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

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