80, 87 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 p₁. Lens 2 is mounted within the farther boxed region, at distanced. 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 distancei₂for the image produced by lens 2 (the final image produced by the system) and (b) the overall lateral magnificationMfor 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,$p_1=+20\text{cm}$.

Distance between lenses 1 and 2, $d=10\text{cm}$.

Lens 1 diverging, focal length, $f_1=-12\text{cm}$.

Lens 2 diverging, focal length,$f_2=-8\text{cm}$.

Using the relation between focal length, image distance, and object distance, find the image distancei₂.

Formulae 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.

For lens 1, focal lengthf₁, object distancep₁;

Using the expression for focal length,

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

Solving further as,

$$\begin{gathered} i_1=\frac{f_1{p}_1}{p_1-f_1}······\left(1\right) \\ {i}_1=\frac{-12\times 20}{20-\left(-12\right)} \\ {i}_1=-7.5\text{cm} \end{gathered}$$

This serves as an object for lens 2, which is diverging, $p_2=d-i_1=10-\left(-7.5\right)=17.\text{5}\text{cm}$and it is given that $f_2=-8\text{cm}$.

Modifying equation 1 for lens 2:

$\begin{array}{rcl}{i}_2& =& \frac{f_2{p}_2}{p_2-f_2}\\ i_2& =& \frac{-8\times 17.5}{17.5-\left(-8\right)}\\ i_2& =& -5.5\text{cm}\end{array}$

Therefore, the image produced by lens 2 is at -5.5 cm.

To find overall magnification use the formula,

$M=m_1{m}_2$

Magnification,

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

$$\begin{gathered} M=-\frac{i_1}{p_1}\times -\frac{i_2}{p_2} \\ M=-\frac{-7.5}{20}\times -\frac{-5.5}{17.5} \\ M=+0.12 \end{gathered}$$

Therefore, the overall magnification for the given lens system is +0.12.

Since the lens 1 and 2 are 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.

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

Hence the image is not inverted.

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.