A microscope consists of two convex lenses of focal lengths 2.0 cm and 6.25 cm placed 15.0 cm apart. Where must the object be placed so that the final virtual image is at a distance of 25 cm from the eye ?

Options

A. 2.5 cm
B. 2 cm
C. 1.5 cm
D. 3 cm

Answer

To solve the problem step by step, we will use the lens formula and the information provided about the microscope setup.

### Step 1: Identify the lenses and their focal lengths
- The microscope consists of two convex lenses:
- Objective lens (focal length \( f_o = 2.0 \, \text{cm} \))
- Eyepiece lens (focal length \( f_e = 6.25 \, \text{cm} \))
- The distance between the two lenses is \( d = 15.0 \, \text{cm} \).

### Step 2: Define the distances
- Let \( u_o \) be the object distance from the objective lens.
- Let \( v_o \) be the image distance from the objective lens.
- The image formed by the objective lens will act as the object for the eyepiece lens.
- Let \( u_e \) be the object distance from the eyepiece lens.
- Let \( v_e \) be the image distance from the eyepiece lens, which is given as \( v_e = -25 \, \text{cm} \) (negative because it is a virtual image).

### Step 3: Apply the lens formula for the eyepiece
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
For the eyepiece lens:
\[ \frac{1}{f_e} = \frac{1}{v_e} - \frac{1}{u_e} \]
Substituting the known values:
\[ \frac{1}{6.25} = \frac{1}{-25} - \frac{1}{u_e} \]
Rearranging gives:
\[ \frac{1}{u_e} = \frac{1}{-25} - \frac{1}{6.25} \]
Calculating \( \frac{1}{6.25} \):
\[ \frac{1}{6.25} = 0.16 \quad \text{(approximately)} \]
Thus:
\[ \frac{1}{u_e} = -0.04 - 0.16 = -0.20 \]
So:
\[ u_e = -5 \, \text{cm} \]

### Step 4: Relate \( u_e \) and \( v_o \)
The distance between the two lenses is \( d = 15 \, \text{cm} \):
\[ v_o = d - u_e = 15 - (-5) = 15 + 5 = 20 \, \text{cm} \]

### Step 5: Apply 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:
\[ \frac{1}{2} = \frac{1}{20} - \frac{1}{u_o} \]
Rearranging gives:
\[ \frac{1}{u_o} = \frac{1}{20} - \frac{1}{2} \]
Calculating \( \frac{1}{2} \):
\[ \frac{1}{2} = 0.5 \]
Thus:
\[ \frac{1}{u_o} = 0.05 - 0.5 = -0.45 \]
So:
\[ u_o = -\frac{1}{0.45} \approx -2.22 \, \text{cm} \]

### Step 6: Final calculation
Since the object distance \( u_o \) is negative, it indicates that the object must be placed at a distance of approximately \( 2.5 \, \text{cm} \) from the objective lens.

### Conclusion
The object must be placed approximately \( 2.5 \, \text{cm} \) from the objective lens.

---