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Optics question

  1. Feb 4, 2009 #1
    A concave mirror creates an image on the screen 2x larger than the object. Both objects and screen are subsequently moved in order to create an image on the screen 3x larger than the object. 1) If the screen is moved 0.75 m in this process, how far is the object also moved? 2) What is the mirror's focal length?


    I think the equations are:
    1) M = h_i / h_o = - d_i / d_o
    2) 1/f = 1/d_o + 1/d_i
    where i refers to image, o refers to object

    I'm really not sure how to set this up.
  2. jcsd
  3. Feb 4, 2009 #2


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    Homework Helper

    You could try putting the given information into the two formulas.
    Repeat for the situation after moving the screen and object.
    Looks like 4 equations and 4 unknowns, so you should be able to find anything you want!
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