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+### Statement
+
+$12.2.3.$ [Insert the problem statement]
+
+### Solution
+
+Huygens–Fresnel Principle (reminder)
+
+Each point on a wavefront becomes an emitter of secondary spherical waves. The envelope of all these waves at a later instant forms the new wavefront.
+
+---
+
+a) Law of reflection (\alpha_1 = \alpha'_1)
+
+Initial setup
+
+· Plane interface between two media (e.g., air–mirror).
+· A plane wave is incident at an angle \alpha_1 with respect to the normal (or equivalently, the wavefront makes an angle \alpha_1 with the interface).
+· Wave speed in medium 1: v_1.
+
+Huygens construction
+
+1. At t = 0, the incident wavefront is segment AB; point A touches the interface, while end B still must travel distance BC to reach the interface at C.
+2. The time it takes for B to reach C is
+ \tau = \frac{BC}{v_1}.
+3. During that interval \tau, point A acts as a secondary source and emits a spherical wave in medium 1 (backward) that reaches a radius
+ AD = v_1 \tau.
+4. At time \tau, all points between A and C have emitted secondary waves. The envelope of those waves (the common tangent) is the reflected wavefront CD, which must be a plane that passes through C and is tangent to the hemisphere centered at A.
+
+Geometry
+
+· \triangle ABC is right‑angled at B. The angle at A is the incidence angle \alpha_1.
+ BC = v_1 \tau.
+· \triangle ADC is right‑angled at D (radius AD is perpendicular to the reflected wavefront CD). The angle at C is the reflection angle \alpha'_1.
+ AD = v_1 \tau.
+· Both triangles share the hypotenuse AC and have one leg of equal length BC = AD = v_1 \tau.
+
+Therefore, right triangles ABC and ADC are congruent. Hence:
+
+\angle BAC = \angle DCA.
+
+But \angle BAC = 90^\circ - \alpha_1 and \angle DCA = 90^\circ - \alpha'_1. Equating:
+
+90^\circ - \alpha_1 = 90^\circ - \alpha'_1 \quad\Rightarrow\quad \boxed{\alpha_1 = \alpha'_1}.
+
+---
+
+b) Law of refraction \left(\frac{\sin\alpha_1}{\sin\alpha_2} = \frac{v_1}{v_2}\right)
+
+Now the wave is transmitted into the second medium with speed v_2 (we assume v_2 < v_1).
+
+Huygens construction
+
+1. Same incident wavefront AB; A touches the interface at t=0; B will arrive at C in \tau = BC / v_1.
+2. Meanwhile, A emits a secondary wave in medium 2 (downward) that in time \tau reaches a radius
+ AD = v_2 \tau.
+3. The points between A and C also emit secondary waves in medium 2. The envelope of all of them is the refracted wavefront CD, a plane that passes through C and is tangent to the hemisphere centered at A.
+
+Geometry
+
+· \triangle ABC: right‑angled at B.
+ \sin\alpha_1 = \frac{BC}{AC} = \frac{v_1 \tau}{AC}. \tag{1}
+· \triangle ADC: right‑angled at D (radius AD is perpendicular to the refracted wavefront). The angle at C is the refraction angle \alpha_2.
+ \sin\alpha_2 = \frac{AD}{AC} = \frac{v_2 \tau}{AC}. \tag{2}
+
+Dividing (1) by (2):
+
+\frac{\sin\alpha_1}{\sin\alpha_2} = \frac{v_1 \tau / AC}{v_2 \tau / AC} = \frac{v_1}{v_2}.
+
+\boxed{\frac{\sin\alpha_1}{\sin\alpha_2} = \frac{v_1}{v_2}}
+
+#### Answer
+
+[Insert a concise answer or boxed result]