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 will use the magnification formula for a compound microscope and the given parameters to find the focal length of the eyepiece lens.

### Step-by-Step Solution:

1. **Understand the Given 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

2. **Use the Magnification Formula:**
The magnification (m) for a compound microscope can be expressed as:
\[ m = \frac{L}{f_0} \times \left(1 + \frac{D}{f_e}\right) \]
Where:
- \(L\) = length of the tube
- \(f_0\) = focal length of the objective lens
- \(D\) = distance of the final image from the eyepiece
- \(f_e\) = focal length of the eyepiece lens

3. **Substitute the Known Values:**
Plugging in the values we have:
\[ 100 = \frac{20}{1} \times \left(1 + \frac{25}{f_e}\right) \]

4. **Simplify the Equation:**
This simplifies to:
\[ 100 = 20 \times \left(1 + \frac{25}{f_e}\right) \]
Dividing both sides by 20 gives:
\[ 5 = 1 + \frac{25}{f_e} \]

5. **Isolate the Fraction:**
Rearranging the equation:
\[ 5 - 1 = \frac{25}{f_e} \]
\[ 4 = \frac{25}{f_e} \]

6. **Solve for \(f_e\):**
Cross-multiplying gives:
\[ 4f_e = 25 \]
Dividing both sides by 4:
\[ f_e = \frac{25}{4} = 6.25 \text{ cm} \]

### Final Answer:
The focal length of the eyepiece lens is \(f_e = 6.25 \text{ cm}\).