A person's near point is `50 cm` and his far point `3m`. answer the following question when
`(i)` Power of the lenses he requires for reading and
`(ii)` Power of the lenses he requires for for seeing distant stars are

Options

A. `-2D` and `-0.33D`
B. `2D` and `-0.33D`
C. `-2D` and `3D`
D. `2D` and `-3D`

Explanation

To solve the problem, we need to calculate the power of the lenses required for two scenarios:

1. For reading (to adjust the near point).
2. For seeing distant stars (to adjust the far point).

### Step-by-step Solution:

**(i) Power of the lens required for reading:**

1. **Identify the near point and normal vision:**
- The person's near point (N) is 50 cm.
- The normal near point (N') for clear vision is 25 cm.

2. **Determine the image distance (V) and object distance (U):**
- We want to make the object at 25 cm appear at the near point of 50 cm.
- Therefore, the image distance (V) = -50 cm (negative because it is a virtual image).
- The object distance (U) = -25 cm (negative as per the sign convention).

3. **Use the lens formula:**
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Substituting the values:
\[ \frac{1}{f} = \frac{1}{-50} - \frac{1}{-25} \]
\[ \frac{1}{f} = -\frac{1}{50} + \frac{1}{25} \]
\[ \frac{1}{f} = -\frac{1}{50} + \frac{2}{50} = \frac{1}{50} \]
Thus, \( f = 50 \) cm.

4. **Calculate the power of the lens:**
\[ P = \frac{100}{f \text{ (in cm)}} \]
\[ P = \frac{100}{50} = 2 \text{ diopters (D)} \]

**(ii) Power of the lens required for seeing distant stars:**

1. **Identify the far point:**
- The person's far point (F) is 3 m (300 cm).

2. **Determine the image distance (V) and object distance (U):**
- We want to adjust the far point to infinity (for distant stars).
- Therefore, the image distance (V) = -300 cm (negative as it is a virtual image).
- The object distance (U) = -∞ (for distant objects).

3. **Use the lens formula:**
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Substituting the values:
\[ \frac{1}{f} = \frac{1}{-300} - \frac{1}{-\infty} \]
\[ \frac{1}{f} = -\frac{1}{300} + 0 = -\frac{1}{300} \]
Thus, \( f = -300 \) cm.

4. **Calculate the power of the lens:**
\[ P = \frac{100}{f \text{ (in cm)}} \]
\[ P = \frac{100}{-300} = -\frac{1}{3} \text{ diopters (D)} \approx -0.33 \text{ D} \]

### Final Answers:
- (i) The power of the lens required for reading is **2 D**.
- (ii) The power of the lens required for seeing distant stars is **-0.33 D**.