Refractive index of a glass convex lens is $1 \cdot 5$. The radius of curvature of each of the two surfaces of the lens is 40 cm . The ratio of the power of the lens when immersed in a liquid of refractive index 1.25 to that when placed in air is

The ratio of the power in the liquid to that in air is determined using the lens maker's formula and the definition of lens power, $P = \frac{1}{f}$.

The relative refractive index for air is $n_{\text{rel,a}} = \frac{n_g}{n_a} = \frac{1.5}{1} = 1.5$.
For a biconvex lens with $R_1 = 40 \text{ cm}$ and $R_2 = -40 \text{ cm}$, the power in air is $P_a = (n_{\text{rel,a}} - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = (0.5)\left(\frac{1}{40} + \frac{1}{40}\right) = \frac{0.5}{20} = \frac{1}{40}$.

In the liquid, $n_{\text{rel,l}} = \frac{n_g}{n_l} = \frac{1.5}{1.25} = 1.2$.
The power becomes $P_l = (n_{\text{rel,l}} - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = (0.2)\left(\frac{1}{40} + \frac{1}{40}\right) = \frac{0.2}{20} = \frac{1}{100}$.

The ratio is $\frac{P_l}{P_a} = \frac{1/100}{1/40} = \frac{40}{100} = \frac{2}{5}$.

The power ratio is $\boxed{\text{C}}$.