The following objects are placed one after each other in given order onto a central axis with a separation of 40 cm each. A point source of light O. a diverging lens of focal length 40 cm. a converging lens of focal length 40 cm and converging mirror of focal length 80 cm. The aperture diameter of lenses and mirror is d=20 cm. If a point source of light is placed at a perpendicular distance of x from central axis then. (You have to consider single optical event at any optical element)
Mark the CORRECT statement(s):-

Options

A. Final image is formed in the plane of converging lens
B. If `x gt 1.5d` the rays will not be reflected by converging mirror
C. For `x gt d` final real images can be captured on a screen after on optical event (i.e refraction or reflection)at every optical system
D. Final image is real and inverted

Explanation

To solve the problem step by step, we will analyze the optical system consisting of a point source of light, a diverging lens, a converging lens, and a converging mirror. We will calculate the image positions and magnifications for each optical element and determine the validity of the statements provided.

### Step-by-Step Solution:

1. **Identify the Optical Elements and Their Positions:**
- Point source of light (O) at position 0 cm.
- Diverging lens (focal length = -40 cm) at position 40 cm.
- Converging lens (focal length = 40 cm) at position 80 cm.
- Converging mirror (focal length = 80 cm) at position 120 cm.

2. **Calculate Image Formed by the Diverging Lens:**
- Object distance (u) for the diverging lens:
\[ u = -40 \text{ cm} \quad (\text{negative as per sign convention}) \]
- Focal length (f) of the diverging lens:
\[ f = -40 \text{ cm} \]
- Using the lens formula:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Rearranging gives:
\[ \frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{-40} + \frac{1}{-40} = \frac{-2}{40} = -\frac{1}{20} \]
Thus,
\[ v = -20 \text{ cm} \]
- The image formed by the diverging lens is at -20 cm (virtual image).

3. **Calculate Image Formed by the Converging Lens:**
- Object distance (u) for the converging lens:
\[ u = -20 \text{ cm} + 40 \text{ cm} = 60 \text{ cm} \quad (\text{distance from the lens to the image of the diverging lens}) \]
- Focal length (f) of the converging lens:
\[ f = 40 \text{ cm} \]
- Using the lens formula again:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Rearranging gives:
\[ \frac{1}{v} = \frac{1}{40} + \frac{1}{-60} = \frac{3 - 2}{120} = \frac{1}{120} \]
Thus,
\[ v = 120 \text{ cm} \]
- The image formed by the converging lens is at 120 cm (real image).

4. **Calculate Image Formed by the Converging Mirror:**
- Object distance (u) for the converging mirror:
\[ u = 120 \text{ cm} - 40 \text{ cm} = 80 \text{ cm} \]
- Focal length (f) of the converging mirror:
\[ f = 80 \text{ cm} \]
- Using the mirror formula:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Rearranging gives:
\[ \frac{1}{v} = \frac{1}{80} - \frac{1}{80} = 0 \]
Thus,
\[ v = -40 \text{ cm} \]
- The final image formed by the converging mirror is at -40 cm (real image).

5. **Determine the Height of the Final Image:**
- The magnification for the diverging lens:
\[ m_1 = \frac{v}{u} = \frac{-20}{-40} = \frac{1}{2} \]
- The magnification for the converging lens:
\[ m_2 = \frac{v}{u} = \frac{120}{60} = 2 \]
- The total magnification (M) is:
\[ M = m_1 \times m_2 = \frac{1}{2} \times 2 = 1 \]
- Therefore, the height of the final image is equal to the height of the object (x).

6. **Check the Conditions for the Statements:**
- **Statement A:** The final image is real and inverted, which is correct.
- **Statement B:** If \( x > 1.5d \) (where \( d = 20 \text{ cm} \)), the rays will not be reflected by the converging mirror. This is correct since \( 1.5d = 30 \text{ cm} \).
- **Statement C:** If \( x > d \), the final real images can be captured on a screen after every optical event. This is incorrect because the diverging lens produces a virtual image.

### Conclusion:
The correct statements are A and B.