A person cannot see objects clearly that are closer that 1 m and farther than 5 m. To correct the eye vision, the person will use

Options

A. Bifocal lenses of power -3 D and +0.2 D
B. Bifocal lenses of power -2.5 D and +3.5 D
C. Bifocal lenses of power -1.5 D and +5D
D. Bifocal lenses of power -0.2 D and +3D

Answer

To solve the problem, we need to determine the type of lenses required for a person who cannot see objects clearly closer than 1 meter and farther than 5 meters.

### Step-by-Step Solution:

1. **Understanding the Vision Problem**:
- The person cannot see objects clearly that are closer than 1 meter (100 cm) and farther than 5 meters (500 cm).
- This indicates that the person is likely suffering from a condition that requires corrective lenses for both near and far vision.

2. **Correcting Near Vision**:
- For objects closer than 1 meter, we need to find a lens that allows the person to see objects at a distance of 25 cm (the least distance of distinct vision for a normal person).
- We will use the lens formula:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
- Here, \( u = -25 \) cm (the object distance, negative as per sign convention) and \( v = -100 \) cm (the image distance, negative since it is virtual and on the same side as the object).

3. **Calculating Focal Length for Near Vision**:
- Substitute the values into the lens formula:
\[ \frac{1}{f} = \frac{1}{-100} - \frac{1}{-25} \]
- Finding a common denominator (100):
\[ \frac{1}{f} = -\frac{1}{100} + \frac{4}{100} = \frac{3}{100} \]
- Therefore, the focal length \( f \) is:
\[ f = \frac{100}{3} \text{ cm} \approx 33.33 \text{ cm} \text{ or } 0.333 \text{ m} \]

4. **Calculating Power for Near Vision**:
- The power \( P \) of the lens is given by:
\[ P = \frac{1}{f} \text{ (in meters)} \]
- Converting focal length to meters:
\[ P = \frac{1}{0.333} \approx 3 \text{ diopters} \]
- This indicates a **convex lens** is needed for correcting near vision.

5. **Correcting Far Vision**:
- For objects farther than 5 meters, we need a lens that allows the person to see objects at infinity.
- Here, \( u = -\infty \) and \( v = -500 \) cm.

6. **Calculating Focal Length for Far Vision**:
- Substitute into the lens formula:
\[ \frac{1}{f} = \frac{1}{-500} - \frac{1}{-\infty} \]
- Since \( \frac{1}{-\infty} = 0 \):
\[ \frac{1}{f} = -\frac{1}{500} \]
- Therefore, the focal length \( f \) is:
\[ f = -500 \text{ cm} = -5 \text{ m} \]

7. **Calculating Power for Far Vision**:
- The power \( P \) of the lens is:
\[ P = \frac{1}{-5} = -0.2 \text{ diopters} \]
- This indicates a **concave lens** is needed for correcting far vision.

8. **Conclusion**:
- The person will need a **bifocal lens** that combines both a convex lens (for near vision with +3 diopters) and a concave lens (for far vision with -0.2 diopters).

### Final Answer:
The person will use a bifocal lens with +3 diopters for near vision and -0.2 diopters for far vision.