Новое решение

Alexphysics правка от 16 июн. 2026 г., 16:58

правка #18929 позже →

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		+	### Statement
		+
		+	$12.2.11.$ [Insert the problem statement]
		+
		+	### Solution
		+
		+	Data:
		+
		+	· Parallel beam of monochromatic light of wavelength $\lambda$
		+	· Observation point A on the axis.
		+	· The plate has rings that coincide with the Fresnel zones.
		+	· It is considered that the plate covers the first m = 9 Fresnel zones.
		+
		+	Reference amplitude (without plate):
		+	With the entire wavefront open, the amplitude at A is half of the first zone:
		+
		+	$A_{\text{free}} \approx \frac{A_1}{2}$
		+
		+	and the corresponding intensity is
		+
		+	$ I_0 \propto \left(\frac{A_1}{2}\right)^2 = \frac{A_1^2}{4}$
		+
		+	---
		+
		+	Case (a): Opaque plate on even zones
		+
		+	The plate has the even zones blackened out (blocked): 2, 4, 6, 8.
		+	The odd zones remain open: 1, 3, 5, 7, 9 (five zones).
		+
		+	Since the amplitudes of the zones are $approximately equal (A_1 \approx A_3 \approx A_5 \approx A_7 \approx A_9)$, the total amplitude at A is:
		+
		+	$A_a = 5 A_1$
		+
		+	Intensity:
		+
		+	$I_a \propto (5 A_1)^2 = 25 A_1^2$
		+
		+	Compared with the intensity without the plate:
		+
		+	$\frac{I_a}{I_0} = \frac{25 A_1^2}{A_1^2/4} = 25 \times 4 = 100$
		+
		+	$\boxed{\text{The intensity increases by a factor of 100.}}$
		+
		+	---
		+
		+	Case (b): Phase plate (phase shift \pi) on even zones
		+
		+	The rings that were previously opaque are replaced by dielectric layers that add a phase shift of \pi to the wave passing through.
		+
		+	· The odd zones (1, 3, 5, 7, 9) pass with no phase change.
		+	· The even zones (2, 4, 6, 8) would normally arrive with a phase shift of $\pi $due to the path difference, but the layer adds another \pi, so they end up in phase with the odd ones.
		+
		+	Now all zones (9 in total) contribute constructively:
		+
		+	$A_b = 9 A_1$
		+
		+	Intensity:
		+
		+	$I_b \propto (9 A_1)^2 = 81 A_1^2$
		+
		+	Compared with the intensity without the plate:
		+
		+	$\frac{I_b}{I_0} = \frac{81 A_1^2}{A_1^2/4} = 81 \times 4 = 324$
		+
		+	$\boxed{\text{The intensity increases by a factor of 324.}}$
		+
		+	#### Answer
		+
		+	[Insert a concise answer or boxed result]

	@@ -0,0 +1,69 @@
			### Statement

			$12.2.11.$ [Insert the problem statement]

			### Solution

			Data:

			· Parallel beam of monochromatic light of wavelength $\lambda$
			· Observation point A on the axis.
			· The plate has rings that coincide with the Fresnel zones.
			· It is considered that the plate covers the first m = 9 Fresnel zones.

			Reference amplitude (without plate):
			With the entire wavefront open, the amplitude at A is half of the first zone:

			$A_{\text{free}} \approx \frac{A_1}{2}$

			and the corresponding intensity is

			$ I_0 \propto \left(\frac{A_1}{2}\right)^2 = \frac{A_1^2}{4}$

			---

			Case (a): Opaque plate on even zones

			The plate has the even zones blackened out (blocked): 2, 4, 6, 8.
			The odd zones remain open: 1, 3, 5, 7, 9 (five zones).

			Since the amplitudes of the zones are $approximately equal (A_1 \approx A_3 \approx A_5 \approx A_7 \approx A_9)$, the total amplitude at A is:

			$A_a = 5 A_1$

			Intensity:

			$I_a \propto (5 A_1)^2 = 25 A_1^2$

			Compared with the intensity without the plate:

			$\frac{I_a}{I_0} = \frac{25 A_1^2}{A_1^2/4} = 25 \times 4 = 100$

			$\boxed{\text{The intensity increases by a factor of 100.}}$

			---

			Case (b): Phase plate (phase shift \pi) on even zones

			The rings that were previously opaque are replaced by dielectric layers that add a phase shift of \pi to the wave passing through.

			· The odd zones (1, 3, 5, 7, 9) pass with no phase change.
			· The even zones (2, 4, 6, 8) would normally arrive with a phase shift of $\pi $due to the path difference, but the layer adds another \pi, so they end up in phase with the odd ones.

			Now all zones (9 in total) contribute constructively:

			$A_b = 9 A_1$

			Intensity:

			$I_b \propto (9 A_1)^2 = 81 A_1^2$

			Compared with the intensity without the plate:

			$\frac{I_b}{I_0} = \frac{81 A_1^2}{A_1^2/4} = 81 \times 4 = 324$

			$\boxed{\text{The intensity increases by a factor of 324.}}$

			#### Answer

			[Insert a concise answer or boxed result]