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Diverging lenses

  1. Mar 25, 2009 #1
    1. The problem statement, all variables and given/known data
    The focal length of a diverging lens is -20 cm. Locate the object, given that the image is real, erect, and 150% of the size of the object.

    2. Relevant equations
    (1/f)=(1/do)+(1/di)
    M=-(di/do)
    3. The attempt at a solution
    i'm not really sure how a diverging lens forms a real image (i thought they could only from virtual since most of the images are on the same side) and a size that is larger than the original object. At first, I thought that the image would be inverted (negative magnification) since a virtual image would have a positive magnification (image is upright)...
     
  2. jcsd
  3. Mar 25, 2009 #2

    mgb_phys

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    Science Advisor
    Homework Helper

    For a diverging lens revealed the image of the object is virtual, upright, reduced in size and located on the same side of the lens as the object.

    You can form a real image from a diverging lens - but only by feeding it a converging beam from another lens (eg in a retrofocus wide angle camera lens)
     
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