Let $t\in\Bbb Q$, $t\
e0,1$ and $$ R_t(z)=z^2+\frac14\,e^{2\pi it}-1. $$ Then $$ p_\pm=\frac{1}{2} \left(-1\pm\sqrt{1-e^{2 i \pi t}}\right) $$ are periodic of period $2$ and $(R^{(2)})'(p_\pm)=e^{2 i \pi t}$.
Note that these points $t = \frac14\,e^{2\pi it}-1$ all lie on the circle of radius $1/4$ centered at the point $c=-1$. This is exactly the boundary of the period 2 disk in the Mandelbrot set. If, for example, we set $t=3/4$, we obtain $z^2-1+i/4$ as shown in yellow below:
![enter image description here](
More generally, the rational values of $t$ produce points where a smaller bulb is attached. The Julia set for this value of $c$ looks like so:
![enter image description here](
The red dots form an orbit of period 2.