Given that the lens is made of a material with refractive index $\mu = 1.6$ and has equal radii of curvature $r_1 = 30\ \text{cm}$ and $r_2 = -30\ \text{cm}$. The lens is placed in three different media with refractive indices $\mu_1 = 1.16$, $\mu_2 = 1.33$, and $\mu_3 = 1.4$. The power of a lens in a medium is given by the lens maker's formula: $$ \begin{aligned} P &= \frac{1}{f} = \left( \frac{\mu}{\mu_m} - 1 \right) \left( \frac{1}{r_1} - \frac{1}{r_2} \right) \end{aligned} $$ Here, $\left( \frac{1}{r_1} - \frac{1}{r_2} \right) = \left( \frac{1}{30} - \frac{1}{-30} \right) \text{cm}^{-1} = \frac{2}{30} \text{cm}^{-1} = \frac{2}{0.3} \text{m}^{-1} = \frac{20}{3} \text{m}^{-1}$. Now, calculating the power for each medium: 1. For medium $\mu_1 = 1.16$: $$ \begin{aligned} P_1 &= \left( \frac{1.6}{1.16} - 1 \right) \times \frac{20}{3} \\ &= \left( \frac{1.6 - 1.16}{1.16} \right) \times \frac{20}{3} \\ &= \left( \frac{0.44}{1.16} \right) \times \frac{20}{3} \approx 2.53\ \text{D} \end{aligned} $$ 2. For medium $\mu_2 = 1.33$: $$ \begin{aligned} P_2 &= \left( \frac{1.6}{1.33} - 1 \right) \times \frac{20}{3} \\ &= \left( \frac{1.6 - 1.33}{1.33} \right) \times \frac{20}{3} \\ &= \left( \frac{0.27}{1.33} \right) \times \frac{20}{3} \approx 1.35\ \text{D} \end{aligned} $$ 3. For medium $\mu_3 = 1.4$: $$ \begin{aligned} P_3 &= \left( \frac{1.6}{1.4} - 1 \right) \times \frac{20}{3} \\ &= \left( \frac{1.6 - 1.4}{1.4} \right) \times \frac{20}{3} \\ &= \left( \frac{0.2}{1.4} \right) \times \frac{20}{3} \approx 0.95\ \text{D} \end{aligned} $$ Comparing the results, we find $P_1 > P_2 > P_3$. Therefore, the lens is used with maximum power in the medium with a refractive index of $1.16$.