of know what the angle 'theta' is, but we're not sure.
It could be the elevation angle of the camera's optical axis, and if so, then
it depends on where the camera is actually aimed.  Is it the center of the
picture ?  Is it one corner ?  Is it a point somewhere along the top of the
frame ?  We don't know.

a).  On the other hand, 'theta' could be the angle subtended by the whole
subject at the camera's focal plane, or at the front surface of its objective
lens.  We don't know.

b).  Here in b)., it sounds like maybe there's some connection between
'theta' and what kind of lens she uses, so it could be the angle of the
camera's field of view. But then, how much of the picture you can capture
also depends on the dimensions of the film frame or the CCD.  We know
that one radian of field of view is one focal length on the film. But we don't
know the size of the 'film', so maybe the image of the subject can completely
fit on it, and maybe it can't.  We don't know.

But, as I said.  It sure is fascinating.


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