Решение до правки #18896 от , автор Alexphysics. Это не текущая версия.

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]