The magnifying power of an astronomical telescope for relaxed vision is 16 and the distance between the objective and eyelens is 34 cm. Then the focal length of objective and eyelens will be respectively

Options

A. 17 cm, 17 cm
B. 20 cm, 14 cm
C. 32 cm, 2 cm
D. 30 cm, 4 cm

Answer

To solve the problem, we need to determine the focal lengths of the objective lens and the eyepiece of an astronomical telescope given the magnifying power and the distance between the two lenses.

### Step-by-Step Solution:

1. **Understanding the Magnifying Power**:
The magnifying power (M) of an astronomical telescope for relaxed vision is given by the formula:
\[ M = \frac{f_o}{f_e} \]
where \( f_o \) is the focal length of the objective lens and \( f_e \) is the focal length of the eyepiece.

2. **Substituting the Given Magnifying Power**:
We know from the problem that the magnifying power \( M = 16 \). Therefore, we can write:
\[ 16 = \frac{f_o}{f_e} \]
Rearranging this gives:
\[ f_o = 16 f_e \]

3. **Using the Length of the Telescope**:
The length of the telescope (L) is the sum of the focal lengths of the objective and the eyepiece:
\[ L = f_o + f_e \]
We are given that the distance between the objective and the eyepiece is 34 cm. Thus, we have:
\[ 34 = f_o + f_e \]

4. **Substituting \( f_o \) in the Length Equation**:
Now, we can substitute \( f_o \) from step 2 into the length equation:
\[ 34 = 16 f_e + f_e \]
This simplifies to:
\[ 34 = 17 f_e \]

5. **Calculating the Focal Length of the Eyepiece**:
To find \( f_e \), we divide both sides by 17:
\[ f_e = \frac{34}{17} = 2 \text{ cm} \]

6. **Calculating the Focal Length of the Objective**:
Now that we have \( f_e \), we can find \( f_o \) using the relation from step 2:
\[ f_o = 16 f_e = 16 \times 2 = 32 \text{ cm} \]

### Final Answer:
The focal lengths of the objective and eyepiece are:
- Focal length of the objective \( f_o = 32 \) cm
- Focal length of the eyepiece \( f_e = 2 \) cm