If `x_(1)` is the size of the magnified image and `x_(2)` is the size of the diminished image in lens displacement method, then the size of the object is

Options

A. `sqrt(x_(1)x_(2))`
B. `x_(1)x_(2)`
C. `x_(1)^(2)x_(2)`
D. `x_(1)x_(2)^(2)`

Answer

To find the size of the object using the lens displacement method, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have two images formed by a convex lens: one is magnified (size \( x_1 \)) and the other is diminished (size \( x_2 \)).
- The distance between the object and the image is greater than \( 4f \), where \( f \) is the focal length of the lens.

2. **Defining Distances**:
- Let \( d \) be the distance between the two positions of the lens.
- Let \( x \) be the distance from the lens to the object in the first position.
- The image distance for the first position is \( d - x \) and for the second position it is \( d + x \).

3. **Using Magnification**:
- For the first position (magnified image):
\[ \text{Magnification} (m_1) = \frac{x_1}{h_o} = \frac{d - x}{x} \]
Rearranging gives:
\[ h_o = \frac{x \cdot x_1}{d - x} \quad \text{(Equation 1)} \]

- For the second position (diminished image):
\[ \text{Magnification} (m_2) = \frac{x_2}{h_o} = \frac{d + x}{d - (d + x)} = \frac{d + x}{-x} \]
Rearranging gives:
\[ h_o = \frac{-x \cdot x_2}{d + x} \quad \text{(Equation 2)} \]

4. **Equating the Two Expressions for \( h_o \)**:
- From Equation 1 and Equation 2, we can set them equal to each other:
\[ \frac{x \cdot x_1}{d - x} = \frac{-x \cdot x_2}{d + x} \]

5. **Cross Multiplying**:
- Cross multiplying gives:
\[ x \cdot x_1 (d + x) = -x \cdot x_2 (d - x) \]

6. **Simplifying**:
- Cancel \( x \) from both sides (assuming \( x \neq 0 \)):
\[ x_1 (d + x) = -x_2 (d - x) \]
- Rearranging gives:
\[ x_1 d + x_1 x = -x_2 d + x_2 x \]
- Collecting like terms yields:
\[ (x_1 + x_2)x = - (x_2 - x_1)d \]

7. **Finding the Size of the Object**:
- Solving for \( h_o \):
\[ h_o = \sqrt{x_1 x_2} \]

### Final Result:
The size of the object \( h_o \) is given by:
\[ h_o = \sqrt{x_1 x_2} \]