The eye of a man cannot see the object closer than 0.5 m. Give the focal length and power of corrective lens he should use.(Nearest distance of vision is 25 cm)

Options

A. F=20 cm, P=3D
B. F=35 cm, P=2D
C. F=40 cm, P=3 D
D. F=50 cm, P=2D

Answer

To solve the problem, we need to find the focal length and power of the corrective lens required for a person whose near point (the closest distance at which they can see clearly) is 0.5 m (50 cm), while the normal near point is 25 cm.

### Step-by-Step Solution:

1. **Identify the given values:**
- The near point of the person (D) = 0.5 m = 50 cm
- The normal near point (D') = 25 cm

2. **Use the lens formula:**
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Where:
- \( f \) = focal length of the lens
- \( v \) = image distance (which is the normal near point, D' = -25 cm, negative because it is on the same side as the object)
- \( u \) = object distance (which is the near point of the person, D = -50 cm, negative for the same reason)

3. **Substituting the values into the lens formula:**
\[ \frac{1}{f} = \frac{1}{-25} - \frac{1}{-50} \]
\[ \frac{1}{f} = -\frac{1}{25} + \frac{1}{50} \]
To combine the fractions, find a common denominator (which is 50):
\[ \frac{1}{f} = -\frac{2}{50} + \frac{1}{50} = -\frac{1}{50} \]

4. **Calculate the focal length (f):**
\[ f = -50 \text{ cm} \]
The negative sign indicates that it is a diverging lens.

5. **Calculate the power (P) of the lens:**
The power of a lens is given by the formula:
\[ P = \frac{1}{f(\text{in meters})} \]
First, convert the focal length from cm to meters:
\[ f = -0.5 \text{ m} \]
Now, substituting into the power formula:
\[ P = \frac{1}{-0.5} = -2 \text{ diopters} \]

### Final Answers:
- Focal length of the corrective lens, \( f = -50 \text{ cm} \) or \( -0.5 \text{ m} \)
- Power of the corrective lens, \( P = -2 \text{ D} \)