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

Answer

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

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

where:
- \( f_o \) = focal length of the objective lens
- \( f_e \) = focal length of the eye lens
- \( D \) = least distance of distinct vision (which is typically 25 cm for a normal eye)

### Step-by-Step Solution:

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

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

3. **Calculate the first part of the equation:**
\[ -\frac{200}{5} = -40 \]

4. **Calculate the second part of the equation:**
\[ \frac{5}{25} = 0.2 \implies 1 + 0.2 = 1.2 \]

5. **Combine the results:**
\[ M = -40 \times 1.2 = -48 \]

6. **Interpret the result:**
The magnifying power is dimensionless, so we can express it as:
\[ M = 48 \text{ (ignoring the negative sign as magnification is often expressed as a positive value)} \]

### Final Answer:
The maximum magnifying power of the telescope is **48**.