The magnification produced by the objective lens of a compound microscope is 25. The focal length of eye piece is 5 cm and it forms find image at least distance of distinct vision. The magnifying power of the compound microscope is

Options

A. 19
B. 31
C. 150
D. `sqrt(150)`

Answer

To solve the problem, we need to calculate the magnifying power of a compound microscope given the magnification produced by the objective lens and the focal length of the eyepiece.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Magnification produced by the objective lens (M₀) = 25
- Focal length of the eyepiece (fₑ) = 5 cm
- Least distance of distinct vision (D) = 25 cm (standard value)

2. **Calculate the Magnification of the Eyepiece (Mₑ):**
The formula for the magnification produced by the eyepiece is given by:
\[ Mₑ = 1 + \frac{D}{fₑ} \]
Substituting the values:
\[ Mₑ = 1 + \frac{25 \text{ cm}}{5 \text{ cm}} = 1 + 5 = 6 \]

3. **Calculate the Total Magnification (M):**
The total magnification of the compound microscope is the product of the magnification of the objective lens and the magnification of the eyepiece:
\[ M = M₀ \times Mₑ \]
Substituting the values:
\[ M = 25 \times 6 = 150 \]

4. **Conclusion:**
The magnifying power of the compound microscope is 150.

### Final Answer:
The magnifying power of the compound microscope is **150**.

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