Solution

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):$\alpha = 20'. \alpha = \frac{20}{60}^\circ = \frac{1}{3}^\circ \approx 0.3333^\circ$

· 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.

now 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.

now 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:

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