A compound microscope having magnifying power 35 with its eye - piece of focal length 10 cm. Assume that the final image is at least distance of distinct vision then the magnification produced by the objective is

Options

A. `-4`
B. 5
C. 10
D. `-10`

Answer

To solve the problem, we need to find the magnification produced by the objective lens of a compound microscope. Here’s a step-by-step solution:

### Step 1: Understand the given data
- Magnifying power of the compound microscope (MP) = 35
- Focal length of the eyepiece (Fe) = 10 cm
- Least distance of distinct vision (D) = 25 cm

### Step 2: Use the formula for magnifying power of a compound microscope
The magnifying power (MP) of a compound microscope can be expressed as:
\[ MP = MO \times (1 + \frac{D}{Fe}) \]
Where:
- \( MO \) = magnifying power of the objective
- \( D \) = least distance of distinct vision
- \( Fe \) = focal length of the eyepiece

### Step 3: Substitute the known values into the formula
We can rearrange the formula to find \( MO \):
\[ MO = \frac{MP}{(1 + \frac{D}{Fe})} \]
Now substituting the known values:
\[ MO = \frac{35}{(1 + \frac{25}{10})} \]

### Step 4: Calculate the term in the denominator
Calculate \( \frac{25}{10} \):
\[ \frac{25}{10} = 2.5 \]
Thus,
\[ 1 + 2.5 = 3.5 \]

### Step 5: Substitute back into the equation for \( MO \)
Now substitute this back into the equation for \( MO \):
\[ MO = \frac{35}{3.5} \]

### Step 6: Perform the division
Now calculate \( \frac{35}{3.5} \):
\[ MO = 10 \]

### Step 7: Conclusion
The magnification produced by the objective lens is:
\[ MO = 10 \]
(Note: The negative sign indicates that the image is inverted, which is typical for a microscope.)

### Final Answer
The magnification produced by the objective lens is **10**.

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