A far sighted person has a near point of 60 cm. What power lens should be used for eye glasses such that the person can read a book at a distance of 25 cm.

Options

A. ` + 1.67` D
B. ` + 2 ` D
C. ` + 2.33` D
D. ` - 1.67 ` D

Explanation

To solve the problem, we need to determine the power of the lens required for a far-sighted person who has a near point of 60 cm to read a book at a distance of 25 cm.

### Step-by-Step Solution:

1. **Understand the Problem**:
- A far-sighted person has a near point (the closest distance at which they can see clearly) of 60 cm.
- They want to read a book at a distance of 25 cm.

2. **Identify the Required Focal Length**:
- The lens needs to create an image of the object (the book) at the near point of the person (60 cm).
- The object distance (u) is -25 cm (the negative sign indicates that the object is in front of the lens).
- The image distance (v) is -60 cm (since the image is formed at the near point of the person).

3. **Use the Lens Formula**:
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
where \( f \) is the focal length of the lens, \( v \) is the image distance, and \( u \) is the object distance.

4. **Substitute the Values**:
\[ \frac{1}{f} = \frac{1}{-60} - \frac{1}{-25} \]

5. **Calculate Each Term**:
- Calculate \( \frac{1}{-60} \) and \( \frac{1}{-25} \):
\[ \frac{1}{-60} = -0.01667 \quad \text{and} \quad \frac{1}{-25} = -0.04 \]

6. **Combine the Terms**:
\[ \frac{1}{f} = -0.01667 + 0.04 = 0.02333 \]

7. **Find the Focal Length**:
\[ f = \frac{1}{0.02333} \approx 42.86 \text{ cm} \]

8. **Convert Focal Length to Meters**:
\[ f = 0.4286 \text{ m} \]

9. **Calculate the Power of the Lens**:
The power \( P \) of a lens is given by:
\[ P = \frac{1}{f} \quad \text{(in meters)} \]
\[ P = \frac{1}{0.4286} \approx 2.33 \text{ diopters} \]

### Conclusion:
The power of the lens required for the far-sighted person to read a book at a distance of 25 cm is approximately **+2.33 diopters**.