In Fig. 34-60, a sand grain is $\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{}\mathit{c}\mathit{m}$from thin lens 1, on the central axis through the two symmetric lenses. The distance between focal point and lens is $\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{}\mathit{c}\mathit{m}$for both lenses; the lenses are separated by $\mathbf{8}\mathbf{.}\mathbf{00}\mathbf{}\mathit{c}\mathit{m}$. (a) What is the distance between lens 2 and the image it produces of the sand grain? Is that image (b) to the left or right of lens 2, (c) real or virtual, and (d) inverted relative to the sand grain or not inverted?

$p_1=3.0cm$

$f_1=4.0cm$

$f_2=4.0cm$

We can use the lens equation to find the distance between lens 2 and the image of the sand grain produced by it. Using sign convention, we can find if the image is to the left or right of lens 2and if it is real or virtual.We can find if the image is inverted or non-inverted by finding the net magnification.

Formula:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{i}$

$m=-\frac{i}{p}$

Net magnification due to two lenses,$m=m_1{m}_2$

We have

$p_1=3.0cm$and$f_1=4.0cm$

By lens equation, we can get the image distance $i_1$

$\begin{array}{c}\frac{1}{f_1}=\frac{1}{p_1}+\frac{1}{i_1}\\ \frac{1}{4.0}=\frac{1}{3}+\frac{1}{i_1}\\ \frac{1}{i_1}=\frac{1}{4.0}-\frac{1}{3}\\ i_1=-12.0cm\end{array}$

Magnification is

$\begin{array}{c}{m}_1=-\frac{i_1}{p_1}\\ =\frac{12.0}{3.0}\\ =+4.0\end{array}$

This image will act as an object for the second lens with

$\begin{array}{c}{p}_2=\left(8.0cm\right)-\left(-12.0cm\right)\\ =20.0cm\end{array}$and$f_2=-4.0cm$

Now the distance of the image $i_2$produced by lens 2 will be

$\begin{array}{c}\frac{1}{f_2}=\frac{1}{p_2}+\frac{1}{i_2}\\ \frac{1}{i_2}=\frac{1}{f_2}-\frac{1}{p_2}\\ \frac{1}{i_2}=\frac{1}{-4}-\frac{1}{20}\\ =-\frac{24}{80}\\ =-0.3c{m}^{-1}\end{array}$

$\begin{array}{c}{i}_2=-\frac{1}{0.3}\\ =-3.33cm\end{array}$

Hence, the distance between lens 2 and the image of the sand grain produced by it is$\left|i_2\right|=3.33cm.$

As $i_2$ is negative, the final image is on the left side of lens 2.

The fact that $i_2$is negative. On the left of lens 2 implies that the image is virtual.

Let’s find the magnification $m_2$.

$\begin{array}{c}{m}_2=-\frac{i_2}{p_2}\\ =-\frac{-3.33}{20.0}\\ =0.1665\end{array}$

The net magnification is

$\begin{array}{c}m=m_1{m}_2\\ =+4.0\times 0.1665\\ =0.66>0\end{array}$

Net magnification is positive; hence the image is non-inverted.