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Two converging lenses with focal lengths of 40 cm and 20 cm are 10 cm apart. A 2.0-cm-tall object is 15 cm in front of the 40-cm-focal-length lens

Below is a diagram of two converging lenses .When tracing the rays in these lenses, the rays are still parallel as it passes through the two lenses.

Below is a diagram of two converging lenses .When tracing the rays in these lenses, the rays are still parallel as it passes through the two lenses.

From figure we have ,

The known values are;-

$$\begin{gathered} f_1=40cm,f_2=20cm \\ d=10cm,h=2cm, \\ {s}_1=15cm,s_2=24+10=34cm \end{gathered}$$

$$\begin{gathered} f_1=40cm,f_2=20cm,d=10cm \\ h=2cm,s_1=15cm,s_2=34cm \end{gathered}$$First ,we will solve for the image distance using the formula;

$s\text{'}=\frac{sf}{s-f}$

Next, we will determine the magnification of the image using the formula;

$m=-\frac{s\text{'}}{s}$

which will be used to determine the height, by substituting the magnification to the formula

find the formula

First ,we will solve for the image distance using the formula;

Next, we will determine the magnification of the image using the formula;

which will be used to determine the height, by substituting the magnification to the formula

Then ,solving for the first lens,

=

=$-24cm$

$$\begin{gathered} m_1=-\frac{s_1\text{'}}{s_1}=-\frac{-24}{15} \\ =1.6cm \end{gathered}$$

Also, solving for the second lens,

Then ,solving for the first lens,

$s_1\text{'}=\frac{s_1{f}_1}{s_{1-}{f}_1}$

=$$\begin{gathered} =\frac{\left(15\right)\left(14\right)}{15-14} \\ =-24cm \end{gathered}$$

$$\begin{gathered} m_1=-\frac{s_1\text{'}}{s_1}=-\frac{-24}{15} \\ =1.6cm \end{gathered}$$

Also, solving for the second lens,

$$\begin{gathered} s_2\text{'}=\frac{s_2{f}_2}{s_2-f_2} \\ =\frac{\left(34\right)\left(20\right)}{34-20} \\ =48.57cm \end{gathered}$$

$$\begin{gathered} m_2=-\frac{s_2\text{'}}{s_2}=-\frac{48.57}{34} \\ =-1.43cm \end{gathered}$$

Solving for the height of the image;

$$\begin{gathered} h\text{'}=m_1{m}_{2}h \\ =(1.6)(-1.43)\left(2\right) \\ =-4.58cm \end{gathered}$$

Thus, the height is 4.58 cm. This result must be identical to the height on the ray tracing.