The focal length of objective and eye lens of a microscope are `4 cm` and `8 cm` respectively. If the least distance of distinct vision is `24 cm` and object distance is `4.5 cm` from the objective lens, then the magnifying power of the microscope will be

Options

A. `18`
B. 32
C. 64
D. 20

Explanation

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

### Step 1: Identify the given values
- Focal length of the objective lens (FO) = 4 cm
- Focal length of the eye lens (FE) = 8 cm
- Least distance of distinct vision (D) = 24 cm
- Object distance from the objective lens (U0) = -4.5 cm (the negative sign indicates that the object is on the same side as the incoming light)

### Step 2: Use the lens formula to find the image distance (V0) for the objective lens
The lens formula is given by:
\[ \frac{1}{F} = \frac{1}{V} - \frac{1}{U} \]
Rearranging gives:
\[ \frac{1}{V} = \frac{1}{F} + \frac{1}{U} \]
Substituting the values:
\[ \frac{1}{V0} = \frac{1}{4} + \frac{1}{-4.5} \]
Calculating the right side:
\[ \frac{1}{V0} = \frac{1}{4} - \frac{1}{4.5} \]
Finding a common denominator (which is 36):
\[ \frac{1}{V0} = \frac{9}{36} - \frac{8}{36} = \frac{1}{36} \]
Thus,
\[ V0 = 36 \text{ cm} \]

### Step 3: Calculate the magnifying power (M) of the microscope
The magnifying power formula for a microscope is given by:
\[ M = \frac{V0}{U0} \left(1 + \frac{D}{FE}\right) \]
Substituting the values we have:
\[ M = \frac{36}{-4.5} \left(1 + \frac{24}{8}\right) \]
Calculating the first part:
\[ \frac{36}{-4.5} = -8 \]
Now calculate the second part:
\[ 1 + \frac{24}{8} = 1 + 3 = 4 \]
Putting it all together:
\[ M = -8 \times 4 = -32 \]
Since magnifying power is usually expressed as a positive value, we take the absolute value:
\[ M = 32 \]

### Final Answer
The magnifying power of the microscope is **32**.

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