(N/A) Linear magnification $(m)$ is defined as the ratio of the height of the image $(h^{\prime})$ to the height of the object $(h)$.
$\therefore m = \frac{h^{\prime}}{h} \quad \dots (1)$
Consider an object $AB$ of height $h$ placed on the principal axis in front of a concave mirror as shown in the figure.
Two rays $AQ$ and $AP$ intersect at $A^{\prime}$ after reflection from the mirror,where $A^{\prime}$ is the image of $A$. Rays from object $AB$ form an image $A^{\prime}B^{\prime}$ after reflection. Let $A^{\prime}B^{\prime} = h^{\prime}$.
From the figure,triangles $\triangle A^{\prime}B^{\prime}P$ and $\triangle ABP$ are similar.
Therefore,$\frac{B^{\prime}A^{\prime}}{BA} = \frac{B^{\prime}P}{BP}$.
According to the sign convention,$B^{\prime}A^{\prime} = -h^{\prime}$ (downward),$BA = h$ (upward),$B^{\prime}P = -v$,and $BP = -u$.
Substituting these values:
$\frac{-h^{\prime}}{h} = \frac{-v}{-u}$
$\therefore \frac{h^{\prime}}{h} = -\frac{v}{u}$
$\therefore m = -\frac{v}{u} \quad (\text{From equation } 1)$.
For a real image,the image formed is inverted,hence its magnification is negative. For a virtual image,the image formed is erect,hence its magnification is positive.