Правка разделов «Statement», «Solution», «Answer»

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		@@ -1,74 +1,96 @@
			### Statement

		−	$12.2.3.$ ~~[Insert the problem 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)
		+	a) Law of reflection $(\alpha_1 = \alpha'_1)$

		−	Initial setup
		+	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.
		+	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.
		+	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.
		+	· $\triangle ABC$ 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.
		+	$\angle BAC = \angle DCA$.

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

		−	~~90^\circ - \alpha_1~~ = 90^\circ - \alpha'_1 ~~\quad\Rightarrow\quad \boxed{\alpha_1 = \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)
		+	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

		−	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.
		+	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}
		+	$\triangle ABC$ : 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}.
		+	$\frac{\sin\alpha_1}{\sin\alpha_2} = \frac{v_1 \tau / AC}{v_2 \tau / AC} = \frac{v_1}{v_2}$.

		−	\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
		−
		−	[Insert a concise answer or boxed result]

	@@ -1,74 +1,96 @@
	### Statement		### Statement

	$12.2.3.$ ~~[Insert the problem 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		### 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 $(\alpha_1 = \alpha'_1)$

	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).
	· Wave speed in medium 1: v_1.

			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		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.		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.

			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		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.		$\angle BAC = \angle DCA$.

	But \angle BAC = 90^\circ - \alpha_1 and \angle DCA = 90^\circ - \alpha'_1. Equating:		But

	~~90^\circ - \alpha_1~~ = 90^\circ - \alpha'_1 ~~\quad\Rightarrow\quad \boxed{\alpha_1 = \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)		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}
	· \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}

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

			\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):		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

	[Insert a concise answer or boxed result]