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 find the focal lengths of the objective and eyepiece of a telescope in normal adjustment, we can follow these steps:

### 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 and the eyepiece lens (denoted as \( D \)) is equal to the sum of their focal lengths:
\[ D = f_o + f_e \]
where \( f_o \) is the focal length of the objective lens and \( f_e \) is the focal length of the eyepiece lens.

### Step 2: Use the given information.
From the problem, we know:
- \( D = 100 \, \text{cm} \)
- The magnifying power \( M \) of the telescope is given by the formula:
\[ M = \frac{f_o}{f_e} \]
Given that \( M = 24 \), we can set up the equation:
\[ 24 = \frac{f_o}{f_e} \]

### Step 3: Express \( f_o \) in terms of \( f_e \).
From the magnifying power equation, we can express \( f_o \) as:
\[ f_o = 24 f_e \]

### Step 4: Substitute \( f_o \) into the distance equation.
Now we substitute \( f_o \) into the distance equation:
\[ 100 = 24 f_e + f_e \]
This simplifies to:
\[ 100 = 25 f_e \]

### Step 5: Solve for \( f_e \).
Now, we can solve for \( f_e \):
\[ f_e = \frac{100}{25} = 4 \, \text{cm} \]

### Step 6: Find \( f_o \).
Now that we have \( f_e \), we can find \( f_o \):
\[ f_o = 24 f_e = 24 \times 4 = 96 \, \text{cm} \]

### Final Result:
Thus, the focal lengths of the lenses are:
- Focal length of the objective lens, \( f_o = 96 \, \text{cm} \)
- Focal length of the eyepiece lens, \( f_e = 4 \, \text{cm} \)

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