When a telescope is in normal adjusment, the distance of the objective from the eyepiece is found to `100 cm`. If the magnifying power of the telescope, at normal adjusment, is `24` focal lengths of the lenses are

Options

A. 96 cm, 4 cm
B. 48 cm, 2 cm
C. 50 cm, 50 cm
D. 80 cm, 20 cm

Explanation

To solve the problem step by step, we will use the given information about the telescope's normal adjustment and its magnifying power.

### Step 1: Understand the relationship between the focal lengths and the distance between the lenses.
In a telescope in normal adjustment, the distance between the objective lens (focal length \( f_0 \)) and the eyepiece lens (focal length \( f_e \)) is given by:
\[ f_0 + f_e = 100 \, \text{cm} \quad \text{(Equation 1)} \]

### Step 2: Use the formula for magnifying power.
The magnifying power \( M \) of a telescope in normal adjustment is given by the ratio of the focal lengths of the objective and the eyepiece:
\[ M = \frac{f_0}{f_e} \]
Given that the magnifying power \( M = 24 \), we can express this as:
\[ f_0 = 24 f_e \quad \text{(Equation 2)} \]

### Step 3: Substitute Equation 2 into Equation 1.
Now, we substitute \( f_0 \) from Equation 2 into Equation 1:
\[ 24 f_e + f_e = 100 \]
This simplifies to:
\[ 25 f_e = 100 \]

### Step 4: Solve for the focal length of the eyepiece.
Now, we can solve for \( f_e \):
\[ f_e = \frac{100}{25} = 4 \, \text{cm} \]

### Step 5: Find the focal length of the objective.
Now that we have \( f_e \), we can find \( f_0 \) using Equation 2:
\[ f_0 = 24 f_e = 24 \times 4 = 96 \, \text{cm} \]

### Final Answer:
The focal lengths of the lenses are:
- Focal length of the eyepiece \( f_e = 4 \, \text{cm} \)
- Focal length of the objective \( f_0 = 96 \, \text{cm} \)

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