An object ABED is placed in front of a concave mirror beyond centre of curvature C as shown in figure. State the shape of the image.

Object is placed beyond C . Hence, the image will be real and it will lie between C and F. Further u. v are negative. hence the mirror Formula will become

$\begin{array}{l}\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}\\ \frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}=\frac{\mathrm{u}-\mathrm{f}}{\mathrm{uf}}\mathrm{or}\mathrm{v}=\frac{\mathrm{f}}{1-\left(\mathrm{f}/\mathrm{u}\right)}\end{array}$

$\begin{array}{l}\mathrm{Now}{\mathrm{u}}_{\mathrm{AB}}>{\mathrm{u}}_{\mathrm{ED}}\\ \therefore {\mathrm{v}}_{\mathrm{AB}}<{\mathrm{v}}_{\mathrm{ED}}\\ \left|{\mathrm{m}}_{\mathrm{AB}}\right|<\left|{\mathrm{m}}_{\mathrm{ED}}\right|\left(\mathrm{m}=-\frac{\mathrm{v}}{\mathrm{u}}\right)\\ \mathrm{Therefore},\mathrm{shape}\mathrm{of}\mathrm{the}\mathrm{image}\mathrm{will}\mathrm{be}\mathrm{as}\mathrm{shown}\mathrm{in}\mathrm{figure}.\\ \mathrm{Also}\mathrm{note}\mathrm{that},{\mathrm{v}}_{\mathrm{AB}}<{\mathrm{u}}_{\mathrm{AB}}\mathrm{and}{\mathrm{v}}_{\mathrm{ED}}<{\mathrm{u}}_{\mathrm{ED}},\\ \mathrm{So},\left|{\mathrm{m}}_{\mathrm{AB}}\right|<1\mathrm{and}\left|{\mathrm{m}}_{\mathrm{ED}}\right|<1\end{array}$