Let $S$ and $T$ be countable complete theories. We assume that $T$ is superstable without the dimensional order property, and $S$ is interpretable in $T$ in such a way that every model of $S$ is coded in a model of $T$. We show that $S$ does not have the dimensional order property, and we discuss the question of whether $\operatorname{Depth}(S) \leq \operatorname{Depth}(T)$. For Mekler's uniform interpretation of arbitrary theories $S$ of finite similarity type into suitable theories $T_s$ of groups we show that $\operatorname{Depth}(S) \leq \operatorname{Depth}(T_S) \leq 1 + \operatorname{Depth}(S)$.