A compound microscope has an objective lens of 1.0 cm focal length and an eye-piece of 2.5 cm of focal length. When the object is placed at 1.1 cm from the objective, the final image is at the least distance of direct vision. Thus, the magnification produced is

Options

A. 10
B. 11
C. 100
D. 110

Answer

To solve the problem of finding the magnification produced by a compound microscope with the given parameters, we will follow these steps:

### Step 1: Identify the given values
- Focal length of the objective lens (f_o) = 1.0 cm
- Focal length of the eyepiece (f_e) = 2.5 cm
- Object distance from the objective lens (u_o) = -1.1 cm (the negative sign indicates that the object is on the same side as the incoming light)

### Step 2: Use the lens formula for the objective lens
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Rearranging gives:
\[ \frac{1}{v_o} = \frac{1}{f_o} + \frac{1}{u_o} \]
Substituting the values:
\[ \frac{1}{v_o} = \frac{1}{1.0} + \frac{1}{-1.1} \]
Calculating:
\[ \frac{1}{v_o} = 1 - \frac{1}{1.1} = 1 - 0.9091 = 0.0909 \]
Thus,
\[ v_o = \frac{1}{0.0909} \approx 11.0 \text{ cm} \]

### Step 3: Calculate the total magnification of the compound microscope
The total magnification (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 = least distance of distinct vision = 25 cm
- u_o = 1.1 cm (considered positive in the magnification formula)

Substituting the values:
\[ M = \frac{11.0}{1.1} \left(1 + \frac{25}{2.5}\right) \]
Calculating:
\[ M = 10 \left(1 + 10\right) = 10 \times 11 = 110 \]

### Final Answer
The magnification produced by the compound microscope is **110**.

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