The image formed by a concave mirror

Using Mirror equation

$\begin{array}{r}\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}\\ \Rightarrow \frac{1}{\mathrm{v}}=\frac{1}{\mathrm{f}}-\frac{1}{\mathrm{u}}\end{array}$

$\mathrm{For} \mathrm{concave} \mathrm{mirror}, \mathrm{focal} \mathrm{length} \mathrm{is} \mathrm{negative}. \mathrm{Hence}, \mathrm{f}<0$

$\mathrm{For} \mathrm{a} \mathrm{virtual} \mathrm{object}, \mathrm{object} \mathrm{distance} \mathrm{is} \mathrm{positive}. \mathrm{Hence} {\mathrm{u}}_{\mathrm{v}}>0$

From above, it can be observed that image distance v is negative when object is virtual. Since image distance is negative, it is formed by real intersection of rays and image is real.

For a real object, object distance is negative.

$\mathrm{Hence} {\mathrm{u}}_{\mathrm{r}}<0$

In this case, image distance v may be positive or negative depending on the location of object.