A far sighted person has his near point 50 cm. Find the power of lens he should use to see at 25 cm, clearly.

Options

A. + 1 D
B. `+` 2 D
C. `-`2D
D. `-`1D

Explanation

To solve the problem of finding the power of the lens for a far-sighted person with a near point of 50 cm who wants to see clearly at 25 cm, we can follow these steps:

### Step-by-step Solution:

1. **Identify the Given Values**:
- Near point of the person (u) = 50 cm (since it's a far-sighted person, this is positive).
- Desired distance for clear vision (v) = 25 cm (this is negative in lens formula as per sign convention).

2. **Apply the Sign Convention**:
- According to the sign convention for lenses:
- Object distance (u) is taken as negative when the object is on the same side as the incoming light (for a virtual object).
- Image distance (v) is taken as negative for virtual images.
- Thus, we have:
- \( u = -50 \, \text{cm} \)
- \( v = -25 \, \text{cm} \)

3. **Use the Lens Formula**:
- The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
- Substituting the values:
\[ \frac{1}{f} = \frac{1}{-25} - \frac{1}{-50} \]

4. **Calculate the Focal Length (f)**:
- Simplifying the right side:
\[ \frac{1}{f} = -\frac{1}{25} + \frac{1}{50} \]
- Finding a common denominator (50):
\[ \frac{1}{f} = -\frac{2}{50} + \frac{1}{50} = -\frac{1}{50} \]
- Thus, we find:
\[ f = -50 \, \text{cm} \]

5. **Calculate the Power of the Lens (P)**:
- The power of a lens is given by the formula:
\[ P = \frac{1}{f(\text{in meters})} \]
- Convert focal length from cm to meters:
\[ f = -0.50 \, \text{m} \]
- Now, substituting into the power formula:
\[ P = \frac{1}{-0.50} = -2 \, \text{diopters} \]

6. **Conclusion**:
- The power of the lens required for the person to see clearly at 25 cm is:
\[ P = -2 \, \text{diopters} \]

### Final Answer:
The power of the lens he should use is **-2 diopters**.