Question

The near point and the far point of a child are at 10 cm and 100 cm. If the retina is 2.0 cm behind the eye-lens, what is the range of the power of the eye-lens?

Answer 1

Given:
Near point of the child = 10 cm
Far point of the child = 100 cm
The retina is at a distance of 2 cm behind the eye lens.
Thus, we have:
v = 2 cm = 0.02 m
When the near point is 10 cm,
u = − 10 cm = − 0.1 m
v = 2 cm = 0.02 m
The lens formula is given by
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
On putting the respective values, we get:
$$\begin{gathered} \frac{1}{f}=\frac{1}{0.02}-\frac{1}{\left(-0.1\right)} \\ =50+10=60\mathrm{m} \end{gathered}$$
∴ Power of the lens, P = $\frac{1}{f}$ = 60 D
Now, consider the near point 100 cm.
Here,
u = − 100 cm = − 1 m
v = 2 cm = 0.02 m
The lens formula is given by
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
Putting the values, we get:
$$\begin{gathered} \frac{1}{f}=\frac{1}{0.02}-\frac{1}{\left(-1\right)} \\ =50+1=51 \end{gathered}$$
∴ Power of the lens = $\frac{1}{f}$ = 51 D
So, the range of the power of the eye lens is from + 60 D to + 51 D.