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

Options

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

Answer

To solve the problem, we need to find the focal length of the objective lens (f₀) of an astronomical telescope given the tube length (L) and the magnifying power (m).

### Step-by-Step Solution:

1. **Identify the given values:**
- Tube length (L) = 105 cm
- Magnifying power (m) = 20

2. **Use the formula for magnifying power of an astronomical telescope:**
\[ m = \frac{f_0}{f_e} \]
where \(f_0\) is the focal length of the objective lens and \(f_e\) is the focal length of the eyepiece lens.

3. **Express \(f_0\) in terms of \(f_e\):**
From the magnifying power formula, we can rearrange it to find:
\[ f_0 = m \cdot f_e \]
Substituting the value of \(m\):
\[ f_0 = 20 \cdot f_e \quad \text{(Equation 1)} \]

4. **Use the relationship between tube length and focal lengths:**
The tube length of the telescope is given by:
\[ L = f_0 + f_e \]
Substituting the given tube length:
\[ 105 = f_0 + f_e \quad \text{(Equation 2)} \]

5. **Substitute Equation 1 into Equation 2:**
Replace \(f_0\) in Equation 2 with the expression from Equation 1:
\[ 105 = 20f_e + f_e \]
This simplifies to:
\[ 105 = 21f_e \]

6. **Solve for \(f_e\):**
\[ f_e = \frac{105}{21} = 5 \text{ cm} \]

7. **Substitute \(f_e\) back into Equation 1 to find \(f_0\):**
Now we can find \(f_0\):
\[ f_0 = 20 \cdot f_e = 20 \cdot 5 = 100 \text{ cm} \]

8. **Conclusion:**
The focal length of the objective lens \(f_0\) is 100 cm.

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

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