A compound microscope has an eye piece of focal length `10 cm` and an objective of focal length `4 cm`. Calculate the magnification, if an object is kept at a distance of `5 cm` from the objective so that final image is formed at the least distance vision `(20 cm)`

Options

A. 12
B. 11
C. 10
D. 13

Answer

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

### Step 1: Identify the given values
- Focal length of the objective lens, \( f_o = 4 \, \text{cm} \)
- Focal length of the eyepiece lens, \( f_e = 10 \, \text{cm} \)
- Object distance from the objective lens, \( u_o = -5 \, \text{cm} \) (negative because the object is on the same side as the incoming light)
- Final image distance (least distance of distinct vision), \( D = 20 \, \text{cm} \)

### Step 2: Use the lens formula to find the image distance from the objective lens, \( v_o \)
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
For the objective lens:
\[ \frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u_o} \]
Substituting the known values:
\[ \frac{1}{4} = \frac{1}{v_o} - \frac{1}{-5} \]
This simplifies to:
\[ \frac{1}{4} = \frac{1}{v_o} + \frac{1}{5} \]
Finding a common denominator (20):
\[ \frac{1}{4} = \frac{5}{20} \quad \text{and} \quad \frac{1}{5} = \frac{4}{20} \]
Thus:
\[ \frac{1}{v_o} = \frac{5}{20} - \frac{4}{20} = \frac{1}{20} \]
So, we find:
\[ v_o = 20 \, \text{cm} \]

### Step 3: Calculate the magnification
The total magnification \( M \) of the compound microscope can be calculated using the formula:
\[ M = \frac{v_o}{u_o} \left(1 + \frac{D}{f_e}\right) \]
Substituting the values we have:
\[ M = \frac{20}{-5} \left(1 + \frac{20}{10}\right) \]
Calculating \( \frac{20}{-5} = -4 \) and \( \frac{20}{10} = 2 \):
\[ M = -4 \left(1 + 2\right) = -4 \times 3 = -12 \]
Thus, the magnification is:
\[ M = -12 \]

### Final Answer
The magnification of the compound microscope is \( 12 \) (the negative sign indicates that the image is inverted).

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