A telescope consists of two lenses of focal lengths `20 cm and 5 cm`. Obtain its magnifying power when final image is
(i) at infinity
(ii) at `25 cm` from the eye.

Explanation

To solve the problem, we will calculate the magnifying power of the telescope in two different scenarios: when the final image is at infinity and when the final image is at 25 cm from the eye.

### Step-by-Step Solution:

**Given:**
- Focal length of the objective lens (F_O) = 20 cm
- Focal length of the eyepiece lens (F_E) = 5 cm

#### Case (i): Final Image at Infinity

1. **Formula for Magnifying Power (M)**:
The magnifying power of a telescope when the final image is at infinity is given by:
\[ M = -\frac{F_O}{F_E} \]

2. **Substituting the values**:
\[ M = -\frac{20 \text{ cm}}{5 \text{ cm}} = -4 \]

3. **Conclusion**:
The magnifying power when the final image is at infinity is -4.

#### Case (ii): Final Image at 25 cm from the Eye

1. **Formula for Magnifying Power (M)**:
The magnifying power when the final image is at a distance D (25 cm in this case) from the eye is given by:
\[ M = -\frac{F_O}{F_E} \left(1 + \frac{F_E}{D}\right) \]

2. **Substituting the values**:
- Here, D = 25 cm
\[ M = -\frac{20 \text{ cm}}{5 \text{ cm}} \left(1 + \frac{5 \text{ cm}}{25 \text{ cm}}\right) \]

3. **Calculating the fraction**:
\[ \frac{5 \text{ cm}}{25 \text{ cm}} = \frac{1}{5} \]
Therefore,
\[ M = -4 \left(1 + \frac{1}{5}\right) = -4 \left(\frac{5}{5} + \frac{1}{5}\right) = -4 \left(\frac{6}{5}\right) \]

4. **Final calculation**:
\[ M = -\frac{24}{5} = -4.8 \]

5. **Conclusion**:
The magnifying power when the final image is at 25 cm from the eye is -4.8.

### Summary of Results:
- Magnifying power when the final image is at infinity: **-4**
- Magnifying power when the final image is at 25 cm from the eye: **-4.8**