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

Answer

To find the magnifying power of the microscope, we can use the following steps:

### Step 1: Understand the formula for magnifying power
The magnifying power (M) of a microscope is given by the formula:
\[ M = \frac{v}{f_o} + 1 \]
where:
- \( v \) is the least distance of distinct vision (D),
- \( f_o \) is the focal length of the objective lens.

### Step 2: Identify the given values
From the question, we have:
- Focal length of the objective lens, \( f_o = 4 \, \text{cm} \)
- Focal length of the eye lens, \( f_e = 8 \, \text{cm} \) (not directly needed for magnifying power calculation)
- Least distance of distinct vision, \( D = 24 \, \text{cm} \)
- Object distance from the objective lens, \( u = -4.5 \, \text{cm} \) (the negative sign is used because the object is on the same side as the incoming light)

### Step 3: Calculate the image distance (v) using the lens formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
For the objective lens:
\[ \frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u} \]
Substituting the values:
\[ \frac{1}{4} = \frac{1}{v_o} - \frac{1}{-4.5} \]
\[ \frac{1}{4} = \frac{1}{v_o} + \frac{1}{4.5} \]

### Step 4: Solve for \( v_o \)
Rearranging the equation:
\[ \frac{1}{v_o} = \frac{1}{4} - \frac{1}{4.5} \]
To find a common denominator (which is 18):
\[ \frac{1}{4} = \frac{4.5}{18}, \quad \frac{1}{4.5} = \frac{4}{18} \]
Thus:
\[ \frac{1}{v_o} = \frac{4.5 - 4}{18} = \frac{0.5}{18} \]
So:
\[ v_o = \frac{18}{0.5} = 36 \, \text{cm} \]

### Step 5: Calculate the magnifying power
Now, substituting \( v_o \) into the magnifying power formula:
\[ M = \frac{D}{f_o} + 1 \]
\[ M = \frac{24}{4} + 1 = 6 + 1 = 7 \]

### Final Answer
The magnifying power of the microscope is \( M = 7 \).

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