Question

Compute the power lens required to correct a hypermetropic eye with its near point at $75\mathrm{cm}$ from the eye.

Answer 1

Step 1: Given data

1. The near point of a hypermetropic eye is given as $75\mathrm{cm}$.
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 $25\mathrm{cm}$. However, 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 objects at the near point of the eye. So, the image distance is $75\mathrm{cm}$.

Step 3: Formulas used:

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

Step 4: Calculating the focal length

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

Image distance, $v=-75cm$

Focal length, $f=?$

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

Substitute the given values in the lens formula.

$$\begin{gathered} \frac{1}{-75}-\frac{1}{-25}=\frac{1}{f} \\ \frac{1}{-75}+\frac{1}{25}=\frac{1}{f} \\ \frac{1}{f}=\frac{-1+3}{75} \\ \frac{1}{f}=\frac{2}{75} \\ f=37.5\mathrm{cm} \end{gathered}$$

$\Rightarrow f=0.375\mathrm{m}$

The convex lens has a positive focal length of $0.375\mathrm{m}$.

Step 5: Calculate the power of the lens

We know the power of the lens,

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

$$\begin{gathered} P=\frac{1}{0.375} \\ P=2.67\mathrm{D} \end{gathered}$$

Hence, the power of the lens required to correct the eye defect is $+2.67\mathrm{D}$.