The two lenses of a compound microscope are of focal lengths 2 cm and 5 cm. If an object is placed at a distance of 2.1 cm from the objective of focal length 2 cm the final image forms at the final image forms at the least distance of distinct vision of a normal eye. Find the magnifying power of the microscope.

Options

A. 20
B. 6
C. 120
D. 60

Answer

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

### Step 1: Identify the given data
- Focal length of the objective lens (\(f_o\)) = 2 cm
- Focal length of the eyepiece lens (\(f_e\)) = 5 cm
- Object distance from the objective lens (\(u_o\)) = -2.1 cm (the negative sign is used because the object is placed on the same side as the incoming light)

### Step 2: Calculate the image distance for the objective lens
Using the lens formula:
\[ \frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u_o} \]
Rearranging gives:
\[ \frac{1}{v_o} = \frac{1}{f_o} + \frac{1}{u_o} \]
Substituting the known values:
\[ \frac{1}{v_o} = \frac{1}{2} + \frac{1}{-2.1} \]
Calculating:
\[ \frac{1}{v_o} = \frac{1}{2} - \frac{1}{2.1} = \frac{2.1 - 2}{4.2} = \frac{0.1}{4.2} \approx 0.0238 \]
Thus,
\[ v_o \approx 42 \text{ cm} \]

### Step 3: Calculate the magnifying power of the microscope
The magnifying power (\(M\)) of a compound microscope is given by the formula:
\[ M = \frac{v_o}{u_o} \left(1 + \frac{D}{f_e}\right) \]
Where \(D\) is the least distance of distinct vision (typically taken as 25 cm for a normal eye).

Substituting the values:
\[ M = \frac{42}{-2.1} \left(1 + \frac{25}{5}\right) \]
Calculating:
\[ M = \frac{42}{-2.1} \left(1 + 5\right) = \frac{42}{-2.1} \times 6 \]
Calculating \(\frac{42}{-2.1} = -20\):
\[ M = -20 \times 6 = -120 \]

### Step 4: Conclusion
The magnifying power of the microscope is \(120\).