For the situation as shown in the figure, find the velocity of image formed by the mirror

For first refraction through lens,  $u=-30cm,f=+20cm$
 $\because \frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
 $\therefore v=+60cm$
 $v_{imagelens}=\frac{v^2}{u^2}\frac{du}{dt}=\frac{{60}^2}{{\left(-30\right)}^2}\left(6-2\right)$
Or    $v_{image}-v_{lens}=16cm{s}^{-1}$
 $\therefore v_{image}=16+2=18cm{s}^{-1}$Image formed by lens will act as object  for the mirror hence the object distance for the mirror is equal to $u^{ }\text{'}$
For reflection through mirror,  $u\text{'}=-(80-60)=-20cm,f\text{'}=-10cm$
 $\therefore v\text{'}=-20cm$
 $\therefore v_{imagemirror}=-\frac{{\left(-20\right)}^2}{{\left(-20\right)}^2}\left(18-4\right)=-14cm{s}^{-1}$
 $\therefore v_{image}=-14+4=-10cm{s}^{-1}$