A telescope has an objective of focal length 100 cm and an eye-piece of focal length 5 cm. What is the magnifying power of the telescope when it is in normal adjustment?

Options

A. 0.2
B. 2
C. 20
D. 200

Explanation

To find the magnifying power of the telescope in normal adjustment, we can follow these steps:

### Step 1: Understand the formula for magnifying power
The magnifying power (m) of a telescope in normal adjustment is given by the formula:
\[ m = \frac{f_o}{f_e} \]
where:
- \( f_o \) = focal length of the objective lens
- \( f_e \) = focal length of the eyepiece lens

### Step 2: Identify the given values
From the question, we have:
- Focal length of the objective lens, \( f_o = 100 \) cm
- Focal length of the eyepiece lens, \( f_e = 5 \) cm

### Step 3: Substitute the values into the formula
Now, we can substitute the given values into the formula:
\[ m = \frac{100 \, \text{cm}}{5 \, \text{cm}} \]

### Step 4: Perform the calculation
Now, we perform the division:
\[ m = \frac{100}{5} = 20 \]

### Step 5: State the final answer
Thus, the magnifying power of the telescope when it is in normal adjustment is:
\[ m = 20 \]