To solve this problem we will apply the concept related to the lens power with which farsightedness can be corrected. Mathematically this value is given by the relationship,
[tex]P = \frac{1}{f}[/tex]
Here,
f =focal length
In turn, said expression can be exposed in terms of the distance of the object and the image as:
[tex]P = \frac{1}{p}+\frac{1}{q}[/tex]
Here,
p = Object Distance ( By convention is 25cm)
q = Image distance
Replacing we have,
[tex]P = \frac{1}{0.25}+\frac{1}{-0.75}[/tex]
[tex]P = +2.67D[/tex]
Therefore the power lens that is needed to correct for farsightedness is +2.67D
