EuPhO 2025, Експеримент — Задача 2. E2 – Hidden pattern

E2 – Solution

Task a.
The light transmitted through the film forms a primary and secondary diffraction pattern as displayed in Fig. 13.

Figure 13: Photo of the diffraction pattern with the primary and secondary pattern marked.

The secondary diffraction pattern is only required for task d. For the primary diffraction pattern, a number of bright interference maxima lying on the same line is visible. This fact implies that locally, within the cross section of the laser spot, the printed lines on the film surface form a diffraction grating consisting of a large number of practically linear parallel fringes. Therefore, the diffraction maxima are situated on a straight line, perpendicular to the tangent to the illuminated sinusoids, like illustrated in Fig. 14.

Figure 14: Qualitative sketch of interference pattern.

By scanning the laser in a horizontal direction, the diffraction pattern will tilt according to the line slope in the illuminated spot. The pattern will be vertical when the laser incidents on crests or valleys of the illuminated sinusoids. Therefore, the distance between two consecutive vertical positions of the diffraction pattern is $l = \Lambda/2$, hence $\Lambda = 2l$.

The experimental setup is designed as follows: As a first step, we fix the foil with two clips to the L-shaped stand and align it carefully vertically. The goniometer is glued to the screen with the $0^{\circ}$-division pointing vertically. Next we fix the laser to another L-shaped stand and align it so that the beam hits the centre of the goniometer. We place the screen (goniometer) as far as possible from the foil to achieve larger displacements of the maxima and hence a better precision, also see Fig. 15.

Figure 15: Setup with optimized usage of the table space for improved precision.

The stand with the foil is being displaced in small steps across the laser beam, and the angle of inclination $\theta$ of the diffraction pattern is being measured as function of the distance $x$ between the laser spot and the edge of the film.

Further we put the ruler on the diagonal scale so that one of its edges is aligned with the alignment line on the scale; we'll be sliding the L-shaped stand with the foil along the edge of the ruler. That way we can focus on observing how the diffraction maxima shift while we slide the stand, without a need for sharing our attention between the diffraction maxima and alignment of the stand. A sample data set is recorded in the first two columns of Table 7, while the corresponding graph of $\theta$ vs. $x$ is shown in Fig. 16.

Figure 16: Graph of $\theta$ vs. $x$.

In what follows we'll be outlining two possible approaches for achieving precise experimental results of the quantities asked in this problem: approach A: graphical; approach B: carefully scanning the diffraction pattern around critical configurations.

Approach A. Points $x_1$ and $x_2$ in Fig. 16 corresponding to $\theta = 0^{\circ}$, i.e. consecutive crest and valley of the sinusoid, can be obtained by linear fits of the four points nearest to the zero-crossing points:

Therefore, $l = x_2 - x_1 = 30.5$ mm with uncertainty $\delta l = \sqrt{\delta x_1^2 + \delta x_2^2} = 0.7$ mm, and

Approach B. Alternatively, we move the stand with the foil so that the diffraction maxima align along a vertical line and take the reading $x_1$ from the diagonal scale. Next we slide the stand along the ruler to find the other position where the diffraction maxima align along a vertical line and take the corresponding reading $x_2$. We can see that achieving a vertical alignment is not easy and incurs an uncertainty, so we need to repeat the measurements. The measurement data are shown in the table below.

According to this we calculate

and $\Delta\Lambda = 0.16$ mm

Task b.
Approach A. It can be easily deduced from Fig. 14 that the period of the diffraction grating, perpendicularly to the printed lines, is

The angle to the $n$-th order diffraction maximum is:

The distance between the 0-th and the $n$-th maxima on the screen is

(9)

where $L = 74.6$ cm is the distance between the film and the screen, and $\sin\varphi \approx \tan\varphi$ since involved angles are much smaller than 1 rad. Therefore, if we choose a specific order maximum and measure the distance $R$ at different points on the screen, i.e. for different angles $\theta$, the distance $d$ can be calculated as a sample average:

While the uncertainty – as a sample standard deviation of $d$. Since the 0-th order could be slightly offset from the goniometer center, we measure the corresponding distances between the two symmetric, $n$-th and $-n$-th, maxima: $D_n = R_n + R_{-n}$ and calculate

The last four columns of Table 7 summarize the measured distances and calculated value of $d$ for the 5-th maximum. By averaging the $d$ values, we obtain:

However, we need to keep in mind the uncertainty of the laser wave length. Adding the relative errors of the measurement data and laser wavelength according to the Pythagorean rule (applicable for uncorrelated error sources), we obtain

We note that if we use the data at $\theta = 0^{\circ}$, the diffraction maxima yield directly the line distance $d$ according to

(10)

Approach B. Alternatively to using many data points for different $\theta$, we can choose the positions $x = x_1$ or $x = x_2$ found in the previous task since they are the laterally points offering the highest precision. Analogously to approach A, we determine the distance between the symmetric diffraction maxima of highest observable order, $n = 6$, to achieve the highest possible precision. The best way to determine the distance $D_n$ is by marking dots onto the screen at the positions of the maxima, and measure the distance between the dots by ruler (to keep the goniometer clean, one can attach another sheet of paper to the stand). Since a single measurement will be very precise if done carefully, repeated measurements are not required for this task. The result of the measurement is $D_6 = 105.5$ mm with $L = 810$ mm, resulting in $d = 59.7\ \mu$m. Estimated error is $\pm 0.7\ \mu$m.

Task c.
Approach A: The point where a sinusoid crosses the $x$-axis corresponds to a maximum angle of inclination of the diffraction pattern $\theta = (27 \pm 1)^{\circ}$ and can be obtained as:

The sinusoid equation can be written as

where:

is the sinusoid wavevector. Since $\tan\theta = dy/dx$, we obtain:

Therefore, the auxiliary variables $z = \cos(k(x - x_0))$ and $t = \tan\theta$ are related by a linear dependence

(11)

and the amplitude of the sinusoid can be calculated by determining the slope coefficient $m$:

Calculated values of $t$ and $z$ are shown in the third and fourth column of Table 7, and the corresponding graph is shown in Fig. 17.

Figure 17: Graph of $\tan\theta$ vs. $\cos(k(x - x_0))$.

The slope coefficient is

Therefore:

Approach B. The amplitude of the sinusoids on the foil can be determined if we know the maximal slope the printed lines $\max_x \frac{dy}{dx} = Ak$ as we already know the wave vector $k = 2\pi/\Lambda$. We notice that the diffraction maxima lay on a line perpendicular to the lines on the foil at the point where the laser beam passes it, hence we can find the maximal slope of the lines as the maximal angle $\theta$ that the array of diffraction maxima form with the vertical axis,

This is done easily by using the same setup as in part A, by sliding the stand of the foil along the edge of the ruler and observing how the maxima move on the screen. As a result we obtain $\theta_{\max} = (27.5 \pm 0.3)^{\circ}$, corresponding to

Task d.

The vertical slabs that come from the printing technique lead to the secondary diffraction pattern as can be seen from the photo in Fig. 13. In this task, there also exist two different approaches that differ in the way the diffraction vectors are modelled but lead to the same and valid result.

Approach A: In the model of Huygen's elementary waves, the elementary wave sources along each of the vertical edges, as shown in thick red dotted lines in Fig. 18, will create wavefronts that propagate towards the screen and create interference pattern.

These edges have a regular distance $g$ of each other that depends on $\theta$, specifically:

Figure 18: Sketch for explanation of the relevant quantities for secondary diffraction pattern

From the diffraction angle $\omega$ of the maxima of the secondary diffraction pattern, $g$ can be expressed via:

Experimentally, we can obtain $\omega$ via

where $R_+ + R_-$ is the distance between the two diffraction orders visible (left and right from the centre) in the secondary pattern. We can combine this knowledge to obtain $g$:

Thus, we get $s$ via the trigonometric relation

It is important to notice that the secondary pattern can only be observed distinctly for large $\theta$ since only in these regions, the slope of the sinusoid barely changes, which in turn means that $\theta$ and the resulting diffraction angle $\omega$ is rather constant. In Tab. 6, the recorded measurement points for the same $L = 74.6$ cm is shown.

Table 6: Measurements of the secondary diffraction order distances for different angles $\theta$

A graphical evaluation has no benefit over computing the average of the point-wise results here, so we use the average of $s$ as the result and its standard deviation as the error estimate. Thus, we get:

Approach B: Alternatively the characteristic diffraction length is the hypotenuse of length $h$ of the blue triangle shown in Fig. 18, which is $h = s/\sin\theta$. Subsequently, the diffraction angle is

Where $\tilde{R}_+$ and $\tilde{R}_-$ are the positions of the secondary diffraction pattern orthogonal to the primary pattern. Thus, the formula for s becomes

which leads to the same outcome as in approach A since $\tilde{R}_+ = \cos\theta R_+$ and $\tilde{R}_- = \cos\theta R_-$.

Table 7: Measurement data for experiment “hidden pattern”