Question

The near point of a hypermetropic eye is $50\mathrm{cm}$. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is $25\mathrm{cm}$.

Answer 1

Step 1: Given data

1. The near point of a hypermetropic eye is given as $50\mathrm{cm}$.
2. A person who is unable to see nearby objects clearly but can see far off or distant objects clearly is suffering from hypermetropia.
3. The near point of the normal eye is $25\mathrm{cm}$. But a hypermetropic eye cannot see the objects kept at a distance of $25\mathrm{cm}$ from the eye distinctly.

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 $50\mathrm{cm}$.

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, $v=-50\mathrm{cm}$

Focal length, $f=?$

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

Substitute the given values in the lens formula.

$$\begin{gathered} \frac{1}{-50}-\frac{1}{-25}=\frac{1}{f} \\ \frac{1}{-50}+\frac{1}{25}=\frac{1}{f} \\ \frac{1}{f}=\frac{-1+2}{50} \\ \frac{1}{f}=\frac{1}{50} \\ f=50\mathrm{cm} \end{gathered}$$

$\Rightarrow f=0.5\mathrm{m}$

A convex lens has a positive focal length.

Step 5: Calculating the power of the lens

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

$$\begin{gathered} P=\frac{1}{f(\mathrm{in}\mathrm{meters})} \\ =\frac{1}{0.5} \\ =\frac{10}{2} \\ =2\mathrm{D} \end{gathered}$$

Thus, the power of the lens required to correct the eye defect is $+2\mathrm{D}$.