4. As shown in Figure 1, the center of a semicircle is $O$. A beam of light travels from point $M$ perpendicular to $PA$ towards the semicircle, and reflects at point $C$ on the circumference. The reflection angle $\angle OCB$ equals the incidence angle $\angle MCO$; then the light intersects the circumference again at point $B$, and reflects in the same manner, finally reaching point $A$. Then $\angle COM=$ $\qquad$
4. $36^{\circ}$.

As shown in Figure 7, a circle is drawn with $PA$ as its diameter, and the trajectory of the light refraction forms a regular pentagon $ABCDE$. Then
$$
\begin{array}{l}
\angle DCB=108^{\circ} \\
\angle OCM=54^{\circ} \\
\angle COM=36^{\circ}
\end{array}
$$