A plano-convex lens having radius of curvature of first surface 2 cm exhibits focal length of \(f_1\) in air. Another plano-convex lens with first surface radius of curvature 3 cm has focal length of \(\mathrm{f}_2\) when it is immersed in a liquid of refractive index 1.2 . If both the lenses are made of same glass of refractive index 1.5 , the ratio of \(f_1\) and \(f_2\) will be _______.

For case (1)

\(\begin{aligned} & \frac{1.5}{v_1}-\frac{1}{\propto}=\frac{(1.5-1)}{R_1} \\ & \frac{1}{f_1}-\frac{1.5}{v_1}=\frac{1-1.5}{\propto}\end{aligned}\)
\(\Rightarrow \frac{1}{f_1}=\frac{(1.5-1)}{R_1}\) ...(1)
\(\Rightarrow \frac{1}{f_1}=\frac{1}{2 R_1}\)
For case (2)
\(\begin{aligned}
& \frac{1.5}{v_1}-\frac{1.2}{\infty}=\frac{(1.5-1.2)}{R_2} \\
& \frac{1.2}{f_2}-\frac{1.5}{v_1}=\frac{(1.2-1.5)}{\propto} \\
& \Rightarrow \frac{1.2}{f_2}=\frac{0.3}{R_2} \\
& \Rightarrow \frac{1}{f_2}=\frac{1}{4 R_2}
\end{aligned}\)
So clearly, \(\frac{f_1}{f_2}=\frac{2 \times 2}{4 \times 3}=\frac{1}{3}\)