As far as I understand, the film inside a photo camera is positioned
in such a way that every light ray that hits some spot X on the film
comes from a single source on the object being photographed. Here's
what I mean:

http://static.howstuffworks.com/gif/camera-diagram3.gif

The various light rays from the top of the pencil, no matter where or
at what angle they hit the lens, will be focused to one point on the
film.

If so, then I have two questions:

1. What about interference effects between the rays of light? Since
they all hit the same point on the film, and come from the same
source, it seems they will be coherent and should produce interferene
at that point.
Yet we don't see that when taking photographs.

2. What about when the source of the rays is not an object (pencil,
etc.) but a mirror? In that case, even if the rays come from the same
point on the mirror, they are still of different colors (frequencies)
since they are reflections of light rays entering the mirror. And yet,
all these various frequencies from the same point in the mirror will
hit one single spot on the film. Shouldn't this produce a mix of
colors on the resulting photograph, instead of a picture of reflection
that we expect?

(The following image shows the situation I have in mind:
http://i2.tinypic.com/wsql2r.jpg The green and blue rays are reflected
off a mirror, at the same point, and then enter the lens and are
focused onto a single point A - and this will happen for every ray
coming from the same point on the mirror. As a result, any
photodector/film/etc. on point A should detect a mix of all colors
(from rays coming from all angles), yet in reality we only see one
color from one ray).

Great questions.  

1) You do see interference.  More specifically, a point source on the
object produces not a point in the image, but an Airy disk. The bigger
the lens, the smaller the resulting Airy disk.  So a pinhole camera
cannot make as sharp an image as a lens can. The interference pattern
you get from light that passes through a single aperature is called a
diffraction pattern, and when the source is far away, it's called
Fraunhofer diffraction.

A second, theoretical point, ignoring diffraction, is that the optical
distance from a point on the object to its corresponding point on the
image is identical regardless of the path.  This is a consequence of
the priciple of least time: in going from point A to point B, a light
ray will always take the path that minimizes the travel time.  A lens
makes all possible paths from A through the lens to B take the same
time to complete, so rays take all the paths.  The result is that rays
that left the source point in phase with each other will still be in
phase upon reaching the image point, and will interfere only
constructively.

2) If a camera is close to a mirror and adjusted to bring the mirror
into focus (that is, so that the edges of the mirror or writing on the
mirror appear sharp), then the image in the mirror will be blurry. 
The effective distance to the source of the light rays that form the
image in the mirror is greater than the distance to the mirror.  So
yes, you do get a 'mix of colors' in the photograph, and the result is
a blurry image.

In xray lithography when they need The best possible image to produce
the smaller and smaller more detailed image to make microprocessors,
etc.

They are now jumping thru many hoops just to get a blury image to make it work.

X-rays can be thought of as light with the frequency turned way up.

rracecarr-ga wrote:

>A second, theoretical point, ignoring diffraction, is that the optical
>distance from a point on the object to its corresponding point on the
>image is identical regardless of the path.  This is a consequence of
>the priciple of least time: in going from point A to point B, a light
>ray will always take the path that minimizes the travel time.  A lens
>makes all possible paths from A through the lens to B take the same
>time to complete, so rays take all the paths.

How can we prove that the lens makes all possible paths from A to B
take the same time (and hence have the same optical distance)? Is
there a proof for this available online, in some text, or elsewhere?

>2) If a camera is close to a mirror and adjusted to bring the mirror
>into focus (that is, so that the edges of the mirror or writing on the
>mirror appear sharp), then the image in the mirror will be blurry. 
>The effective distance to the source of the light rays that form the
>image in the mirror is greater than the distance to the mirror.

But it's not just the effective distance that's different - the
frequencies of the various light rays that come from a single point in
the mirror are different, because they originally came from different
sources.

My question is how come we are, in fact, able to take photographs (or
see with our eyes) images reflected in a mirror, no matter how we
adjust the focus?

Fermat's principle of least time is discussed, for example, in the
Feynman Lectures on Physics, Volume 1, Chapter 26, sections 3-6.  Here
is an exerpt (hopefully not a copyright violation) where Feynman
starts to discuss lenses:

"As another important example of the principle of least time, suppose
that we would like to arrange a situation where we have all the light
that comes out of one point, P, collected back together at another
point, P'.  That means, of course, that the light can go in a straight
line from P to P'.  That is all right.  But how can we arrange that
not only does it go straight, but also so that the light starting out
from P toward Q also ends up at P'? We want to bring all the light
back to what we call a focus.  How?  If the light always takes the
path of least time, then certainly it should not want to go over all
these other paths.  The only way that the light can be perfectly
satisfied to take several adjacent paths is to make those times
exactly equal!  Otherwise, it would select the one of least time. 
Therefore the problem of making a fousing system is merely to arrange
a device so that it takes the same time for the light to go on all the
different paths!
This is east to do.  Supose that we had a pice of glass in which light
goes slower than it does in the air. Now consider a ray which goes in
air in the path PQP'.  That is a longer path the from P directly to P'
and no doubt takes a longer time.  But if we were to insert a pice of
glass of just the right thickness (we shall later figure out how
thick), it might exactly compensate the excess time that it would take
th elight to go at an angle!  In those circumstances we can arrange
that the time the light takes to go straight through is the same as
the time it takes to go in the path PQP'..."

All a flat mirror does is make it appear as though the light bouncing
off the mirror comes from behind the mirror.  So if you focus your
camera for the optical distance to the thing you are photographing
(which is the distance from the camera to the mirror plus the distance
from the mirror to the object, all the light entering the camera from
a given point on the object will end up at the same spot on the film. 
It does not matter that light from two different locations on the
object bounce off the same spot on the mirror.  The reflection is
specular, and if light from two different points on the object hit the
same spot on the mirror, they hit with different angels, and so hit
the lens in different locations, and ultimately end up at different
spots on the film.  That is for a camera focused to take a sharp
picture of an image in a mirror.  If you focus the camera for just the
distance to the mirror (so that the edges of the mirror are in focus)
then all light rays coming from the same spot on the mirror will
indeed end up at the same spot on the film.  That causes blurring of
the image.  It doesn't completely wipe out the image, because only
light from a limited area on the object can bounce of a given spot on
the mirror and still enter the camera aperture.  So a given point on
the film is only 'contaminated' by light from parts of the object
close to the 'correct' point on the object.

1. There is no reason to suppose the light coming from your object is
coherent. Coherence is a property of the source of the light, which in
this case is not the object but whatever is illuminating it. In
general it is highly unlikely that the light will be coherent.

However, any aperture will diffract light. The smaller the aperture,
the greater the diffraction it will cause. So a monochromatic point
source of light taken through your lens would produce a set of rings
known as an Airy Disc on the film.

In reality, of course, your object is not a point source, and is not
illuminated monchromatically. Hence the effect of diffraction will be
to reduce the sharpness of the image, and indeed a larger lens will
produce a sharper image for this reason.

This is also why telescopes need to be large (it also helps them to
collect a lot of light). Some of the best radio telescope results come
from linking telescopes on opposite sides of the Earth - this gives
the "effect" of a lens the diameter of the Earth, allowing very
"sharp" results.

2. Your second points seems a little confused. The diagram cannot
possibly represent the setup you describe; trace the incident rays
back - they cannot come from the same point on the object. You are
persuading yourself that a mirror can separate different wavelengths
of light, which it cannot - it will faithfully reflect them to the
lens which will then behave in much the same way as in part 1.

However, a prism can separate wavelengths of light, and a lens is a
lot like a prism. In fact for a single glass lens, the focal length
(the point at which the lens focuses parallel rays of light) will be
different for different wavelengths of light. As a result, different
wavelengths of light coming from the same point on your object will
arrive in slightly different places on the film. This leads to
coloured fringes around objects in pictures - it is called chromatic
abberation, and it is a problem compound lenses attempt to fix.

The thing to remember is that colour is not a simple matter of
wavelength of light. If you lit your object with white light (equal
intensity of each wavelength across the visible spectrum) the point on
your object would reflect back a range of wavelengths of different
intensity, and would absorb other wavelengths to various degrees. This
is what makes up its colour.

The visual perception of colour is even more complex - the eye is no
good at picking out wavelength of light - it sees colour effectively
by comparison with what is around. This is important because light
conditions change. For instance, in a room lit by bulbs there is
almost no light outside the red end of the spectrum, yet a bit of
paper still looks white.

But the main issue you need to think about is what is colour?