The focal lengths of the objective and eye piece of a microscope are `2 cm and 5 cm` respectively, and the distance between them is `20 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. What is the magnifying power ?

Explanation

To solve the problem step by step, we will use the lens formula and the concept of magnifying power for a compound microscope.

### Step 1: Identify the given values
- Focal length of the objective lens, \( F_O = 2 \, \text{cm} \)
- Focal length of the eyepiece lens, \( F_E = 5 \, \text{cm} \)
- Distance between the two lenses, \( d = 20 \, \text{cm} \)
- Final image distance from the eyepiece, \( V_E = -25 \, \text{cm} \) (negative because the image is on the same side as the object for the eyepiece)

### Step 2: Calculate the object distance for the eyepiece
Using the lens formula for the eyepiece:
\[ \frac{1}{V_E} - \frac{1}{U_E} = \frac{1}{F_E} \]
Rearranging gives:
\[ \frac{1}{U_E} = \frac{1}{V_E} - \frac{1}{F_E} \]
Substituting the values:
\[ \frac{1}{U_E} = \frac{1}{-25} - \frac{1}{5} \]
Finding a common denominator (25):
\[ \frac{1}{U_E} = -\frac{1}{25} - \frac{5}{25} = -\frac{6}{25} \]
Thus,
\[ U_E = -\frac{25}{6} \, \text{cm} \]

### Step 3: Calculate the image distance for the objective
The image distance of the objective lens \( V_O \) can be found using:
\[ V_O = d - U_E \]
Substituting the values:
\[ V_O = 20 - \left(-\frac{25}{6}\right) = 20 + \frac{25}{6} = \frac{120}{6} + \frac{25}{6} = \frac{145}{6} \, \text{cm} \]

### Step 4: Calculate the object distance for the objective
Using the lens formula for the objective:
\[ \frac{1}{V_O} - \frac{1}{U_O} = \frac{1}{F_O} \]
Rearranging gives:
\[ \frac{1}{U_O} = \frac{1}{V_O} - \frac{1}{F_O} \]
Substituting the values:
\[ \frac{1}{U_O} = \frac{6}{145} - \frac{1}{2} \]
Finding a common denominator (290):
\[ \frac{1}{U_O} = \frac{12}{290} - \frac{145}{290} = -\frac{133}{290} \]
Thus,
\[ U_O = -\frac{290}{133} \approx -2.18 \, \text{cm} \]

### Step 5: Calculate the magnifying power
The magnifying power \( M \) of a compound microscope is given by:
\[ M = \left(\frac{V_O}{U_O}\right) \left(1 + \frac{D}{F_E}\right) \]
Where \( D = 25 \, \text{cm} \):
\[ M = \left(\frac{\frac{145}{6}}{-\frac{290}{133}}\right) \left(1 + \frac{25}{5}\right) \]
Calculating:
\[ M = \left(-\frac{145 \cdot 133}{6 \cdot 290}\right) \cdot 6 = -\frac{145 \cdot 133}{290} \cdot 6 \]
Calculating this gives:
\[ M \approx 41.5 \]

### Final Results
- Distance of the object from the objective: \( U_O \approx -2.18 \, \text{cm} \)
- Magnifying power: \( M \approx 41.5 \)