A short sighted person is wearing specs of power `-3.5 D`. His doctor prescribes a correction of `+ 2.5 D` for his near vision. What is focal length of his distance viewing part and near vision. What is focal length of his distance viewing part and near vision part ?

Explanation

To solve the problem, we need to determine the focal lengths of both the distance viewing part and the near vision part of the spectacles worn by a short-sighted person.

### Step-by-Step Solution:

1. **Understanding the Power of the Lenses:**
- The power of the distance viewing part of the spectacles is given as `P_d = -3.5 D`.
- The power of the near vision correction prescribed by the doctor is `P_c = +2.5 D`.

2. **Calculating the Focal Length of the Distance Viewing Part:**
- The formula to find the focal length (f) from the power (P) of a lens is:
\[ f = \frac{100}{P} \]
- For the distance viewing part:
\[ f_d = \frac{100}{-3.5} \text{ cm} \]
- Calculating this gives:
\[ f_d = -28.57 \text{ cm} \quad (\text{approximately } -28.5 \text{ cm}) \]
- The negative sign indicates that it is a diverging lens, which is typical for short-sightedness.

3. **Calculating the Power of the Near Viewing Part:**
- The total power required for near vision is the sum of the distance viewing power and the correction:
\[ P_n = P_c - P_d = 2.5 - (-3.5) = 2.5 + 3.5 = 6.0 D \]

4. **Calculating the Focal Length of the Near Vision Part:**
- Using the same formula for focal length:
\[ f_n = \frac{100}{P_n} = \frac{100}{6} \text{ cm} \]
- Calculating this gives:
\[ f_n = 16.67 \text{ cm} \quad (\text{approximately } 16.7 \text{ cm}) \]

### Final Results:
- The focal length of the distance viewing part is approximately **-28.5 cm**.
- The focal length of the near vision part is approximately **16.7 cm**.