The objective of a small telescope has focal length 120 cm and diameter 5 cm. The focal length of the eye piece is 2 cm. The magnifying power of the telescope for distant object is -

Options

A. 12
B. 24
C. 60
D. 300

Answer

To find the magnifying power of the telescope for distant objects, we can use the formula for the magnifying power of an astronomical telescope in normal adjustment:

\[ M = \frac{f_o}{f_e} \]

where:
- \(M\) is the magnifying power,
- \(f_o\) is the focal length of the objective lens,
- \(f_e\) is the focal length of the eyepiece.

### Step-by-Step Solution:

1. **Identify the given values**:
- Focal length of the objective lens, \(f_o = 120 \, \text{cm}\)
- Focal length of the eyepiece, \(f_e = 2 \, \text{cm}\)

2. **Substitute the values into the magnifying power formula**:
\[ M = \frac{f_o}{f_e} = \frac{120 \, \text{cm}}{2 \, \text{cm}} \]

3. **Calculate the magnifying power**:
\[ M = \frac{120}{2} = 60 \]

4. **State the final answer**:
The magnifying power of the telescope for distant objects is \(M = 60\).

### Final Answer:
The magnifying power of the telescope for distant objects is **60**.

---