The focal length of the objective and the eye piece of a compound microscope are 2.0 cm and 3.0 cm, respectively. The distance between the objective and the eye piece is 15.0 cm. The final image formed by the eye piece is at infinity. The two lenses are thin. The distance in cm of the object and the image produced by the objective, measured from the objective lens, are respectively

Options

A. 2.4 and 12.0
B. 2.4 and 15.0
C. 2.0 and 12.0
D. 2.0 and 3.0

Explanation

To solve the problem, we need to find the object distance (u) and the image distance (v) for the objective lens of the compound microscope. We will use the lens formula and the information provided.

### Step 1: Understand the given data
- Focal length of the objective lens (f_o) = 2.0 cm
- Focal length of the eyepiece lens (f_e) = 3.0 cm
- Distance between the objective and eyepiece (d) = 15.0 cm
- The final image formed by the eyepiece is at infinity.

### 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, since the final image is at infinity, we have:
- v_e = ∞ (image distance for eyepiece)
- u_e = distance of the image formed by the objective lens from the eyepiece.

Using the lens formula for the eyepiece:
\[ \frac{1}{f_e} = \frac{1}{v_e} - \frac{1}{u_e} \]
Since \( v_e = \infty \), \( \frac{1}{v_e} = 0 \):
\[ \frac{1}{f_e} = -\frac{1}{u_e} \]
This gives:
\[ u_e = -f_e \]
Substituting the value of f_e:
\[ u_e = -3.0 \text{ cm} \]

### Step 3: Calculate the image distance from the objective lens
The distance between the objective and eyepiece is given as 15 cm. Therefore, the image distance (v_o) from the objective lens can be calculated as:
\[ v_o = d + u_e \]
Substituting the known values:
\[ v_o = 15.0 \text{ cm} - 3.0 \text{ cm} = 12.0 \text{ cm} \]

### Step 4: Use the lens formula for the objective lens
Now we can use the lens formula for the objective lens:
\[ \frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u_o} \]
Rearranging gives:
\[ \frac{1}{u_o} = \frac{1}{v_o} - \frac{1}{f_o} \]
Substituting the known values:
- v_o = 12.0 cm
- f_o = 2.0 cm

Calculating:
\[ \frac{1}{u_o} = \frac{1}{12.0} - \frac{1}{2.0} \]
\[ \frac{1}{u_o} = \frac{1}{12} - \frac{6}{12} = -\frac{5}{12} \]
Thus:
\[ u_o = -\frac{12}{5} = -2.4 \text{ cm} \]

### Step 5: Conclusion
The object distance (u_o) is -2.4 cm (the negative sign indicates that the object is on the same side as the incoming light). The image distance (v_o) is 12.0 cm.

Therefore, the distances of the object and the image produced by the objective, measured from the objective lens, are:
- Object distance (u_o) = 2.4 cm (measured from the objective lens)
- Image distance (v_o) = 12.0 cm (measured from the objective lens)

### Final Answer:
The distances of the object and the image produced by the objective, measured from the objective lens, are respectively:
- Object distance: 2.4 cm
- Image distance: 12.0 cm