An astronomical telescope has angular magnification m=7 for distant object. Distance between objective and eyepeice is 40 cm. final image is formed at infinitely focal length of objective and eyepeice respectively are

Options

A. 35 cm, 5 cm
B. 28 cm, 14 cm
C. 5 cm, 35 cm
D. 70 cm, 10 cm

Explanation

To solve the problem, we need to find the focal lengths of the objective (fo) and the eyepiece (fe) of an astronomical telescope given the angular magnification (m) and the distance between the objective and the eyepiece.

### Step-by-Step Solution:

1. **Understanding Angular Magnification**:
The angular magnification (m) of an astronomical telescope is given by the formula:
\[ m = \frac{f_o}{f_e} \]
where \( f_o \) is the focal length of the objective and \( f_e \) is the focal length of the eyepiece.

2. **Given Values**:
- Angular magnification \( m = 7 \)
- Distance between the objective and eyepiece \( D = 40 \, \text{cm} \)

3. **Setting Up the Equation**:
From the magnification formula, we can express \( f_o \) in terms of \( f_e \):
\[ f_o = m \cdot f_e = 7 \cdot f_e \]

4. **Using the Distance Between the Lenses**:
The distance between the objective and the eyepiece is given by:
\[ D = f_o + f_e \]
Substituting \( f_o \) from the previous step:
\[ 40 = 7f_e + f_e \]
This simplifies to:
\[ 40 = 8f_e \]

5. **Solving for \( f_e \)**:
Now, we can solve for \( f_e \):
\[ f_e = \frac{40}{8} = 5 \, \text{cm} \]

6. **Finding \( f_o \)**:
Now that we have \( f_e \), we can find \( f_o \):
\[ f_o = 7 \cdot f_e = 7 \cdot 5 = 35 \, \text{cm} \]

### Final Answer:
- Focal length of the objective \( f_o = 35 \, \text{cm} \)
- Focal length of the eyepiece \( f_e = 5 \, \text{cm} \)