b
(b) Convex lens will form image \({I_1}\) at it’s focus which acts like a virtual object for concave lens.

Hence for concave lens \(u = + 4\,cm\), \(f = 20\)\(cm\,.\)

So by lens formula \(\frac{1}{{ - 20}} = \frac{1}{v} - \frac{1}{4} \Rightarrow v = 5\,cm\)

i.e. distance of final image \(({I_2})\) from concave lens \(v = 5\,cm\) by using

\(\frac{v}{u} = \frac{I}{O} \Rightarrow \frac{5}{4} = \frac{I}{2} \Rightarrow ({I_2}) = 2.5\,cm\)