I want a picture of the insides of a photocopier.  I realize that this
is not as easy as just putting a mirror on the machine.  I doubt that
the reflection of the inside will come out as an image.  I am thinking
that it may come out all black.  Or will it? I was hoping that you
could explain exactly what does happen.

Back to subcatalog.

---

OK here is the deal.  If you want the photocopier to take a picture
of itself you have to be able to make a real image of the camera lens
on the glass that you would normaly place something to be copied.
You can estimate by eye the depth of the camera lens in the copy
machine by eye. Lets call this distance x.  Now suppose that you
have a concave mirror of focal length f.  What you want to do is
position the mirror so that a real focused image of the camera lens
is formed at the glass plate. To do this you need to calculate the 
distance i from the mirror to the glass that will produce a focused 
image.

>From the optics equation for mirrors

        1   1   1
(1)     - = - + -
        f   o   i
     
Where f is the focal length of the mirror, o is the distance from the
object (in this case the camera) to the mirror and i is the distance
from the image (glass) to the mirror.  In our case (see the figure)

(2)          o = x + i 



                    |
  -----             |                |
      |_       x    |      i          |
       _O-----------|-----------------|
      |             |                 |
  -----             |                |
  camera            |               convex mirror
                  glass
                  plate


 Thus substitution of (2) into (1) gives


        1     1     1
(3)     - = ----- + -
        f   x + i   i                          

Now solve for i


              f     f
(4)     1 = ----- + -
            x + i   i    

              fi + f(x+i)
(5)     1 =   ------------
              (x + i)i    

              
(6)     (x + i)i = fi + f(x+i)
 
(7)      xi + i^2 = fi + fx+fi

(8)      xi + i^2 - fi - fx - fi = 0

(9)      xi + i^2 - 2fi - fx = 0

(10)     i^2 + xi - 2fi - fx = 0

(11)     i^2 + (x - 2f)i - fx = 0 

Thus by the quadratic formula

(12)           - (x -2f) +/- sqrt( (x-2f)^2 + 4fx )
          i = --------------------------------------
                           2   

               (2f - x) +/- sqrt( (x-2f)^2 + 4fx )
            = --------------------------------------
                           2 

               (2f - x) +/- sqrt(x^2 + 4f^2)
            = -------------------------------
                           2
                           
There are two roots and you will need the root that gives a positive
valie since i has been definesd as positive. 

Example: Suppose you happen to have a mirror of focal length 40cm
and the camera lens depth is estimated to be 30cm then


              (2(40cm) - 30cm) +/- sqrt((30cm)^2 + 4(40cm)^2)
        i   = -----------------------------------------------
                           2   
                             
                 50cm +/- sqrt(900cm^2 +6400cm^2)
            = -----------------------------------
                             2 
                             
                 50cm +/- 85.4cm
            = -------------------
                      2 
                      
            = (-17.7cm, 67.7cm)
            
The distance you place the mirror then is 67.7 cm away from the glass.
I suggest that you shine bright light into the camera well to illuminate
it but not on the lens itself to prevent glare.  Unfortunately the
light in the photocopier points in the wrong direction so you cant use it.

Good Luck Tommy!