91Ó°ÊÓ

From mirror equation,

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

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

$M=\frac{\text{image height}}{\text{object height}}=\frac{h^{\mathrm{′}}}{h}=−\frac{s^{\mathrm{′}}}{s}$

The object height is localid="1649264664326" $h=1.0cm.$

The distance lens object islocalid="1649264668623" $s=20cm.$

The focal length is localid="1649264673679" $f=-60cm.$

The mirror is convex.

From mirror equation is

$\begin{array}{l}\frac{1}{s}+\frac{1}{s^{\mathrm{′}}}=\frac{1}{f}\\ \frac{1}{s^{\mathrm{′}}}=\frac{1}{f}−\frac{1}{s}\\ \frac{1}{s^{\mathrm{′}}}=\frac{s−f}{sf}\\ s^{\mathrm{′}}=\frac{sf}{s−f}\\ s^{\mathrm{′}}=\frac{\left(20\mathrm{cm}\right)×(−60\mathrm{cm})}{20\mathrm{cm}−(−60\mathrm{cm})}\\ s^{\mathrm{′}}=−15\mathrm{cm}.\end{array}$

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

The lateral magnification lens by

$\begin{array}{l}M=\frac{h^{\mathrm{′}}}{h}=−\frac{s^{\mathrm{′}}}{s}\\ h^{\mathrm{′}}=−h\frac{s^{\mathrm{′}}}{s}\\ h^{\mathrm{′}}=−\left(1.0\mathrm{cm}\right)×\frac{−15\mathrm{cm}}{20\mathrm{cm}}\\ h^{\mathrm{′}}=0.75\mathrm{cm}.\end{array}$

The positive sign,

$$\begin{gathered} M=−(−15\mathrm{cm})/20\mathrm{cm} \\ \mathrm{M}=3/4. \end{gathered}$$

shows that image is upright.