A compound microscope has objective of focal length 3.2 mm and eyepiece of 12 mm focal length. If the objective forms image of the object 16 cm beyond its pole, then total magnification achieved is

Options

A. `1040x`
B. `-1040x`
C. `-1020x`
D. `1020x`

Answer

To find the total magnification achieved by the compound microscope, we will follow these steps:

### Step 1: Identify the given values
- Focal length of the objective lens, \( f_0 = 3.2 \, \text{mm} = 0.32 \, \text{cm} \)
- Focal length of the eyepiece lens, \( f_e = 12 \, \text{mm} = 1.2 \, \text{cm} \)
- Image distance from the objective lens, \( V_0 = 16 \, \text{cm} \)

### Step 2: Use the lens formula to find the object distance \( U_0 \)
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{V} - \frac{1}{U} \]
Rearranging this for \( U_0 \):
\[ \frac{1}{U_0} = \frac{1}{f_0} - \frac{1}{V_0} \]
Substituting the values:
\[ \frac{1}{U_0} = \frac{1}{0.32} - \frac{1}{16} \]
Calculating each term:
\[ \frac{1}{0.32} = 3.125 \quad \text{and} \quad \frac{1}{16} = 0.0625 \]
Thus,
\[ \frac{1}{U_0} = 3.125 - 0.0625 = 3.0625 \]
Now, taking the reciprocal to find \( U_0 \):
\[ U_0 = \frac{1}{3.0625} \approx 0.327 \, \text{cm} \]

### Step 3: Calculate the total magnification \( M \)
The total magnification of a compound microscope is given by:
\[ M = \frac{V_0}{U_0} \left(1 + \frac{d}{f_e}\right) \]
Where \( d \) (the least distance of distinct vision) is approximately \( 25 \, \text{cm} \).

Substituting the values:
\[ M = \frac{16}{0.327} \left(1 + \frac{25}{1.2}\right) \]

### Step 4: Calculate \( 1 + \frac{25}{1.2} \)
Calculating \( \frac{25}{1.2} \):
\[ \frac{25}{1.2} \approx 20.8333 \]
Thus,
\[ 1 + \frac{25}{1.2} \approx 21.8333 \]

### Step 5: Calculate \( M \)
Now substituting back into the magnification formula:
\[ M = \frac{16}{0.327} \times 21.8333 \]
Calculating \( \frac{16}{0.327} \):
\[ \frac{16}{0.327} \approx 48.92 \]
Now multiplying:
\[ M \approx 48.92 \times 21.8333 \approx 1067.73 \]

### Final Result
The total magnification achieved by the compound microscope is approximately:
\[ M \approx 1068.29 \]