The length of the tube of a compound microscope 15 cm. The focal length of objective and eye lenses are 1 cm and 5 cm respectively. The magnifying power of microscope for relaxed vision is

Options

A. 50
B. 75
C. 25
D. 100

Explanation

To find the magnifying power of a compound microscope for relaxed vision, we can use the formula:

\[ M = \frac{L \cdot D}{F_o \cdot F_e} \]

Where:
- \( M \) = magnifying power
- \( L \) = length of the tube of the microscope
- \( D \) = distance of distinct vision (typically taken as 25 cm)
- \( F_o \) = focal length of the objective lens
- \( F_e \) = focal length of the eyepiece lens

### Step-by-Step Solution

1. **Identify the given values:**
- Length of the tube \( L = 15 \, \text{cm} \)
- Focal length of the objective lens \( F_o = 1 \, \text{cm} \)
- Focal length of the eyepiece lens \( F_e = 5 \, \text{cm} \)
- Distance of distinct vision \( D = 25 \, \text{cm} \)

2. **Substitute the values into the magnifying power formula:**
\[ M = \frac{L \cdot D}{F_o \cdot F_e} \]
\[ M = \frac{15 \, \text{cm} \cdot 25 \, \text{cm}}{1 \, \text{cm} \cdot 5 \, \text{cm}} \]

3. **Calculate the numerator:**
\[ 15 \cdot 25 = 375 \, \text{cm}^2 \]

4. **Calculate the denominator:**
\[ 1 \cdot 5 = 5 \, \text{cm}^2 \]

5. **Divide the numerator by the denominator to find \( M \):**
\[ M = \frac{375 \, \text{cm}^2}{5 \, \text{cm}^2} = 75 \]

6. **Conclusion:**
The magnifying power of the microscope for relaxed vision is \( M = 75 \).

### Final Answer:
The magnifying power of the microscope for relaxed vision is **75**.