A thin converging lens is placed between as fixed object and a screen. There are two positions of the lens for which a sharp image is formed on the screen. The height of one of the image is 2 cm while the magnification of the other image is 3. What is the height of the object?

Options

A. 6 cm
B. 9 cm
C. 10 cm
D. 4 cm

Answer

To solve the problem, we need to find the height of the object given the height of one image and the magnification of the other image formed by a thin converging lens.

### Step-by-Step Solution:

1. **Understanding the Problem:**
- We have a thin converging lens placed between a fixed object and a screen.
- There are two positions of the lens where a sharp image is formed on the screen.
- The height of one image (let's call it \( h' \)) is 2 cm.
- The magnification of the other image (let's call it \( m_2 \)) is 3.

2. **Using the Magnification Formula:**
- The magnification \( m \) of a lens is given by the formula:
\[ m = \frac{h'}{h} \]
where \( h' \) is the height of the image and \( h \) is the height of the object.

3. **Setting Up the Equations:**
- For the first image:
\[ m_1 = \frac{h'}{h} \]
- For the second image, we know \( m_2 = 3 \):
\[ m_2 = \frac{h''}{h} \quad \text{(where \( h'' \) is the height of the second image)} \]

4. **Relating the Two Magnifications:**
- The magnifications from the two positions of the lens are related by:
\[ m_1 \cdot m_2 = 1 \]
- Therefore, we can express \( m_1 \) in terms of \( m_2 \):
\[ m_1 = \frac{1}{m_2} = \frac{1}{3} \]

5. **Substituting Known Values:**
- We know \( h' = 2 \) cm, so we can substitute this into the equation for \( m_1 \):
\[ \frac{h'}{h} = \frac{1}{3} \]
- Substituting \( h' \):
\[ \frac{2}{h} = \frac{1}{3} \]

6. **Solving for the Height of the Object:**
- Cross-multiplying gives:
\[ 2 \cdot 3 = h \implies h = 6 \text{ cm} \]

### Final Answer:
The height of the object is \( 6 \) cm.