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 p₁. Lens 2 is mounted within the middle boxed region, at distance d₁₂ from lens 1. Lens 3 is mounted in the farthest boxed region, at distance d₂₃ 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 i₃ 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 (R) or virtual (V), (d) inverted (I) from object O or non-inverted (NI), and (e) on the same side of lens 3 as object O or on the opposite side.

• The focal length of the first lens, $f_1=8.0\text{cm}$.
• The focal length of the second lens, $f_2=6.0\text{cm}$.
• The focal length of the third lens, $f_3=6.0\text{cm}$.
• The object distance from the first lens, $p_1=12.0\text{cm}$.
• The distance between lens 1 and 2, $d_{12}=28.0\text{cm}$.
• The distance between lens 2 and 3, $d_{23}=8.0\text{cm}$.

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,

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

The magnification is,

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

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

All lenses converge so that the focal length of each lens is positive.

For lens 1:

Write the equation of lens as below for lens 1.

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

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

Rearrange the above equation for the image distance for lens 1.

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

Substituting the known values in the above equation.

$\begin{array}{rcl}{i}_1& =& \frac{12\text{cm}\times 8\text{cm}}{\left(12-8\right)\text{cm}}\\ & =& \frac{96{\text{cm}}^2}{4\text{cm}}\\ & =& 12\mathrm{cm}\end{array}$

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

$p_2=d_{12}-i_1$

Putting the value of I₁ and d₁₂ in the above equation, you get

$\begin{array}{rcl}{p}_2& =& 28\text{cm}-24\text{cm}\\ & =& 4\text{cm}\\ & & \end{array}$

From p₂, calculate the image distance I₂ of the image produced by the lens 2 is,

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

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

Substitute known values in the above equation.

$\begin{array}{rcl}{i}_2& =& \frac{4\text{cm}\times 6\text{cm}}{\left(4-6\right)\text{cm}}\\ & =& \frac{24{\text{cm}}^2}{-2\text{cm}}\\ & =& -12\text{cm}\\ & & \end{array}$

Now, calculate the object distance p₃ as,

$p_3=d_{23}-i_2$

Putting the value of I₂ and

$\begin{array}{rcl}{p}_3& =& 8.0\text{cm}-\left(-12\text{cm}\right)\\ & =& 20\text{cm}\\ & & \end{array}$

The image distance due to lens 3 is

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

After rearranging the above equation for image distance, you have

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

Substituting the known numerical values in the above equation.

$\begin{array}{rcl}{i}_3& =& \frac{20\text{cm}\times 6\text{cm}}{\left(20-6\right)\text{cm}}\\ & =& \frac{120{\text{cm}}^2}{14\text{cm}}\\ & =& \text{8.6 cm}\\ & & \end{array}$

Hence, the image distance I₃ due to the image produced by lens 3 is $i_3=+8.6\text{cm}$.

The lateral magnification is given by,

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

For lens 1, the magnification is,

$\begin{array}{rcl}{m}_1& =& -\frac{i_1}{p_1}\\ & =& -\frac{24\text{cm}}{12\text{cm}}\\ & =& -2\\ & & \end{array}$

For lens 2, the magnification is,

localid="1663017835554" $\begin{array}{rcl}{m}_2& =& -\frac{i_2}{p_2}\\ & =& -\frac{\left(-12\text{cm}\right)}{4\text{cm}}\\ & =& 3\end{array}$

For lens 3, the magnification is,

$\begin{array}{rcl}{m}_3& =& -\frac{i_3}{p_3}\\ & =& -\frac{\left(8.6\text{cm}\right)}{20\text{cm}}\\ & =& -0.43\\ & & \end{array}$

Thus, the total magnification is as below.

$\begin{array}{rcl}m& =& m_1{m}_2{m}_3\\ & =& -2\times 3\times \left(-0.43\right)\\ & =& +2.6\\ & & \end{array}$

Hence, the overall lateral magnification islocalid="1663018349890" $m=+2.6$.

If the image distance due to the last lens is positive, then the image is real, and if the image distance due to the last lens is negative, then the image is virtual. So, in this case, the i₃is positive, so the image is real.

Hence, 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 our case, the total magnification is positive.

Therefore, the final image is non-inverted.

From the results (c) and (d) we can conclude that the image is on the opposite side of lens 3 from the object.

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