A concave mirror with a focal length of $\mathrm{f}$ provides a picture that is multiplied by the size of the item If the image is accurate, the object's distance is as follows:

 

The ratio of image distance to object distance is known as linear magnification.

$\text{m=}\frac{\text{image distance (v)}}{\text{object distance (u)}}\text{=}\frac{-\text{v}}{\text{u}}$

Given $\mathrm{f}$$=$ focal length and $\text{m=-n}$ (The minus sign denotes that the image will be inverted in reference to the item.)

The true image in the concave mirror is always inverted to the object, as we know.

Since,

$\text{m=}\frac{{\text{h}}^{\text{'}}}{\text{h}}\text{=}\frac{-\text{v}}{\text{u}}\text{=-n}$

$\text{v-nu --- (1)}$

The mirror formula is as follows:

$\frac{1}{\text{f}}=\frac{1}{\text{u}}+\frac{1}{\text{v}} --- \left(2\right)$

Using equations 1 and 2,

$\text{f=}\frac{1}{\text{nu}}\text{+}\frac{1}{\text{u}}$

$\frac{1}{\text{f}}=\frac{\text{n+1}}{\text{nu}}$

$\text{u=}\frac{\left(\text{n+1}\right)}{\text{n}}\text{f}$

Hence option (C) is correct.