A farsighted woman breaks her current eyeglasses and is using an old pair whose refractive power is 1.655 diopters. Since these eyeglasses do not completely correct her vision, she must hold a newspaper 39.4 cm from her eyes in order to read it. She wears the eyeglasses 1.90 cm from her eyes. How far is her near point from her eyes?
1/f = 1/(do) + 1/(di)
di < 0 for virtual images (what this is)
The Attempt at a Solution
do is .394m and 1/f=1.655 diopters
so 1/(1/.394m - 1.655) = di
di = 1.13m
Near Point = 1.13m + .019m = 1.15m
Where did I go wrong? Thanks!