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			### Statement

			$12.2.3.$

			Using the Huygens — Fresnel principle, prove that when a plane wave falls
			on the interface of two media: a) the angle of incidence is equal to the angle
			of reflection (α1= α − 3); b) the ratio of the sine of the angle of incidence to
			the sine of the angle of refraction is equal to the ratio of the wave velocity in
			=v1
			the first medium to the wave velocity in the second medium (sin α1).v2

			### 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:

			$\angle BAC = \angle DCA$.
		@@ -54,7 +54,7 @@Solution

			But

		−	$\angle BAC = 90^\circ - \alpha_1 and \angle DCA = 90^\circ - \alpha'_1$.
		+	$\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):

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

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

			#### Answer
		unchanged lines 36

	### Statement		### Statement

	$12.2.3.$		$12.2.3.$

	Using the Huygens — Fresnel principle, prove that when a plane wave falls		Using the Huygens — Fresnel principle, prove that when a plane wave falls
	on the interface of two media: a) the angle of incidence is equal to the angle		on the interface of two media: a) the angle of incidence is equal to the angle
	of reflection (α1= α − 3); b) the ratio of the sine of the angle of incidence to		of reflection (α1= α − 3); b) the ratio of the sine of the angle of incidence to
	the sine of the angle of refraction is equal to the ratio of the wave velocity in		the sine of the angle of refraction is equal to the ratio of the wave velocity in
	=v1		=v1
	the first medium to the wave velocity in the second medium (sin α1).v2		the first medium to the wave velocity in the second medium (sin α1).v2

	### Solution		### Solution

	Huygens–Fresnel Principle (reminder)		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.		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 $(\alpha_1 = \alpha'_1)$		a) Law of reflection $(\alpha_1 = \alpha'_1)$

	configurate a Initial setup		configurate a Initial setup

	Plane interface between two media (e.g., air–mirror).		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).		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: $v_1$.		· Wave speed in medium 1: $v_1$.

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

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

	· $\triangle ABC$ is right‑angled at B. The angle at A is the incidence angle $\alpha_1$.		· $\triangle ABC$ is right‑angled at B. The angle at A is the incidence angle $\alpha_1$.
	$BC = v_1 \tau$.		$BC = v_1 \tau$.
	·$ \triangle ADC$ is right‑angled at D (radius AD is perpendicular to the reflected wavefront CD). The angle at C is the reflection $angle \alpha'_1$.		·$ \triangle ADC$ is right‑angled at D (radius AD is perpendicular to the reflected wavefront CD). The angle at C is the reflection $angle \alpha'_1$.
	$AD = v_1 \tau$.		$AD = v_1 \tau$.


	Both triangles share the hypotenuse AC and have one leg of equal length $ BC = 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:		Therefore, right triangles ABC and ADC are congruent. Hence:

	$\angle BAC = \angle DCA$.		$\angle BAC = \angle DCA$.
	@@ -54,7 +54,7 @@Solution

	But		But

	$\angle BAC = 90^\circ - \alpha_1 and \angle DCA = 90^\circ - \alpha'_1$.		$\angle BAC = 90^\circ - \alpha_1 $and, $\angle DCA = 90^\circ - \alpha'_1$.

	Equating:		Equating:

	$90^\circ - \alpha_1 = 90^\circ - \alpha'_1 \quad\Rightarrow\quad \boxed{\alpha_1 = \alpha'_1}$.		$90^\circ - \alpha_1 = 90^\circ - \alpha'_1 \quad\Rightarrow\quad \boxed{\alpha_1 = \alpha'_1}$.

	---		---

	b) Law of refraction $\left(\frac{\sin\alpha_1}{\sin\alpha_2} = \frac{v_1}{v_2}\right)$		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)$.		Now the wave is transmitted into the second medium with speed $v_2$ (we assume$ v_2 < v_1)$.

	Huygens construction		Huygens construction

	1. Same incident wavefront AB; A touches the interface at t=0; B will arrive at C in$ \tau = BC / v_1$.		1. Same incident wavefront AB; A touches the interface at t=0; B will arrive at C in$ \tau = BC / v_1$.
	2. Meanwhile, A emits a secondary wave in medium 2 (downward) that in time$ \tau" reaches a radius		2. Meanwhile, A emits a secondary wave in medium 2 (downward) that in time$ \tau" reaches a radius
	$AD = v_2 \tau$		$AD = v_2 \tau$


	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.		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		Geometry

	$\triangle ABC$ : right‑angled at B.		$\triangle ABC$ : right‑angled at B.

	$\sin\alpha_1 = \frac{BC}{AC} = \frac{v_1 \tau}{AC}. \tag{1}$		$\sin\alpha_1 = \frac{BC}{AC} = \frac{v_1 \tau}{AC}. \tag{1}$

	\triangle ADC:		\triangle ADC:

	The angle at C is the refraction angle$\alpha_2$.		The angle at C is the refraction angle$\alpha_2$.

	$\sin\alpha_2 = \frac{AD}{AC} = \frac{v_2 \tau}{AC}. \tag{2}$		$\sin\alpha_2 = \frac{AD}{AC} = \frac{v_2 \tau}{AC}. \tag{2}$

	Dividing (1) by (2):		Dividing (1) by (2):

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

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

	#### Answer		#### Answer
	unchanged lines 36