The magnifying power of a microscope with an objective of `5 mm` focal length is 400. The length of its tube is `20cm.` Then the focal length of the eye`-` piece is

Options

A. `200 cm`
B. `160 cm`
C. `2.5 cm`
D. `0.1 cm`

Explanation

To find the focal length of the eyepiece (FE) of the microscope, we can use the formula for the magnifying power (M) of a microscope, which is given by:

\[ M = \frac{L - f_o - f_e}{f_o \cdot f_e} \cdot D \]

Where:
- \( M \) = magnifying power
- \( L \) = length of the tube
- \( f_o \) = focal length of the objective lens
- \( f_e \) = focal length of the eyepiece
- \( D \) = least distance of distinct vision (usually taken as 25 cm)

Given:
- \( M = 400 \)
- \( L = 20 \, \text{cm} \)
- \( f_o = 5 \, \text{mm} = 0.5 \, \text{cm} \)
- \( D = 25 \, \text{cm} \)

### Step 1: Rearranging the formula
We can rearrange the formula to isolate \( f_e \):

\[ M = \frac{(L - f_o - f_e) \cdot D}{f_o \cdot f_e} \]

Multiplying both sides by \( f_o \cdot f_e \):

\[ M \cdot f_o \cdot f_e = (L - f_o - f_e) \cdot D \]

### Step 2: Substitute known values
Substituting the known values into the equation:

\[ 400 \cdot (0.5) \cdot f_e = (20 - 0.5 - f_e) \cdot 25 \]

### Step 3: Simplifying the equation
This simplifies to:

\[ 200 f_e = (19.5 - f_e) \cdot 25 \]

Expanding the right side:

\[ 200 f_e = 19.5 \cdot 25 - 25 f_e \]

### Step 4: Combine like terms
Combining like terms gives:

\[ 200 f_e + 25 f_e = 19.5 \cdot 25 \]

\[ 225 f_e = 19.5 \cdot 25 \]

### Step 5: Calculate \( 19.5 \cdot 25 \)
Calculating \( 19.5 \cdot 25 \):

\[ 19.5 \cdot 25 = 487.5 \]

### Step 6: Solve for \( f_e \)
Now, we can solve for \( f_e \):

\[ f_e = \frac{487.5}{225} \]

Calculating this gives:

\[ f_e = 2.1667 \, \text{cm} \approx 2.5 \, \text{cm} \]

### Conclusion
Thus, the focal length of the eyepiece \( f_e \) is approximately \( 2.5 \, \text{cm} \).