An astronomical telescope has an objective of focal length 100 cm and an eye piece of focal length 5 cm. The final image of a star is seen 25 cm from the eyepiece. The magnifying power of the telescope is

Options

A. 20
B. 22
C. 24
D. 26

Answer

To find the magnifying power of the astronomical telescope, we can use the formula for magnifying power (M) of a telescope, which is given by:

\[ M = \frac{F_0}{F_e} \left(1 + \frac{F_e}{D}\right) \]

Where:
- \( F_0 \) = focal length of the objective lens
- \( F_e \) = focal length of the eyepiece
- \( D \) = distance of the final image from the eyepiece

### Step-by-Step Solution:

1. **Identify the given values:**
- Focal length of the objective lens \( F_0 = 100 \, \text{cm} \)
- Focal length of the eyepiece \( F_e = 5 \, \text{cm} \)
- Distance of the final image from the eyepiece \( D = 25 \, \text{cm} \)

2. **Substitute the values into the magnifying power formula:**
\[ M = \frac{F_0}{F_e} \left(1 + \frac{F_e}{D}\right) \]
Substituting the values:
\[ M = \frac{100}{5} \left(1 + \frac{5}{25}\right) \]

3. **Calculate \( \frac{F_0}{F_e} \):**
\[ \frac{100}{5} = 20 \]

4. **Calculate \( \frac{F_e}{D} \):**
\[ \frac{5}{25} = \frac{1}{5} \]

5. **Substitute back into the formula:**
\[ M = 20 \left(1 + \frac{1}{5}\right) \]

6. **Simplify the expression inside the parentheses:**
\[ 1 + \frac{1}{5} = \frac{5}{5} + \frac{1}{5} = \frac{6}{5} \]

7. **Final calculation of magnifying power:**
\[ M = 20 \times \frac{6}{5} = 20 \times 1.2 = 24 \]

### Conclusion:
The magnifying power of the telescope is \( M = 24 \).