A person has near point at 100 cm. What power of lens is needed to read at 20 cm if he/she uses (a) contact lens, (b) spectacles having glasses 2.0 cm separated from the eyes?

Given: A person has near point of 100 cm.
According to the question, it is needed to read at a distance of 20 cm.
(a) When the contact lens is used:
u = – 20 cm = – 0.2 m,
v = – 100 cm = – 1 m
The lens formula is given by
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
On substituting the values in the above formula, we get:
$\frac{1}{f}=\frac{1}{\left(-1\right)}-\frac{1}{\left(-0.2\right)}=4$
∴ Power of the lens, P = $\frac{1}{f}$ = 4 D
(b) When the person wears spectacles at a distance of 2 cm from the eyes:
u = – (20 – 2) = – 18 cm = – 0.18 m,
v = – 100 cm = –1 m
The lens formula is given by
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
On substituting the values, we get:
$\frac{1}{f}=\frac{1}{\left(-1\right)}-\frac{1}{\left(-0.18\right)}=4.53$
∴ Power of the lens, P = $\frac{1}{f}$ = 4.53 D