Edits to “Statement”, “Solution”, “Answer”

Alexphysics edited Jun 16, 2026, 2:25 AM

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			### Statement

		−	$12.2.6.$ ~~[Insert the problem statement]~~
		+	$12.2.6.$

		+	A very narrow beam of Kα-radiation from copper (λ = 1.541 · 10−10m) falls
		+	on a glass diffraction grating having 200 lines per 1 mm and covered with a
		+	thin layer of gold at an angle of 20′to its surface. Determine the difference in
		+	reflection angles between the first-and zero-order beams.
		+
			### Solution

		−	Data and geometry
		+	first remark data and geometry

		−	· Wavelength: \lambda = 1.541 \times 10^{-10}\,\text{m}.
		−	· Diffraction grating: N = 200 lines/mm.
		−	d = \frac{1}{N} = \frac{1}{200\,\text{mm}^{-1}} = 5 \times 10^{-6}\,\text{m}.
		−	· Grazing incidence angle (relative to the surface): \alpha = 20'.
		−	\alpha = \frac{20}{60}^\circ = \frac{1}{3}^\circ \approx 0.3333^\circ.
		+	· Wavelength: $\lambda = 1.541 \times 10^{-10}\,\text{m}$
		+
		+	· Diffraction grating:$ N = 200 lines/mm$.
		+	$d = \frac{1}{N} = \frac{1}{200\,\text{mm}^{-1}} = 5 \times 10^{-6}\,\text{m}$.
		+
		+	· Grazing incidence angle (relative to the surface):$ \alpha = 20'.
		+	\alpha = \frac{20}{60}^\circ = \frac{1}{3}^\circ \approx 0.3333^\circ$
		+
			· The grating works in reflection, and the gold layer ensures high reflectivity without modifying the diffraction condition.

			When working with grazing angles (measured from the surface), the diffraction condition for a reflection grating is:

		−	d(\cos\alpha - \cos\beta_m) = m\lambda, \qquad m = 0, 1, 2, \dots
		+	$d(\cos\alpha - \cos\beta_m) = m\lambda, \qquad m = 0, 1, 2, \dots$

		−	· \alpha: incident angle relative to the surface.
		−	· \beta_m: diffracted beam angle of order m relative to the surface.
		+	· $\alpha$: incident angle relative to the surface.
		+
		+	· $\beta_m$: diffracted beam angle of order m relative to the surface.
		+
			· d: grating period.
		+
			· m: diffraction order.

		−	For m = 0:

		−	d(\cos\alpha - \cos\beta_0) = 0 \;\Longrightarrow\; \cos\beta_0 = \cos\alpha.
		+	now For m = 0:

		−	Since angles are between 0 and \pi/2, we obtain:
		+	$d(\cos\alpha - \cos\beta_0) = 0 \;\Longrightarrow\; \cos\beta_0 = \cos\alpha$.

		−	\beta_0 = \alpha = 20'.
		+	Since angles are between 0 and $\pi/2$, we obtain:

		+	$\beta_0 = \alpha = 20'$.
		+
			This is specular reflection.

		−	For m = 1:
		+	now For m = 1:

		−	\cos\beta_1 = \cos\alpha - \frac{\lambda}{d}.
		+	$\cos\beta_1 = \cos\alpha - \frac{\lambda}{d}$.

			Compute each term:

		−	· \cos\alpha = \cos(20') = \cos(0.33333^\circ).
		−	Using a calculator or series: \cos(0.33333^\circ) \approx 0.9999831.
		−	· \lambda / d = \dfrac{1.541 \times 10^{-10}}{5 \times 10^{-6}} = 3.082 \times 10^{-5} = 0.00003082.
		+	· $\cos\alpha = \cos(20') = \cos(0.33333^\circ)$.
		+
		+	Using a calculator or series: $\cos(0.33333^\circ) \approx 0.9999831$.

		+	· $\lambda / d = \dfrac{1.541 \times 10^{-10}}{5 \times 10^{-6}} = 3.082 \times 10^{-5} = 0.00003082$.
		+
			Then:

		−	\cos\beta_1 = 0.9999831 - 0.00003082 = 0.99995228.
		+	$\cos\beta_1 = 0.9999831 - 0.00003082 = 0.99995228$.

			Thus:

		−	\beta_1 = \arccos(0.99995228).
		+	$\beta_1 = \arccos(0.99995228)$.

		−	For small angles we can use \cos\theta \approx 1 - \theta^2/2:
		+	For small angles we can use $\cos\theta \approx 1 - \theta^2/2$:

		−	1 - \frac{\beta_1^2}{2} \approx 0.99995228 \;\Longrightarrow\; \frac{\beta_1^2}{2} \approx 0.00004772 \;\Longrightarrow\; \beta_1 \approx \sqrt{0.00009544} \approx 0.00977\,\text{rad}.
		+	$1 - \frac{\beta_1^2}{2} \approx 0.99995228 \;\Longrightarrow\; \frac{\beta_1^2}{2} \approx 0.00004772 \;\Longrightarrow\; \beta_1 \approx \sqrt{0.00009544} \approx 0.00977\,\text{rad}$.

		−	In degrees: 0.00977\,\text{rad} \times \dfrac{180^\circ}{\pi} \approx 0.5597^\circ.
		+	In degrees:

		−	~~Exact calculation with arccos: \beta_1~~ \approx 0.5594^\circ.
		+	$0.00977\,\text{rad} \times \dfrac{180^\circ}{\pi} \approx 0.5597^\circ$.

		+	Exact calculation with arccos: $\beta_1 \approx 0.5594^\circ$.
		+
			The desired difference is:

		−	\Delta\beta = \beta_1 - \beta_0 = \beta_1 - \alpha.
		+	$\Delta\beta = \beta_1 - \beta_0 = \beta_1 - \alpha$.

			Substituting:

		−	\Delta\beta \approx 0.5594^\circ - 0.3333^\circ = 0.2261^\circ.
		+	$\Delta\beta \approx 0.5594^\circ - 0.3333^\circ = 0.2261^\circ$.

			In minutes of arc:

		−	0.2261^\circ \times 60'/\text{degree} \approx 13.57'.
		+	$0.2261^\circ \times 60'/\text{degree} \approx 13.57'$.

		−	---

		−	Final result
		+	#### Answer

		−	\boxed{\Delta\beta \approx 0.226^\circ \;\;(\text{about } 13.6\text{ minutes of arc})}.
		+	The Final result

		−	Interpretation:
		−	The zero-order beam is specularly reflected (grazing angle of 20′), while the first-order beam emerges at a slightly larger angle (≈ 33.6′). The difference is small because the wavelength is much smaller than the grating period. The gold layer does not affect the diffraction condition; it merely ensures reflection of the X‑rays.
		+	$\boxed{\Delta\beta \approx 0.226^\circ \;\;(\text{about } 13.6\text{ minutes of arc})}$.

		−
		−	#### Answer
		−
		−	[Insert a concise answer or boxed result]
		+	Interpretation:
		+	The zero-order beam is specularly reflected (grazing angle of 20′), while the first-order beam emerges at a slightly larger angle (≈ 33.6′).
		+	The difference is small because the wavelength is much smaller than the grating period. The gold layer does not affect the diffraction condition; it merely ensures reflection of the X‑rays.

	@@ -1,85 +1,98 @@
	### Statement		### Statement

	$12.2.6.$ ~~[Insert the problem statement]~~		$12.2.6.$

			A very narrow beam of Kα-radiation from copper (λ = 1.541 · 10−10m) falls
			on a glass diffraction grating having 200 lines per 1 mm and covered with a
			thin layer of gold at an angle of 20′to its surface. Determine the difference in
			reflection angles between the first-and zero-order beams.

	### Solution		### Solution

	Data and geometry		first remark data and geometry

	· Wavelength: \lambda = 1.541 \times 10^{-10}\,\text{m}.		· Wavelength: $\lambda = 1.541 \times 10^{-10}\,\text{m}$
	· Diffraction grating: N = 200 lines/mm.
	d = \frac{1}{N} = \frac{1}{200\,\text{mm}^{-1}} = 5 \times 10^{-6}\,\text{m}.		· Diffraction grating:$ N = 200 lines/mm$.
	· Grazing incidence angle (relative to the surface): \alpha = 20'.		$d = \frac{1}{N} = \frac{1}{200\,\text{mm}^{-1}} = 5 \times 10^{-6}\,\text{m}$.
	\alpha = \frac{20}{60}^\circ = \frac{1}{3}^\circ \approx 0.3333^\circ.
			· Grazing incidence angle (relative to the surface):$ \alpha = 20'.
			\alpha = \frac{20}{60}^\circ = \frac{1}{3}^\circ \approx 0.3333^\circ$

	· The grating works in reflection, and the gold layer ensures high reflectivity without modifying the diffraction condition.		· The grating works in reflection, and the gold layer ensures high reflectivity without modifying the diffraction condition.

	When working with grazing angles (measured from the surface), the diffraction condition for a reflection grating is:		When working with grazing angles (measured from the surface), the diffraction condition for a reflection grating is:

	d(\cos\alpha - \cos\beta_m) = m\lambda, \qquad m = 0, 1, 2, \dots		$d(\cos\alpha - \cos\beta_m) = m\lambda, \qquad m = 0, 1, 2, \dots$

	· \alpha: incident angle relative to the surface.		· $\alpha$: incident angle relative to the surface.
	· \beta_m: diffracted beam angle of order m relative to the surface.
			· $\beta_m$: diffracted beam angle of order m relative to the surface.

	· d: grating period.		· d: grating period.

	· m: diffraction order.		· m: diffraction order.

	For m = 0:

	d(\cos\alpha - \cos\beta_0) = 0 \;\Longrightarrow\; \cos\beta_0 = \cos\alpha.		now For m = 0:

	Since angles are between 0 and \pi/2, we obtain:		$d(\cos\alpha - \cos\beta_0) = 0 \;\Longrightarrow\; \cos\beta_0 = \cos\alpha$.

	\beta_0 = \alpha = 20'.		Since angles are between 0 and $\pi/2$, we obtain:

			$\beta_0 = \alpha = 20'$.

	This is specular reflection.		This is specular reflection.

	For m = 1:		now For m = 1:

	\cos\beta_1 = \cos\alpha - \frac{\lambda}{d}.		$\cos\beta_1 = \cos\alpha - \frac{\lambda}{d}$.

	Compute each term:		Compute each term:

	· \cos\alpha = \cos(20') = \cos(0.33333^\circ).		· $\cos\alpha = \cos(20') = \cos(0.33333^\circ)$.
	Using a calculator or series: \cos(0.33333^\circ) \approx 0.9999831.
	· \lambda / d = \dfrac{1.541 \times 10^{-10}}{5 \times 10^{-6}} = 3.082 \times 10^{-5} = 0.00003082.		Using a calculator or series: $\cos(0.33333^\circ) \approx 0.9999831$.

			· $\lambda / d = \dfrac{1.541 \times 10^{-10}}{5 \times 10^{-6}} = 3.082 \times 10^{-5} = 0.00003082$.

	Then:		Then:

	\cos\beta_1 = 0.9999831 - 0.00003082 = 0.99995228.		$\cos\beta_1 = 0.9999831 - 0.00003082 = 0.99995228$.

	Thus:		Thus:

	\beta_1 = \arccos(0.99995228).		$\beta_1 = \arccos(0.99995228)$.

	For small angles we can use \cos\theta \approx 1 - \theta^2/2:		For small angles we can use $\cos\theta \approx 1 - \theta^2/2$:

	1 - \frac{\beta_1^2}{2} \approx 0.99995228 \;\Longrightarrow\; \frac{\beta_1^2}{2} \approx 0.00004772 \;\Longrightarrow\; \beta_1 \approx \sqrt{0.00009544} \approx 0.00977\,\text{rad}.		$1 - \frac{\beta_1^2}{2} \approx 0.99995228 \;\Longrightarrow\; \frac{\beta_1^2}{2} \approx 0.00004772 \;\Longrightarrow\; \beta_1 \approx \sqrt{0.00009544} \approx 0.00977\,\text{rad}$.

	In degrees: 0.00977\,\text{rad} \times \dfrac{180^\circ}{\pi} \approx 0.5597^\circ.		In degrees:

	~~Exact calculation with arccos: \beta_1~~ \approx 0.5594^\circ.		$0.00977\,\text{rad} \times \dfrac{180^\circ}{\pi} \approx 0.5597^\circ$.

			Exact calculation with arccos: $\beta_1 \approx 0.5594^\circ$.

	The desired difference is:		The desired difference is:

	\Delta\beta = \beta_1 - \beta_0 = \beta_1 - \alpha.		$\Delta\beta = \beta_1 - \beta_0 = \beta_1 - \alpha$.

	Substituting:		Substituting:

	\Delta\beta \approx 0.5594^\circ - 0.3333^\circ = 0.2261^\circ.		$\Delta\beta \approx 0.5594^\circ - 0.3333^\circ = 0.2261^\circ$.

	In minutes of arc:		In minutes of arc:

	0.2261^\circ \times 60'/\text{degree} \approx 13.57'.		$0.2261^\circ \times 60'/\text{degree} \approx 13.57'$.

	---

	Final result		#### Answer

	\boxed{\Delta\beta \approx 0.226^\circ \;\;(\text{about } 13.6\text{ minutes of arc})}.		The Final result

	Interpretation:		$\boxed{\Delta\beta \approx 0.226^\circ \;\;(\text{about } 13.6\text{ minutes of arc})}$.
	The zero-order beam is specularly reflected (grazing angle of 20′), while the first-order beam emerges at a slightly larger angle (≈ 33.6′). The difference is small because the wavelength is much smaller than the grating period. The gold layer does not affect the diffraction condition; it merely ensures reflection of the X‑rays.

			Interpretation:
	#### Answer		The zero-order beam is specularly reflected (grazing angle of 20′), while the first-order beam emerges at a slightly larger angle (≈ 33.6′).
			The difference is small because the wavelength is much smaller than the grating period. The gold layer does not affect the diffraction condition; it merely ensures reflection of the X‑rays.
	[Insert a concise answer or boxed result]