The focal length of the eyepiece and the objective of an astronomical telescope are 2cm and 100cm respectively. Find the magnifying power of the telescope for normal adjustment and the length of the telescope.

Options

A. 50, 102 cm
B. 100, 204 cm
C. 25, 62 cm
D. 75, 125 cm

Answer

To solve the problem, we need to calculate the magnifying power of the astronomical telescope for normal adjustment and the length of the telescope.

### Step-by-step Solution:

1. **Identify the Focal Lengths**:
- Focal length of the eyepiece (\(f_e\)) = 2 cm
- Focal length of the objective (\(f_o\)) = 100 cm

2. **Calculate the Magnifying Power**:
- The formula for magnifying power (\(M\)) of an astronomical telescope in normal adjustment is given by:
\[ M = \frac{f_o}{f_e} \]
- Substituting the values:
\[ M = \frac{100 \, \text{cm}}{2 \, \text{cm}} = 50 \]

3. **Calculate the Length of the Telescope**:
- The length of the telescope (\(L\)) in normal adjustment is the sum of the focal lengths of the objective and the eyepiece:
\[ L = f_o + f_e \]
- Substituting the values:
\[ L = 100 \, \text{cm} + 2 \, \text{cm} = 102 \, \text{cm} \]

### Final Answers:
- Magnifying Power (\(M\)) = 50
- Length of the Telescope (\(L\)) = 102 cm