Съдържание

IPhO 2014, Експеримент — Задача 3. Part 2.2 Light transmission through plastic rulers

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от IPhO 2014

Experimental competition. Thursday, 17 July 2014 — Experiment. To see invisible! (20 points) · 17 юли 2014 г. · 5,4 т.

Условие

Place the plastic ruler between the polarizers. You can move the ruler with your hands. Then, secure it with the clamps on the screen with a slit (see Fig. 2B). The lower edge of the ruler should coincide with a line drawn on the screen, and its scale should be at the top. Make sure that both of the rulers provided demonstrate the birefringence phenomenon. Observe a picture emerging when you put one ruler on the top of the other so that light is to pass through both of them.

In this part use an optical scheme shown in Fig. 4B. Make sure that the rulers are fixed on the screen at the position described in Part 1.2.

2.2.1 Measure the intensity of the transmitted light (in $mV$ ) as a function of the coordinate $x$ of the point of the light incidence on the ruler in the range from 0 to 10 cm.. Measurements must be carried out for each ruler provided and for the two rulers stacked together. In each case, measure the maximum value of the voltage. Plot the corresponding graphs in the same figure. (2.0 points)

2.2.2 For each of the two rules calculate the values of the phase shift $\Delta\varphi$ between the ordinary and extraordinary waves in the range of $x$ from 0 to 7 cm. Plot the corresponding graphs $\Delta\varphi(x)$ . Put down the formula you have used for calculations. (1.2 points)

Note that the phase difference cannot unambiguously be determined from formula (1),additional physical assumptions should be applied to determine it correctly. [1,2 т.]

2.2.3 Assuming that $\Delta\varphi(x)$ is linear for each ruler

\Delta\varphi_1 = a_1x + b_1,

\Delta\varphi_2 = a_2x + b_2,

calculate the numerical values of the above coefficients for rulers 1 and 2. (1.0 point)

2.2.4 Using those data obtained in parts 2.2.1-2.2.3, calculate the theoretical values of the intensity of light passing through the two rulers stacked together. Put down the formula that you have used for calculations. Plot the theoretical dependence in the same figure from part 2.2.1. (1.2 point)

Отговори

Решение

Часть 2.2 Light transition through a plastic ruler

2.2.1 Results of measurements of the light intensity as a function of coordinates of transmission points through ruler #1, #2 and both rulers, are shown in table 3 and in the graph below.

Table 3.

№1			№2			Both rulers
X,mm	U,mV	$\Delta\varphi$	X,mm	U,mV	$\Delta\varphi$	X,mm	U,mV	$U_{calc}$
5	23	0,778	5	14	0,601	5	1	0,0
10	6	0,390	10	22	0,760	10	2	2,1
15	0	0,000	15	30	0,896	15	8	7,7
20	3	0,275	20	43	1,090	20	18	15,7
25	12	0,555	25	55	1,253	25	31	24,7
30	30	0,896	30	67	1,408	30	39	33,0
35	50	1,186	35	78	1,546	35	41	39,2
40	71	1,458	40	90	1,696	40	42	41,9
45	93	1,734	45	99	1,811	45	41	40,8
50	113	1,996	50	107	1,915	50	36	36,0
55	128	2,214	55	116	2,038	55	27	28,5
60	150	2,636	60	123	2,138	60	21	19,6
65	156	2,824	65	129	2,230	65	12	10,9
70	153	2,720	70	133	2,295	70	7	4,1
75	160	3,142	75	130	2,246	75	2	0,4
80	146	2,541	80	134	2,312	80	1	0,5
85	146	2,541	85	143	2,478	85	1	4,4
90	140	2,419	90	144	2,498	90	3	11,4
95	131	2,262	95	146	2,541	95	6	20,1
100	113	1,996	100	145	2,519	100	11	29,0

Graph U, mV (0–160) versus x, mm (0–100): data and curves for ruler №1 (blue, oscillating with a maximum near 70 mm), ruler №2 (green, slowly rising) and both rulers (orange triangles with the calculated curve).

2.2.2 To calculate a phase shift, we use equation (1), mentioned in the problems formulation, which can be represented as

U = U_{max}\sin^2\frac{\Delta\varphi}{2}, \qquad (1)

where $U_{max}$ is the largest value of voltage. But we have to be sure that this value actually corresponds to the maximum of function (1), not just another boundary point. According to measurements, (see the graph) for each ruler the most suitable value for $U_{max}$ is $U_{max} = 160\ mV$ .

The following equation

y_0 = \sin^2\frac{\Delta\varphi}{2} \qquad (1)

has multiple roots and it is not easy to find actual values of the phase shift, even if it’s possible to calculate certain value for $U_{max}$ , roots of the equation mentioned above are shown in the figure below

Graph of y = sin²(Δφ/2) versus Δφ from −π to about 4π with the horizontal line y0 ≈ 0,7; its intersections (roots) are marked with circles and diamonds.

Formally we can represent the roots in different forms, for example,

\Delta\varphi = \pm 2(\arcsin\sqrt{y_0} + k\pi),

\Delta\varphi = \pm 2(\pi - \arcsin\sqrt{y_0} + k\pi), \qquad (2)

k = 0,1,2\ ....

Choosing a correct root should depend on a function obtained experimentally.

Values for the phase shifts calculated by the equation

\Delta\varphi = 2\arcsin\sqrt{\frac{U}{U_{max}}}. \qquad (3)

are shown in table 3

It is clear that the function $\Delta\varphi(x)$ has to be monotonous, that is why the signs of roots for the first two points must be changed, which is a mathematically correct operation (just reflecting the graph). Note that the phase shifts are calculated with uncertainty of $\pm 2\pi k$ .

2.2.3 Obtained functions are close to linear, using MLS we get

\Delta\varphi_1 = 0.059x - 0.94,

\Delta\varphi_2 = 0.028x + 0.52.

Graphs of those functions are shown below.

Graph Δφ (−1,0 to 3,0) versus x, mm (0–70): straight lines for ruler №1 (steeper) and ruler №2; an arrow shows the first two points of ruler №1 reflected to negative values.

2.2.4 If two rulers are stacked together, then phase shifts simply add, and, theoretically, the intensity as function of phase shifts can be written as

U = U_{max}\sin^2\frac{\Delta\varphi_1 + \Delta\varphi_2}{2}. \qquad (5)

Here $U_{max}$ is the largest value of voltage at the light transition through both rulers and can be obtained from experimental data.

Graph U, mV (0–160) versus x, mm (0–70): rulers №1 and №2 data, and for both rulers the experimental points U_exp (orange) and the theoretical points U_theor (red).

Results of the calculations are shown in table 3 and in the graph above. Consistency of theoretical calculations and experimental data can clearly be seen.

Оригинал в Архива: 2014_IPhO_Experimenta_Competition.pdf · официални решения: 2014_IPhO_Solution_to_Experimental_Competition.pdf