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

Explanation

To solve the problem, we need to find the focal length of the objective lens (FO) of an astronomical telescope given the tube length and magnifying power. Here’s a step-by-step solution:

### Step 1: Understand the given information
- Tube length (L) = 105 cm
- Magnifying power (m) = 20

### Step 2: Use the formula for magnifying power in normal adjustment
The magnifying power (m) for an astronomical telescope in normal adjustment is given by:
\[ m = \frac{F_O}{F_E} \]
Where:
- \( F_O \) = Focal length of the objective lens
- \( F_E \) = Focal length of the eyepiece lens

### Step 3: Relate the focal lengths
From the magnifying power formula, we can express the focal length of the objective in terms of the focal length of the eyepiece:
\[ F_O = m \cdot F_E \]
Substituting the given magnifying power:
\[ F_O = 20 \cdot F_E \]

### Step 4: Use the tube length formula
The tube length (L) of the telescope in normal adjustment is given by:
\[ L = F_O + F_E \]
Substituting the expression for \( F_O \):
\[ 105 = 20 \cdot F_E + F_E \]
This simplifies to:
\[ 105 = 21 \cdot F_E \]

### Step 5: Solve for the focal length of the eyepiece
Now, we can solve for \( F_E \):
\[ F_E = \frac{105}{21} = 5 \, \text{cm} \]

### Step 6: Calculate the focal length of the objective lens
Now that we have \( F_E \), we can find \( F_O \):
\[ F_O = 20 \cdot F_E = 20 \cdot 5 = 100 \, \text{cm} \]

### Conclusion
The focal length of the objective lens \( F_O \) is 100 cm.

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

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