Answer
a) farsighted
b) $ N=40cm $
Work Step by Step
(a) We know that
$\frac{1}{f_{near}}=\frac{1}{N}+\frac{1}{d_i}$
$\implies \frac{1}{f_{near}}=\frac{1}{0.25m}+\frac{1}{0.0240m}=45.67m^{-1}$
Now the power of the lens is given as
$ P=\frac{1}{f_{near}}$
$ P=\frac{1}{45.67m^{-1}}=45.67diopters $
We can see that the greatest refractive power of the patient is less than the refractive power of a normal person; hence, the patient can not focus on closer objects and the patient is farsighted.
(b) We know that
$ f_{max}=\frac{1}{P}$
$\implies f_{max}=\frac{1}{44.1diopters}=0.02267m $
Now $ N=\frac{f_{max}d_i}{d_i-f_{max}}$
We plug in the known values to obtain:
$ N=\frac{(0.02267m)(0.024m)}{0.024m-0.02267m}$
$ N=0.40m=40cm $
