Question

A person can see clearly up to a distance of $30\mathrm{cm}$. He wants to read a book kept at a distance of $40\mathrm{cm}$. Find the nature, focal length and, power of lens he requires for his spectacles.

Answer 1

Step 1: Given data

Object distance, $u=-40\mathrm{cm}$

Image distance, $v=-30\mathrm{cm}$

Step 2: Formula used

1. Lens formula, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
2. Power of lens, $P=\frac{1}{f(\mathrm{in}\mathrm{meters})}$

Here, $f$ is the focal length of the lens.

Step 3: Calculating the focal length and nature of the lens

Object distance, $u=-40\mathrm{cm}$

Image distance, $v=-30\mathrm{cm}$

Focal length, $f=?$

We know, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ (lens formula)

Substitute the given values in the lens formula as follows:

$$\begin{gathered} \frac{1}{-30}-\frac{1}{-40}=\frac{1}{f} \\ \frac{1}{-30}+\frac{1}{40}=\frac{1}{f} \\ \frac{-4+3}{120}=\frac{1}{f} \\ \frac{1}{f}=\frac{-1}{120} \\ f=-120\mathrm{cm} \\ f=-1.2\mathrm{m} \end{gathered}$$

To correct this defect, the person has to use spectacles with a concave lens having a focal length of $-1.2\mathrm{m}$.

Step 4: Calculating the power of the lens

We know the power of the lens,

$P=\frac{1}{f(\mathrm{in}\mathrm{meters})}$

$$\begin{gathered} P=\frac{1}{-1.2} \\ P=\frac{10}{-12} \\ P=-0.833\mathrm{D} \end{gathered}$$

Hence, the power of the lens required to correct the eye defect is $-0.833\mathrm{D}$.