The near point of a farsighted person's uncorrected eyes is 80 cm. what power contact lens should be used to move the near point to 25 cm from this person's eyes?

To solve this problem, we will get f and then we will use it to calculate the power.

So, for this farsighted person,
do = 25 cm and di = -80
Therefore:
1/f = (1/25) + (1/-80) = 0.00275 = 0.275 m

Power = 1/f = 1/0.275 = +3.6363 Diopeters.
This means that the lens is converging.

Try c ? that’s what i got