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}}