(a) Calculation of distance of an object and location of image (b) Reason for virtual image, through convex mirror (a) Given $R=-20\,\text{cm}$ , and magnification $m=-2$ Focal length of the mirror $f=\frac{R}{2}=-10\,\text{cm}$ $\Rightarrow -\frac{1}{10} = \frac{1}{2u} + \frac{1}{u}$ $\Rightarrow u=-15\,\text{cm}$ $\therefore v=2 \times -15\,\text{cm} = -30\,\text{cm}$ $\text{Magnification}\,(m)=-\frac{v}{u}$ $-2=-\frac{v}{u}$ $\Rightarrow v=2u$ Using mirror formula $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$ (b) $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$Using sign convention, for convex mirror, we have $f>0, u<0$ From the formula $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$ $\because f$ is positive and $u$ is negative, $\Rightarrow v$ is always positive, hence image is always virtual.