The focal lengths of obj ective and eyepiece of a compound microscope are 5 cm, 6.25 cm respectively. When an object is placed infyrnt of the objective at a distance of 6.25 cm, the final image is formed at least distance of distinct vision. The length of microscope is

Options

A. 22.5cm
B. 25cm
C. 30cm
D. 31.25cm

Explanation

To find the length of the compound microscope given the focal lengths of the objective and eyepiece, we can follow these steps:

### Step 1: Identify the given data
- Focal length of the objective (Fo) = 5 cm
- Focal length of the eyepiece (Fe) = 6.25 cm
- Object distance from the objective (Uo) = -6.25 cm (negative as per sign convention)

### Step 2: Use the lens formula for the objective
The lens formula is given by:
\[ \frac{1}{V} - \frac{1}{U} = \frac{1}{F} \]
For the objective lens, we can write:
\[ \frac{1}{Vo} - \frac{1}{Uo} = \frac{1}{Fo} \]
Substituting the known values:
\[ \frac{1}{Vo} - \frac{1}{-6.25} = \frac{1}{5} \]

### Step 3: Solve for Vo
Rearranging the equation gives:
\[ \frac{1}{Vo} = \frac{1}{5} - \frac{1}{-6.25} \]
Calculating the right side:
\[ \frac{1}{Vo} = \frac{1}{5} + \frac{1}{6.25} \]
Finding a common denominator (which is 31.25):
\[ \frac{1}{5} = \frac{6.25}{31.25}, \quad \frac{1}{6.25} = \frac{5}{31.25} \]
So,
\[ \frac{1}{Vo} = \frac{6.25 + 5}{31.25} = \frac{11.25}{31.25} \]
Thus,
\[ Vo = \frac{31.25}{11.25} = 2.78 \text{ cm} \]

### Step 4: Calculate the distance of the first image (I1)
Now we know that the distance of the image formed by the objective (I1) is:
\[ Vo = 25 \text{ cm} \]

### Step 5: Use the lens formula for the eyepiece
For the eyepiece, we can write:
\[ \frac{1}{Ve} - \frac{1}{Ue} = \frac{1}{Fe} \]
Where:
- Ve = -25 cm (the final image distance, as it is at the least distance of distinct vision)
- Ue = distance of the first image from the eyepiece (which is the object for the eyepiece)

Rearranging gives:
\[ \frac{1}{Ue} = \frac{1}{Fe} + \frac{1}{Ve} \]
Substituting the known values:
\[ \frac{1}{Ue} = \frac{1}{6.25} + \frac{1}{-25} \]
Finding a common denominator (which is 25):
\[ \frac{1}{6.25} = \frac{4}{25}, \quad \frac{1}{-25} = -\frac{1}{25} \]
Thus,
\[ \frac{1}{Ue} = \frac{4 - 1}{25} = \frac{3}{25} \]
So,
\[ Ue = \frac{25}{3} \approx 8.33 \text{ cm} \]

### Step 6: Calculate the length of the microscope
The length of the microscope (L) is given by:
\[ L = Vo + Ue \]
Substituting the values:
\[ L = 25 + 8.33 = 33.33 \text{ cm} \]

### Final Answer
The length of the microscope is approximately **33.33 cm**.