In a compound microscope, the objective and eye piece have focal lengths `0.95 cm and 5 cm` respectively, and are kept at a distance of `20 cm`. The final image is formed at a distance of `25 cm` from the eye piece. Calculate the position of the object and the total magnification.

Options

A. 94
B. 84
C. 75
D. 88

Answer

To solve the problem, we need to find the position of the object and the total magnification of the compound microscope. We'll follow these steps:

### Step 1: Identify the given data
- Focal length of the objective lens, \( f_o = 0.95 \, \text{cm} \)
- Focal length of the eyepiece lens, \( f_e = 5 \, \text{cm} \)
- Distance between the objective and eyepiece, \( d = 20 \, \text{cm} \)
- Final image distance from the eyepiece, \( v_e = 25 \, \text{cm} \)

### Step 2: Use the lens formula for the eyepiece
The lens formula is given by:
\[ \frac{1}{v_e} - \frac{1}{u_e} = \frac{1}{f_e} \]
Where:
- \( v_e \) is the image distance from the eyepiece (25 cm)
- \( u_e \) is the object distance from the eyepiece (which we need to find)

Rearranging the lens formula gives:
\[ \frac{1}{u_e} = \frac{1}{v_e} - \frac{1}{f_e} \]

Substituting the known values:
\[ \frac{1}{u_e} = \frac{1}{25} - \frac{1}{5} \]
\[ \frac{1}{u_e} = \frac{1}{25} - \frac{5}{25} = -\frac{4}{25} \]
Thus,
\[ u_e = -\frac{25}{4} = -6.25 \, \text{cm} \]

### Step 3: Calculate the object distance from the objective lens
The distance between the objective and eyepiece is 20 cm. Therefore, the object distance from the objective lens \( u_o \) can be calculated as:
\[ u_o = d - |u_e| = 20 - 6.25 = 13.75 \, \text{cm} \]

### Step 4: Use the lens formula for the objective
Using the lens formula for the objective lens:
\[ \frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o} \]
Where \( v_o \) is the image distance from the objective lens. Rearranging gives:
\[ \frac{1}{v_o} = \frac{1}{f_o} + \frac{1}{u_o} \]
Substituting the values:
\[ \frac{1}{v_o} = \frac{1}{0.95} + \frac{1}{13.75} \]
Calculating:
\[ \frac{1}{v_o} = 1.0526 + 0.0727 = 1.1253 \]
Thus,
\[ v_o = \frac{1}{1.1253} \approx 0.888 \, \text{cm} \]

### Step 5: Calculate the total magnification
The total magnification \( M \) of a compound microscope is given by:
\[ M = \left( \frac{v_o}{u_o} \right) \left( 1 + \frac{D}{f_e} \right) \]
Where \( D \) is the least distance of distinct vision (approximately 25 cm).

Calculating \( M \):
\[ M = \left( \frac{0.888}{13.75} \right) \left( 1 + \frac{25}{5} \right) \]
Calculating the first part:
\[ \frac{0.888}{13.75} \approx 0.0646 \]
Calculating the second part:
\[ 1 + \frac{25}{5} = 1 + 5 = 6 \]
Thus,
\[ M \approx 0.0646 \times 6 \approx 0.3876 \]

### Final Result
The position of the object from the objective lens is approximately \( 13.75 \, \text{cm} \) and the total magnification is approximately \( 0.3876 \).