The focal lengths of the objective and eyepiece of a microscope are 0.6 cm and 5 cm respectively and the distance between them is 12 cm. Find the distance of the object from the objective when the final image seen by the eye is 25 cm from the eye-piece. Also find the magnifying power.

Answer

To solve the problem step by step, we will follow these calculations:

### Given Data:
- Focal length of the objective lens, \( f_o = 0.6 \, \text{cm} \)
- Focal length of the eyepiece lens, \( f_e = 5 \, \text{cm} \)
- Distance between the objective and eyepiece, \( d = 12 \, \text{cm} \)
- Final image distance from the eyepiece, \( v_e = 25 \, \text{cm} \)

### Step 1: Calculate the image distance from the objective lens \( v_o \)
Using the lens formula for the eyepiece:
\[ \frac{1}{f_e} = \frac{1}{v_e} - \frac{1}{u_e} \]
We rearrange this to find \( u_e \):
\[ \frac{1}{u_e} = \frac{1}{v_e} - \frac{1}{f_e} \]
Substituting the known values:
\[ \frac{1}{u_e} = \frac{1}{25} - \frac{1}{5} \]
Calculating the right-hand side:
\[ \frac{1}{u_e} = \frac{1 - 5}{25} = \frac{-4}{25} \]
Thus,
\[ u_e = -\frac{25}{4} = -6.25 \, \text{cm} \]
The negative sign indicates that the image formed by the eyepiece is virtual and on the same side as the object.

### Step 2: Calculate the object distance from the objective lens \( u_o \)
Using the relation between the distances:
\[ u_e + v_o = d \]
Substituting \( u_e \) and \( d \):
\[ -6.25 + v_o = 12 \]
Thus,
\[ v_o = 12 + 6.25 = 18.25 \, \text{cm} \]

### Step 3: Use the lens formula for the objective lens to find \( u_o \)
Using the lens formula:
\[ \frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u_o} \]
Rearranging gives:
\[ \frac{1}{u_o} = \frac{1}{v_o} - \frac{1}{f_o} \]
Substituting the values:
\[ \frac{1}{u_o} = \frac{1}{18.25} - \frac{1}{0.6} \]
Calculating the right-hand side:
\[ \frac{1}{u_o} = \frac{1}{18.25} - \frac{1.6667}{1} \]
Calculating \( \frac{1}{18.25} \):
\[ \frac{1}{18.25} \approx 0.0548 \]
Thus:
\[ \frac{1}{u_o} = 0.0548 - 1.6667 \approx -1.6119 \]
So,
\[ u_o \approx -\frac{1}{1.6119} \approx -0.620 \, \text{cm} \]
The negative sign indicates that the object is on the same side as the incoming light.

### Step 4: Calculate the magnifying power \( M \)
The total magnification \( M \) is given by:
\[ M = M_o \times M_e \]
Where:
- \( M_o = \frac{v_o}{u_o} \)
- \( M_e = 1 + \frac{d}{f_e} \)

Calculating \( M_o \):
\[ M_o = \frac{18.25}{-0.620} \approx -29.39 \]

Calculating \( M_e \):
\[ M_e = 1 + \frac{12}{5} = 1 + 2.4 = 3.4 \]

Thus, the total magnification:
\[ M = (-29.39) \times (3.4) \approx -100.84 \]

### Final Answers:
- Distance of the object from the objective lens: \( u_o \approx -0.620 \, \text{cm} \)
- Magnifying power: \( M \approx -100.84 \)