$$ \frac{1}{f} = \left(\frac{n}{n'} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) $$

For a thin plano-convex lens, one surface is flat, meaning $$R_2 = \infty$$ and $$\frac{1}{R_2} = 0.$$ The formula reduces to:

$$ \frac{1}{f} = \left(\frac{n}{n'} - 1\right)\frac{1}{R_1} $$

Thus, the focal length is:

$$ f = \frac{R_1}{\left(\frac{n}{n'} - 1\right)} $$

For the First Lens (in Air)

Refractive index of lens, $$n = 1.5$$

Surrounding medium (air), $$n' = 1$$

Radius of curvature, $$R_1 = 2 \text{ cm}$$

Substitute into the formula:

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

For the Second Lens (in Liquid)

Refractive index of lens, $$n = 1.5$$

Surrounding medium (liquid), $$n' = 1.2$$

Radius of curvature, $$R_1 = 3 \text{ cm}$$

Substitute into the formula:

$$ f_2 = \frac{3}{\left(\frac{1.5}{1.2} - 1\right)} $$

First, calculate the term inside the parentheses:

$$ \frac{1.5}{1.2} = 1.25 \quad \Rightarrow \quad 1.25 - 1 = 0.25 $$

Now, compute $$f_2$$:

$$ f_2 = \frac{3}{0.25} = 12 \text{ cm} $$

Ratio of the Focal Lengths

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

Thus, the ratio of $$f_1$$ to $$f_2$$ is:

$$ 1 : 3 $$