$13 \cdot 64$ As shown in the figure, a beam of light emitted from point $A$ propagates in a plane, reflecting between the lines $AD$ and $CD$ several times before hitting point $B$ (point $B$ may be on $AD$ or $CD$) perpendicularly, and then reflecting back to point $A$ along the same path (at each reflection point, as shown in the figure, the angle of incidence equals the angle of reflection. The figure only represents the case of $n=3$). If $\angle CDA=8^{\circ}$, then the maximum value that $n$ can take is
(A) 6.
(B) 10.
(C) 38.
(D) 98.
(E) There is no maximum value for $n$.
[Solution] Let $\angle D A R_{1}=\theta$, and let $\theta_{i}$ be the angle of incidence (which is also the angle of reflection) at the $i$-th reflection point.

For $\triangle A R_{1} D$ and successively for $\triangle R_{i-1} R_{i} D, 2 \leqslant i \leqslant n$, up to $\triangle R_{n} B D$, using the exterior angle theorem of a triangle, we have the following results:
$$
\begin{array}{l}
\theta_{1}=\theta+8^{\circ}, \\
\theta_{2}=\theta_{1}+8^{\circ}=\theta+16^{\circ}, \\
\theta_{3}=\theta_{2}+8^{\circ}=\theta+24^{\circ}, \\
\cdots \cdots \cdots \cdots \\
\theta_{n}=\theta_{n-1}+8^{\circ}=\theta+(8 n)^{\circ}, \\
90^{\circ}=\theta_{n}+8^{\circ}=\theta+(8 n+8)^{\circ} .
\end{array}
$$

Since $\theta$ is positive, we have $0^{\circ} \leqslant \theta=90^{\circ}-(8 n+8)^{\circ}$, which gives
$$
n \leqslant \frac{82}{8}<11 \text {. }
$$

If $\theta=2^{\circ}$, $n$ achieves its maximum value of 10.
Therefore, the answer is $(B)$.