An object 2 cm high is placed at a distance of 14 cm formed below the principal axis that is , it is a real and inverted image:1) what is the focal length of the mirror
2) find the position of the image

Here given, distance of the object from the mirror $u=-14\ cm$

 size $h=2\ cm$ and image size $h'=-3\ cm$

As known $m=\frac{h}{h'}=-\frac{3}{2}$

And we know $m=-\frac{v}{u}$

Therefore, $-\frac{v}{u}=-\frac{3}{2}$

Or $v=\frac{3}{2}u$

Or $v=\frac{3}{2}\times(-14)$

Or $v=3\times(-7)$

Or $v=-21$

Let $f$ be the focal length of the mirror.

Therefore, $\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

Or $\frac{1}{f}=\frac{1}{-14}+\frac{1}{-21}$

Or $\frac{1}{f}=-( \frac{3+2}{42})$

Or $\frac{1}{f}=-\frac{5}{42}$

Or $f=-\frac{42}{5}$

Or $f=-8.4\ cm$

Therefore, The position of the mirror is v=-21\ cm$

And the focal length of the mirror is $f=-8.4\ cm$