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

Options

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

Answer

To solve the problem, we need to find the power of the lenses required for a person whose near point is 50 cm and far point is 3 m. We will tackle this in two parts:

### Part (i): Finding the Power of the Lens for Reading

1. **Identify the Near Point**: The near point of the person is given as 50 cm, which we convert to meters for consistency.
\[ \text{Near Point (D)} = 50 \text{ cm} = 0.5 \text{ m} \]

2. **Determine the Focal Length Required**: The lens needs to be able to bring the near point closer. The formula for the focal length (f) of a lens for a near point is given by:
\[ \frac{1}{f} = \frac{1}{D} - \frac{1}{d} \]
where \(D\) is the near point (0.5 m) and \(d\) is the desired near point (which we can take as 25 cm or 0.25 m for reading comfortably).

3. **Calculate the Focal Length**:
\[ \frac{1}{f} = \frac{1}{0.25} - \frac{1}{0.5} = 4 - 2 = 2 \text{ m}^{-1} \]
Therefore,
\[ f = \frac{1}{2} = 0.5 \text{ m} \]

4. **Calculate the Power of the Lens**: The power (P) of a lens is given by the formula:
\[ P = \frac{1}{f} \text{ (in meters)} \]
Thus,
\[ P = \frac{1}{0.5} = 2 \text{ D} \]

### Part (ii): Finding the Power of the Lens for Seeing Distant Stars

1. **Identify the Far Point**: The far point of the person is given as 3 m.
\[ \text{Far Point (D)} = 3 \text{ m} \]

2. **Determine the Focal Length Required**: For distant vision, we want the far point to be at infinity. The formula for the focal length (f) for the far point is:
\[ \frac{1}{f} = \frac{1}{D} - \frac{1}{d} \]
where \(D\) is the far point (3 m) and \(d\) is infinity (which we can consider as 0).

3. **Calculate the Focal Length**:
\[ \frac{1}{f} = \frac{1}{\infty} - \frac{1}{3} = 0 - \frac{1}{3} = -\frac{1}{3} \text{ m}^{-1} \]
Therefore,
\[ f = -3 \text{ m} \]

4. **Calculate the Power of the Lens**: Using the power formula:
\[ P = \frac{1}{f} = \frac{1}{-3} = -0.33 \text{ D} \]

### Final Answers:
- (i) Power of the lens for reading: **+2 D**
- (ii) Power of the lens for seeing distant stars: **-0.33 D**