Edits to “Statement”, “Solution”

Alexphysics edited Jun 16, 2026, 5:03 PM

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		@@ -1,16 +1,30 @@
			### Statement

		−	$12.2.11.$ ~~[Insert the problem statement]~~
		+	$12.2.11.$

		+	The drawing shows a flat glass plate with blackened ring sections. This
		+	plate overlapped a parallel beam of monochromatic light with a wavelength
		+	λ. It turned out that the blackened rings of the plate coincided with even
		+	Fresnel zones for the axial point A. How did the light intensity change at this
		+	point?
		+	b. A parallel beam of monochromatic light was blocked by a plate in which the
		+	blackened ring sections were replaced by layers of a dielectric that changes the
		+	phase of the passing wave by π. How did the light intensity change at point
		+	A in this case?
		+	[For problem $12.2.11$\|720x224, 50%](../../img/12.2.11/1000125861.jpg)!
		+
			### 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.
		+	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
		@@ -20,14 +34,15 @@Solution

			$ 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:
		+	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]
		unchanged lines 36

	@@ -1,16 +1,30 @@
	### Statement		### Statement

	$12.2.11.$ ~~[Insert the problem statement]~~		$12.2.11.$

			The drawing shows a flat glass plate with blackened ring sections. This
			plate overlapped a parallel beam of monochromatic light with a wavelength
			λ. It turned out that the blackened rings of the plate coincided with even
			Fresnel zones for the axial point A. How did the light intensity change at this
			point?
			b. A parallel beam of monochromatic light was blocked by a plate in which the
			blackened ring sections were replaced by layers of a dielectric that changes the
			phase of the passing wave by π. How did the light intensity change at point
			A in this case?
			[For problem $12.2.11$\|720x224, 50%](../../img/12.2.11/1000125861.jpg)!

	### Solution		### Solution

	Data:		Data:

	· Parallel beam of monochromatic light of wavelength $\lambda$		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.

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

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

	and the corresponding intensity is		and the corresponding intensity is
	@@ -20,14 +34,15 @@Solution

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

	---

	Case (a): Opaque plate on even zones		Case (a): Opaque plate on even zones

	The plate has the even zones blackened out (blocked): 2, 4, 6, 8.		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).		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:		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:

	$A_a = 5 A_1$		$A_a = 5 A_1$

	Intensity:		Intensity:

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

	Compared with the intensity without the plate:		Compared with the intensity without the plate:

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

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

	---		---

	Case (b): Phase plate (phase shift \pi) on even zones		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 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 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.		· 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:		Now all zones (9 in total) contribute constructively:

	$A_b = 9 A_1$		$A_b = 9 A_1$

	Intensity:		Intensity:

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

	Compared with the intensity without the plate:		Compared with the intensity without the plate:

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

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

	#### Answer		#### Answer

	[Insert a concise answer or boxed result]		[Insert a concise answer or boxed result]
	unchanged lines 36