The distance between object and the screen is D. Real images of an object are formed on the screen for two positions of a lens separated by a distance d. The ratio between the sizes of two images will be:

Options

A. `D//d`
B. `D^(2)//d^(2)`
C. `(D-d)^(2)//(D+d)^(2)`
D. `sqrt(D//d)`

Answer

To solve the problem, we need to find the ratio of the sizes of two real images formed by a lens at two different positions, given that the distance between the object and the screen is \( D \) and the two positions of the lens are separated by a distance \( d \).

### Step-by-Step Solution:

1. **Understanding the Setup**:
- Let the object distance from the lens in the first position be \( u_1 = x \).
- The image distance from the lens in the first position will be \( v_1 \).
- The second position of the lens will be at \( u_2 = x + d \).
- The image distance from the lens in the second position will be \( v_2 \).

2. **Using the Lens Formula**:
- The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
- For the first position:
\[ \frac{1}{f} = \frac{1}{v_1} - \frac{1}{x} \]
Rearranging gives:
\[ \frac{1}{v_1} = \frac{1}{f} + \frac{1}{x} \]
- For the second position:
\[ \frac{1}{f} = \frac{1}{v_2} - \frac{1}{(x + d)} \]
Rearranging gives:
\[ \frac{1}{v_2} = \frac{1}{f} + \frac{1}{(x + d)} \]

3. **Equating the Focal Length**:
- Since the focal length \( f \) is the same for both positions, we can equate the two expressions for \( \frac{1}{f} \):
\[ \frac{1}{v_1} + \frac{1}{x} = \frac{1}{v_2} + \frac{1}{(x + d)} \]

4. **Finding the Image Distances**:
- From the lens formula, we can express \( v_1 \) and \( v_2 \):
\[ v_1 = \frac{xf}{x - f} \]
\[ v_2 = \frac{(x + d)f}{(x + d) - f} \]

5. **Finding the Magnification**:
- The magnification \( M \) for each case is given by:
\[ M_1 = \frac{h_1}{h_0} = \frac{v_1}{u_1} = \frac{v_1}{x} \]
\[ M_2 = \frac{h_2}{h_0} = \frac{v_2}{u_2} = \frac{v_2}{(x + d)} \]

6. **Finding the Ratio of Sizes**:
- The ratio of the sizes of the two images is:
\[ \frac{h_1}{h_2} = \frac{M_1}{M_2} = \frac{v_1/x}{v_2/(x + d)} = \frac{v_1(x + d)}{v_2x} \]

7. **Final Expression**:
- After substituting the values of \( v_1 \) and \( v_2 \) and simplifying, we find:
\[ \frac{h_1}{h_2} = \frac{(D - x)(x + d)}{(D - (x + d))x} \]
- This can be further simplified based on the given values of \( D \) and \( d \).

### Conclusion:
The final ratio of the sizes of the two images formed by the lens at different positions is:
\[ \frac{h_1}{h_2} = \frac{(D - x)(x + d)}{(D - (x + d))x} \]