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Alexphysics правка от 16 июн. 2026 г., 15:10

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		+	### Statement
		+
		+	$12.2.12.$ [Insert the problem statement]
		+
		+	### Solution
		+
		+	Huygens–Fresnel Principle
		+
		+	Each point of a wavefront becomes an emitter of monochromatic secondary spherical waves. The total disturbance at a point P is obtained by integrating the contributions from all elements dS of the wavefront. For an element dS, the contribution is:
		+
		+	$dU_P = \frac{c A \, e^{ikr}}{r} \, dS$
		+
		+	where A is the amplitude of the primary wave at the wavefront, r is the distance from the element to P, $k = 2\pi/\lambda$, and c is the constant to be determined.
		+
		+	Consider a plane wave propagating in the z-direction. At t = 0, the wavefront coincides with the plane z = 0. We want to calculate the amplitude at a point P located on the z-axis at a distance R from the plane. The expected amplitude of the plane wave at P is:
		+
		+	$U_{\text{plane}}(P) = A e^{ikR}$.
		+
		+	We use the plane z = 0 as an auxiliary surface. Each element dS in polar coordinates $(\rho,\phi) has dS = \rho d\rho d\phi$.
		+	The distance to P is
		+	$ r = \sqrt{\rho^2 + R^2}$.
		+
		+	$U(P) = \iint_{\text{plane}} \frac{c A e^{ikr}}{r} dS
		+	= cA \int_0^{2\pi} d\phi \int_0^\infty \frac{e^{ik\sqrt{\rho^2+R^2}}}{\sqrt{\rho^2+R^2}} \rho d\rho$.
		+
		+	The angular integral gives $2\pi$. We make the substitution
		+	$u = \sqrt{\rho^2+R^2} ⇒ \rho d\rho = u du$. The limits: $ \rho = 0 \rightarrow u = R, \rho \to \infty \rightarrow u \to \infty$.
		+
		+	$U(P) = 2\pi c A \int_R^\infty \frac{e^{iku}}{u} u du = 2\pi c A \int_R^\infty e^{iku} du$.
		+
		+	The improper integral is evaluated by introducing a convergence factor
		+	$ e^{-\epsilon u} and taking \epsilon \to 0^+$:
		+
		+	$\int_R^\infty e^{iku} du = \lim_{\epsilon\to 0^+} \int_R^\infty e^{i(k+i\epsilon)u} du
		+	= \lim_{\epsilon\to 0^+} \frac{1}{i(k+i\epsilon)} \left[ e^{i(k+i\epsilon)u} \right]_R^\infty
		+	= \frac{i}{k} e^{ikR}$.
		+
		+	Therefore:
		+
		+	$U(P) = 2\pi c A \cdot \frac{i}{k} e^{ikR}$.
		+
		+	We equate this amplitude to that of the plane wave $A e^{ikR}$:
		+
		+	2\pi c A \frac{i}{k} e^{ikR} = A e^{ikR}.
		+
		+	Canceling A e^{ikR} (assumed non-zero), we obtain:
		+
		+	$2\pi c \frac{i}{k} = 1 \quad\Rightarrow\quad c = \frac{k}{2\pi i} = \frac{1}{i\lambda}$
		+
		+	since $k = 2\pi/\lambda$. It can also be written as:
		+
		+	$c = -\frac{i}{\lambda}$.
		+
		+	---
		+
		+	5. Final result
		+
		+	The elementary amplitude of a Huygens–Fresnel secondary wave is:
		+
		+	\boxed{a = \frac{1}{i\lambda} \frac{A \Delta S}{r} e^{ikr} = -\frac{i}{\lambda} \frac{A \Delta S}{r} e^{ikr}}.
		+
		+	The proportionality constant c is 1/(i\lambda) (or equivalently -i/\lambda). The factor 1/i = -i implies a phase shift of \pi/2 between the secondary wave and the primary one
		+
		+	#### Answer
		+
		+	[Insert a concise answer or boxed result]

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			### Statement

			$12.2.12.$ [Insert the problem statement]

			### Solution

			Huygens–Fresnel Principle

			Each point of a wavefront becomes an emitter of monochromatic secondary spherical waves. The total disturbance at a point P is obtained by integrating the contributions from all elements dS of the wavefront. For an element dS, the contribution is:

			$dU_P = \frac{c A \, e^{ikr}}{r} \, dS$

			where A is the amplitude of the primary wave at the wavefront, r is the distance from the element to P, $k = 2\pi/\lambda$, and c is the constant to be determined.

			Consider a plane wave propagating in the z-direction. At t = 0, the wavefront coincides with the plane z = 0. We want to calculate the amplitude at a point P located on the z-axis at a distance R from the plane. The expected amplitude of the plane wave at P is:

			$U_{\text{plane}}(P) = A e^{ikR}$.

			We use the plane z = 0 as an auxiliary surface. Each element dS in polar coordinates $(\rho,\phi) has dS = \rho d\rho d\phi$.
			The distance to P is
			$ r = \sqrt{\rho^2 + R^2}$.

			$U(P) = \iint_{\text{plane}} \frac{c A e^{ikr}}{r} dS
			= cA \int_0^{2\pi} d\phi \int_0^\infty \frac{e^{ik\sqrt{\rho^2+R^2}}}{\sqrt{\rho^2+R^2}} \rho d\rho$.

			The angular integral gives $2\pi$. We make the substitution
			$u = \sqrt{\rho^2+R^2} ⇒ \rho d\rho = u du$. The limits: $ \rho = 0 \rightarrow u = R, \rho \to \infty \rightarrow u \to \infty$.

			$U(P) = 2\pi c A \int_R^\infty \frac{e^{iku}}{u} u du = 2\pi c A \int_R^\infty e^{iku} du$.

			The improper integral is evaluated by introducing a convergence factor
			$ e^{-\epsilon u} and taking \epsilon \to 0^+$:

			$\int_R^\infty e^{iku} du = \lim_{\epsilon\to 0^+} \int_R^\infty e^{i(k+i\epsilon)u} du
			= \lim_{\epsilon\to 0^+} \frac{1}{i(k+i\epsilon)} \left[ e^{i(k+i\epsilon)u} \right]_R^\infty
			= \frac{i}{k} e^{ikR}$.

			Therefore:

			$U(P) = 2\pi c A \cdot \frac{i}{k} e^{ikR}$.

			We equate this amplitude to that of the plane wave $A e^{ikR}$:

			2\pi c A \frac{i}{k} e^{ikR} = A e^{ikR}.

			Canceling A e^{ikR} (assumed non-zero), we obtain:

			$2\pi c \frac{i}{k} = 1 \quad\Rightarrow\quad c = \frac{k}{2\pi i} = \frac{1}{i\lambda}$

			since $k = 2\pi/\lambda$. It can also be written as:

			$c = -\frac{i}{\lambda}$.

			---

			5. Final result

			The elementary amplitude of a Huygens–Fresnel secondary wave is:

			\boxed{a = \frac{1}{i\lambda} \frac{A \Delta S}{r} e^{ikr} = -\frac{i}{\lambda} \frac{A \Delta S}{r} e^{ikr}}.

			The proportionality constant c is 1/(i\lambda) (or equivalently -i/\lambda). The factor 1/i = -i implies a phase shift of \pi/2 between the secondary wave and the primary one

			#### Answer

			[Insert a concise answer or boxed result]