If the focal length of the objective lens and the eye lens are 4 mm and 25 mm respectively in a compound microscope. The length of the tube is 16 cm . Find its magnifying power for relaxed eye position

Options

A. 32.75
B. 327.5
C. 0.3275
D. None of the above

Answer

To find the magnifying power of a compound microscope for a relaxed eye position, we can follow these steps:

### Step 1: Understand the Given Data
- Focal length of the objective lens (fo) = 4 mm = 0.4 cm
- Focal length of the eye lens (fe) = 25 mm = 2.5 cm
- Length of the tube (L) = 16 cm
- Distance of distinct vision (D) = 25 cm

### Step 2: Write the Formula for Magnifying Power
The magnifying power (M) of a compound microscope for a relaxed eye position is given by the formula:

\[ M = \frac{(L - fe - fo) \cdot D}{fe \cdot fo} \]

### Step 3: Substitute the Values into the Formula
Now, we substitute the values into the formula:

- L = 16 cm
- fe = 2.5 cm
- fo = 0.4 cm
- D = 25 cm

Substituting these values gives:

\[ M = \frac{(16 - 2.5 - 0.4) \cdot 25}{2.5 \cdot 0.4} \]

### Step 4: Calculate the Numerator
First, calculate the numerator:

\[ 16 - 2.5 - 0.4 = 16 - 2.9 = 13.1 \]

Now multiply by D:

\[ 13.1 \cdot 25 = 327.5 \]

### Step 5: Calculate the Denominator
Now calculate the denominator:

\[ fe \cdot fo = 2.5 \cdot 0.4 = 1.0 \]

### Step 6: Calculate the Magnifying Power
Now, we can find M:

\[ M = \frac{327.5}{1.0} = 327.5 \]

### Final Answer
The magnifying power of the compound microscope for a relaxed eye position is **327.5**.

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