Question

A person with defective eyesight is unable to see objects nearer than $150\mathrm{cm}$. He wants to read a book placed at a distance of $25\mathrm{cm}$. Find the nature, focal length, and power of lens he requires for his spectacles.

Answer 1

Step 1: Understanding the question

1. A person who is unable to see nearby objects clearly but can see far off or distant objects clearly is suffering from hypermetropia.
2. Therefore, the person is suffering from hypermetropia as he is not able to see nearby objects.
3. The near point of a hypermetropic eye is $150\mathrm{cm}$.
4. It is known that the near point of a normal eye is $25\mathrm{cm}$. But a hypermetropic eye is unable to see objects kept at a distance of $25\mathrm{cm}$ clearly.

Step 2: Nature of the lens

To correct this defect, the person has to use spectacles with a convex lens of suitable focal length. The convex lens forms a virtual image of the nearby object at the near point of the eye. So, the image distance is $150\mathrm{cm}$.

Step 3: Formula used

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

Step 4: Calculating the focal length

Object distance $u=-25\mathrm{cm}$

Image distance, $v=-150\mathrm{cm}$

Focal length, $f=?$

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

Putting the given values in the lens formula, we get:

$$\begin{gathered} \frac{1}{-150}-\frac{1}{-25}=\frac{1}{f} \\ \frac{1}{-150}+\frac{1}{25}=\frac{1}{f} \\ \frac{1}{f}=\frac{-1+6}{150} \\ \frac{1}{f}=\frac{5}{150} \\ f=\frac{150}{5} \\ f=30\mathrm{cm} \end{gathered}$$

Thus, the convex lens with a positive focal length of $30\mathrm{cm}$ is required in the given case.

Step 5: Conversion

$$\begin{gathered} 1\mathrm{m}=100\mathrm{cm} \\ \Rightarrow 1\mathrm{cm}=\frac{1}{100}\mathrm{m} \\ \therefore 30\mathrm{cm}=30\times \frac{1}{100}\mathrm{m} \\ =0.3\mathrm{m} \end{gathered}$$

Step 5: Calculating the power of the lens

We know the power of the lens,

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

$$\begin{gathered} P=\frac{1}{{\displaystyle 0.3}} \\ P=\frac{10}{3} \\ P=3.33\mathrm{D} \end{gathered}$$

Hence, the power of the lens required to correct the eye defect is $+3.33\mathrm{D}$.