A plano-convex lens having radius of curvature of first surface $2\ \text{cm}$ exhibits focal length of $f_1$ in air. Another plano-convex lens with first surface radius of curvature $3\ \text{cm}$ has focal length of $f_2$ when it is immersed in a liquid of refractive index $1.2$. If both the lenses are made of the same glass of refractive index $1.5$, the ratio of $f_1$ and $f_2$ will be
Answer: (B) $1 : 3$
Plano-convex lens: $\dfrac{1}{f} = \left(\dfrac{\mu_g}{\mu_m} - 1\right)\dfrac{1}{R}$.
In air: $\dfrac{1}{f_1} = (1.5 - 1)\dfrac{1}{2} = \dfrac{1}{4} \Rightarrow f_1 = 4\ \text{cm}$.
In the liquid: $\dfrac{1}{f_2} = \left(\dfrac{1.5}{1.2} - 1\right)\dfrac{1}{3} = \dfrac{0.25}{3} \Rightarrow f_2 = 12\ \text{cm}$.
$f_1 : f_2 = 4 : 12 = 1 : 3$
Solution by Sreeraj P, M.Sc Physics