In compound microscope final image formed at 25 cm from eyepiece lens. Length of tube is 20 cm. Given that `f_0 = 1 cm` , m= 100. Find focal length of eyepiece lens

Answer

To solve the problem, we need to find the focal length of the eyepiece lens (FE) in a compound microscope given the following information:

- Final image distance from the eyepiece (D) = 25 cm
- Length of the tube (L) = 20 cm
- Focal length of the objective lens (F0) = 1 cm
- Magnification (m) = 100

### Step-by-Step Solution:

1. **Understand the relationship between magnification, length of the tube, and focal lengths:**
The magnification (m) for a compound microscope can be expressed as:
\[ m = \frac{L}{F_0} \left(1 + \frac{D}{F_E}\right) \]
where:
- L = Length of the tube
- F0 = Focal length of the objective lens
- D = Image distance from the eyepiece
- FE = Focal length of the eyepiece lens

2. **Substitute the known values into the magnification formula:**
Given:
- m = 100
- L = 20 cm
- F0 = 1 cm
- D = 25 cm

Substitute these values into the magnification formula:
\[ 100 = \frac{20}{1} \left(1 + \frac{25}{F_E}\right) \]

3. **Simplify the equation:**
\[ 100 = 20 \left(1 + \frac{25}{F_E}\right) \]
Divide both sides by 20:
\[ 5 = 1 + \frac{25}{F_E} \]

4. **Isolate the term involving FE:**
Subtract 1 from both sides:
\[ 4 = \frac{25}{F_E} \]

5. **Solve for FE:**
Rearranging gives:
\[ F_E = \frac{25}{4} \]
Calculate the value:
\[ F_E = 6.25 \text{ cm} \]

### Final Answer:
The focal length of the eyepiece lens (FE) is **6.25 cm**.