80 through 87 80, 87 SSM WWW 83 Two-lens systems. In Fig. 34-45, stick figure O (the object) stands on the common central axis of two thin, symmetric lenses, which are mounted in the boxed regions. Lens 1 is mounted within the boxed region closer to O, which is at object distance ${\mathit{p}}_{\mathbf{1}}$. Lens 2 is mounted within the farther boxed region, at distance $\mathit{d}$. Each problem in Table 34-9 refers to a different combination of lenses and different values for distances, which are given in centimeters. The type of lens is indicated by for converging and for diverging; the number after or is the distance between a lens and either of its focal points (the proper sign of the focal distance is not indicated). Find (a) the image distance ${\mathit{i}}_{\mathbf{2}}$for the image produced by lens 2 (the final image produced by the system) and (b) the overall lateral magnification $\mathit{M}$for the system, including signs. Also, determine whether the final image is (c) real(R) or virtual (V), (d) inverted(I) from object or non inverted (NI), and (e) on the same side of lens 2 as object O or on the opposite side.

The object stands on the common central axis of two thin symmetric lenses.

Distance between object and lens 1,$p_1=+4\text{cm}$

Distance between lenses 1 and 2,$d=8\text{cm}$

Lens 1 converging, focal length$f_1=6\text{cm}$

Lens 2 diverging, focal length$f_2=6\text{cm}$

Using the relation between focal length, image distance and object distance find the image distance $i_2$. Then using the formula for overall magnification find its value.

From the solution of part a and b answer part c, d and e.

Formula are as follows:

The formula for focal length,$\frac{1}{f}=\frac{1}{p}+\frac{1}{i}$

Overall magnification,$M=m_1{m}_2$

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

Where,$m$is the magnification, $p$is the pole,$f$is the focal length, and$i$is the image distance.

(a)

For lens 1, focal length $f_1$, object distance $p_1$

Using the expression for focal length,

$$\begin{gathered} \frac{1}{f_1}=\frac{1}{p_1}+\frac{1}{i_1} \\ \frac{1}{i_1}=\frac{1}{f_1}-\frac{1}{p_1} \end{gathered}$$

$\begin{array}{rcl}\frac{1}{i_1}& =& \frac{p_1-f_1}{f_1{p}_1}\\ i_1& =& \frac{f_1{p}_1}{p_1-f_1}\end{array}$

$i_1=\frac{f_1{p}_1}{p_1-f_1}\dots \dots \dots \dots \dots \dots ..\left(1\right)$

$\begin{array}{rcl}{i}_1& =& \frac{6\times 4}{4-6}\\ i_1& =& -12\text{cm}\end{array}$

This serves as an object for lens 2 which is diverging, $\therefore p_2=d-i_1=8-\left(-12\right)=20\text{cm}$and it is given that$f_2=-6\text{cm}$

Modifying equation 1 for lens 2,

$$\begin{gathered} i_2=\frac{f_2{p}_2}{p_2-f_2} \\ {i}_2=\frac{-6\times 20}{20-\left(-6\right)} \\ {i}_2=-4.6\text{cm} \end{gathered}$$

Therefore, the image produced by lens 2 is at$-4.6cm$

(b)

To find overall magnification use the formula,

$M=m_1{m}_2$

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

$\begin{array}{rcl}\therefore M& =& -\frac{i_1}{p_1}\times -\frac{i_2}{p_2}\\ M& =& -\frac{\left(-12\right)}{4}\times -\frac{-4.6}{20}\\ M& =& +0.69\end{array}$

Therefore, the overall magnification for the given lens system is$+0.69$ .

(c)

Since the lens 2 is diverging, the object for lens 2 is inside the focal point. The final image distance is negative.

Hence, the image formed by this lens system is virtual.

(d)

Overall magnification for this lens system is the positive which shows that the image and the object have the same orientation.

Hence, the image is not inverted.

(e)

The final image distance is negative, which is on the same side of the object relative to lens 2, which is diverging.

Hence, the image is diverging.

The focal length and overall magnification of the two-lens system can be found using corresponding formulae. The nature of the image can be predicted from the characteristics of the image formed due to the given two-lens system.