1. Set up the lens condition using the mirror

The object is placed $12\,\text{cm}$ to the left of the convex lens. A convex mirror is placed $8\,\text{cm}$ to the right of the lens. The final image after reflection and second refraction coincides with the object.

This means that after reflection from the mirror, the rays must retrace such that for the lens on the return path, the object for the lens is at the same position as the first image formed by the lens. A standard way to use this condition is:

• Let the lens alone form its first image at distance $v$ to the right of the lens.
• This image acts as object for the convex mirror.
• Since the final image comes back to coincide with the original object, the image formed by the mirror must be at the same point as the first image itself, so that after passing back through the lens the rays retrace their path.

Thus, for the mirror, the object and image are at the same point.

2. Apply mirror formula

Take the convex mirror at $8\,\text{cm}$ to the right of the lens. If the first image by lens is at distance $v$ to the right of the lens, then its distance from mirror is $u_m = v-8$ (with sign convention handled via magnitude relation).

For a convex mirror, if object and image coincide, that point must be at the center of curvature in the mirror geometry extension sense, giving distance from mirror equal to radius of curvature: $R = 2f = 30\,\text{cm}$ So the point must be $30\,\text{cm}$ from the mirror.

Hence, $v-8 = 30$ $v = 38\,\text{cm}$

So the lens alone forms image at $38\,\text{cm}$ to the right of the lens.

3. Find focal length of the lens

Using lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ For the object at $12\,\text{cm}$ to the left, $u=-12\,\text{cm}, \qquad v=+38\,\text{cm}$ Therefore, $\frac{1}{f} = \frac{1}{38} - \left(-\frac{1}{12}\right) = \frac{1}{38}+\frac{1}{12}$ $\frac{1}{f} = \frac{12+38}{456} = \frac{50}{456} = \frac{25}{228}$ $f = \frac{228}{25} = 9.12\,\text{cm}$

4. Now remove the mirror

With only the convex lens remaining, object distance is still $u=-12\,\text{cm}$ and focal length is $f=\frac{228}{25}\,\text{cm}$

Again use lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{v}+\frac{1}{12}$ So, $\frac{1}{v} = \frac{1}{f} - \frac{1}{12} = \frac{25}{228} - \frac{1}{12}$ $\frac{1}{v} = \frac{25}{228} - \frac{19}{228} = \frac{6}{228} = \frac{1}{38}$ Thus, $v = 38\,\text{cm}$

5. Distance between object and image

The object is $12\,\text{cm}$ to the left of the lens and the image is $38\,\text{cm}$ to the right of the lens. Hence separation is $12+38 = 50\,\text{cm}$

Final Answer

$\boxed{50}$