The focal lengths of the eye piece and objective of a compound microscope are `5 cm and 1 cm` respectively, and the length of the tube is `20 cm`. Calculate magnifying power of microscope when the final image is formed at inifinity. The least distance of distinct cision is `25 cm`.

Explanation

To calculate the magnifying power of a compound microscope when the final image is formed at infinity, we can follow these steps:

### Step 1: Identify the given values
- Focal length of the objective lens (F_o) = 1 cm
- Focal length of the eyepiece lens (F_e) = 5 cm
- Length of the tube (L) = 20 cm
- Least distance of distinct vision (D) = 25 cm

### Step 2: Determine the image distance for the objective lens (V_o)
Since the final image is formed at infinity, we can use the relationship between the length of the tube, the focal length of the eyepiece, and the image distance of the objective lens:

\[ V_o = L - F_e \]

Substituting the values:

\[ V_o = 20 \, \text{cm} - 5 \, \text{cm} = 15 \, \text{cm} \]

### Step 3: Use the lens formula to find the object distance for the objective lens (U_o)
The lens formula is given by:

\[ \frac{1}{F_o} = \frac{1}{V_o} - \frac{1}{U_o} \]

Rearranging gives:

\[ \frac{1}{U_o} = \frac{1}{V_o} - \frac{1}{F_o} \]

Substituting the known values:

\[ \frac{1}{U_o} = \frac{1}{15} - \frac{1}{1} \]

Calculating the right-hand side:

\[ \frac{1}{U_o} = \frac{1 - 15}{15} = \frac{-14}{15} \]

Thus,

\[ U_o = -\frac{15}{14} \, \text{cm} \]

### Step 4: Calculate the magnifying power (M)
The magnifying power of the microscope when the final image is at infinity is given by:

\[ M = \frac{V_o}{U_o} \]

Substituting the values we found:

\[ M = \frac{15}{-\frac{15}{14}} = 15 \times \frac{14}{15} = 14 \]

### Step 5: Calculate the overall magnifying power using the least distance of distinct vision
Since the final image is at infinity, the magnifying power can also be expressed as:

\[ M = \frac{D}{F_e} \]

Substituting the values:

\[ M = \frac{25}{5} = 5 \]

### Final Calculation of Magnifying Power
Now, we combine the two results for the overall magnifying power:

\[ M = M_{objective} \times M_{eyepiece} = 14 \times 5 = 70 \]

Thus, the magnifying power of the compound microscope is **70**.