Given the reaction $\ce{A->B}$, where the absorbance of the product $\ce{B}$ has been measured (see the table below), and the initial concentration of $\ce{A}$ is $\pu{10^{-4} M},$ calculate the concentration of the reactant at the time $t.$
$$ \begin{array}{rr} \hline t/\pu{min} & A \\ \hline 0 & 0.000 \\ 1 & 0.115 \\ 2 & 0.188 \\ 3 & 0.237 \\ 4 & 0.273 \\ 5 & 0.301 \\ 6 & 0.321 \\ 7 & 0.339 \\ 8 & 0.353 \\ 9 & 0.365 \\ 10 & 0.375 \\ \infty & 0.500 \\ \hline \end{array}$$
From the answer it is stated that one should use the following equation to calculate the concentration of $\ce{A}$:
$$[\ce{A}]_t = [\ce{A}]_0\left(1 – \frac{A_t}{A_\infty}\right)\label{eqn:1q}\tag{1}$$
I don't understand where this comes from since I have only ever converted absorbance to concentration using the equation:
$$A = \varepsilon lc\tag{2}$$
Could someone please explain the reasoning behind the equation \eqref{eqn:1q}?