The magnifying power of an astronomical telescope is `5`. When it is set for normal adjustment, the distance between the two lenses is `24 cm`. Calculate the focal lengths of eye piece and objective lens.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between the focal lengths and 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.

### Step 2: Use the given magnifying power
From the problem, we know that the magnifying power \( M = 5 \). Therefore, we can write:
\[ \frac{F_O}{F_E} = 5 \]
This implies:
\[ F_O = 5 \times F_E \]

### Step 3: Use the distance between the lenses
In normal adjustment, the distance between the two lenses is equal to the sum of their focal lengths:
\[ F_O + F_E = 24 \, \text{cm} \]

### Step 4: Substitute \( F_O \) in the distance equation
Now, substituting \( F_O \) from Step 2 into the distance equation:
\[ 5F_E + F_E = 24 \]
This simplifies to:
\[ 6F_E = 24 \]

### Step 5: Solve for \( F_E \)
Now, we can solve for \( F_E \):
\[ F_E = \frac{24}{6} = 4 \, \text{cm} \]

### Step 6: Calculate \( F_O \)
Now that we have \( F_E \), we can find \( F_O \):
\[ F_O = 5 \times F_E = 5 \times 4 = 20 \, \text{cm} \]

### Final Result
Thus, the focal lengths of the eyepiece and the objective lens are:
- Focal length of the eyepiece \( F_E = 4 \, \text{cm} \)
- Focal length of the objective \( F_O = 20 \, \text{cm} \)

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