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)`

Explanation

To find the magnifying power of the compound microscope, we can follow these steps:

### Step 1: Understand the components of the compound microscope
A compound microscope consists of two lenses: the objective lens and the eyepiece. The total magnification of the microscope is the product of the magnifications produced by each lens.

### Step 2: Identify the given values
- Magnification produced by the objective lens (M₀) = 25
- Focal length of the eyepiece (Fₑ) = 5 cm
- Distance of distinct vision (d) = 25 cm

### Step 3: Calculate the magnification produced by the eyepiece (Mₑ)
The formula for the magnification produced by the eyepiece is given by:
\[ Mₑ = 1 + \frac{d}{Fₑ} \]

Substituting the known values:
\[ Mₑ = 1 + \frac{25 \text{ cm}}{5 \text{ cm}} \]
\[ Mₑ = 1 + 5 = 6 \]

### Step 4: Calculate the total magnification (M)
The total magnification (M) of the compound microscope is the product of the magnifications of the objective lens and the eyepiece:
\[ M = M₀ \times Mₑ \]

Substituting the values we have:
\[ M = 25 \times 6 \]
\[ M = 150 \]

### Conclusion
The magnifying power of the compound microscope is 150.

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