A telescope has an objective of focal length 200 cm and eyepiece of focal length 5 cm. Calculate its magnifying power when the final image is formed (a) at infinity and (b) at distance of distinct vision.

Answer

To solve the problem, we will calculate the magnifying power of the telescope in two scenarios: when the final image is formed at infinity and when it is formed at the distance of distinct vision.

### Given Data:
- Focal length of the objective lens, \( F_O = 200 \, \text{cm} \)
- Focal length of the eyepiece lens, \( F_E = 5 \, \text{cm} \)

### (a) Magnifying Power when the Final Image is Formed at Infinity:

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

2. **Substituting the Values**:
\[ M = \frac{200 \, \text{cm}}{5 \, \text{cm}} = 40 \]

3. **Conclusion**:
The magnifying power when the final image is formed at infinity is \( M = 40 \).

### (b) Magnifying Power when the Final Image is Formed at the Distance of Distinct Vision:

1. **Formula for Magnifying Power**:
The magnifying power \( M \) when the final image is formed at the distance of distinct vision (usually taken as \( D = 25 \, \text{cm} \)) is given by:
\[ M = \frac{F_O}{F_E} \left( 1 + \frac{F_E}{D} \right) \]

2. **Substituting the Values**:
- First, calculate \( \frac{F_E}{D} \):
\[ \frac{F_E}{D} = \frac{5 \, \text{cm}}{25 \, \text{cm}} = \frac{1}{5} \]

- Now substitute this into the magnifying power formula:
\[ M = \frac{200 \, \text{cm}}{5 \, \text{cm}} \left( 1 + \frac{1}{5} \right) \]
\[ M = 40 \left( 1 + 0.2 \right) = 40 \times 1.2 = 48 \]

3. **Conclusion**:
The magnifying power when the final image is formed at the distance of distinct vision is \( M = 48 \).

### Final Answers:
- (a) Magnifying power when the final image is at infinity: **40**
- (b) Magnifying power when the final image is at the distance of distinct vision: **48**