intuition for the normal bundle to a divisor

Question

I understand that if $i:D\hookrightarrow \mathbb{P}^n$ is a divisor of degree $d$, then the normal bundle to $D$ is $O_D(d)=i^*(O(d))$. The only way I know to prove this is through the conormal bundle, i.e., to show that $O_D(d)^v=I_D/I_D^2$. It would be great if someone explain this in geometric terms, i.e., how do you actually see $O(D)$ as sth normal to $D$.