= Solution
A <plane mirror> sends each incident ray into its specularly reflected direction, with equal incidence and reflection <angles> measured from the surface normal. For a point object in front of the mirror, the backward extensions of the reflected rays meet at the reflected point behind the plane. The image is therefore \b[virtual, upright, equal in size and as far behind the mirror as the object is in front]. The actual rays do not travel behind the mirror. In coordinates with mirror plane $x=0$, the object point $(x,y,z)$ is mapped to $(-x,y,z)$.
A converging <convex lens> refracts a paraxial pencil according to the <thin-lens equation>
$$
\boxed{\frac1u+\frac1v=\frac1f,\qquad m=-\frac vu,}
$$
where $u>0$ is the distance to a real object, $f>0$ is the <focal length>, and $v>0$ denotes an image on the opposite side. Two constructed rays locate the image: a ray parallel to the <optical axis> emerges towards the far <optical focus>, and a ray through the ideal thin lens's optical centre proceeds undeviated. Their intersection is a real image; if they diverge, their backward extensions locate a virtual one.
For $u>2f$ the real inverted image is reduced; at $u=2f$ it is equal in size; for $f<u<2f$ it is enlarged. At $u=f$ the emergent pencil is parallel and the image is at infinity in the ideal model. For $0<u<f$, the image is virtual, upright and enlarged on the object's side. A distant object gives an image near the far focal plane. The <lensmaker's equation> explains why shape alone does not guarantee convergence: the surrounding medium and <refractive index> also matter. <Spherical aberration>, <chromatic aberration> and <diffraction> limit the ideal point-image construction. The statement that the centre ray is undeviated is the thin-lens approximation, rather than a general rule for a thick compound optical instrument.
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