Geometrical construction for Snell's law? Snell's law from geometrical optics states that the ratio of the angles of incidence $\theta_1$ and of the angle of refraction $\theta_2$ as shown in figure1, is the same as the opposite ratio of the indices of refraction $n_1$ and $n_2$. 
$$
\frac{\sin\theta_1}{\sin \theta_2} = \frac{n_2}{n_1}
$$

(figure originally from wikimedia)
Now let $P$ be a point in one medium (with refraction index $n_1$) and $Q$ a point in the other one as in the figure. My question is, is there is a nice geometrical construction (at best using only ruler and compass) to find the point $O$ in the figure such that Snell's law is satisfied. (Suppose you know the interface and $n_2/n_1$)?
Edit
A long time ago user17762 announced to post a construction. However until now no simple construction was given by anybody. So, does anybody know how to do this?

Yes, right. When entering a denser medium light slows down..  Adding another answer keeping only to the construction method. Drawn on Geogebra, removed the axes and grid to trace a ray inside a medium of higher refractive index $\mu$.
Choose an arbitrary point X such that
$$ \mu= \dfrac{XP}{XQ} \left( =\dfrac{n_2}{n_1}=\dfrac{v_1}{v_2} \right) $$
which is  possible using Ruler & Compass using suitable segment lengths.
From $X$ draw bisector to angle $PXQ$ cutting interface at $O,$ which is the required point of incidence. Due to refraction, the point of incidence $O$ always shifts to the right, when stat point $P$ is at top left, compared to the straight unrefracted ray $PQ$ point of impingement on interface.
The ratio of sides and times spent in each medium is maintained due to the constant bisector property of Apollonian Circle ( no need to draw it).
The diagram represents a lightray refraction into  a medium of refractive index 1.5

