Question

A $1.0-cm$-tall object is $75cm$in front of a converging lens that has a $30cm$focal length.

$\left(a\right)$ Use ray tracing to find the position and height of the image. To do this accurately, use a ruler or paper with a grid. Determine the image distance and image height by making measurements on your diagram.
$\left(b\right)$Calculate the image position and height. Compare with your ray-tracing answers in part $a$.

Short answer

Part $\left(a\right)$

$\left(a\right)$The image distance is localid="1649234286834" $s^{\text{'}}=50cm$and image height islocalid="1649234293008" $h^{\text{'}}=0.67cm.$

Part $\left(b\right)$

$\left(a\right)$The image distance is $s^{\text{'}}=50cm$and image height is$h^{\text{'}}=0.67cm.$

The both part agrees.

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

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

object height is$h=1.0cm.$

The lens is converging.

The distance lens object is$s=75cm.$

The focal length is$f=30cm.$

Step: 2 Finding image distance: (part a) 

From lens equation is

$\begin{array}{l}\frac{1}{s}+\frac{1}{s^{\mathrm{\prime }}}=\frac{1}{f}\\ \frac{1}{s^{\mathrm{\prime }}}=\frac{1}{f}-\frac{1}{s}\\ \frac{1}{s^{\mathrm{\prime }}}=\frac{s-f}{sf}\\ s^{\mathrm{\prime }}=\frac{sf}{s-f}\\ s^{\mathrm{\prime }}=\frac{\left(75\mathrm{cm}\right)\left(30\mathrm{cm}\right)}{75\mathrm{cm}-30\mathrm{cm}}\\ s^{\mathrm{\prime }}=50\mathrm{cm}.\end{array}$

The ray trace diagram is

Step: 3 Finding image height: (part b)   

The lateral magnification lens by

$\begin{array}{l}M=-\frac{s^{\mathrm{\prime }}}{s}\\ M=-\frac{50\mathrm{cm}}{75\mathrm{cm}}\\ M=-\frac{2}{3}.\\ \begin{array}{l}\left|M\right|=\frac{h^{\mathrm{\prime }}}{h}\\ h^{\mathrm{\prime }}=h\left|M\right|\\ h^{\mathrm{\prime }}=\left(1.0\mathrm{cm}\right)\left(\frac{2}{3}\right)\\ h^{\mathrm{\prime }}=0.67\mathrm{cm}.\end{array}\end{array}$

By comparing, the both parts are agree.