A point object is placed at a distance of 60 cm from a convex lens of focal length 30 cm. If a plane mirror were put perpendicular to the principal axis of the lens and at a distance of 40 cm from it, the final image would be formed at a distance of :

The correct option is 20 cm from the plane mirror, it would be a virtual image.

Let us analyze the optical system step-by-step by tracing the light rays:

Step 1: First refraction through the convex lens (light traveling from left to right)
The object is placed at a distance of 60 cm to the left of the lens. According to the sign convention:
Object distance, $u_1=-60cm$
Focal length of the convex lens, $f=+30cm$
Using the lens formula:
$\frac{1}{v_1}-\frac{1}{u_1}=\frac{1}{f}$
Substitute the values into the formula:
$\frac{1}{v_1}-\frac{1}{-60}=\frac{1}{30}$
$\frac{1}{v_1}+\frac{1}{60}=\frac{1}{30}$
$\frac{1}{v_1}=\frac{1}{30}-\frac{1}{60}=\frac{1}{60}$
$v_1=+60cm$
This means the first image $I_1$ is formed at a distance of 60 cm to the right of the lens.

Step 2: Reflection from the plane mirror
The plane mirror is placed at a distance of 40 cm to the right of the lens.
Since the image $I_1$ is located 60 cm to the right of the lens, it lies behind the plane mirror.
Distance of $I_1$ from the plane mirror $=60cm-40cm=20cm$.
The image $I_1$ acts as a virtual object for the plane mirror. The plane mirror forms a real image $I_2$ at an equal distance of 20 cm in front of it (to the left of the mirror).
The position of this reflected image $I_2$ from the lens is:
$40cm-20cm=20cm$ (to the right of the lens).

Step 3: Second refraction through the convex lens (light traveling from right to left)
After reflection from the mirror, the rays travel from right to left back towards the lens.
Taking the leftward direction of light propagation as positive:
The source of light $I_2$ is 20 cm to the right of the lens, which is in the direction opposite to the light propagation.
Object distance, $u_2=-20cm$
Focal length of the lens, $f=+30cm$
Using the lens formula again:
$\frac{1}{v_2}-\frac{1}{u_2}=\frac{1}{f}$
Substitute the values:
$\frac{1}{v_2}-\frac{1}{-20}=\frac{1}{30}$
$\frac{1}{v_2}+\frac{1}{20}=\frac{1}{30}$
$\frac{1}{v_2}=\frac{1}{30}-\frac{1}{20}=\frac{2-3}{60}=-\frac{1}{60}$
$v_2=-60cm$
Since $v_2$ is negative, the final image is formed 60 cm to the right of the lens (opposite to the direction of light propagation), making it a virtual image.

Conclusion:
The distance of this final virtual image from the plane mirror is:
$60cm-40cm=20cm$ (behind the mirror).
Thus, the final image is formed at a distance of 20 cm from the plane mirror, and it is a virtual image.