Suppose your wave equation reads
$$
u_{tt}=a^2u_{xx}.
$$
Let
$$
u_j^n\approx u(t_n,x_j).
$$
Let $k$ be your time step, and $h$ be your spatial step. Let
$$
0=x_0<x_1<\cdots<x_N=1,
$$
with $x_j=jh$. Your finite difference scheme could be
$$
\frac{u_j^{n+1}-2u_j^n+u_j^{n-1}}{k^2}=a^2\frac{u_{j+1}^n-2u_j^n+u_{j-1}^n}{h^2},\quad j=1,2,\cdots,N-1.
$$
If you expect that the wave could propagate beyond $x=0$ and $x=1$, i.e., $u$ is defined for $x<0$ and $x>1$. Then you are assuming that $u_{tt}=a^2u_{xx}$, and hence the above discretized scheme, applies at $x=0$ and $x=1$ as well.
Take $x=1$ for instance. Since the scheme holds at $x=1$, you must put $j=N$, i.e.,
$$
\frac{u_N^{n+1}-2u_N^n+u_N^{n-1}}{k^2}=a^2\frac{u_{N+1}^n-2u_N^n+u_{N-1}^n}{h^2}.
$$
In this last scheme, everything is fine except the "ghost" $u_{N+1}^n$ term. Nevertheless, it suffices to approximate $u_{N+1}^n$ by using $u_N^n$, $u_{N-1}^n$, etc. For example,
$$
u_{N+1}^n=2u_N^n-u_{N-1}^n,
$$
which is second-order accurate, because
\begin{align}
u(t_n,x_{N+1})&=u(t_n,x_N+h)\\
&=2u(t_n,x_N)-u(t_n,x_N-h)+O(h^2)\\
&=2u(t_n,x_N)-u(t_n,x_{N-1})+O(h^2).
\end{align}
You may check this by Taylor expansion. You could make use of $u_{N-2}^n$, $u_{N-3}^n$, ..., as well if you wish to get a higher order of accuracy.
Similar trick also applies to the other "ghost" $u_{-1}^n$.
