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

For the first planoconvex lens in air:

$$\mu_g = 1.5, \quad \mu_{\text{air}} = 1, \quad R_1 = 2\text{ cm}, \quad R_2 = \infty$$

$$\frac{1}{f_1} = (1.5 - 1)\left(\frac{1}{2} - \frac{1}{\infty}\right) \implies \frac{1}{f_1} = 0.5 \times \frac{1}{2} = \frac{1}{4}\text{ cm}^{-1} \implies f_1 = 4\text{ cm}$$

For the second planoconvex lens in liquid:

$$\mu_g = 1.5, \quad \mu_l = 1.2, \quad R_3 = 3\text{ cm}, \quad R_4 = \infty$$

$$\frac{1}{f_2} = \left(\frac{1.5}{1.2} - 1\right)\left(\frac{1}{3} - \frac{1}{\infty}\right) \implies \frac{1}{f_2} = (1.25 - 1) \times \frac{1}{3} = 0.25 \times \frac{1}{3} = \frac{1}{12}\text{ cm}^{-1} \implies f_2 = 12\text{ cm}$$

Ratio of focal lengths:

$$\frac{f_1}{f_2} = \frac{4}{12} = \frac{1}{3}$$