Question

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?

Short answer

1. Distance between the lens and the image of the sand grain it produces is $\left|i_2\right|=3.33cm$.
2. The image is to the left of the lens 2.
3. The image is virtual.
4. The image is non-inverted relative to the sand grain.

Step-by-step solution

Listing the given quantities

$p_1=3.0cm$

$f_1=4.0cm$

$f_2=4.0cm$

Understanding the concepts of lens equation

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,$m=m_1{m}_2$

(a Find the distance between lens 2 and the image of the sand grain 

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$\left|i_2\right|=3.33cm.$

(b) Determine whether the image is to the left or to the right of lens 2 

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

(c) Determine whether the image is real or virtual

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

(d) Determine whether the image is inverted or non-inverted relative to the sand grain

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.