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*Question:* How far must an object be
placed in front of a diverging lens
of focal length cm in order to ensure that the
size of the image is fifteen times less than the
size of the object? How far in front of the lens is
the image located?

*Answer:* The focal length of a diverging
lens is *negative* by convention,
so cm, in this case.
If the image is fifteen times smaller than the object then the magnification
is
. We can be sure that , as opposed to , because
we know that images formed in diverging lenses are always virtual and
upright.
According to Eq. (364), the image distance
is given by

where is the object distance. This can be combined with Eq. (367)
to give

Thus, the object must be placed cm in front of the lens.
The image distance is given by

Thus, the image is located cm *in front* of the lens.

** Next:** Wave Optics
** Up:** Paraxial Optics
** Previous:** Example 13.3: Converging lenses
Richard Fitzpatrick
2007-07-14