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 step by step, we will use the given information about the compound microscope and the relationships between its components.

### Step 1: Understand the given data
- Magnifying power (M) of the microscope = 35
- Focal length (f) of the eyepiece = 10 cm = 0.1 m
- Least distance of distinct vision (d) = 25 cm = 0.25 m

### Step 2: Write the formula for magnifying power of a compound microscope
The magnifying power (M) of a compound microscope when the final image is formed at the least distance of distinct vision is given by the formula:
\[ M = -\frac{1}{f_{objective}} \left( 1 + \frac{d}{f_{eyepiece}} \right) \]

### Step 3: Substitute the known values into the formula
We can substitute the values of M, d, and f_eyepiece into the formula:
\[ 35 = -\frac{1}{f_{objective}} \left( 1 + \frac{0.25}{0.1} \right) \]

### Step 4: Calculate the term inside the parentheses
Calculate \( \frac{d}{f_{eyepiece}} \):
\[ \frac{0.25}{0.1} = 2.5 \]
Thus,
\[ 1 + \frac{d}{f_{eyepiece}} = 1 + 2.5 = 3.5 \]

### Step 5: Rewrite the equation
Now we can rewrite the equation with the calculated value:
\[ 35 = -\frac{1}{f_{objective}} \cdot 3.5 \]

### Step 6: Solve for \( \frac{1}{f_{objective}} \)
Rearranging the equation to solve for \( \frac{1}{f_{objective}} \):
\[ \frac{1}{f_{objective}} = -\frac{35}{3.5} \]

### Step 7: Calculate \( \frac{1}{f_{objective}} \)
Now, calculate \( -\frac{35}{3.5} \):
\[ \frac{1}{f_{objective}} = -10 \]

### Step 8: Conclusion
The magnification produced by the objective lens is:
\[ \text{Magnification by objective} = -10 \]

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

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