Question

A $1.0-cm-$tall candle flame is $60cm$from a lens with a focal length of $20cm$. What are the image distance and the height of the flame's image?

Short answer

The distance image is$S^{\text{'}}=30cm$and the flames's image height is$h^{\text{'}}=0.5cm.$

Step-by-step solution

Step: 1 Finding lateral magnification:

From thin lens equation,

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

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

The lateral magnification is

localid="1649225203836" $$\begin{gathered} \left|M\right|=\frac{\text{image height}}{\text{object height}}=\frac{h^{\mathrm{\prime }}}{h} \\ M=-\frac{s^{\mathrm{\prime }}}{s}. \end{gathered}$$

The candle height flame is localid="1649225208567" $h=1.0cm.$

The candle flame distance to the lens is localid="1649225213777" $s=60cm.$

The focal lens length is localid="1649225220303" $f=20cm.$

Step: 2 Finding image flame's distance:

By thin lens equation,

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

Step: 3 Finding image height:

The lateral magnification lens by

$$\begin{gathered} \begin{array}{l}M=-\frac{s^{\mathrm{\prime }}}{s}\\ M=-\frac{30\mathrm{cm}}{60\mathrm{cm}}\\ M=-0.5.\end{array} \\ \begin{array}{l}\left|M\right|=\frac{h^{\mathrm{\prime }}}{h}\\ h^{\mathrm{\prime }}=h\left|M\right|\\ h^{\mathrm{\prime }}=\left(1.0\mathrm{cm}\right)\left(0.5\right)\\ h^{\mathrm{\prime }}=0.5\mathrm{cm}.\end{array} \end{gathered}$$