A compound microscope has tube length 300 mm and an objective of focal length 7.5 mm and an eyepiece of certain focal length. When eyepiece is replaced by another lens of focal length 20 mm, the magnifying power of compound microscope is increased by 100. The focal length of previous eyepiece is

Options

A. 10 mm
B. 15 mm
C. 22 mm
D. 25 mm

Explanation

To find the focal length of the previous eyepiece in a compound microscope, we can follow these steps:

### Step 1: Understand the magnifying power formula
The magnifying power \( M \) of a compound microscope is given by the formula:
\[ M = \frac{L}{F_o} \left(1 + \frac{D}{F_e}\right) \]
where:
- \( L \) = tube length
- \( F_o \) = focal length of the objective lens
- \( D \) = distance of distinct vision (typically taken as 25 cm or 250 mm)
- \( F_e \) = focal length of the eyepiece

### Step 2: Set up the equations for magnifying power
Let \( F_{e1} \) be the focal length of the original eyepiece and \( F_{e2} = 20 \) mm be the focal length of the new eyepiece. The magnifying power with the original eyepiece is:
\[ M_1 = \frac{L}{F_o} \left(1 + \frac{D}{F_{e1}}\right) \]
The magnifying power with the new eyepiece is:
\[ M_2 = \frac{L}{F_o} \left(1 + \frac{D}{F_{e2}}\right) \]

### Step 3: Relate the magnifying powers
According to the problem, the magnifying power increases by 100 when the eyepiece is changed:
\[ M_2 = M_1 + 100 \]

### Step 4: Substitute the magnifying power equations
Substituting the expressions for \( M_1 \) and \( M_2 \):
\[ \frac{L}{F_o} \left(1 + \frac{D}{F_{e2}}\right) = \frac{L}{F_o} \left(1 + \frac{D}{F_{e1}}\right) + 100 \]

### Step 5: Simplify the equation
Since \( \frac{L}{F_o} \) is common, we can cancel it out:
\[ 1 + \frac{D}{F_{e2}} = 1 + \frac{D}{F_{e1}} + \frac{100 F_o}{L} \]
This simplifies to:
\[ \frac{D}{F_{e2}} - \frac{D}{F_{e1}} = \frac{100 F_o}{L} \]

### Step 6: Substitute known values
Given:
- \( L = 300 \) mm
- \( F_o = 7.5 \) mm
- \( D = 250 \) mm
- \( F_{e2} = 20 \) mm

Substituting these values into the equation:
\[ \frac{250}{20} - \frac{250}{F_{e1}} = \frac{100 \times 7.5}{300} \]

### Step 7: Calculate the right side
Calculating the right side:
\[ \frac{100 \times 7.5}{300} = \frac{750}{300} = 2.5 \]

### Step 8: Rearranging the equation
Now we have:
\[ 12.5 - \frac{250}{F_{e1}} = 2.5 \]
Rearranging gives:
\[ \frac{250}{F_{e1}} = 12.5 - 2.5 = 10 \]

### Step 9: Solve for \( F_{e1} \)
Now, solving for \( F_{e1} \):
\[ F_{e1} = \frac{250}{10} = 25 \text{ mm} \]

### Final Answer
The focal length of the previous eyepiece is **25 mm**.

---