A $1.0-cm$-tall object is $20cm$in front of a concave mirror that has a$60cm$focal length. Calculate the position and height of the image. State whether the image is in front of or behind the mirror, and whether the image is upright or inverted.

From mirror equation,

$\frac{1}{s}+\frac{1}{s^{\mathrm{\prime }}}=\frac{2}{R}=\frac{1}{f}$

The convex lens of focal lens is negative and concave lens is positive:the image virutal is negative and real image distance is positive.

The lateral magnification is

$M=\frac{\text{image height}}{\text{object height}}=\frac{h^{\mathrm{\prime }}}{h}=-\frac{s^{\mathrm{\prime }}}{s}$

The object height is localid="1649264457469" $h=1.0cm.$

The distance lens object is localid="1649264464609" $s=20cm.$

The focal length is localid="1649264469803" $f=60cm.$

The mirror is concave.

From mirror equation is

$\begin{array}{l}\frac{1}{s}+\frac{1}{s^{\mathrm{\prime }}}=\frac{1}{f}\\ \frac{1}{s^{\mathrm{\prime }}}=\frac{1}{f}-\frac{1}{s}\\ \frac{1}{s^{\mathrm{\prime }}}=\frac{s-f}{sf}\\ s^{\mathrm{\prime }}=\frac{sf}{s-f}\\ s^{\mathrm{\prime }}=\frac{\left(20\mathrm{cm}\right)\times \left(60\mathrm{cm}\right)}{20\mathrm{cm}-60\mathrm{cm}}\\ s^{\mathrm{\prime }}=-30\mathrm{cm}.\end{array}$

where negative sign denotes virtual image and behind the mirror is located.

The lateral magnification lens by

$\begin{array}{l}M=\frac{h^{\mathrm{\prime }}}{h}=-\frac{s^{\mathrm{\prime }}}{s}\\ h^{\mathrm{\prime }}=-h\frac{s^{\mathrm{\prime }}}{s}\\ h^{\mathrm{\prime }}=-\left(1.0\mathrm{cm}\right)\times \frac{-30\mathrm{cm}}{20\mathrm{cm}}\\ h^{\mathrm{\prime }}=1.5\mathrm{cm}.\end{array}$

The positive sign,

$$\begin{gathered} M=-(-30\mathrm{cm})/20\mathrm{cm} \\ \mathrm{M}=3/2. \end{gathered}$$

shows that image is upright.