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+### Statement
+
+$12.2.6.$ [Insert the problem statement]
+
+### Solution
+
+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.
+· 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
+
+· \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.
+
+Since angles are between 0 and \pi/2, we obtain:
+
+\beta_0 = \alpha = 20'.
+
+This is specular reflection.
+
+For m = 1:
+
+\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.
+
+Then:
+
+\cos\beta_1 = 0.9999831 - 0.00003082 = 0.99995228.
+
+Thus:
+
+\beta_1 = \arccos(0.99995228).
+
+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}.
+
+In degrees: 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.
+
+Substituting:
+
+\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'.
+
+---
+
+Final result
+
+\boxed{\Delta\beta \approx 0.226^\circ \;\;(\text{about } 13.6\text{ minutes of arc})}.
+
+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.
+
+
+#### Answer
+
+[Insert a concise answer or boxed result]