The magnifying power of an astronomical telescope in the normal adjustment position is `100`. The distance between the objective and eye piece is `101 cm` . Calculate the focal lengths of objective and eye piece.

Explanation

To solve the problem, we will use the information provided about the magnifying power of the astronomical telescope and the distance between the objective and eyepiece.

### Step-by-Step Solution:

1. **Understanding the Magnifying Power**:
The magnifying power (M) of an astronomical telescope in normal adjustment 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. According to the problem, the magnifying power is 100:
\[ M = 100 \]

2. **Setting Up the Equation**:
From the magnifying power formula, we can express the focal length of the objective in terms of the focal length of the eyepiece:
\[ F_O = 100 \times F_E \]

3. **Using the Distance Between the Lenses**:
The distance between the objective and the eyepiece is given as 101 cm. In normal adjustment, this distance is equal to the sum of the focal lengths of the two lenses:
\[ F_O + F_E = 101 \, \text{cm} \]

4. **Substituting the Expression for \( F_O \)**:
Substitute \( F_O \) from step 2 into the equation from step 3:
\[ 100 \times F_E + F_E = 101 \]
This simplifies to:
\[ 101 \times F_E = 101 \]

5. **Solving for \( F_E \)**:
Divide both sides by 101:
\[ F_E = 1 \, \text{cm} \]

6. **Finding \( F_O \)**:
Now, substitute \( F_E \) back into the equation for \( F_O \):
\[ F_O = 100 \times F_E = 100 \times 1 = 100 \, \text{cm} \]

### Final Results:
- Focal length of the objective lens \( F_O = 100 \, \text{cm} \)
- Focal length of the eyepiece lens \( F_E = 1 \, \text{cm} \)