Obtain the mirror equation for the real image formed by a concave mirror.

(N/A) Consider an object $AB$ placed perpendicular to the principal axis beyond the center of curvature $C$ of a concave mirror.
$A$ ray $AM$ from point $A$ incident on the mirror at $M$ reflects through the principal focus $F$.
$A$ ray $AP$ from point $A$ incident on the pole $P$ reflects back following the law of reflection,such that $\angle APB = \angle A'PB'$.
These reflected rays intersect at $A'$,forming a real image $A'B'$.
Let $FP = f$ (focal length),$CP = R$ (radius of curvature),$BP = u$ (object distance),and $B'P = v$ (image distance).
For paraxial rays,$MP$ can be considered a straight line perpendicular to the principal axis.
The right-angled triangles $\triangle A'B'F$ and $\triangle MPF$ are similar.
Therefore,$\frac{B'A'}{PM} = \frac{B'F}{FP}$. Since $PM = AB$,we have $\frac{B'A'}{AB} = \frac{B'F}{FP} = \frac{B'P - FP}{FP} = \frac{v - f}{f} \quad \dots (1)$
Similarly,$\triangle A'B'P$ and $\triangle ABP$ are similar.
Therefore,$\frac{B'A'}{AB} = \frac{B'P}{BP} = \frac{v}{u} \quad \dots (2)$
Equating $(1)$ and $(2)$,we get $\frac{v - f}{f} = \frac{v}{u}$.
Dividing by $v$,we get $\frac{1}{f} - \frac{1}{v} = \frac{1}{u}$,which simplifies to the mirror equation: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.