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. (a)10mm
B. (b)15mm
C. (c)22mm
D. (d)25mm

Explanation

To solve the problem, we need to determine the focal length of the previous eyepiece of a compound microscope given the following information:

- Tube length (L) = 300 mm
- Focal length of the objective lens (F₀) = 7.5 mm
- The new eyepiece focal length (Fₑ') = 20 mm
- The increase in magnifying power (ΔM) = 100

### Step-by-Step Solution:

1. **Understanding the Magnifying Power of a Compound Microscope**:
The magnifying power (M) of a compound microscope is given by the formula:
\[ M = \frac{L}{F₀} \left(1 + \frac{D}{Fₑ}\right) \]
where \(D\) is the near point distance (usually taken as 25 cm or 250 mm for the human eye).

2. **Calculate the Magnifying Power with the Original Eyepiece**:
Let the focal length of the original eyepiece be \(Fₑ\). The magnifying power with the original eyepiece is:
\[ M = \frac{300}{7.5} \left(1 + \frac{250}{Fₑ}\right) \]

3. **Calculate the Magnifying Power with the New Eyepiece**:
The magnifying power with the new eyepiece (focal length = 20 mm) is:
\[ M' = \frac{300}{7.5} \left(1 + \frac{250}{20}\right) \]

4. **Finding the Increase in Magnifying Power**:
According to the problem, the increase in magnifying power when replacing the eyepiece is 100:
\[ M' - M = 100 \]

5. **Substituting the Magnifying Powers**:
Substitute the expressions for \(M\) and \(M'\):
\[ \frac{300}{7.5} \left(1 + \frac{250}{20}\right) - \frac{300}{7.5} \left(1 + \frac{250}{Fₑ}\right) = 100 \]

6. **Simplifying the Equation**:
Factor out \(\frac{300}{7.5}\):
\[ \frac{300}{7.5} \left(\left(1 + \frac{250}{20}\right) - \left(1 + \frac{250}{Fₑ}\right)\right) = 100 \]
This simplifies to:
\[ \frac{300}{7.5} \left(\frac{250}{20} - \frac{250}{Fₑ}\right) = 100 \]

7. **Calculating the Coefficients**:
Calculate \(\frac{300}{7.5} = 40\):
\[ 40 \left(\frac{250}{20} - \frac{250}{Fₑ}\right) = 100 \]
This simplifies to:
\[ \frac{250}{20} - \frac{250}{Fₑ} = \frac{100}{40} = 2.5 \]

8. **Rearranging the Equation**:
Rearranging gives:
\[ \frac{250}{Fₑ} = \frac{250}{20} - 2.5 \]

9. **Finding \(Fₑ\)**:
Calculate \(\frac{250}{20} = 12.5\):
\[ \frac{250}{Fₑ} = 12.5 - 2.5 = 10 \]
Therefore:
\[ Fₑ = \frac{250}{10} = 25 \text{ mm} \]

### Conclusion:
The focal length of the previous eyepiece is **25 mm**.