If `I_(1)` and `I_(2)` be the size of the images respectively for the two positions of lens in the displacement method, then the size of the object is given by

Options

A. `I_(1)//I_(2)`
B. `I_(1)xxI_(2)`
C. `sqrt(I_(1)xxI_(2))`
D. `sqrt(I_(1)//I_(2))`

Answer

To find the size of the object (O) in the displacement method using the sizes of the images (I1 and I2), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Magnification**:
We know that magnification (m) is defined as the ratio of the size of the image (I) to the size of the object (O). For the first position of the lens, we can write:
\[ m_1 = \frac{I_1}{O} \]
where \(I_1\) is the size of the first image.

2. **Using the Second Position**:
For the second position of the lens, the magnification can be expressed as:
\[ m_2 = \frac{I_2}{O} \]
where \(I_2\) is the size of the second image.

3. **Relating the Two Positions**:
In the displacement method, the image formed by the first position acts as an object for the second position. Therefore, we can relate the two magnifications:
\[ \frac{I_1}{O} = \frac{O}{I_2} \]

4. **Cross Multiplying**:
Cross-multiplying the equation gives:
\[ I_1 \cdot I_2 = O^2 \]

5. **Solving for the Object Size**:
To find the size of the object (O), we take the square root of both sides:
\[ O = \sqrt{I_1 \cdot I_2} \]

Thus, the size of the object is given by:
\[ O = \sqrt{I_1 \cdot I_2} \]

### Final Answer:
The size of the object is \(O = \sqrt{I_1 \cdot I_2}\).

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