Question

At what distance from a concave mirror of focal length 10 cm should an object be placed so that:

(a) its real image is formed 20 cm from the mirror?
(b) its virtual image is formed 20 cm from the mirror?

Answer 1

(a)
$$\begin{gathered} Given, \\ Itisaconcavemirror. \\ \mathrm{f}=-10\mathrm{cm} \\ \mathrm{v}=-20\mathrm{cm} \\ S\mathrm{ince}the\mathrm{image}\mathrm{is}\mathrm{real}and\mathrm{form}s\mathrm{in}\mathrm{front}\mathrm{of}\mathrm{the}\mathrm{mirror},theequationwillbe \\ \frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}} \\ \frac{1}{-10}=\frac{1}{-20}+\frac{1}{\mathrm{u}} \\ \frac{1}{\mathrm{u}}=\frac{1}{-10}+\frac{1}{20} \\ \frac{1}{\mathrm{u}}=\frac{-20+10}{200} \\ \frac{1}{\mathrm{u}}=\frac{-10}{200} \\ \frac{1}{\mathrm{u}}=-\frac{1}{20} \\ \mathrm{u}=-20\mathrm{cm} \\ Therefore,t\mathrm{he}\mathrm{object}\mathrm{should}\mathrm{be}\mathrm{placed}\mathrm{at}\mathrm{a}\mathrm{distance}\mathrm{of}20\mathrm{cm}fromthemirror\mathrm{to}\mathrm{form}a\mathrm{real}\mathrm{image}. \end{gathered}$$

(b)
$$\begin{gathered} Given, \\ \mathrm{f}=-10\mathrm{cm} \\ \mathrm{v}=+20\mathrm{cm} \\ S\mathrm{ince}the\mathrm{image}\mathrm{is}\mathrm{virtual}and\mathrm{form}s\mathrm{behind}\mathrm{the}\mathrm{mirror},theequationwillbe \\ \frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}} \\ \frac{1}{-10}=\frac{1}{20}+\frac{1}{\mathrm{u}} \\ -\frac{1}{10}-\frac{1}{20}=\frac{1}{\mathrm{u}} \\ \frac{1}{\mathrm{u}}=\frac{-20-10}{200} \\ \frac{1}{\mathrm{u}}=\frac{-30}{200} \\ \frac{1}{\mathrm{u}}=\frac{-3}{20} \\ \mathrm{u}=-\frac{20}{3} \end{gathered}$$
Therefore, the object must be placed at a distance of $\frac{20}{3}$cm from the mirror to form the virtual image.