Question

An eye has a near point at a distance of $75cm$. What kind of lens should he get fixed in spectacles to decrease his near point distance to $25cm$? Also, calculate the power of the lens.

Answer 1

Step 1: Given data

Near point distance for an eye is $d=-75cm$
Near point distance for a normal eye is $D=-25cm$

Step 2: Finding the focal length of the lens

According to the lens formula,

$\frac{1}{f}=\frac{1}{\left(d\right)}-\frac{1}{\left(D\right)}$

Substituting the known values

$$\begin{gathered} \frac{1}{f}=\frac{1}{-75cm}-\frac{1}{\left(-25cm\right)} \\ \frac{1}{f}=-\frac{1}{75}+\frac{1}{25} \\ \frac{1}{f}=\frac{2}{75} \\ f=37.5cm \end{gathered}$$

As the sign of the focal length of the lens is positive, the lens used is convex.
Therefore, a person should get a convex lens fixed in spectacles to decrease his near point distance to $25cm$.

Step 3: Calculating the power of the lens

According to the formula for the power of the lens,

$P=\frac{100}{f\left(incm\right)}D$

where $D$ is the power of the lens in dioptre,

Substituting known values

$$\begin{gathered} P=\frac{100}{37.5cm}D \\ P=\frac{100}{37.5}D \\ P=2.67D \end{gathered}$$

Therefore, the power of the lens is $2.67D$.