Question

A mirror forms an image which is 30 cm from an object and twice its height.

(a) Where must the mirror be situated?
(b) What is the radius of curvature?
(c) Is the mirror convex or concave?

Answer 1

(a) Given, v - u = 30 cm
Magnification 'm' = 2
$$\begin{gathered} m=\frac{-v}{u} \\ \Rightarrow 2=\frac{-\mathrm{v}}{\mathrm{u}} \\ \Rightarrow v=-2u \\ \mathrm{By}\mathrm{putting}\mathrm{the}\mathrm{va}\mathrm{lue}\mathrm{of}\text{'}\mathrm{v}\text{'}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{equation},weget \\ -2u-u=30 \\ \Rightarrow -3u=30 \\ u=\frac{-30}{3}=-10\mathrm{cm} \\ \mathrm{Therefore},\mathrm{the}\mathrm{object}\mathrm{is}\mathrm{situated}\mathrm{at}\mathrm{a}\mathrm{distance}\mathrm{of}10\mathrm{cm}\mathrm{from}\mathrm{the}\mathrm{lens}. \\ \mathrm{Hence},\mathrm{the}\mathrm{mirror}\mathrm{is}\mathrm{at}\mathrm{a}\mathrm{distance}\mathrm{of}10\mathrm{cm}\mathrm{from}\mathrm{the}\mathrm{object}. \\ \mathrm{Therefore},\mathrm{distance}\mathrm{of}\mathrm{the}\mathrm{image}\text{'}\mathrm{v}\text{'}=-2\times -10=20\mathrm{cm} \\ \left(\mathrm{b}\right)U\sin gthe\mathrm{mirror}\mathrm{formula},weget \\ \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \\ \Rightarrow \frac{1}{f}=\frac{1}{20}+\frac{1}{-10}\Rightarrow \frac{1}{f}=\frac{-10}{200} \\ \Rightarrow \frac{1}{f}=\frac{10-20}{200} \\ \Rightarrow \frac{1}{f}=\frac{-1}{20} \\ \Rightarrow f=-20\mathrm{cm} \\ \mathrm{Radius}\mathrm{of}\mathrm{the}\mathrm{curvature}R=2f \\ R=2\times -20 \\ R=-40\mathrm{cm} \\ \mathrm{Thus},\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{curvature}\mathrm{of}\mathrm{the}\mathrm{lens}\mathrm{is}40\mathrm{cm}. \\ \left(\mathrm{c}\right)\mathrm{Since}\mathrm{the}\mathrm{focal}\mathrm{length}\mathrm{shows}\mathrm{a}\mathrm{negative}\mathrm{value},\mathrm{the}\mathrm{given}\mathrm{mirror}\mathrm{is}\mathrm{concave}. \end{gathered}$$