The focal lengths of the objective and eye lenses of a telescope are respectively 200 cm and 5 cm . The maximum magnifying power of the telescope will be

Options

A. `-40`
B. `-48`
C. `-60`
D. `-100`

Explanation

To find the maximum magnifying power of the telescope, we can use the formula for the magnifying power (M) of an astronomical telescope:

\[ M = \frac{f_o}{f_e} \left(1 + \frac{D}{f_e}\right) \]

Where:
- \( f_o \) = focal length of the objective lens
- \( f_e \) = focal length of the eyepiece lens
- \( D \) = distance of distinct vision (typically taken as 25 cm for a normal human eye)

### Step-by-Step Solution:

1. **Identify the given values:**
- Focal length of the objective lens, \( f_o = 200 \) cm
- Focal length of the eyepiece lens, \( f_e = 5 \) cm
- Distance of distinct vision, \( D = 25 \) cm

2. **Substitute the values into the magnifying power formula:**
\[ M = \frac{f_o}{f_e} \left(1 + \frac{D}{f_e}\right) \]
\[ M = \frac{200}{5} \left(1 + \frac{25}{5}\right) \]

3. **Calculate \( \frac{f_o}{f_e} \):**
\[ \frac{200}{5} = 40 \]

4. **Calculate \( \frac{D}{f_e} \):**
\[ \frac{25}{5} = 5 \]

5. **Now substitute back into the equation:**
\[ M = 40 \left(1 + 5\right) \]
\[ M = 40 \times 6 \]

6. **Calculate the final magnifying power:**
\[ M = 240 \]

7. **Note on sign convention:**
The magnification in telescopes is typically negative because the image formed is inverted. Thus, we write:
\[ M = -240 \]

### Final Answer:
The maximum magnifying power of the telescope is \( -240 \).