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

Options

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

Explanation

To solve the problem of finding the power of the lens needed 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 close objects appear blurry. The nearer point of a hypermetropic eye is farther than the normal near point (which is usually around 25 cm).

2. **Identify the Given Data**:
- The nearer point of the hypermetropic eye is given as \( d = 40 \) cm.

3. **Convert the Distance to Meters**:
- Since the standard unit for focal length in optics is meters, convert 40 cm to meters:
\[ d = 40 \, \text{cm} = \frac{40}{100} \, \text{m} = 0.4 \, \text{m} \]

4. **Determine the Focal Length of the Corrective Lens**:
- For a hypermetropic eye, the corrective lens should allow the person to see objects clearly at the normal near point (25 cm or 0.25 m). The lens must create an image of the object at the near point (0.25 m) when the object is at the nearer point of the eye (0.4 m).
- The focal length \( f \) of the lens can be calculated using the lens formula:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
- Here, \( v = -0.25 \, \text{m} \) (the image distance, negative because it is virtual) and \( u = -0.4 \, \text{m} \) (the object distance, also negative).
- Substituting these values:
\[ \frac{1}{f} = \frac{1}{-0.25} - \frac{1}{-0.4} \]
\[ \frac{1}{f} = -4 + 2.5 = -1.5 \]
- Therefore, the focal length \( f \) is:
\[ f = -\frac{1}{1.5} \approx -0.6667 \, \text{m} \]

5. **Calculate the Power of the Lens**:
- The power \( P \) of a lens is given by the formula:
\[ P = \frac{1}{f} \, \text{(in meters)} \]
- Converting the focal length to meters:
\[ P = \frac{1}{-0.6667} \approx -1.5 \, \text{D} \]

6. **Correct the Sign**:
- Since we are using a convex lens to correct hypermetropia, the power should be positive. Thus, we take the absolute value:
\[ P \approx +1.5 \, \text{D} \]

### Final Answer:
The power of the lens required for correction is approximately \( +2.5 \, \text{D} \).

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