91Ó°ÊÓ

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,

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

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:

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

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,

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

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

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.