The length of the tube of a microscope is 10 cm . The focal lengths of the objective and eye lenses are 0.5 cm and 1.0 cm . The magnifying power of the microscope is about

Options

A. 5
B. 23
C. 166
D. 500

Answer

To find the magnifying power of the microscope, we can use the formula for the magnification of a compound microscope, which is given by:

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

Where:
- \(M\) is the magnifying power.
- \(L\) is the length of the tube (distance between the objective and eye lens).
- \(F_o\) is the focal length of the objective lens.
- \(D\) is the least distance of distinct vision (typically taken as 25 cm).
- \(F_e\) is the focal length of the eye lens.

### Step-by-step Solution:

1. **Identify the given values:**
- Length of the tube, \(L = 10 \, \text{cm}\)
- Focal length of the objective lens, \(F_o = 0.5 \, \text{cm}\)
- Focal length of the eye lens, \(F_e = 1.0 \, \text{cm}\)
- Least distance of distinct vision, \(D = 25 \, \text{cm}\)

2. **Substitute the values into the magnification formula:**
\[ M = \frac{L}{F_o} \times \frac{D}{F_e} \]
Substituting the values:
\[ M = \frac{10 \, \text{cm}}{0.5 \, \text{cm}} \times \frac{25 \, \text{cm}}{1.0 \, \text{cm}} \]

3. **Calculate the magnification:**
- First, calculate \(\frac{10}{0.5}\):
\[ \frac{10}{0.5} = 20 \]
- Next, calculate \(\frac{25}{1.0}\):
\[ \frac{25}{1.0} = 25 \]
- Now multiply these two results:
\[ M = 20 \times 25 = 500 \]

4. **Conclusion:**
The magnifying power of the microscope is approximately \(500\).

### Final Answer:
The magnifying power of the microscope is about **500**.