Question

A diverging and converging lens with equal focal lengths of f = 0.5 m are placed so that their "fake" focal points overlap. If you place an object 2 m to the left of the left most lens, where is the final image formed? What are its attributes? Draw the ray diagram to double-check your work.Suppose you are allowed to use these the same type of lenses with what ever focal lengthyou need. What type of devise you can build with these two lenses.

Answer 1

Note that      
      
1/f = 1/do + 1/di      
      
where      
      
f = focal length =    0.5   m
do = object distance =    2   m
di = image distance      
      
Thus, solving for di,      
      
di =    0.666666667   m
      
Thus, the magnification, m = -di/do,      
      
m =    -0.333333333  

This will be 0.3333 m from the second diverging lens.

Thus, for the second lens, note that      
      
1/f = 1/do + 1/di      
      
where      
      
f = focal length =    -0.5   m
do = object distance =    0.333333   m
di = image distance      
      
Thus, solving for di,      
      
di =    -0.19999988   m
      
Thus, the magnification, m = -di/do,      
      
m =    0.60000024  

Thus, the total magnification is

mtot = -0.3333*0.6000

mtot= -0.200

Thus, the final image is [INVERTED, MINIFIED, VIRTUAL] and formed 0.200 m behind to the left of the second lens.