Explanation

(b) Lens-maker’s formula is given by
$\frac{1}{f} = \left( \ _{a}\mu_{g} - 1 \right)\left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right)\ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
If the lens is immersed in a liquid of refractive index $\mu_{1}$ then
$\frac{1}{f_{1}} = \left( \ _{l}\mu_{g} - 1 \right)\left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right)\ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
Here, $\ _{l}\mu_{g}\ $is refractive index of glass w.r.t liquid
Dividing Eq. (i) by Eq. (ii), we have
$\frac{f_{1}}{f} = \ \frac{\left( \ _{a}\mu_{g} - 1 \right)}{\left( \ _{l}\mu_{g} - 1 \right)}$
$\Rightarrow \frac{f_{1}}{f} = \left( \frac{1.5 - 1}{\frac{1.5}{1.25} - 1} \right)$
$\Longrightarrow \frac{f_{1}}{f} = \frac{0.5 \times 1.25}{0.25} = 2.5$
Hence, focal length increases by a factor of 2.5.