In compound microscope the magnification is 95, and the distance of object from objective lens 1/3.8 cm and focal length of objective is ¼ cm. What is the magnification of eye pieces when final image is formed at least distance of distinct vision :

Options

A. 5
B. 10
C. 100
D. None

Explanation

To solve the problem step by step, we will use the principles of optics related to a compound microscope.

### Step 1: Understand the given data
- Total magnification (M) = 95
- Object distance from the objective lens (u) = -1/3.8 cm (negative because it is on the same side as the incoming light)
- Focal length of the objective lens (f₀) = 1/4 cm

### Step 2: Use the lens formula to find the image distance (v₀) for the objective lens
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Substituting the values we have:
\[ \frac{1}{\frac{1}{4}} = \frac{1}{v₀} - \frac{1}{-\frac{1}{3.8}} \]
This simplifies to:
\[ 4 = \frac{1}{v₀} + \frac{1}{\frac{1}{3.8}} \]
\[ 4 = \frac{1}{v₀} + 3.8 \]
Now, rearranging gives:
\[ \frac{1}{v₀} = 4 - 3.8 = 0.2 \]
Thus,
\[ v₀ = \frac{1}{0.2} = 5 \text{ cm} \]

### Step 3: Calculate the magnification of the objective lens (M₀)
The magnification of the objective lens is given by:
\[ M₀ = \frac{v₀}{u} \]
Substituting the values:
\[ M₀ = \frac{5}{-\frac{1}{3.8}} = 5 \times 3.8 = 19 \]

### Step 4: Relate total magnification to the magnification of the eyepiece (Mₑ)
The total magnification (M) of the compound microscope is the product of the magnification of the objective (M₀) and the magnification of the eyepiece (Mₑ):
\[ M = M₀ \times Mₑ \]
Substituting the known values:
\[ 95 = 19 \times Mₑ \]
Now, solving for Mₑ:
\[ Mₑ = \frac{95}{19} = 5 \]

### Final Answer
The magnification of the eyepiece (Mₑ) is **5**.

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