The magnifying power of terrestrial telescope is 25 when it is in normal adjustment and the length of the telescope is 124 cm. If the focal length of the erecting lens is 5 cm, the focal lengths of the objective and the eye-piece are respectively.

Options

A. ) 50 cm, 2 cm
B. 50 cm, 2.5 cm
C. 100 cm, 4 cm
D. 100 cm, 5 cm

Explanation

To solve the problem, we need to find the focal lengths of the objective and eyepiece of a terrestrial telescope given the magnifying power, the length of the telescope, and the focal length of the erecting lens.

### Step-by-Step Solution:

1. **Understanding the Magnifying Power**:
The magnifying power (M) of a terrestrial telescope in normal adjustment 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. Given that \( M = 25 \), we can write:
\[ 25 = \frac{f_o}{f_e} \]
This implies:
\[ f_o = 25 f_e \quad \text{(1)} \]

2. **Using the Length of the Telescope**:
The total length (L) of the telescope is given by:
\[ L = f_o + f_e + 4f_r \]
where \( f_r \) is the focal length of the erecting lens. We know:
- \( L = 124 \, \text{cm} \)
- \( f_r = 5 \, \text{cm} \)

Substituting these values into the equation gives:
\[ 124 = f_o + f_e + 4 \times 5 \]
Simplifying this, we have:
\[ 124 = f_o + f_e + 20 \]
Thus:
\[ f_o + f_e = 124 - 20 = 104 \quad \text{(2)} \]

3. **Substituting Equation (1) into Equation (2)**:
Now, substitute \( f_o \) from equation (1) into equation (2):
\[ 25 f_e + f_e = 104 \]
This simplifies to:
\[ 26 f_e = 104 \]
Therefore:
\[ f_e = \frac{104}{26} = 4 \, \text{cm} \]

4. **Finding the Focal Length of the Objective**:
Now that we have \( f_e \), we can find \( f_o \) using equation (1):
\[ f_o = 25 f_e = 25 \times 4 = 100 \, \text{cm} \]

### Final Answer:
The focal lengths of the objective and eyepiece are:
- Focal length of the objective \( f_o = 100 \, \text{cm} \)
- Focal length of the eyepiece \( f_e = 4 \, \text{cm} \)