Question

The far point of a myopic person is $75\mathrm{cm}$ in front of the eye. What is the nature and power of the lens required to correct the problem?

Answer 1

Step 1: Given data

The far point of a myopic person, that is the image distance is given as $75\mathrm{cm}$.

Step 2: Nature of the lens

A person with a myopic eye should use spectacles fitted with a concave lens or diverging lens of suitable focal length to enable him to see very distant objects clearly.

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

To see the object at infinity, Object distance $u=\infty \left(\mathrm{infinity}\right)$

The corrective concave lens should form an image of a distant object placed at infinity at the far point of the myopic eye.

Image distance, $v=-75\mathrm{cm}$

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

$\frac{1}{-75}-\frac{1}{\infty }=\frac{1}{f}$

$$\begin{gathered} \frac{1}{f}=\frac{1}{-75} \\ f=-75\mathrm{cm} \end{gathered}$$

$\Rightarrow f=0.75\mathrm{m}$

A negative sign of the focal length shows that the lens required to correct the eye defect (myopia) is a concave lens.

Step 5: Calculating the power of the lens

We know the power of the lens can be calculated by using the formula given as,

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

$$\begin{gathered} P=\frac{1}{{\displaystyle -0.75}} \\ P=-\frac{100}{75} \\ P=-1.33\mathrm{D} \end{gathered}$$

Power of lens required to correct the eye defect is $-1.33\mathrm{D}$.