\(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\) \( = \frac{1}{{12}} + \frac{1}{{240}}\) \( = \frac{{20 + 1}}{{240}} = \frac{{21}}{{240}}\)

\(f=\frac{240}{21}\, \mathrm{cm}\)

Shift \( = t\left( {1 - \frac{1}{\mu }} \right)\) \( \Rightarrow 1\left( {1 - \frac{1}{{3/2}}} \right) = 1 \times \frac{1}{3}\)

Now \(v^{\prime}=12-\frac{1}{3}=\frac{35}{3} \,\mathrm{cm}\)

Now the object distance \(u\).

\(\frac{1}{u} = \frac{3}{{35}} - \frac{{21}}{{240}}\) \( = \frac{1}{5}\left[ {\frac{3}{7} - \frac{{21}}{{48}}} \right]\)

\(\frac{1}{u}=\frac{1}{5}\left[\frac{48-49}{7 \times 16}\right]\)

\(u=-7 \times 16 \times 5=-560\, \mathrm{cm}=-5.6\, \mathrm{m}\)

shift \(=5.6-2.4=3.2 \;cm\)