The nearer point of hypermetropic eye is `40 cm`. The lens to be used for its correction should have the power

Options

A. `+1.5 D`
B. `-1.5 D`
C. `+2.5 D`
D. `+0.5 D`

Explanation

To solve the problem of finding the power of the lens required to correct a hypermetropic eye with a nearer point of 40 cm, we can follow these steps:

### Step-by-Step Solution:

1. **Understand Hypermetropia**: Hypermetropia, or farsightedness, is a condition where distant objects can be seen clearly, but nearby objects appear blurry. The nearest point of clear vision is farther than normal.

2. **Identify the Given Data**:
- The nearest point (D) of the hypermetropic eye is given as 40 cm.

3. **Determine the Focal Length**:
- For a hypermetropic eye, we need to bring the nearer point closer to the normal near point of the human eye, which is typically taken as 25 cm.
- The focal length (f) of the lens required can be calculated using the formula:
\[ f = D - d \]
where \(d\) is the normal near point (25 cm).
- Here, \(D = 40 \, \text{cm}\) and \(d = 25 \, \text{cm}\).
- Thus,
\[ f = 25 \, \text{cm} - 40 \, \text{cm} = -15 \, \text{cm} \]
- However, since we need a lens to correct the vision, we actually consider the focal length as positive for a convex lens.

4. **Convert Focal Length to Meters**:
- The focal length in meters is:
\[ f = +40 \, \text{cm} = +0.4 \, \text{m} \]

5. **Calculate the Power of the Lens**:
- The power (P) of a lens is given by the formula:
\[ P = \frac{1}{f} \quad \text{(in meters)} \]
- Substituting the value of \(f\):
\[ P = \frac{1}{0.4} = 2.5 \, \text{diopters} \]

6. **Conclusion**:
- The power of the lens required to correct the hypermetropic eye is \(+2.5 \, \text{D}\).

### Final Answer:
The lens to be used for its correction should have the power of **+2.5 D**.

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