Magnification of compound microscope is 375. Length of tube is 150mm. Given that focal length of objective lens is 5mm, then value of focal length of eyepiece is:

Options

A. 2mm
B. 22mm
C. 12mm
D. 33mm

Answer

To find the focal length of the eyepiece (Fe) in a compound microscope, we can use the formula for magnification (M) of a compound microscope:

\[ M = \frac{L}{f_o} \left(1 + \frac{d}{f_e}\right) \]

Where:
- \(M\) = Magnification
- \(L\) = Length of the tube
- \(f_o\) = Focal length of the objective lens
- \(d\) = Least distance of distinct vision (approximately 25 cm or 250 mm)
- \(f_e\) = Focal length of the eyepiece

Given:
- \(M = 375\)
- \(L = 150 \, \text{mm}\)
- \(f_o = 5 \, \text{mm}\)

### Step 1: Convert the least distance of distinct vision to mm
Since \(d\) is given in cm, we convert it to mm:
\[ d = 25 \, \text{cm} = 250 \, \text{mm} \]

### Step 2: Substitute the known values into the magnification formula
Substituting the known values into the magnification formula:
\[ 375 = \frac{150}{5} \left(1 + \frac{250}{f_e}\right) \]

### Step 3: Simplify the equation
First, calculate \(\frac{150}{5}\):
\[ \frac{150}{5} = 30 \]
Now substitute this back into the equation:
\[ 375 = 30 \left(1 + \frac{250}{f_e}\right) \]

### Step 4: Divide both sides by 30
\[ \frac{375}{30} = 1 + \frac{250}{f_e} \]
Calculating \(\frac{375}{30}\):
\[ 12.5 = 1 + \frac{250}{f_e} \]

### Step 5: Rearrange the equation to isolate \(\frac{250}{f_e}\)
Subtract 1 from both sides:
\[ 12.5 - 1 = \frac{250}{f_e} \]
\[ 11.5 = \frac{250}{f_e} \]

### Step 6: Solve for \(f_e\)
Cross-multiply to solve for \(f_e\):
\[ f_e = \frac{250}{11.5} \]

### Step 7: Calculate the value of \(f_e\)
Perform the division:
\[ f_e \approx 21.7391 \, \text{mm} \]
Rounding this to two decimal places gives:
\[ f_e \approx 22 \, \text{mm} \]

### Final Answer
The focal length of the eyepiece is approximately \(22 \, \text{mm}\).

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