Explanation

(a) Focal length in air is given by
$\frac{1}{f_{a}ₐ} = \left( \ _{a}\mu_{g} - 1 \right)\left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right)\ $
The focal length of lens immersed in water is given by
$\frac{1}{f_{1}} = \left( \ _{l}n_{g} - 1 \right)\left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right)$
When, $R_{1},R_{2}$ are radii of curvatures of the two surfaces of lens and $\ _{l}n_{g}\ $is refractive index of glass with respect to liquid.
Also, $\ _{l}\mu_{g}\ = \frac{\ _{a}n_{g}}{ₐ\ _{a}n_{l}}$
Given, $ₐ\ _{a}n_{g} = 1.5,\ f_{a}ₐ = 12\ cm,\ ₐ\ _{a}n_{l} = \frac{4}{3}$
$\therefore\frac{f_{l}}{f_{a}ₐ} = \frac{\left( ₐ\ _{a}n_{g} - 1 \right)}{\left( \ _{l}n_{g} - 1 \right)}$
$\frac{f_{1}}{12} = \frac{(1.5 - 1)}{\left( \frac{1.5}{4/3} - 1 \right)} = \frac{0.5 \times 4}{0.5}$
$\Rightarrow f_{1}\ = 4 \times 12 = 48\ cm$