95 through 100. Three-lens systems. In Fig. 34-49, stick figure O (the object) stands on the common central axis of three thin, symmetric lenses, which are mounted in the boxed regions. Lens 1 is mounted within the boxed region closest to O, which is at object distance $p_1$. Lens 2 is mounted within the middle boxed region, at distance ${\mathbf{d}}_{\mathbf{12}}$from lens 1. Lens 3 is mounted in the farthest boxed region, at distance ${\mathbf{d}}_{\mathbf{23}}$from lens 2. Each problem in Table 34-10 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 the focal points (the proper sign of the focal distance is not indicated). Find (a) the image distance $i_3$ \for the (final) image produced by lens 3 (the final image produced by the system) and (b) the overall lateral magnification M for the system, including signs. Also, determine whether the final image is (c) real $\left(\mathrm{R}\right)$or virtual $\left(\mathrm{V}\right)$ , (d) inverted $\left(\mathrm{I}\right)$ from object O or non-inverted $\left(Nl\right)$, and (e) on the same side of lens 3 as object O or on the opposite side.

$f_1=6.0\mathrm{cm}$

$f_2=4.0\mathrm{cm}$

width="87" height="25" role="math">f3=12.0cm

• Objects distance; $p_1=4.0\mathrm{cm}$
• Distance between the lens 1 and 2; $d_{12}=8.0\text{cm}$
• Distance between the lens 2 and 3; $d_{23}=5.7\text{cm}$

First, find out the image distance for each lens using the lens formula. After that, find the magnification of each lens using the corresponding formula. Using this, find the total magnification. From the sign of the total magnification, conclude whether the image is inverted or non-inverted. Also from the sign of the final image distance for the last lens, conclude whether the image is real or virtual and on which side of the object the final image is present.

Formulae are as follows:

$\frac{1}{f}=\left(\frac{1}{p}+\frac{1}{i}\right)$

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

According to the equation 34-4, first, find the image distance $i_1$ as
$\frac{1}{f}=\left(\frac{1}{p}+\frac{1}{i}\right)$

The lens 1 is converging, so the focal length of the length is positive.

$\frac{1}{f_1}=\frac{1}{p_1}+\frac{1}{i_1}$

$\frac{1}{i_1}=\frac{1}{f_1}-\frac{1}{p_1}$

So,

$i_1=\frac{f_1{p}_1}{p_1-f_1}$

Substituting the value,

$$\begin{gathered} i_1=\frac{4\times \left(6\right)}{4-6} \\ =-\frac{24}{2} \\ =-12.0\text{cm} \end{gathered}$$

So now find the$p_2$i.e., the object distance due to the second lens.

role="math" localid="1663216199144" $p_2=d_{12}-i_1$

Putting the value of$p_2=8.0-\left(-12.0\right)$

$p_2=20.0\mathrm{cm}$

From $p_2$ calculate the image distance$i_2$produced by the lens 2:

$\frac{1}{f_2}=\frac{1}{p_2}+\frac{1}{i_2}$

Here, lens 2 is diverging, so the focal length is negative.

$\frac{1}{i_2}=\frac{1}{f_2}-\frac{1}{p_2}$

So,

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

Substituting the value,

$$\begin{gathered} i_2=\frac{20.0\times \left(-4\right)}{20+4} \\ =-\frac{80}{24} \\ =-3.33\text{cm} \end{gathered}$$

Now calculate the object distance $p_3$as,

$p_3=d_{23}-i_2$

Putting the value of$i_2$,

$$\begin{gathered} p_3=5.7-\left(-3.33\right) \\ =9.3\text{cm} \end{gathered}$$

Now calculate the image distance due to the lens 3 as,

$\frac{1}{f_3}=\frac{1}{p_3}+\frac{1}{i_3}$

Lens three is the diverging lens, so the focal length is negative so that,

$\frac{1}{i_3}=\frac{1}{f_3}-\frac{1}{p_3}$

So,

$i_3=\frac{f_3{p}_3}{p_3-f_3}$

Substituting the value,

$$\begin{gathered} i_3=\frac{9.3\times \left(-12.0\right)}{9.3-\left(-12.0\right)} \\ =-\frac{111.6}{21.3} \\ =-5.23 \end{gathered}$$

So,

$i_3\approx -5.2\mathrm{cm}$

Therefore, the image distance $i_3$ due to the image produced by the lens 3 is $i_3=-5.2\mathrm{cm}$.

Now calculate the lateral magnification. The total magnification is the product of the magnification due to each lens.

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

Magnification for the lens 1;

$m_1=-\frac{i_1}{p_1}$

Substituting the values,

$$\begin{gathered} m_1=-\frac{\left(-12.0\right)}{4} \\ =3.0 \end{gathered}$$$$\begin{gathered} m_1=-\frac{\left(-12.0\right)}{4} \\ =3.0 \end{gathered}$$

Magnification for the lens 2;

$m_2=-\frac{i_2}{p_2}$

Substituting the values,

$$\begin{gathered} m_2=-\frac{\left(-3.33\right)}{20.0} \\ =0.17 \end{gathered}$$

Magnification for the lens 3;

$m_3=-\frac{i_3}{p_3}$

Substituting the values,

$$\begin{gathered} m_3=-\frac{\left(-5.2\right)}{9.3} \\ =0.56 \end{gathered}$$

So, the total magnification is,

$m=m_1{m}_2{m}_3$

Substitute the above calculated values of the magnification;

$$\begin{gathered} m=\left(3\right)\times \left(0.17\right)\times \left(0.56\right) \\ =0.285 \\ \approx +0.29 \end{gathered}$$

Therefore, the overall lateral magnification is $m=+0.29$.

If the image distance fromthelast lens is positive, then the image is real, and if the image distance due tothelast lens is negative, then the image is virtual.

In this case$i_3$is negative, so the image is virtual.

Therefore, the final image is virtual.

If the total magnification is positive, then the image is non-inverted, and if the total magnification is negative, then the image is inverted.

In our case, the total magnification is positive.

Hence, the image is non-inverted.

The result c), concludes that the image is virtual. So, the image is on the same side of lens 3 from the object.

Therefore, the final image is on the same side of the object.

The image distance for any lens and overall magnification of a three-lens system can be found using the corresponding formulae. The nature of the image can be predicted from the characteristics of the image formed due to the given three-lens system.