When the distance between the object and the screen is more than `4 f`. We can obtain image of the object on the screen for the two positions of the lens. If is called displacement method. In one case, the image is magnified and in the other case, it is diminished. If `I_(1)` and `I_(2)` be the sized of the two images, then the size of the object is

Options

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

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the setup
We have a fixed object and a fixed screen, and we can place a lens in two different positions. The distance between the object and the screen is more than \(4f\) (where \(f\) is the focal length of the lens). When the lens is in the first position, it produces a magnified image, and in the second position, it produces a diminished image.

**Hint:** Visualize the arrangement with the object, lens, and screen to understand how the lens positions affect the image formation.

### Step 2: Define magnification for both positions
For the first position of the lens, the magnification \(M_1\) is given by:
\[ M_1 = \frac{I_1}{O} = \frac{V}{u} \]
where \(I_1\) is the size of the image, \(O\) is the size of the object, \(V\) is the image distance, and \(u\) is the object distance.

For the second position of the lens, the magnification \(M_2\) is given by:
\[ M_2 = \frac{I_2}{O} = \frac{U}{v} \]
where \(I_2\) is the size of the second image, \(U\) is the image distance, and \(v\) is the object distance.

**Hint:** Remember that magnification is the ratio of the size of the image to the size of the object.

### Step 3: Relate the two magnifications
From the definitions of magnification, we can express:
\[ M_1 = \frac{I_1}{O} \quad \text{and} \quad M_2 = \frac{I_2}{O} \]
Now, if we multiply the two magnifications, we have:
\[ M_1 \cdot M_2 = \left(\frac{I_1}{O}\right) \cdot \left(\frac{I_2}{O}\right) = \frac{I_1 \cdot I_2}{O^2} \]

**Hint:** This step combines the two magnification equations into one, which will help us find the relationship between the sizes of the images and the object.

### Step 4: Set the product of magnifications equal to 1
From the lens formula and the properties of lenses, we know that:
\[ M_1 \cdot M_2 = 1 \]
Thus, we can write:
\[ \frac{I_1 \cdot I_2}{O^2} = 1 \]

**Hint:** This relationship indicates that the product of the magnifications for the two positions of the lens is equal to one.

### Step 5: Solve for the size of the object
Rearranging the equation gives us:
\[ I_1 \cdot I_2 = O^2 \]
Taking the square root of both sides, we find:
\[ O = \sqrt{I_1 \cdot I_2} \]

**Hint:** This final equation gives us the size of the object in terms of the sizes of the two images formed by the lens.

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