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 C for converging and D for diverging; the number after C or D 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 O 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,$p_1=+15\text{cm}$
• Distance between lenses 1 and 2,$d=67\text{cm}$
• Lens 1 is converging, focal length$f_1=12\text{cm}$
• Lens 2 is converging, focal length$f_1=12\text{cm}$

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.

Formulae are as follows:

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

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

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

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

For lens1 focal length $f_1$, object distance $p_1$

Using the expression for focal length,

$$\begin{gathered} \frac{1}{f_1}=\frac{1}{p_1}+\frac{1}{i_1} \\ \frac{1}{i_1}=\frac{1}{f_1}-\frac{1}{p_1} \\ \frac{1}{i_1}=\frac{p_1-f_1}{f_1{p}_1} \end{gathered}$$

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

$\begin{array}{rcl}{i}_1& =& \frac{12\times 15}{15-12}\\ & =& +60\text{cm}\\ & & \end{array}$

This serves as an object for lens 2,

$\begin{array}{rcl}{p}_2& =& d-i_1\\ & =& 67-60\\ & =& 7\text{cm}\\ & & \end{array}$

It is given that$f_2=10\text{cm}$

Modifying equation (1) for lens 2,

$i_2=\frac{f_2{p}_2}{p_2-f_2}$

$i_2=\frac{10\times 7}{7-10}$

$i_2=-23\text{cm}$

Therefore, the image produced by lens 2 is at -23 cm.

To find overall magnification use the formula,

$M=m_1{m}_2$

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

$\therefore M=-\frac{i_1}{p_1}\times -\frac{i_2}{p_2}$

$M=\frac{60}{15}\times -\frac{-23}{7}$

$M=-13$

Therefore, the overall magnification for the given lens system is -13.

Since the lens 1 and 2 are converging, 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.

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

Hence the image is inverted.

The final image distance is negative which shows that it is on the negative side of lens 2 that is on the same side of the object.

Hence, the image is on the same side of the object.

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.