A thin lens made of glass (refractive index =1.5)

(2) Using lens maker's formula

$\frac{1}{f}=\left(\frac{\mu_{g}}{\mu_{a}}-1\right)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]$

Here, $\mu_{g}$ and $\mu_{a}$ are the refractive index of glass and air respectively                   ....(i)

$\Rightarrow \frac{1}{f}=(1.5-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$

When immersed in liquid

$\frac{1}{f_{l}}=\left(\frac{\mu_{g}}{\mu_{l}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$

Dividing (i) by (ii)

$\Rightarrow \quad \frac{f_{l}}{f}=\frac{(1.5-1) 1.42}{0.08}=\frac{1.42}{0.16}=\frac{142}{16} \approx 9$