An upright object is placed at a distance of \(40\, cm\) in front of a convergent lens of focal length \(20\, cm\). A convergent mirror of focal length \(10\, cm\) is placed at a distance of \(60\, cm\) on the other side of the lens. The position and size of the final image will be

\((i)\) Refraction from lens

\((ii)\) Reflection from mirror

\((iii)\) Refraction from lens

After these phenomena. Image will be on object and will have same size. None of the option depicts so this question is Bonus.

\(1^{\text {st }}\) refraction \(\mathrm{u}=-40\,\mathrm{cm} ; \mathrm{f}=+20\, \mathrm{cm}\)

\( \Rightarrow {\text{v}} =  + 40\,{\text{cm}}\)  (image \(I_1\)) and \(m_1\,=\,-\,1\) for reflction

\({\text{u}} =  - 20\,{\mkern 1mu} {\text{cm}}\,;{\text{f}} =  - 10\,{\mkern 1mu} {\text{cm}}\)

\( \Rightarrow v =  - 20{\mkern 1mu} \,{\text{cm}}\) (image \(I_2\)) and \(m_2\,=\,-\,1\) \(2^{\text {nd }}\) refraction 

\(\mathrm{u}=-40\, \mathrm{cm}\) ;    \(\mathrm{f}=+20\, \mathrm{cm}\)

\( \Rightarrow v =  + 40{\mkern 1mu} \,{\text{cm}}\) (image \(I_3\)) and \(m_3=-\,1\)

Total magnification \(=m_{1} \times m_{2} \times m_{3}=-1\) and final image is formed at distance \(40\,\mathrm{cm}\) from convergent lens and is of same size as the object.