Mirror placed horizontally between two half convex lenses

[Null_Void]

Homework Statement

See pic

Relevant Equations

None

So I was able to find out quite easily the image distance after the first refraction and the corresponding magnification.

Employing Cartesian convention
$1/v - 1/u = 1/f$

Substituting the values gives us:
$v_1 = +60cm$
$m_1 = -3$

Thus the height of the image below the principal axis must be $6mm$

I was also able to find the horizontal position of the final image.
Now the problem arises when calculating the final image height for the refraction by the second half-lens.

The distance between the plane mirror and the principal axis is $4cm$ while the image is formed $6cm$ below the principal axis (height). I can see that $2cm$ of the image is formed as a virtual object and here's where I'm stuck.

1) Can the real image formed by the first lens on the plane mirror give rise to a virtual image of the same size?
This is my first time encountering a scenario like this and I'm a bit skeptical

2) how do I establish a relationship between the virtual object and the corresponding final image formed? I'm confused as there's a lot of stuff going on.

3) Will the virtual image formed by the plane mirror of the first real image, act as an object for the second lens? I feel that it shouldn't as it isn't a virtual object but the book solution seems to assume it does.

The height of the final image is given as 8/3mm

 

[kuruman]

I think it will be instructive for you to do part (b) first in order to get your bearings.

First make a ray diagram as if the mirror is not there. Try drawing it to scale as best as possible. As you have already found, you will get a real, inverted image at 60 cm from the first half-lens. Now put the mirror in place. Does it look like the image is still formed at 60 cm or does it look like the converging rays that would have formed the image at 60 cm diverge upon reflection from the mirror?

 

[haruspex]

It is very unclear what is meant by the "size" of the final image. The shown object consists of a 2mm long stick. As you found, only the top $\frac 43$mm form an image below the mirror, which folds that to be above the mirror. So the final image is of a folded stick. What is meant by its size?