Answer:
The final image relative to the diverging lens at 9.0 cm.
Step-by-step explanation:
Given that,
Focal length of diverging lens = -6.18 cm
Focal length of converging lens = 12.5 cm
Distance of object = 36.9 cm
We need to calculate the image distance of converging lens
Using formula of lens
![(1)/(v)=(1)/(f)-(1)/(u)](https://img.qammunity.org/2020/formulas/physics/college/8syfwgrfiff7l3fejesh6gnv25gvk6h87v.png)
![(1)/(v)=(1)/(12.5)-(1)/(-36.9)](https://img.qammunity.org/2020/formulas/physics/college/3dfnbozkmpwf5bspx6wa0rob24xqx5p5yr.png)
![(1)/(v)=(988)/(9225)](https://img.qammunity.org/2020/formulas/physics/college/qnu25bsnsw7ds1m3bahfgvbmdc7w3tttk9.png)
![v=9.33\ cm](https://img.qammunity.org/2020/formulas/physics/college/33gmtvowjxbth62xmkphwbwutec4bv88xm.png)
We need to calculate the image distance of diverging lens
Now, object distance is
![u=28.8-9.33=19.47 cm](https://img.qammunity.org/2020/formulas/physics/college/vl9ert9jzcf5k69ax0cgmmb8gx3lw53m5v.png)
Using formula of lens
![(1)/(v)=(1)/(-6.18)-(1)/(-19.47)](https://img.qammunity.org/2020/formulas/physics/college/68pdneisc5dmpwevu789glkbt2ub38inc6.png)
![(1)/(v)=-(22150)/(200541)](https://img.qammunity.org/2020/formulas/physics/college/mifbxdaovu0s4h6jribfz8do6vbk4kzy4h.png)
![v=-9.0\ cm](https://img.qammunity.org/2020/formulas/physics/college/mqdccmf1wiosihukc07mnbxpy4muy96rls.png)
The image is formed 9.0 cm to the left of the diverging lens.
Hence, The final image relative to the diverging lens at 9.0 cm.