A telescope, when in normal adjustment, has a magnifying power of 6 and the objective and eye-piece are 14 cm apart. The focal lengths of the eye-piece and the objective, respectively,

Options

A. 3 cm and 14 cm
B. 3 cm and 12 cm
C. 2 cm and 14 cm
D. 2 cm and 12 cm

Answer

To solve the problem, we need to find the focal lengths of the eyepiece (Fe) and the objective (Fo) of a telescope that has a magnifying power of 6 and a total length of 14 cm.

### Step-by-Step Solution:

1. **Understanding the Magnifying Power:**
The magnifying power (M) of a 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.

2. **Using Given Values:**
We know from the problem that:
\[ M = 6 \]
Therefore, we can write:
\[ \frac{F_o}{F_e} = 6 \quad \text{(Equation 1)} \]

3. **Total Length of the Telescope:**
The total length of the telescope is the sum of the focal lengths of the objective and the eyepiece:
\[ F_o + F_e = 14 \quad \text{(Equation 2)} \]

4. **Substituting Equation 1 into Equation 2:**
From Equation 1, we can express \( F_o \) in terms of \( F_e \):
\[ F_o = 6F_e \]
Now, substitute this expression for \( F_o \) into Equation 2:
\[ 6F_e + F_e = 14 \]

5. **Solving for \( F_e \):**
Combine the terms:
\[ 7F_e = 14 \]
Now, divide both sides by 7:
\[ F_e = 2 \, \text{cm} \]

6. **Finding \( F_o \):**
Now that we have \( F_e \), we can find \( F_o \) using Equation 1:
\[ F_o = 6F_e = 6 \times 2 = 12 \, \text{cm} \]

### Final Result:
Thus, the focal lengths of the eyepiece and the objective are:
- Focal length of the eyepiece, \( F_e = 2 \, \text{cm} \)
- Focal length of the objective, \( F_o = 12 \, \text{cm} \)