CATAGORY: Light
 QUESTION: I was given this problem, and did not know how to solve it:
An object is placed 35 cm in frount of a covex mirror which has a Radius
of curvature of 78 cm. If the object is 10cm high, where is the image
formed, and how tall is it.

Back to subcatalog.

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We use the fact that the focal length is one half the radius of 
curvature.

                      1
(1)               f = - r
                      2

And the optics formula


(2)          1   1   1
             - = - + -
             f   o   i
             
             
Where f is the focal length, o is the object distance from the mirror
and i is the image distance from the mirror. From (1) and using r = 78cm
we have 
                     1
(3)              f = - 78 cm = 39 cm
                     2
                     
since o =35 cm we can find i from equation (2).  But first notice that 
the object is closer to the mirror than the focal point so we anticipate
that the image will be a virtual erect image appearing "in reflection"
to be "behind the mirror".


(4)          1       1       1
             -     = -    +  -
             39cm    35cm    i
             
so             
             
              (35cm)(39cm)      1365
(5)       i =  ----------- = -  ---- cm = -341.25 cm
               35cm - 39cm        4
                 
                  
The negative answer we get indicates a virtual image behind the mirror
as predicted above.  The image height ho relative to the
object height ho is given by

            hi     i
(6)         -- = - -
            ho     o
            
so with ho = 10 cm
 
                   i        -341.25 cm
(7)         hi = - - ho = - ----------- 10 cm = 97.5cm           
                   o           35cm
                   
The image is magnified and erect since hi > 0 

               ********************
               *                  *
               *  i = -341.25 cm  *
               * hi =    97.5 cm  *
               *                  *   
               ********************