The linear magnification for a spherical mirror is the ratio of the size of the image to the size of the object and is denoted by $m$ . Then $m$ is equal to (symbols have their usual meanings).
A) $\dfrac{u}{{u - f}}$
B) $\dfrac{{uf}}{{u - f}}$
C) $\dfrac{f}{{u - f}}$
D) None of these

Hint: The linear magnification can also be defined as the ratio of the distance corresponding to the position of the image from the mirror to the distance corresponding to the position of the object from the mirror. The image distance can be expressed in terms of the object distance and focal length of the mirror using the mirror equation.

Formulas used:
-The mirror equation is given by, $\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$ where $v$ is the image distance, $u$ is the object distance and $f$ is the focal length of the mirror.
-The magnification of a mirror is given by, $m = \dfrac{v}{u}$ where $v$ is the image distance and $u$ is the object distance.

Complete step by step solution:
Step 1: Express the mirror equation to obtain a relation for the image distance.
The mirror equation is represented as $\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$ --------- (1) where $v$ is the image distance, $u$ is the object distance and $f$ is the focal length of the mirror.
Simplifying equation (1) we get, $\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} = \dfrac{{u - f}}{{uf}}$
Then taking the reciprocal we obtain the required relation for the image distance as
$v = \dfrac{{uf}}{{u - f}}$ -------- (2)

Step 2: Express the relation for the magnification of the spherical mirror.
The magnification of the spherical mirror can be expressed as $m = \dfrac{v}{u}$ -------- (3)
Substituting equation (2) in (3) we get, $m = \dfrac{{uf}}{{u\left( {u - f} \right)}} = \dfrac{f}{{u - f}}$
$\therefore $ the magnification of the mirror is obtained to be $m = \dfrac{f}{{u - f}}$ .

Hence the correct option is C.

Note: The relation for the linear magnification actually contains an additional negative sign when it is expressed as the ratio of the image distance to the object distance i.e., $m = \dfrac{{ - v}}{u}$ . This is because the size of the image and that of the object are always considered to be positive by sign convention whereas the image distance and object distance can be negative depending on the type of the mirror used. A positive value for $m$ indicates that the nature of the image is erect and if the value was greater than one then the image is a magnified one.