A compound microscope has an objective of focal length `2.0 cm` and an eye-piece of focal length `6.25cm` and distance between the objective and eye-piece is `15cm`. If the final image is formed at the least distance vision `(25 cm)`, the distance of the object form the objective is

Explanation

To solve the problem of finding the distance of the object from the objective in a compound microscope, we can follow these steps:

### Step 1: Identify the Given Data
- Focal length of the objective lens, \( f_o = 2.0 \, \text{cm} \)
- Focal length of the eyepiece lens, \( f_e = 6.25 \, \text{cm} \)
- Distance between the objective and eyepiece, \( L = 15 \, \text{cm} \)
- Final image distance (least distance of distinct vision), \( v_e = -25 \, \text{cm} \) (negative because it is on the same side as the object)

### Step 2: Use the Lens Formula for the Eyepiece
The lens formula is given by:
\[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \]
For the eyepiece, we can write:
\[ \frac{1}{v_e} - \frac{1}{u_e} = \frac{1}{f_e} \]
Substituting the known values:
\[ \frac{1}{-25} - \frac{1}{u_e} = \frac{1}{6.25} \]

### Step 3: Solve for \( u_e \)
Rearranging the equation gives:
\[ -\frac{1}{u_e} = \frac{1}{6.25} + \frac{1}{25} \]
Calculating the right-hand side:
\[ \frac{1}{6.25} = 0.16 \quad \text{and} \quad \frac{1}{25} = 0.04 \]
Thus,
\[ -\frac{1}{u_e} = 0.16 + 0.04 = 0.20 \]
Taking the reciprocal:
\[ u_e = -5 \, \text{cm} \]

### Step 4: Calculate the Image Distance from the Objective
Using the relationship between the distances:
\[ L = v_o + u_e \]
Where \( v_o \) is the image distance from the objective. Rearranging gives:
\[ v_o = L - u_e = 15 - (-5) = 15 + 5 = 20 \, \text{cm} \]

### Step 5: Use the Lens Formula for the Objective
Now, apply the lens formula for the objective:
\[ \frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o} \]
Substituting the known values:
\[ \frac{1}{20} - \frac{1}{u_o} = \frac{1}{2} \]

### Step 6: Solve for \( u_o \)
Rearranging gives:
\[ -\frac{1}{u_o} = \frac{1}{2} - \frac{1}{20} \]
Finding a common denominator (20):
\[ \frac{1}{2} = \frac{10}{20} \quad \text{and} \quad \frac{1}{20} = \frac{1}{20} \]
Thus,
\[ -\frac{1}{u_o} = \frac{10}{20} - \frac{1}{20} = \frac{9}{20} \]
Taking the reciprocal:
\[ u_o = -\frac{20}{9} \approx -2.22 \, \text{cm} \]

### Conclusion
The distance of the object from the objective is approximately \( 2.22 \, \text{cm} \) (the negative sign indicates the direction of the object).