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 $(\alpha_1 = \alpha'_1)$

configurate a 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

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.

The time it takes for B to reach C is
$\tau = \frac{BC}{v_1}$

During that interval $\tau$, point A acts as a secondary source and emits a spherical wave in medium 1 (backward) that reaches a radius
$AD = v_1 \tau$
.
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:

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):

$\frac{\sin\alpha_1}{\sin\alpha_2} = \frac{v_1 \tau / AC}{v_2 \tau / AC} = \frac{v_1}{v_2}$.

$\boxed{\frac{\sin\alpha_1}{\sin\alpha_2} = \frac{v_1}{v_2}}$