A point object $O$ is placed in front of two thin symmetrical coaxial convex lenses $L_1$ and $L_2$ with focal length $24\ \text{cm}$ and $9\ \text{cm}$ respectively. The distance between two lenses is $10\ \text{cm}$ and the object is placed $6\ \text{cm}$ away from lens $L_1$ as shown in the figure. The distance between the object and the image formed by the system of two lenses is ______ cm.
Numerical value type. Enter your answer.
Answer: 34
$L_1$: $\dfrac1v=\dfrac1{24}-\dfrac16=-\dfrac18\Rightarrow v=-8\ \text{cm}$ (virtual, $8\ \text{cm}$ before $L_1$).
For $L_2$ this image is $18\ \text{cm}$ away: $\dfrac1v=\dfrac19-\dfrac1{18}=\dfrac1{18}\Rightarrow v=18\ \text{cm}$ beyond $L_2$.
Object to image: $6+10+18=34\ \text{cm}$.
Solution by Sreeraj P, M.Sc Physics