= Solution
Take the positive horizontal direction along B. The initial direction is $(\alpha-90^\circ)-\theta$. Reflection at the horizontal face B reverses the vertical direction. At C, symmetry of the equilateral geometry gives internal incidence $r$, so <Snell's law> makes the emerging angle to C's normal equal to $\theta$. Its direction is therefore $(90^\circ-\alpha)+\theta$. Subtracting the incoming direction gives
$$
\boxed{\delta=180^\circ-2\alpha+2\theta=60^\circ+2\theta.}
$$
This is the smaller angle between the incident and emergent propagation directions in the requested reflection interval. The cancellation of $n$ occurs because the entry and exit <refraction angles> have equal magnitudes; <refractive index> still determines whether the reflected route exists and where it exits.
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