Posted by Kyle on Monday, May 28, 2012 at 8:27pm.
A converging or positive lens is required, with
a power such that when an object is placed 25 cm in front of the
lens, the lens forms a virtual image.
The image distance di= 60 cm, so that this image serves as an
object for the eye at the eye’s near point.
Taking into account the distance 2 cm
di =0.62 m, do =0.27m
Applying the thin lens equation, we find
1/F =1/do +1/di = 1/0.27 +1/0.62 =3.7+1.61 =5.31.
F = 0.188 m
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