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Mirrors and images

  1. May 25, 2007 #1
    A virtual image is formed by a convex mirror that has a 30cm focal length. Where must the object be placed if the image is to be half the size of the object?



    I know that since the mirror is convex F is negative (F is the focal length). Useful equations are 1/Do + 1/Di = 1/F, where Do is the objects distance and Di is the images distance. I/O = Di/Do, where I is the height of the image and O is the height of the object. And O/I = F/F-Di.



    I'm not really sure on where to start could some one give me some pointers. :confused:

    I = Ox(1/2) or 2I = O

    F= -30cm
     
  2. jcsd
  3. May 25, 2007 #2
    Hello,

    The amplification I/O also equls to [tex]\frac{|D_i|}{|D_o|}[/tex] when we talk about near-axis light.
    Be careful and recall that the image is a virtual one and therefore [tex]D_i[/tex] is a negative value in Gauss' formula.


    Best regards
     
  4. May 25, 2007 #3
    Thanks for the tip :D
     
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