In a compound microscope , the magnificent virtual image is found at a distance of 25 cm from the eye-piece . The focal length of its objective lens is 1 cm. If the microspore is 20 cm , then the focal length of the eye-piece lens ( in cm) is .................

Answer

To solve the problem, we need to use the formula for magnification in a compound microscope. The magnification (M) is given by the formula:

\[ M = \frac{L}{F_O} \times \left(1 + \frac{D}{F_E}\right) \]

Where:
- \( L \) = Length of the tube (20 cm)
- \( F_O \) = Focal length of the objective lens (1 cm)
- \( D \) = Distance from the eyepiece to the virtual image (25 cm)
- \( F_E \) = Focal length of the eyepiece lens (which we need to find)

Given that the magnification \( M \) is 100, we can substitute the known values into the equation.

### Step 1: Write the magnification formula
\[ M = \frac{L}{F_O} \times \left(1 + \frac{D}{F_E}\right) \]

### Step 2: Substitute the known values into the formula
\[ 100 = \frac{20}{1} \times \left(1 + \frac{25}{F_E}\right) \]

### Step 3: Simplify the equation
\[ 100 = 20 \times \left(1 + \frac{25}{F_E}\right) \]

### Step 4: Divide both sides by 20
\[ 5 = 1 + \frac{25}{F_E} \]

### Step 5: Isolate the term with \( F_E \)
\[ 5 - 1 = \frac{25}{F_E} \]
\[ 4 = \frac{25}{F_E} \]

### Step 6: Cross-multiply to solve for \( F_E \)
\[ 4F_E = 25 \]

### Step 7: Divide both sides by 4
\[ F_E = \frac{25}{4} \]
\[ F_E = 6.25 \, \text{cm} \]

Thus, the focal length of the eyepiece lens is **6.25 cm**.