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 $d_{12}$from lens 1. Lens 3 is mounted in the farthest boxed region, at distance 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.

The focal length of lens 1, $f_1=6.0\mathrm{cm}$

The focal length of lens 2, $f_2=6.0\mathrm{cm}$

The focal length of lens 3, $f_3=5.0\mathrm{cm}$

The object distance from lens 1, $p_1=2.0\mathrm{cm}$

The distance between lens 1 and 2, $d_{12}=15.0\text{cm}$$d_{12}=15.0\text{cm}$

The distance between lens 2 and 3,$d_{23}=19.0\text{cm}$

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:

The lens formula is,

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

The magnification is define by,

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

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

All lenses are converging so that the focal length due to each lens is positive.

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

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

Rearrange the above equation for image distance from lens 1 as below.

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

Substituting the known values in the above equation, and you have

$$\begin{gathered} i_1=\frac{2\text{cm}\times 6\text{cm}}{\left(2-6\right)\text{cm}} \\ =\frac{12{\text{cm}}^2}{-4\text{cm}} \\ =-3.0\text{cm} \end{gathered}$$

So now find the$p_2$, i.e.,theobject distance due tothesecond lens.

$p_2=d_{12}-i_1$

Putting the value of$i_1$ and $d_{12}$ in the above expression.

$$\begin{gathered} p_2=15.0\text{cm}-\left(-3.0\text{cm}\right) \\ =18\text{cm} \end{gathered}$$

From role="math" localid="1663212671171" $P_2$, calculate the image distance$i_2$produced by the lens 2 as,

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

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

So, the image distance from lens 2 is,

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

Substituting the known value in the above equation, and you get

$$\begin{gathered} i_2=\frac{18.0\text{cm}\times 6\text{cm}}{\left(18-6\right)\text{cm}} \\ =\frac{108{\text{cm}}^2}{12\text{cm}} \\ =9.0\text{cm} \end{gathered}$$

Now calculate the object distance$p_3$as,

$p_3=d_{23}-i_2$

Putting the value of $i_2$and $d_{23}$, you get

$$\begin{gathered} p_3=19.0\text{cm}-9.0\text{cm} \\ =10.0\text{cm} \end{gathered}$$

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

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

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

So, the image distance is,

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

Substituting known the value in the above equation.

$$\begin{gathered} i_3=\frac{10.0\text{cm}\times 5.0\text{cm}}{\left(10.0-5.0\right)\text{cm}} \\ =\frac{50.0{\text{cm}}^2}{5.0\text{cm}} \\ =10.0\text{cm} \end{gathered}$$

Hence, the image distance $i_3$ due to the image produced by lens 3 is $i_3=+10\mathrm{cm}$.

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

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

The magnification of lens 1 is,

$$\begin{gathered} m_1=-\frac{i_1}{p_1} \\ =-\frac{\left(-3.0\text{cm}\right)}{2\text{cm}} \\ =1.5 \end{gathered}$$

The magnification of lens 2 is,

$$\begin{gathered} m_2=-\frac{i_2}{p_2} \\ =-\frac{9.0\text{cm}}{18.0\text{cm}} \\ =-0.5 \end{gathered}$$

The magnification of lens 3 is,

$$\begin{gathered} m_3=-\frac{i_3}{p_3} \\ =-\frac{10\text{cm}}{10\text{cm}} \\ =-1 \end{gathered}$$

So the total magnification is,

$$\begin{gathered} m=m_1{m}_2{m}_3 \\ =\left(1.5\right)\times \left(-0.5\right)\times \left(-1\right) \\ =+0.75 \end{gathered}$$

Hence, the overall lateral magnification is $m=+0.75$

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

In this case, $i_3$is positive, therefore, the final image is real.

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 this case, the total magnification is positive, thus, the final image is non-inverted.

From the result (c), conclude that the image is real so that the image is on the opposite side of the lens 3 from the object.

Hence, the final image is on the opposite side of the object.