If tube length Of astronomical telescope is 105cm and magnifying power is 20 for normal setting, the focal length of the objective is:

Options

A. 100 cm
B. 10 cm
C. 20 cm
D. 25 cm

Answer

To find the focal length of the objective lens in an astronomical telescope given the tube length and magnifying power, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Magnifying Power**:
The magnifying power (M) of an astronomical telescope in normal setting 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 lens.

2. **Using Given Magnifying Power**:
We know from the problem that the magnifying power \(M\) is 20. Therefore, we can write:
\[ 20 = \frac{f_o}{f_e} \]
Rearranging this gives:
\[ f_e = \frac{f_o}{20} \]

3. **Using the Tube Length**:
The total 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 tube length \(L\) is 105 cm. Substituting \(f_e\) from the previous step into this equation gives:
\[ 105 = f_o + \frac{f_o}{20} \]

4. **Combining Terms**:
To combine the terms on the right side, we can express \(f_o\) in a common fraction:
\[ 105 = f_o + \frac{f_o}{20} = \frac{20f_o}{20} + \frac{f_o}{20} = \frac{21f_o}{20} \]

5. **Solving for \(f_o\)**:
Now, we can solve for \(f_o\):
\[ 105 = \frac{21f_o}{20} \]
Multiplying both sides by 20 to eliminate the fraction:
\[ 2100 = 21f_o \]
Dividing both sides by 21 gives:
\[ f_o = \frac{2100}{21} = 100 \text{ cm} \]

### Final Answer:
The focal length of the objective lens \(f_o\) is **100 cm**.