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. Height of final image is `(x)/(2)`
B. Height of final image is `(x)/(3)`
C. Image formed by converging lens is real and inverted
D. Image formed by converging mirror is real

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 heights and their nature as the light passes through each optical element.

### Step 1: Analyze the Diverging Lens
- **Given**: Focal length \( f_1 = -40 \) cm (negative for diverging lens).
- **Object Distance**: \( u_1 = -40 \) cm (the object is placed at the focal point).

Using the lens formula:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
We rearrange it to find \( v_1 \):
\[ \frac{1}{v_1} = \frac{1}{f_1} + \frac{1}{u_1} \]
Substituting the values:
\[ \frac{1}{v_1} = \frac{1}{-40} + \frac{1}{-40} = -\frac{1}{20} \]
Thus,
\[ v_1 = -20 \text{ cm} \]
- **Nature of Image**: Virtual (since \( v_1 \) is negative).
- **Magnification**:
\[ M_1 = \frac{v_1}{u_1} = \frac{-20}{-40} = \frac{1}{2} \]

### Step 2: Analyze the Converging Lens
- **Given**: Focal length \( f_2 = 40 \) cm.
- **Object Distance**: The image from the diverging lens acts as the object for the converging lens. The distance from the diverging lens to the converging lens is 40 cm, so:
\[ u_2 = -20 + 40 = -60 \text{ cm} \]

Using the lens formula:
\[ \frac{1}{v_2} = \frac{1}{f_2} + \frac{1}{u_2} \]
Substituting the values:
\[ \frac{1}{v_2} = \frac{1}{40} + \frac{1}{-60} \]
Finding a common denominator:
\[ \frac{1}{v_2} = \frac{3 - 2}{120} = \frac{1}{120} \]
Thus,
\[ v_2 = 120 \text{ cm} \]
- **Nature of Image**: Real (since \( v_2 \) is positive).
- **Magnification**:
\[ M_2 = \frac{v_2}{u_2} = \frac{120}{-60} = -2 \]

### Step 3: Analyze the Converging Mirror
- **Given**: Focal length \( f_3 = -80 \) cm (negative for a converging mirror).
- **Object Distance**: The image from the converging lens acts as the object for the mirror. The distance from the converging lens to the mirror is 40 cm, so:
\[ u_3 = 120 - 40 = 80 \text{ cm} \]

Using the mirror formula:
\[ \frac{1}{f_3} = \frac{1}{v_3} + \frac{1}{u_3} \]
Rearranging gives:
\[ \frac{1}{v_3} = \frac{1}{f_3} - \frac{1}{u_3} \]
Substituting the values:
\[ \frac{1}{v_3} = \frac{1}{-80} - \frac{1}{80} = -\frac{2}{80} = -\frac{1}{40} \]
Thus,
\[ v_3 = -40 \text{ cm} \]
- **Nature of Image**: Virtual (since \( v_3 \) is negative).
- **Magnification**:
\[ M_3 = \frac{v_3}{u_3} = \frac{-40}{80} = -\frac{1}{2} \]

### Step 4: Calculate Total Magnification
The total magnification \( M \) is given by:
\[ M = M_1 \times M_2 \times M_3 \]
Substituting the values:
\[ M = \frac{1}{2} \times (-2) \times \left(-\frac{1}{2}\right) = \frac{1}{2} \]

### Step 5: Final Image Height
If the height of the object is \( h_0 = x \):
\[ h = M \times h_0 = \frac{1}{2} \times x = \frac{x}{2} \]

### Conclusion
The height of the final image is \( \frac{x}{2} \).