Question

A $20-cm$-tall object is $40cm$in front of a converging lens that has a$20cm$ 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 $s^{\text{'}}=40cm$and image height is$h^{\text{'}}=2.0cm.$

Part $\left(b\right)$

$\left(b\right)$The image distance is $s^{\text{'}}=40cm$and image height is $h^{\text{'}}=2.0cm.$

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$h=2.0cm.$

The lens is converging.

The distance lens object is$s=40cm.$

The focal length islocalid="1649230311637" $f=20cm.$

Step: 2 Finding image distance: (part a) 

The ray trace diagram is

From lens equation is

$\begin{array}{l}\frac{1}{s}+\frac{1}{s^{\mathrm{\prime }}}=\frac{1}{f}\\ \frac{1}{s^{\mathrm{\prime }}}=\frac{1}{f}-\frac{1}{s}\\ \frac{1}{s^{\mathrm{\prime }}}=\frac{s-f}{sf}\\ s^{\mathrm{\prime }}=\frac{\left(40\mathrm{cm}\right)\times \left(20\mathrm{cm}\right)}{40\mathrm{cm}-20\mathrm{cm}}\\ s^{\mathrm{\prime }}=40\mathrm{cm}.\end{array}$

Step: 3 Finding image height: (part b) 

The lateral magnification lens by

$\begin{array}{l}M=-\frac{s^{\mathrm{\prime }}}{s}\\ M=-\frac{40\mathrm{cm}}{40\mathrm{cm}}\\ M=-1.\\ \begin{array}{l}\left|M\right|=\frac{h^{\mathrm{\prime }}}{h}\\ h^{\mathrm{\prime }}=h\left|M\right|\\ h^{\mathrm{\prime }}=\left(2.0\mathrm{cm}\right)\times \left(1\right)\\ h^{\mathrm{\prime }}=2.0\mathrm{cm}.\end{array}\end{array}$

By comparing, the both parts are agree.