\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\) ; \(\frac{1}{v}+\frac{1}{30}=\frac{1}{20}\)
\(v=+60\, \mathrm{cm}\)
If image position does not change even when mirror is removed it means image formed by lens is formed at centre of curvature of spherical mirror.
Radius of curvature of mirror \(=80-60=20\,{cm}\)
\(\Rightarrow\) Focal length of mirror \(f=10\, \mathrm{cm}\) for virtual image, object is to be kept between focus and pole.
\(\Rightarrow\) Maximum distance of object from spherical mirror for which virtual image is formed, is \(10\,\mathrm{cm} .\)