Find the maximum magnifying power of a compound microscope having a 25 diopter lens as the objective, a 5 diopter lens as the eyepiece and the separation 30 cm between the two lenses. The least distance for clear vision is 25 cm.

Explanation

To solve the problem of finding the maximum magnifying power of a compound microscope, we will follow these steps:

### Step 1: Determine the Focal Lengths of the Lenses
The power \( P \) of a lens is related to its focal length \( F \) by the formula:
\[ P = \frac{100}{F} \]
where \( P \) is in diopters and \( F \) is in centimeters.

- For the **objective lens** with a power of 25 diopters:
\[ 25 = \frac{100}{F_o} \implies F_o = \frac{100}{25} = 4 \text{ cm} \]

- For the **eyepiece lens** with a power of 5 diopters:
\[ 5 = \frac{100}{F_e} \implies F_e = \frac{100}{5} = 20 \text{ cm} \]

### Step 2: Use the Lens Formula for the Eyepiece
The lens formula is given by:
\[ \frac{1}{F} = \frac{1}{V} - \frac{1}{U} \]
For the eyepiece, we know that the least distance of distinct vision \( D \) is 25 cm, and we will consider the image distance \( V_e \) to be -25 cm (since it is virtual).

Substituting the values:
\[ \frac{1}{20} = \frac{1}{-25} - \frac{1}{U_e} \]
Rearranging gives:
\[ \frac{1}{U_e} = \frac{1}{-25} - \frac{1}{20} \]
Calculating the right-hand side:
\[ \frac{1}{U_e} = -\frac{4}{100} - \frac{5}{100} = -\frac{9}{100} \implies U_e = -\frac{100}{9} \approx -11.11 \text{ cm} \]

### Step 3: Calculate the Image Distance for the Objective Lens
The total separation between the lenses is given as 30 cm. Hence, the image distance \( V_o \) for the objective lens can be calculated as:
\[ V_o = 30 - U_e = 30 - (-11.11) = 30 + 11.11 = 41.11 \text{ cm} \]

### Step 4: Use the Lens Formula for the Objective Lens
Now, we apply the lens formula for the objective lens:
\[ \frac{1}{F_o} = \frac{1}{V_o} - \frac{1}{U_o} \]
Substituting the known values:
\[ \frac{1}{4} = \frac{1}{41.11} - \frac{1}{U_o} \]
Rearranging gives:
\[ \frac{1}{U_o} = \frac{1}{41.11} - \frac{1}{4} \]
Calculating the right-hand side:
\[ \frac{1}{U_o} = \frac{1}{41.11} - \frac{10.25}{41.11} = -\frac{9.25}{41.11} \implies U_o \approx -4.43 \text{ cm} \]

### Step 5: Calculate the Magnifying Power
The magnifying power \( M \) of the compound microscope is given by:
\[ M = \frac{V_o}{U_o} \left(1 + \frac{D}{F_e}\right) \]
Substituting the values:
\[ M = \frac{41.11}{-4.43} \left(1 + \frac{25}{20}\right) \]
Calculating:
\[ M = \frac{41.11}{-4.43} \left(1 + 1.25\right) = \frac{41.11}{-4.43} \times 2.25 \]
Calculating the magnification:
\[ M \approx -8.376 \]

### Final Answer
The maximum magnifying power of the compound microscope is approximately **-8.376**.