If an object is placed at 0.25 m ( the normal near point), where will the image be? This is the place where the object must be placed in the absence of the correcting lens.
1/p + 1/q = P
1/0.25 + 1/q = 3
-1/p - 1/q = 3 - 1/0.25 = 3 - 4 = -1
q = -1 m
Answer is A