Question

The near point of a hypermetropic eye is $1m$. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is $25cm$.

Answer 1

Step 1: Hypermetropia

1. Hypermetropia is also termed long-sightedness which is a type of defect in vision.
2. In this defect objects closer to the eye cannot be seen clearly.
3. Objects further away from the eye can be seen clearly.
4. When the size of the eyeball reduces, the distance between the retina and lens of the eye is reduced, which causes this defect to arise.
5. Resulting which the focal length of the eye lens increases.
6. For greater focal length converging power reduces.
7. The convergence of the eye lens is increased by using the converging lens i.e. convex lens.

Step 2: Calculate Power of Lens

Let $y=$distance of the $N\text{'}$ from the defective eye.

Now $N$ of the normal eye is at distance $D=25cm$.

The object placed at $N$ forms its virtual image at $N\text{'}$ due to the convex lens.

$\therefore u=-D,v=-y$

By lens formula

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}=\frac{1}{-y}-\frac{1}{-D}=\frac{-1}{y}+\frac{1}{D}=\frac{y-D}{yD}$

$\therefore$ Required focal length, $f=\frac{yD}{y-D}$

Required power, $P=\frac{1}{f}=\frac{y-D}{yD}$

Step 3: Substitute the Values

We have $y=1m=100cm$

$D=25cm$

$\therefore$Required power,

$$\begin{gathered} P=\frac{100-25}{100\times 25} \\ P=0.03c{m}^{-1} \\ P=0.03\times 100m^{-1} \\ P=+3\mathrm{dioptre} \end{gathered}$$

Hence, the power of the lens required is $+3\mathrm{dioptre}$.