Answer to Question #16566 in Optics for tariq
A diverging lens is to be used to produce a virtual image one-third as tall as the object. Where the object should be placed
All virtual images formed by diverging lenses are upright images. Thus,& M > 0 and the magnification gives
M = -p/q = 1/3,
q = -p/3
Then, from the thin lens equation,
1/p - 3/p = -2/p = 1/f,
p = -2f = 2|f|.
So, the object should be placed at distance& 2|f| in front of the lens.