95, 96, 99 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 $d_{23}$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 Mfor the system, including signs. Also, determine whether the final image is (c) real (R) or virtual (V), (d) inverted (I) from object Oor non-inverted (NI), and (e) on the same side of lens 3 as object Oor on the opposite side.

• Three thin symmetrical lenses are mounted in a straight line as shown in figure 34-49.
• Object distance from lens 1,$p_1=+2.0\mathrm{cm}$
• Lens 1 is a converging lens with focal length,$f_1=6.0\mathrm{cm}$
• Distance of lens 2 from lens 1,$d_1=15\mathrm{cm}$
• Lens 2 is a converging lens with focal length,$f_2=6.0\mathrm{cm}$
• Distance of lens 3 from lens 2,$d_{2}3=19\mathrm{cm}$
• Lens 3 is a converging lens with focal length,$f_3=5.0\mathrm{cm}$

If parallel rays are sent through a converging lens parallel to the central axis, the refracted rays pass through a common point (a real focus F) at a focal distance f (a positive quantity) from the lens. A converging lens can form a real image (if the object is outside the focal point) or a virtual image (if the object is inside the focal point). For a system of lenses with a common central axis, the image produced by the first lens acts as the object for the second lens, and so on, and the overall magnification is the product of the individual magnifications.

Formulae:

The lens equation,$\frac{1}{f}=\frac{1}{p}+\frac{1}{i}$ (i)

The lateral magnification of an object, $m=\frac{-i}{p}$ (ii)

The overall lateral magnification of the lens system, $M=\prod _i^n{m}_i$ (iii)

The focal length of a converging lens is always positive. Thus, the focal lengths of te converging lens is given as:$f_1=+6.0\mathrm{cm}$$f_2=+6.0\mathrm{cm}$role="math" localid="1663210990436" $f_3=+5.0\mathrm{cm}$

Now, the image distance of lens 1 is given using equation (i) as follows:

$$\begin{gathered} \frac{1}{i_1}=\frac{1}{f_1}-\frac{1}{p_1} \\ =\frac{1}{+6}-\frac{1}{+2} \\ =\frac{1-3}{6} \\ =\frac{-1}{\text{3cm}} \\ {i}_1=-3\text{cm} \end{gathered}$$

This image from lens 1 will act as an object to lens 2, thus the object distance of lens 2 is given by:

$$\begin{gathered} p_2=d_{12}-i_1 \\ =15\text{cm}-(-3\text{cm}) \\ =+18\text{cm} \end{gathered}$$

Now, the image distance of lens 2 can be given using the given data in equation (i) as follows:

$$\begin{gathered} \frac{1}{i_2}=\frac{1}{f_2}-\frac{1}{p_2} \\ =\frac{1}{+6}-\frac{1}{+1} \\ =\frac{3-1}{18} \\ =\frac{1}{\text{9cm}} \end{gathered}$$

So,

$i_2=9\mathrm{cm}$

Now, this image from lens 2 will act as an object to lens 3, thus the object distance of lens 3 is given by:

$$\begin{gathered} p_3=d_{23}-i_2 \\ =19\text{cm}-\left(9\text{cm}\right) \\ =+10\text{cm} \end{gathered}$$

Now, the image distance of lens 3 can be given using the given data in equation (i) as follows:

$$\begin{gathered} \frac{1}{i_3}=\frac{1}{f_3}-\frac{1}{p_3} \\ =\frac{1}{+5}-\frac{1}{+10} \\ =\frac{2-1}{10} \\ =\frac{1}{\text{10cm}} \end{gathered}$$

So,$i_3=10\mathrm{cm}$

Hence, the final image distance is $+10\mathrm{cm}$.

Using the given data with equation (ii) in equation (i), the overall magnification of the lens system can be given as follows:

$$\begin{gathered} \backslash M=m_1{m}_2{m}_3 \\ =\left(\frac{-i_1}{p_1}\right)\left(\frac{-i_2}{p_2}\right)\left(\frac{-i_3}{p_3}\right) \\ =\left(\frac{-\left(-3\text{cm}\right)}{+2\text{cm}}\right)\left(\frac{-\left(\text{9cm}\right)}{+18\text{cm}}\right)\left(\frac{-\left(\text{10cm}\right)}{+10\text{cm}}\right) \\ =+\frac{3}{4} \\ =+0.75 \end{gathered}$$

Hence, the overall magnification is $+0.75$.

The image formed by the final lens has image distance greater than its focal length.

Hence, the image formed is real.

The overall magnification of the image formed is positive.

Hence, the image formed is non-inverted.

The final image formed has positive value that means right to the lens 3.

Hence, the image is on the opposite side of lens 3 from the object (which is expected for a real final image).