A lake is covered with a layer of ice under a steady sub-zero...

The rate of heat conduction through an ice layer of thickness $x$ is given by $\dfrac{dQ}{dt} = \dfrac{KA\theta}{x}$.

The heat released when a layer of thickness $dx$ freezes is $dQ = \rho A L dx$.

Equating the two, we get $\rho A L \dfrac{dx}{dt} = \dfrac{KA\theta}{x} \Rightarrow dt = \dfrac{\rho L}{KA\theta} x dx$.

Integrating both sides, the time taken for the thickness to increase from $x_i$ to $x_f$ is $t = \dfrac{\rho L}{2KA\theta} (x_f^2 - x_i^2)$.

Thus, the time taken is proportional to the difference of the squares of the thicknesses.

For the first interval: $t_1 \propto (2^2 - 1^2) = 4 - 1 = 3$.

For the second interval: $t_2 \propto (3^2 - 2^2) = 9 - 4 = 5$.

Therefore, the ratio $t_1 : t_2$ is $3 : 5$.

Answer: $3 : 5$