The focal length of objective and eyelens of a astonomical telescope are respectively 20 cm and 5 cm. Final image is formed at least distance of distinct vision. The magnifying power will be

Options

A. -4.8
B. -4
C. 4.8
D. 4

Explanation

To find the magnifying power of an astronomical telescope, we can use the formula for magnification when the final image is formed at the least distance of distinct vision (D). The formula is:

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

Where:
- \( F_0 \) = focal length of the objective lens
- \( F_E \) = focal length of the eyepiece lens
- \( D \) = least distance of distinct vision (typically 25 cm)

### Step-by-Step Solution:

1. **Identify the given values:**
- Focal length of the objective lens, \( F_0 = 20 \, \text{cm} \)
- Focal length of the eyepiece lens, \( F_E = 5 \, \text{cm} \)
- Least distance of distinct vision, \( D = 25 \, \text{cm} \)

2. **Substitute the values into the magnification formula:**
\[ M = -\frac{F_0}{F_E} \left(1 + \frac{F_E}{D}\right) \]
\[ M = -\frac{20}{5} \left(1 + \frac{5}{25}\right) \]

3. **Calculate the first part of the equation:**
\[ -\frac{20}{5} = -4 \]

4. **Calculate the second part of the equation:**
\[ \frac{5}{25} = 0.2 \]
\[ 1 + 0.2 = 1.2 \]

5. **Combine the results:**
\[ M = -4 \times 1.2 = -4.8 \]

6. **Interpret the result:**
The magnifying power of the telescope is \( |M| = 4.8 \).

### Final Answer:
The magnifying power of the astronomical telescope is \( 4.8 \).

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