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, we will follow these steps:

### Step 1: Understand the given data
We are given:
- Total magnification of the compound microscope (M) = 95
- Distance of the object from the objective lens (u₀) = -1/3.8 cm (negative as per sign convention)
- Focal length of the objective lens (f₀) = 1/4 cm (positive for a convex lens)

### Step 2: Use the lens formula to find the image distance (v₀) for the objective lens
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Rearranging gives:
\[ \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \]
Substituting the values:
\[ \frac{1}{v₀} = \frac{1}{(1/4)} + \frac{1}{(-1/3.8)} \]
Calculating:
\[ \frac{1}{v₀} = 4 - \frac{3.8}{-1} = 4 + 3.8 = 7.8 \]
Thus,
\[ v₀ = \frac{1}{7.8} \text{ cm} \]

### Step 3: Calculate the magnification of the objective lens (M₀)
The magnification of the objective lens is given by:
\[ M₀ = \frac{v₀}{u₀} \]
Substituting the values:
\[ M₀ = \frac{1/7.8}{-1/3.8} = \frac{3.8}{-7.8} \]
Calculating:
\[ M₀ = -\frac{3.8}{7.8} = -0.487 \]

### Step 4: Relate the total magnification to the eyepiece magnification
The total magnification (M) of the compound microscope is given by:
\[ M = M₀ \times M_E \]
Where \(M_E\) is the magnification of the eyepiece. We know:
\[ 95 = (-0.487) \times M_E \]
Thus,
\[ M_E = \frac{95}{-0.487} \]
Calculating:
\[ M_E \approx -195.3 \]

### Step 5: Interpret the result
The negative sign indicates that the image formed by the eyepiece is inverted relative to the object. The absolute value indicates the magnification factor.

### Final Answer
The magnification of the eyepiece when the final image is formed at the least distance of distinct vision is approximately **-195.3**.

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