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Alexphysics edited Jun 16, 2026, 4:09 AM

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		+	### Statement
		+
		+	$12.2.8.$ [Insert the problem statement]
		+
		+	### Solution
		+
		+	Definition of Fresnel zones
		+
		+	Consider a spherical wavefront passing through a circular hole of variable radius. The observation point A is on the axis, at a distance b from the diaphragm. The wavefront is divided into Fresnel zones: the m-th zone is a ring bounded by radii r_{m-1} and r_m such that the distance from the edge of the m-th zone to point A is b + m\lambda/2.
		+
		+	The radius of the m-th zone (for m not too large) is:
		+
		+	r_m \approx \sqrt{m \lambda \frac{ab}{a+b}},
		+
		+	where a is the distance from the source to the diaphragm.
		+
		+	The area of each zone is approximately constant:
		+
		+	\Delta S_m \approx \frac{\pi \lambda a b}{a+b} \quad \text{(independent of } m\text{)}.
		+
		+	Amplitude of the resultant wave at A
		+
		+	Each zone contributes an amplitude A_m at point A. Due to the path difference of \lambda/2 between consecutive zones, the contributions alternate in phase:
		+
		+	· Zone 1: amplitude A_1, reference phase.
		+	· Zone 2: amplitude A_2, phase shifted by \pi (opposite).
		+	· Zone 3: amplitude A_3, in phase with zone 1
		+
		+	The total amplitude (considering an aperture with N complete zones) is the alternating sum:
		+
		+	A = A_1 - A_2 + A_3 - A_4 + \cdots \pm A_N.
		+
		+	The A_m decrease slowly as m increases due to inclination (obliquity factor), but for the first few zones they are practically equal: A_1 \approx A_2 \approx A_3 \dots
		+
		+	Case of 1 Fresnel zone (radius r_1)
		+
		+	Only the first zone is uncovered:
		+
		+	A^{(1)} \approx A_1.
		+
		+	Intensity at A:
		+
		+	I^{(1)} \propto A_1^2.
		+
		+	This is the maximum possible intensity for that source and distance (about 4 times the intensity without the screen).
		+
		+	Case of 2 zones (radius r_2)
		+
		+	Now the first two complete zones are uncovered:
		+
		+	A^{(2)} = A_1 - A_2.
		+
		+	Since A_1 \approx A_2, it follows that:
		+
		+	A^{(2)} \approx 0 \quad\Longrightarrow\quad \boxed{I^{(2)} \approx 0}.
		+
		+	Destructive interference between the two zones produces darkness at the center (point A).
		+
		+	The power passing through the hole is proportional to the total open area. The area of one zone is \Delta S. For 1 zone, area \approx \Delta S; for 2 zones, area \approx 2\Delta S.
		+	Thus the flux roughly doubles when going from 1 to 2 zones:
		+
		+	\Phi^{(2)} \approx 2\,\Phi^{(1)}.
		+
		+	Energy is not obliged to concentrate at the central point A.
		+
		+	· With 1 zone, almost all the energy that passes through is focused at A (single diffraction maximum).
		+	· With 2 zones, interference cancels the intensity on the axis, but the energy is redistributed into lateral directions (a bright ring appears around the dark center, corresponding to the secondary maxima of Fresnel diffraction).
		+
		+	Conservation of energy: The integral of the intensity over a closed surface must equal the total power passing through the hole. The energy that is "missing" at A is found in the side lobes.
		+
		+	---
		+
		+	Conclusion with equations
		+
		+	The total amplitude on the axis is the alternating sum of the zone contributions. For N = 2, one obtains A \approx 0 and therefore I_{\text{center}} \approx 0. At the same time, the uncovered area is double, so the total flux is also double. The energy does not disappear; rather, it is diffracted away from the central axis. Both facts are perfectly compatible and are described by the same Fresnel formalism.
		+
		+	#### Answer
		+
		+	[Insert a concise answer or boxed result]

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			### Statement

			$12.2.8.$ [Insert the problem statement]

			### Solution

			Definition of Fresnel zones

			Consider a spherical wavefront passing through a circular hole of variable radius. The observation point A is on the axis, at a distance b from the diaphragm. The wavefront is divided into Fresnel zones: the m-th zone is a ring bounded by radii r_{m-1} and r_m such that the distance from the edge of the m-th zone to point A is b + m\lambda/2.

			The radius of the m-th zone (for m not too large) is:

			r_m \approx \sqrt{m \lambda \frac{ab}{a+b}},

			where a is the distance from the source to the diaphragm.

			The area of each zone is approximately constant:

			\Delta S_m \approx \frac{\pi \lambda a b}{a+b} \quad \text{(independent of } m\text{)}.

			Amplitude of the resultant wave at A

			Each zone contributes an amplitude A_m at point A. Due to the path difference of \lambda/2 between consecutive zones, the contributions alternate in phase:

			· Zone 1: amplitude A_1, reference phase.
			· Zone 2: amplitude A_2, phase shifted by \pi (opposite).
			· Zone 3: amplitude A_3, in phase with zone 1

			The total amplitude (considering an aperture with N complete zones) is the alternating sum:

			A = A_1 - A_2 + A_3 - A_4 + \cdots \pm A_N.

			The A_m decrease slowly as m increases due to inclination (obliquity factor), but for the first few zones they are practically equal: A_1 \approx A_2 \approx A_3 \dots

			Case of 1 Fresnel zone (radius r_1)

			Only the first zone is uncovered:

			A^{(1)} \approx A_1.

			Intensity at A:

			I^{(1)} \propto A_1^2.

			This is the maximum possible intensity for that source and distance (about 4 times the intensity without the screen).

			Case of 2 zones (radius r_2)

			Now the first two complete zones are uncovered:

			A^{(2)} = A_1 - A_2.

			Since A_1 \approx A_2, it follows that:

			A^{(2)} \approx 0 \quad\Longrightarrow\quad \boxed{I^{(2)} \approx 0}.

			Destructive interference between the two zones produces darkness at the center (point A).

			The power passing through the hole is proportional to the total open area. The area of one zone is \Delta S. For 1 zone, area \approx \Delta S; for 2 zones, area \approx 2\Delta S.
			Thus the flux roughly doubles when going from 1 to 2 zones:

			\Phi^{(2)} \approx 2\,\Phi^{(1)}.

			Energy is not obliged to concentrate at the central point A.

			· With 1 zone, almost all the energy that passes through is focused at A (single diffraction maximum).
			· With 2 zones, interference cancels the intensity on the axis, but the energy is redistributed into lateral directions (a bright ring appears around the dark center, corresponding to the secondary maxima of Fresnel diffraction).

			Conservation of energy: The integral of the intensity over a closed surface must equal the total power passing through the hole. The energy that is "missing" at A is found in the side lobes.

			---

			Conclusion with equations

			The total amplitude on the axis is the alternating sum of the zone contributions. For N = 2, one obtains A \approx 0 and therefore I_{\text{center}} \approx 0. At the same time, the uncovered area is double, so the total flux is also double. The energy does not disappear; rather, it is diffracted away from the central axis. Both facts are perfectly compatible and are described by the same Fresnel formalism.

			#### Answer

			[Insert a concise answer or boxed result]