Question

An object is placed 15 cm from (a) a converging mirror, and (b) a diverging mirror, of radius of curvature 20 cm. Calculate the image position and magnification in each case.

Answer 1

Case -1

It is a converging mirror, i.e. concave mirror.

Distance of the object 'u' = -15 cm

Radius of curvature of the mirror 'R' = -20 cm

Focal length of the mirror 'f' = $\frac{\mathrm{R}}{2}$ = -10 cm

We have to find the position of the image 'v' and its magnification 'M'.

Using the mirror formula, we get
$$\begin{gathered} \frac{1}{f}=\frac{1}{u}+\frac{1}{v} \\ \Rightarrow \frac{1}{-10}=\frac{1}{-15}+\frac{1}{\mathrm{v}} \\ \Rightarrow \frac{1}{\mathrm{v}}=\frac{1}{-10}+\frac{1}{15} \\ \Rightarrow \frac{1}{\mathrm{v}}=-\frac{1}{10}+\frac{1}{15} \\ \Rightarrow \frac{1}{\mathrm{v}}=\frac{-15}{150}+\frac{10}{150} \\ \Rightarrow \frac{1}{\mathrm{v}}=\frac{-5}{150} \\ \Rightarrow \mathrm{v}=-30\mathrm{cm} \\ \mathrm{the}\mathrm{image}\mathrm{will}\mathrm{be}\mathrm{form}\mathrm{at}\mathrm{a}\mathrm{distance}\mathrm{of}30\mathrm{cm}\mathrm{in}\mathrm{front}\mathrm{of}\mathrm{converging}\mathrm{mirror}. \\ \mathrm{magnification}=\frac{-v}{u} \\ m=\frac{-(-30)}{-15} \\ m=-2 \\ \mathrm{magnification}=-2 \end{gathered}$$
Thus the image is real, inverted and large in size.

Case -2

Mirror is converging mirror i.e. convex mirror.

Distance of the object u = -15 cm

Radius of curvature of the mirror R = 20 cm

Focal length of the mirror f = $\frac{\mathrm{R}}{2}$ = 10 cm

We have to find the position of the image v = ?

Magnification = ?
$$\begin{gathered} U\mathrm{sing}the\mathrm{mirror}\mathrm{formula},weget \\ \frac{1}{f}=\frac{1}{u}+\frac{1}{v} \\ \Rightarrow \frac{1}{f}=\frac{1}{-15}+\frac{1}{v} \\ \Rightarrow \frac{1}{10}=\frac{1}{-15}+\frac{1}{v} \\ \Rightarrow \frac{1}{v}=\frac{1}{10}+\frac{1}{15} \\ \Rightarrow \frac{1}{v}=\frac{15}{150}+\frac{10}{150} \\ \Rightarrow \frac{1}{v}=\frac{25}{150} \\ \Rightarrow \frac{1}{v}=\frac{1}{6} \\ \Rightarrow v=6\mathrm{cm} \\ There\mathrm{fore},\mathrm{the}\mathrm{image}\mathrm{willl}\mathrm{form}6\mathrm{cm}\mathrm{behind}\mathrm{the}\mathrm{mirror}. \\ \mathrm{Using}\mathrm{the}\mathrm{magnification}\mathrm{formula},\mathrm{we}\mathrm{get} \\ \mathrm{m}=\frac{-v}{u} \\ m=\frac{-6}{-15} \\ m=0.4 \end{gathered}$$

Thus, the image is virtual, erect and small in size.