Does there exist a if and only if condition so that arc length of a convergent sequence of functions to converge to the arc length of the limit.

A sufficient condition is that $f_n'\to f'$ in $L^1$, that is $\int_a^b |f_n'-f'|\to 0$. The justification for this (unexciting) observation is that the function $t\mapsto \sqrt{1+t^2}$ has derivative bounded by $1$, hence $|\sqrt{1+t^2}-\sqrt{1+s^2}|\le |t-s|$ for all $t,s\in\mathbb R$, which implies $$ \left| \int_a^b \sqrt{1+{f_n'}^2} - \int_a^b \sqrt{1+{f'}^2}\right| \le \int_a^b |f_n'-f'| $$

Within some classes of functions one can give weaker sufficient conditions. For example, if every $f_n$ is convex, and $f_n\to f$ uniformly, then the lengths of graphs converge.

Another relevant fact is that mere pointwise convergence $f_n\to f$ suffices to conclude that $$\int_a^b \sqrt{1+{f'}^2}\le\liminf_{n\to\infty} \int_a^b \sqrt{1+{f_n'}^2}$$ (the upper semicontinuity of length). To see this, pick a partition $t_j$ of $[a,b]$ that almost realizes the length of the graph of $f$. For sufficiently large $n$, the differences $|f_n(t_j)-f(t_j)|$ are as small as we want. Since the length of the graph of $f_n$ between $(t_{j-1}, f_{n}(t_{j-1}))$ and $(t_j, f_n(t_j))$ is no less than the length of line segment connecting these points, the claim follows.

All that said, I don't think there is any non-tautological necessary and sufficient condition for the general case.