Съдържание

IPhO 2014, Експеримент — Задача 5. Part 2.4.Light transmission through a curved strip

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Всички материали от IPhO 2014

Experimental competition. Thursday, 17 July 2014 — Experiment. To see invisible! (20 points) · 17 юли 2014 г. · 3,4 т.

Условие

Secure the plastic strip on the screen as described in part 1.3.

2.4.1 Measure the intensity of light, transmitted through the optical system, as a function of the coordinate $x$ of the point of light penetration into the strip in the range of $\pm 20\ mm$ from its center. Plot the corresponding graph. (1.2 point)

2.4.2 Calculate the phase shift between the ordinary and extraordinary waves $\Delta\varphi_0$ , passing through an uncurved strip. It is known that $\Delta\varphi_0$ lies in the range of $10\pi$ to $12\pi$ . (1.2 points)

Near the center of the strip its shape may be approximated a circular arc of radius R. The theoretical dependence of the phase shift $\Delta\varphi$ on the distance z from the center of the strip, $z \ll R$ , has the form:

\Delta\varphi = \Delta\varphi_0\left(1 + \frac{z^2}{2n^2R^2}\right),

where $n = 1.4$ , is the refractive index of the strip material

2.4.3 Using the data obtained in the previous parts, calculate the radius of strip curvature $R$ near its center. The refractive index of the strip material is equal to $n = 1.4$ . (1.0 points)

Отговори

Решение

Section 2.4 Light transmission through a curved strip

2.4.1 Results of the measurements of the light intensity as a function of coordinate $z$ of the point of light penetration into the strip are presented in table 6 and plotted below.

Table 6.

x, mm	U, mV	$\Delta\varphi'$	$\Delta\varphi$	z, mm
40	16	0,794	5,489	-22
41	8	0,554	5,729	-21
42	3	0,336	5,947	-20
43	1	0,194	6,090	-19
44	1	0,194	6,090	-18
45	1	0,194	6,090	-17
46	1	0,194	6,090	-16
47	1	0,194	6,090	-15
48	1	0,194	6,090	-14
49	4	0,389	5,894	-13
50	11	0,653	5,630	-12
51	23	0,964	5,319	-11
52	41	1,335	4,948	-10
53	61	1,711	4,572	-9
54	78	2,046	4,237	-8
55	90	2,322	3,962	-7
56	99	2,588	3,696	-6
57	102	2,706	2,706	-5
58	100	2,624	2,624	-4
59	98	2,553	2,553	-3
60	96	2,489	2,489	-2
61	95	2,459	2,459	-1
62	93	2,401	2,401	0
63	96	2,489	2,489	1
64	99	2,588	2,588	2
65	104	2,805	2,805	3
66	107	3,142	3,142	4
67	107	3,142	3,142	5
68	98	2,553	3,730	6
69	81	2,111	4,173	7
70	65	1,787	4,496	8
71	44	1,392	4,891	9
72	24	0,987	5,296	10
73	10	0,621	5,662	11
74	3	0,336	5,947	12
75	1	0,194	6,090	13
76	1	0,194	6,090	14
77	1	0,194	6,090	15
78	1	0,194	6,090	16
79	1	0,194	6,090	17
80	2	0,274	6,009	18

Graph U, mV (0–120) versus z, mm (−25 to 20): near zero for |z| above about 13 mm, rising to two maxima of about 102 and 107 mV near z = −5 and z = 5 mm with a shallow dip of 93 mV at z = 0.

2.4.2 The shape of the curve indicates that $\Delta\varphi_0$ lies at the ascending part of the relation between the intensity and the phase shift, which can be calculated as

\Delta\varphi_0 = 10\pi + 2\arcsin\sqrt{\frac{U_0}{U_{max}}} \approx 33.9.

Graph of the phase shift is drawn in the figure on right. Because we are interested in the central part of the graph, “reflection” parts are not shown. (This graph is not required from participants)

Graph Δφ (0–3,5) versus z, mm (−25 to 20): points with a parabola fitted to the central part around z = 0.

2.4.3 The graph shows that the central part is approximately parabolic function of $z$

\Delta\varphi = az^2 + b. \qquad (10)

In order to determine the coefficients of the function we can draw graph of $\Delta\varphi$ as a function of $z^2$ (see figure on the right). Using MLS, we can determine the parameters

Graph Δφ (2,0–3,0) versus z², mm² (0–25): five points on a straight rising line from about 2,45 to 2,7.

$a = 0.0104$ мм $^{-1}$ ,

$b = 2.45$ .

It is necessary to add $10\pi$ to the obtained value of $b$ .

Comparing equations (9) and (10), we conclude that the parameters can be represented by the strip characteristics as:

\alpha = \frac{\Delta\varphi_0}{2n^2R^2}, \quad b = \Delta\varphi_0. \qquad (11)

From those equations we get the radius of curvature of the strip

R = \frac{1}{n}\sqrt{\frac{b}{2a}}. \qquad (12)

Substitution of the obtained results leads us to $R = 29\ mm$ . Note that the obtained result is quite rough, due to uncertainties in measuring.

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