If we need a magnification of 400 from compound microscope of tube length 150 mm and an objective of focal length 5 mm, the focal length of the eye-piece, should be dose to : (final Image is at infinity).

Options

A. 22mm
B. 19mm
C. 33mm
D. 2 mm

Explanation

To solve the problem of finding the focal length of the eyepiece in a compound microscope, we can follow these steps:

### Step 1: Understand the Magnification Formula
The total magnification (M) of a compound microscope can be expressed as:
\[ M = \frac{L}{f_o} \cdot \frac{D}{f_e} \]
where:
- \(L\) = tube length (150 mm)
- \(f_o\) = focal length of the objective (5 mm)
- \(D\) = least distance of distinct vision (typically taken as 25 cm or 250 mm)
- \(f_e\) = focal length of the eyepiece (what we need to find)

### Step 2: Substitute Known Values
We know:
- \(M = 400\)
- \(L = 150 \text{ mm}\)
- \(f_o = 5 \text{ mm}\)
- \(D = 250 \text{ mm}\)

Substituting these values into the magnification formula gives:
\[ 400 = \frac{150}{5} \cdot \frac{250}{f_e} \]

### Step 3: Simplify the Equation
First, calculate \(\frac{150}{5}\):
\[ \frac{150}{5} = 30 \]
Now, substituting this back into the equation:
\[ 400 = 30 \cdot \frac{250}{f_e} \]

### Step 4: Rearranging the Equation
Rearranging the equation to solve for \(f_e\):
\[ f_e = 30 \cdot \frac{250}{400} \]

### Step 5: Calculate \(f_e\)
Now, simplify the right side:
\[ f_e = 30 \cdot \frac{250}{400} = 30 \cdot \frac{25}{40} = 30 \cdot \frac{5}{8} = \frac{150}{8} = 18.75 \text{ mm} \]

### Step 6: Conclusion
The focal length of the eyepiece should be approximately 19 mm.

### Final Answer
The focal length of the eyepiece should be close to **19 mm**.

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