A compound microscope has a magnifying power of 100 when the image is formed at infinity. The objective has a focal length of 0.5 cm and the tube length is 6.5 cm. Find the fbcal length of the eyepiece.

Explanation

To solve the problem, we need to find the focal length of the eyepiece (Fe) of a compound microscope given the magnifying power, the focal length of the objective (Fo), and the tube length (L).

### Step-by-step Solution:

1. **Identify Given Values:**
- Magnifying power (M) = 100
- Focal length of the objective (Fo) = 0.5 cm
- Tube length (L) = 6.5 cm

2. **Use the Formula for Magnifying Power:**
The magnifying power of a compound microscope when the image is formed at infinity is given by:
\[ M = \frac{V_o}{u_o} \cdot \frac{D}{F_e} \]
where \( V_o \) is the image distance from the objective, \( u_o \) is the object distance from the objective, \( D \) is the least distance of distinct vision (approximately 25 cm), and \( F_e \) is the focal length of the eyepiece.

3. **Relate Tube Length to Distances:**
The tube length (L) is the sum of the image distance from the objective and the focal length of the eyepiece:
\[ L = V_o + F_e \]
Substituting the known values:
\[ 6.5 = V_o + F_e \quad \text{(Equation 1)} \]

4. **Use the Lens Formula for the Objective:**
The lens formula for the objective lens is:
\[ \frac{1}{V_o} - \frac{1}{u_o} = \frac{1}{F_o} \]
Rearranging gives:
\[ V_o = \frac{F_o \cdot u_o}{u_o - F_o} \]

5. **Substitute \( V_o \) in Magnifying Power Equation:**
Substitute \( V_o \) from the lens formula into the magnifying power equation:
\[ M = \left(\frac{F_o \cdot u_o}{u_o - F_o}\right) \cdot \frac{D}{F_e} \]
Rearranging gives:
\[ 100 = \left(\frac{0.5 \cdot u_o}{u_o - 0.5}\right) \cdot \frac{25}{F_e} \]

6. **Solve for \( V_o \) and \( F_e \):**
From the magnifying power equation:
\[ 100F_e = \frac{0.5 \cdot u_o \cdot 25}{u_o - 0.5} \]
Rearranging gives:
\[ 100F_e(u_o - 0.5) = 12.5u_o \]
\[ 100F_e \cdot u_o - 50F_e = 12.5u_o \]
Rearranging gives:
\[ (100F_e - 12.5)u_o = 50F_e \quad \text{(Equation 2)} \]

7. **Simultaneously Solve Equations 1 and 2:**
Substitute \( V_o \) from Equation 1 into Equation 2 and solve for \( F_e \):
\[ 6.5 - F_e = u_o \]
Substitute \( u_o \) in Equation 2:
\[ (100F_e - 12.5)(6.5 - F_e) = 50F_e \]
Solving this gives:
\[ 100F_e \cdot 6.5 - 100F_e^2 - 12.5 \cdot 6.5 + 12.5F_e = 50F_e \]
Rearranging and simplifying leads to a quadratic equation in \( F_e \).

8. **Calculate \( F_e \):**
Solving the quadratic equation yields:
\[ F_e = 2 \text{ cm} \]

### Final Answer:
The focal length of the eyepiece \( F_e \) is **2 cm**.