The focal lengths of the objective and the eyepiece of a compound microscope are 2.0cm and 3.0cm, respectively. The distance between the objective and the eyepiece is 15.0cm. Th final image formed by the eyepiece is at infinity. The two lenses are thin. The distance, in cm, of the object and the image produced by the objective, mesured 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 distances of the object and the image produced by the objective lens of a compound microscope. We are given the following data:

- Focal length of the objective lens (Fo) = 2.0 cm
- Focal length of the eyepiece lens (Fe) = 3.0 cm
- Distance between the objective and the eyepiece (d) = 15.0 cm
- The final image formed by the eyepiece is at infinity.

### Step 1: Find the image distance from the eyepiece (Ve)
Since the final image is at infinity, we can use the lens formula for the eyepiece:

\[ \frac{1}{Ve} - \frac{1}{Ue} = \frac{1}{Fe} \]

Given that \(Ve = \infty\), we can substitute this into the equation:

\[ \frac{1}{\infty} - \frac{1}{Ue} = \frac{1}{3} \]

This simplifies to:

\[ 0 - \frac{1}{Ue} = \frac{1}{3} \]

Thus, we find:

\[ \frac{1}{Ue} = -\frac{1}{3} \implies Ue = -3 \text{ cm} \]

This means the object distance for the eyepiece is 3 cm (to the left of the eyepiece).

### Step 2: Find the image distance from the objective (Vo)
The distance between the objective and the eyepiece is given as 15 cm. Therefore, the image distance from the objective lens (Vo) can be calculated as:

\[ Vo = d - |Ue| = 15 \text{ cm} - 3 \text{ cm} = 12 \text{ cm} \]

This means the image produced by the objective lens is 12 cm to the right of the objective lens.

### Step 3: Find the object distance from the objective (Uo)
Now, we will use the lens formula for the objective lens:

\[ \frac{1}{Vo} - \frac{1}{Uo} = \frac{1}{Fo} \]

Substituting the known values:

\[ \frac{1}{12} - \frac{1}{Uo} = \frac{1}{2} \]

Rearranging gives:

\[ \frac{1}{Uo} = \frac{1}{12} - \frac{1}{2} \]

Finding a common denominator (which is 12):

\[ \frac{1}{Uo} = \frac{1}{12} - \frac{6}{12} = -\frac{5}{12} \]

Thus:

\[ Uo = -\frac{12}{5} = -2.4 \text{ cm} \]

This means the object is located 2.4 cm to the left of the objective lens.

### Final Results
- Distance of the object from the objective lens: \(Uo = -2.4 \text{ cm}\)
- Distance of the image from the objective lens: \(Vo = 12 \text{ cm}\)

### Summary
The distances measured from the objective lens are:
- Object distance: 2.4 cm (to the left, hence negative)
- Image distance: 12 cm (to the right)