Question

A $1.0-cm$-tall object is $20cm$in front of a convex 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.

Short answer

The picture is $15cm$behind the mirror, upright, and it's tall at $0.75cm.$

Step-by-step solution

Step: 1 Finding lateral magnification:   

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="1649264664326" $h=1.0cm.$

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

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

The mirror is convex.

Step: 2 Finding image distance: (part a) 

From mirror equation is

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

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

Step: 3 Finding image height: (part b) 

The lateral magnification lens by

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

The positive sign,

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

shows that image is upright.