CBSE Class 10 Science 2017 · Paper 31/1/3 · Question 4

Here $f=\frac{R}{2}=-15$ cm and $u=-12$ cm, so $\frac{1}{v}=-\frac{1}{15}+\frac{1}{12}=\frac{1}{60}$, giving $v=+60$ cm, and $m=-\frac{v}{u}=+5$. The object lies between the pole and the focus, and the four characteristics of the image are: (1) Virtual: it is formed behind the mirror and cannot be obtained on a screen. (2) Erect: $m$ is positive. (3) Enlarged: five times the size of the object. (4) Position: 60 cm behind the mirror.