91Ó°ÊÓ

We need to find the focal length and distance from screen from second lens.

First, Let us find the image of diverging lens:

$$\begin{gathered} \frac{1}{S_1}=\frac{1}{S_1^{\text{'}}}=\frac{1}{f_1} \\ \frac{1}{20}+\frac{1}{S_1^{\text{'}}}=\frac{1}{-20} \\ {S}_1^{\text{'}}=-10cm \end{gathered}$$

Here, $S_1$is the distance of the object from first lens, $f_1$is the focal length of the first lens and ${S^{\text{'}}}_1$is the distance of the image from first lens.

Than we Let us find the magnification of the diverging lens:

localid="1648745456574" $m_1=-\frac{S_1^{\text{'}}}{S_1}=-\frac{(-10)}{20}=\frac{1}{2}$

Converging the lens magnification is;

$m_2=-\frac{S_2^{\text{'}}}{S_2}=-4$

Here, $S{\text{'}}_2$is the distance of the image from the second lens,$S_2$is the distance of the object from the second lens and $m_2$is the magnification of second image.

And from the previous relation, we get the value for $S_2^{\text{'}}$

$S_2^{\text{'}}=4S_2$

Total magnification is

$M=m_1{m}_2=\frac{1}{2}.(-4)=-2$.

Let us find the value of $S_2$and $S_2^{\text{'}}$:

$$\begin{gathered} S_2+S_2^{\text{'}}=100cm \\ {S}_2+4S_2=100cm \\ 5S_2=100cm \\ {S}_2=20cm \\ ⇒S_2^{\text{'}}=4S_2 \\ {S}_2^{\text{'}}=4.20cm \\ {S}_2^{\text{'}}=80cm \end{gathered}$$.

Finally, we got the focal length:

$$\begin{gathered} \frac{1}{S_2}+\frac{1}{S{\text{'}}_2}=\frac{1}{f_2} \\ \frac{1}{f_2}=\frac{1}{20}+\frac{1}{80} \\ \frac{1}{f_2}=\frac{1}{16} \\ {f}_2=16cm \end{gathered}$$.