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Условие MATERIAL DESCRIPTION OF THE PHENOMENON EXPERIMENTAL SETUP (1 т.)MEASURING INTENSITIES (4,25 т.)DATA ANALYSIS. (4,75 т.)Отговори Решение

IPhO 2009, Експеримент — Задача 2. BIREFRINGENCE OF MICA

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EXPERIMENTAL PROBLEM 2 BIREFRINGENCE OF MICA

Условие

In this experiment you will measure the birefringence of mica (a crystal widely used in polarizing optical components).

MATERIAL

In addition to items 1), 2) and 3), you should use,

Two polarizing films mounted in slide holders, each with an additional acrylic support (LABEL J). See photograph for mounting instructions.
A thin mica plate mounted in a plastic cylinder with a scale with no numbers; acrylic support for the cylinder (LABEL K). See photograph for mounting instructions.
Photodetector equipment. A photodetector in a plastic box, connectors and foam support. A multimeter to measure the voltage of the photodetector (LABEL L). See photograph for mounting and connecting instructions.
Calculator.
White index cards, masking tape, stickers, scissors, triangle squares set.
Pencils, paper, graph paper.

Photograph of a polarizing film in a slide holder marked 'Polarized' and 'Linear Polarized Filter', standing in a clear acrylic support with a screw at the bottom. — Polarizer mounted in slide holder with acrylic support (LABEL J).

Photograph of a transparent cylinder with a black and red line scale holding a thin mica plate, mounted in a clear acrylic block with a screw at the bottom; a sticker on the cylinder reads 43. — Thin mica plate mounted in cylinder with a scale with no numbers, and acrylic support (LABEL K).

Photograph of a yellow digital multimeter connected by red and black leads to a grey photodetector box standing on a black foam support on the optical table. — A photodetector in a plastic box, connectors and foam support. A multimeter to measure the voltage of the photodetector (LABEL L). Set the connections as indicated.

DESCRIPTION OF THE PHENOMENON

Light is a transverse electromagnetic wave, with its electric field lying on a plane perpendicular to the propagation direction and oscillating in time as the light wave travels.

If the direction of the electric field remains in time oscillating along a single line, the wave is said to be linearly polarized, or simply, polarized. See Figure 2.1.

Axes x, y, z with a sinusoidal wave along the y axis; vertical arrows show the electric field oscillating along z. — Figure 2.1 A wave travelling in the y-direction and polarized in the z-direction.

A polarizing film (or simply, a polarizer) is a material with a privileged axis parallel to its surface, such that, transmitted light emerges polarized along the axis of the polarizer. Call (+) the privileged axis and (-) the perpendicular one.

A thick arrow labelled 'unpolarized light' hits a square polarizer; behind it a sine wave labelled 'polarized light'; arrows mark the '(+) axis' (vertical) and the '(-) axis' (horizontal) of the polarizer. — Figure 2.2 Unpolarized light normally incident on a polarizer. Transmitted light is polarized in the (+) direction of the polarizer.

Common transparent materials (such as window glass), transmit light with the same polarization as the incident one, because its index of refraction does not depend on the direction and/or polarization of the incident wave. Many crystals, including mica, however, are sensitive to the direction of the electric field of the wave. For propagation perpendicular to its surface, the mica sheet has two characteristic orthogonal axes, which we will call Axis 1 and Axis 2. This leads to the phenomenon called birefringence.

A grey disc with a vertical red line labelled Axis 1 and a horizontal green line labelled Axis 2. — Figure 2.3 Thin slab of mica with its two axes, Axis 1 (red) and Axis 2 (green).

Let us analyze two simple cases to exemplify the birefringence. Assume that a wave polarized in the vertical direction is normally incident on one of the surfaces of the thin slab of mica.

Case 1) Axis 1 or Axis 2 is parallel to the polarization of the incident wave. The transmitted wave passes without changing its polarization state, but the propagation is characterized as if the material had either an index of refraction $n_1$ or $n_2$ . See Figs. 2.4 and 2.5.

A vertically polarized wave passes through a mica disc whose red Axis 1 is vertical (labelled Axis 1); the transmitted wave keeps vertical polarization. — Figure 2.4 Axis 1 is parallel to polarization of incident wave. Index of refraction is $n_1$ .

A vertically polarized wave passes through a mica disc whose green Axis 2 is vertical (labelled Axis 2); the transmitted wave keeps vertical polarization. — Figure 2.5 Axis 2 is parallel to polarization of incident wave. Index of refraction is $n_2$ .

Case 2) Axis 1 makes an angle $\theta$ with the direction of polarization of the incident wave. The transmitted light has a more complicated polarization state. This wave, however, can be seen as the superposition of two waves with different phases, one that has polarization parallel to the polarization of the incident wave (i.e. "vertical") and another that has polarization perpendicular to the polarization of the incident wave (i.e. "horizontal").

A vertically polarized wave hits a mica disc whose red Axis 1 is tilted by 'angle θ' from the dashed vertical; the transmitted light is drawn as two components, one vertical and one horizontal. — Figure 2.6 Axis 1 makes and angle $\theta$ with polarization of incident wave

Call $I_P$ the intensity of the wave transmitted parallel to the polarization of the incident wave, and $I_O$ the intensity of the wave transmitted perpendicular to polarization of the incident wave. These intensities depend on the angle $\theta$ , on the wavelength $\lambda$ of the light source, on the thickness $L$ of the thin plate, and on the absolute value of the difference of the refractive indices, $|n_1 - n_2|$ . This last quantity is called the birefringence of the material. The measurement of this quantity is the goal of this problem. Together with polarizers, birefringent materials are useful for the control of light polarization states.

We point out here that the photodetector measures the intensity of the light incident on it, independent of its polarization.

The dependence of $I_P(\theta)$ and $I_O(\theta)$ on the angle $\theta$ is complicated due to other effects not considered, such as the absorption of the incident radiation by the mica. One can obtain, however, approximated but very simple expressions for the normalized intensities $\bar{I}_P(\theta)$ and $\bar{I}_O(\theta)$ , defined as,

\bar{I}_P(\theta) = \frac{I_P(\theta)}{I_P(\theta) + I_O(\theta)} \qquad (2.1)

and

\bar{I}_O(\theta) = \frac{I_O(\theta)}{I_P(\theta) + I_O(\theta)} \qquad (2.2)

It can be shown that the normalized intensities are (approximately) given by,

\bar{I}_P(\theta) = 1 - \frac{1}{2}\left(1 - \cos\Delta\phi\right)\sin^2(2\theta) \qquad (2.3)

and

\bar{I}_O(\theta) = \frac{1}{2}\left(1 - \cos\Delta\phi\right)\sin^2(2\theta) \qquad (2.4)

where $\Delta\phi$ is the difference of phases of the parallel and perpendicular transmitted waves. This quantity is given by,

\Delta\phi = \frac{2\pi L}{\lambda}|n_1 - n_2| \qquad (2.5)

where $L$ is the thickness of the thin plate of mica, $\lambda$ the wavelength of the incident radiation and $|n_1 - n_2|$ the birefringence.

EXPERIMENTAL SETUP (1 т.)

Task 2.1 Experimental setup for measuring intensities. Design an experimental setup for measuring the intensities $I_P$ and $I_O$ of the transmitted wave, as a function of the angle $\theta$ of any of the optical axes, as shown in Fig. 2.6. Do this by writing the LABELS of the different devices on the drawing of the optical table. Use the convention (+) and (-) for the direction of the polarizers. You can make additional simple drawings to help clarify your design.

Task 2.1 a) Setup for $I_P$ (0.5 points). [0,5 т.]

Task 2.1 b) Setup for $I_O$ (0.5 points). [0,5 т.]

Laser beam alignment. Align the laser beam in such a way that it is parallel to the table and is incident on the center of the cylinder holding the mica. You may align by using one the white index cards to follow the path. Small adjustments can be made with the movable mirror.

Photodetector and the multimeter. The photodetector produces a voltage as light impinges on it. Measure this voltage with the multimeter provided. The voltage produced is linearly proportional to the intensity of the light. Thus, report the intensities as the voltage produced by the photodetector. Without any laser beam incident on the photodetector, you can measure the background light intensity of the detector. This should be less than 1 mV. Do not correct for this background when you perform the intensity measurements.

WARNING: The laser beam is partially polarized but it is not known in which direction. Thus, to obtain polarized light with good intensity readings, place a polarizer with either its (+) or (–) axes vertically in such a way that you obtain the maximum transmitted intensity in the absence of any other optical device.

MEASURING INTENSITIES (4,25 т.)

Task 2.2 The scale for angle settings. The cylinder holding the mica has a regular graduation for settings of the angles. Write down the value in degrees of the smallest interval (i.e. between two black consecutive lines). (0.25 points). [0,25 т.]

Finding (approximately) the zero of $\theta$ and/or the location of the mica axes. To facilitate the analysis, it is very important that you find the appropriate zero of the angles. We suggest that, first, you identify the location of one of the mica axes, and call it Axis 1. It is almost sure that this position will not coincide with a graduation line on the cylinder. Thus, consider the nearest graduation line in the mica cylinder as the provisional origin for the angles. Call $\bar{\theta}$ the angles measured from such an origin. Below you will be asked to provide a more accurate location of the zero of $\theta$ .

Task 2.3 Measuring $I_P$ and $I_O$ . Measure the intensities $I_P$ and $I_O$ for as many angles $\bar{\theta}$ as you consider necessary. Report your measurements in Table I. Try to make the measurements for $I_P$ and $I_O$ for the same setting of the cylinder with the mica, that is, for a fixed angle $\bar{\theta}$ . (3.0 points). [3 т.]

Task 2.4 Finding an appropriate zero for $\theta$ . The location of Axis 1 defines the zero of the angle $\theta$ . As mentioned above, it is mostly sure that the location of Axis 1 does not coincide with a graduation line on the mica cylinder. To find the zero of the angles, you may proceed either graphically or numerically. Recognize that the relationship near a maximum or a minimum may be approximated by a parabola where:

I(\bar{\theta}) \approx a\bar{\theta}^2 + b\bar{\theta} + c

and the minimum or maximum of the parabola is given by,

\bar{\theta}_m = -\frac{b}{2a}.

Either of the above choices gives rise to a shift $\delta\bar{\theta}$ of all your values of $\bar{\theta}$ given in Table I of Task 2.3, such that they can now be written as angles $\theta$ from the appropriate zero, $\theta = \bar{\theta} + \delta\bar{\theta}$ . Write down the value of the shift $\delta\bar{\theta}$ in degrees. (1.0 points). [1 т.]

DATA ANALYSIS. (4,75 т.)

Task 2.5 Choosing the appropriate variables. Choose $\bar{I}_P(\theta)$ or $\bar{I}_O(\theta)$ to make an analysis to find the difference of phases $\Delta\phi$ . Identify the variables that you will use. (0.5 point). [0,5 т.]

Task 2.6 Data analysis and the phase difference.

Use Table II to write down the values of the variables needed for their analysis. Make sure that you use the corrected values for the angles $\theta$ . Include uncertainties. Use graph paper to plot your variables. (1.0 points).
Perform an analysis of the data needed to obtain the phase difference $\Delta\phi$ . Report your results including uncertainties. Write down any equations or formulas used in the analysis. Plot your results. (1.75 points).
Calculate the value of the phase difference $\Delta\phi$ in radians, including its uncertainty. Find the value of the phase difference in the interval $[0, \pi]$ . (0.5 points). [3,25 т.]

Task 2.7 Calculating the birefringence $|n_1 - n_2|$ . You may note that if you add $2N\pi$ to the phase difference $\Delta\phi$ , with N any integer, or if you change the sign of the phase, the values of the intensities are unchanged. However, the value of the birefringence $|n_1 - n_2|$ would change. Thus, to use the value $\Delta\phi$ found in Task 2.6 to correctly calculate the birefringence, you must consider the following:

\Delta\phi = \frac{2\pi L}{\lambda}|n_1 - n_2| \qquad \text{if} \qquad L \lt 82\times 10^{-6}\ \mathrm{m}

2\pi - \Delta\phi = \frac{2\pi L}{\lambda}|n_1 - n_2| \qquad \text{if} \qquad L \gt 82\times 10^{-6}\ \mathrm{m}

where the value $L$ of the thickness of the slab of mica you used is written on the cylinder holding it. This number is given in micrometers (1 micrometer = $10^{-6}$ m). Assign $1\times 10^{-6}\mathrm{m}$ as the uncertainty for $L$ . For the laser wavelength, you may use the value you found in Problem 1 or the average value between $620\times 10^{-9}$ m and $750\times 10^{-9}$ m, the reported range for red in the visible spectrum. Write down the values of $L$ and $\lambda$ as well as the birefringence $|n_1 - n_2|$ with its uncertainty. Include the formulas that you used to calculate the uncertainties. (1.0 points). [1 т.]

Отговори

Решение

Answer Form

Experimental Problem No. 2

Birefringence of mica

Task 2.1 a) Experimental setup for $I_P$ . (0.5 points)

Top view of the optical table (grid of holes) with the red beam path: laser A, mirror B, polarizer + (J+), mica K, Polarizer + (J+), detector L; labels laser, mirror, polarizer +, mica, Polarizer +, detector. — Task 2.1 a) Experimental setup for $I_P$ .

Task 2.1 b) Experimental setup for $I_O$ . (0.5 points)

Top view of the optical table with the red beam path: laser A, mirror B, polarizer + (J+), mica K, polarizer - (J-), detector L; labels laser, mirror, polarizer +, mica, polarizer -, detector. — Task 2.1 b) Experimental setup for $I_O$ .

2.1 — 1.0

Photograph of the black optical table with the mirror, a polarizer, the mica cylinder, a second polarizer and the photodetector connected to a multimeter. — Experimental setup for measurement of mica birefringence

Experimental setup for measurement of mica birefringence

Task 2.2 The scale for angles.

2.2 — 0.25

The angle between two consecutive black lines is

$\theta_{\mathrm{int}} = 3.6$ degrees because there are 100 lines.

Tasks 2.3 Measuring $I_P$ and $I_O$ .Use additional sheets if necessary.

TABLE I (3 points)

$\bar{\theta}$ (degrees)	$(I_P \pm 1)\times 10^{-3}\ \mathrm{V}$	$(I_O \pm 1)\times 10^{-3}\ \mathrm{V}$
-3.6	46.4	1.1
0	48.1	0.2
3.6	47.0	0.6
7.2	46.0	2.0
10.8	42.3	4.9
14.4	38.2	9.0
18.0	33.9	12.5
21.6	27.7	17.9
25.2	23.4	22.0
28.8	17.8	27.0
32.4	12.5	31.7
36.0	8.8	34.8
39.6	5.2	38.0
43.2	3.6	39.4
46.8	3.2	39.6
50.4	4.5	38.7
54.0	6.9	36.6
57.6	10.3	33.6
61.2	14.7	29.4
64.8	20.1	24.7
68.4	25.4	19.7
72.0	30.5	14.7
75.6	36.6	10.2
79.2	40.7	6.1
82.8	44.3	3.2
86.4	46.9	1.0
90.0	47.8	0.2
93.6	47.0	0.4
97.2	45.7	2.0

Parallel $I_P$ and perpendicular $I_O$ intensities vs angle $\bar{\theta}$ .

GRAPH NOT REQUIRED!

Task 2.4 Finding an appropriate zero for $\theta$ .

2.4 — 1.0

a) Graphical analysis

The value for the shift is $\delta\bar{\theta} = -1.0$ degrees.

Add the graph paper with the analysis of this Task.

b) Numerical analysis

From Table I choose the first three points of $\bar{\theta}$ and $I_O(\bar{\theta})$ :
(intensities in millivolts)

(x_1, y_1) = (-3.6, 1.1) \qquad (x_2, y_2) = (0, 0.2) \qquad (x_3, y_3) = (3.6, 0.6)

We want to fit $y = ax^2 + bx + c$ . This gives three equations:

1.1 = a(3.6)^2 - b(3.6) + c

0.2 = c

0.6 = a(3.6)^2 + b(3.6) + c

second in first $\Rightarrow \quad b = \dfrac{-0.9 + a(3.6)^2}{3.6}$

in third $\Rightarrow \quad 0.6 = a\left((3.6)^2 + (3.6)^2\right) - 0.9 + 0.2$

$\Rightarrow \quad a = 0.050 \qquad b = -0.069$

The minimum of the parabola is at:

\bar{\theta}_{\min} = -\frac{b}{2a} \approx 0.7\ \text{degrees}

Therefore, $\delta\bar{\theta} = -0.7$ degrees.

Task 2.5 Choosing the appropriate variables.

2.5 — 0.5

Equation (2.4) for the perpendicular intensity is

\bar{I}_O(\theta) = \frac{1}{2}\left(1 - \cos\Delta\phi\right)\sin^2(2\theta)

This can be cast as a straight line $y = mx + b$ , with

y = \bar{I}_O(\theta) \quad , \quad x = \sin^2(2\theta) \quad \text{and} \quad m = \frac{1}{2}\left(1 - \cos\Delta\phi\right)

from which the phase may be obtained.

NOTE: This is not the only way to obtain the phase difference. One may, for instance, analyze the 4 maxima of either $\bar{I}_P(\theta)$ or $\bar{I}_O(\theta)$ .

Task 2.6 Statistical analysis and the phase difference.

2.6 — 1.0

To perform the statistical analysis, we shall then use

y = \bar{I}_O(\theta) \quad \text{and} \quad x = \sin^2(2\theta) \ .

Since for $\theta : 0 \to \frac{\pi}{4}, \quad x : 0 \to 1$ , we use only 12 pairs of data points to cover this range, as given in Table II.

x may be left without uncertainty since it is a setting. The uncertainty in y may be calculated as

\Delta\bar{I}_O = \sqrt{\left(\frac{\partial \bar{I}_O}{\partial I_O}\right)^2 \Delta I_O^2 + \left(\frac{\partial \bar{I}_P}{\partial I_P}\right)^2 \Delta I_P^2} \quad \text{and one gets}

\Delta\bar{I}_O = \frac{\sqrt{I_O^2 + I_P^2}}{(I_O + I_P)^2}\Delta I_O \approx 0.018,

approximately the same for all values.

TABLE II

$\bar{\theta}$ (degrees)	$x = \sin^2(2\theta)$	$y = \bar{I}_O \pm 0.018$
2.9	0.010	0.013
6.5	0.051	0.042
10.1	0.119	0.104
13.7	0.212	0.191
17.3	0.322	0.269
20.9	0.444	0.392
24.5	0.569	0.484
28.1	0.690	0.603
31.7	0.799	0.717
35.3	0.890	0.798
38.9	0.955	0.880
42.5	0.992	0.916

2.6 — 1.75

We now perform a least square analysis for the variables y vs x in Table II. The slope and y-intercept are:

m \pm \Delta m = 0.913 \pm 0.012

b \pm \Delta b = -0.010 \pm 0.008

The formulas for this analysis are:

m = \frac{N\sum_{n=1}^{N} x_n y_n - \sum_{n=1}^{N} x_n \sum_{n'=1}^{N} y_{n'}}{\Delta}

b = \frac{\sum_{n=1}^{N} x_n^2 \sum_{n'=1}^{N} y_{n'} - \sum_{n=1}^{N} x_n \sum_{n'=1}^{N} x_{n'} y_{n'}}{\Delta}

where

\Delta = N\sum_{n=1}^{N} x_n^2 - \left(\sum_{n=1}^{N} x_n\right)^2

with N the number of data points.
The uncertainty is calculated as

(\Delta m)^2 = N\frac{\sigma^2}{\Delta} \quad , \quad (\Delta b)^2 = \frac{\sigma^2}{\Delta}\sum_{n=1}^{N} x_n^2 \qquad \text{with,}

\sigma^2 = \frac{1}{N-2}\sum_{n=1}^{N}\left(y_n - b - m x_n\right)^2

with N = 12 in this example.

Include the accompanying plot or plots.

Handwritten graph on IPhO 2009 graph paper: Ī_O against sen²(2θ) from 0 to 1.0 with error bars and a fitted straight line; notes 'fit y=mx+b', 'm = 0.913 ± 0.012', 'b = -0.010 ± 0.008'.

2.6 — 0.5

Calculate the value of the phase $\Delta\phi$ in radians in the interval $[0, \pi]$ .

From the slope $m = \frac{1}{2}(1 - \cos\Delta\phi)$ , one finds

\Delta\phi \pm \Delta(\Delta\phi) = 2.54 \pm 0.04

Write down the formulas for the calculation of the uncertainty.

We see that,

\Delta m = \left|\frac{\partial m}{\partial \Delta\phi}\right|\Delta(\Delta\phi) = \frac{1}{2}\sin(\Delta\phi)\Delta(\Delta\phi), \ \text{therefore,} \quad \Delta(\Delta\phi) = \frac{2\Delta m}{\sin(\Delta\phi)} \ .

Task 2.7 Calculating the birefringence $|n_1 - n_2|$ .

2.7 — 1.0

Write down the width of the slab of mica you used,

L \pm \Delta L = (100 \pm 1)\times 10^{-6}\ \mathrm{m}

Write down the wavelength you use,

\lambda \pm \Delta\lambda = (663 \pm 25)\times 10^{-9}\ \mathrm{m}\ \text{(from Problem 1)}

Calculate the birefringence

|n_1 - n_2| \pm \Delta|n_1 - n_2| = (3.94 \pm 0.16)\times 10^{-3}

The birefringence is between 0.003 and 0.005. Nominal value 0.004

Write down the formulas you used for the calculation of the uncertainty of the birefringence.

Since the width $L \gt 82$ micrometers, we use

2\pi - \Delta\phi = \frac{2\pi L}{\lambda}|n_1 - n_2|

The error is

\Delta|n_1 - n_2| = \sqrt{\left(\frac{\partial|n_1 - n_2|}{\partial\lambda}\right)^2 \Delta\lambda^2 + \left(\frac{\partial|n_1 - n_2|}{\partial L}\right)^2 \Delta L^2 + \left(\frac{\partial|n_1 - n_2|}{\partial\Delta\phi}\right)^2 \Delta(\Delta\phi)^2}

\Delta|n_1 - n_2| = \sqrt{\left(\frac{|n_1 - n_2|}{\lambda}\right)^2 \Delta\lambda^2 + \left(\frac{|n_1 - n_2|}{L}\right)^2 \Delta L^2 + \left(\frac{\lambda}{2\pi L}\right)^2 \Delta(\Delta\phi)^2}

Since the data may appear somewhat disperse and/or the errors in the intensities may be large, a graphical analysis may be performed.

In the accompanying plot, it is exemplified a simple graphical analysis: first the main slope is found, then, using the largest deviations one can find two extreme slopes.

The final result is,

m = 0.91 \pm 0.08 \qquad \text{and} \qquad b = -0.01 \pm 0.04

The calculation of the birefringence and its uncertainty follows as before. One now finds,

|n_1 - n_2| \pm \Delta|n_1 - n_2| = (3.94 \pm 0.45)\times 10^{-3}.

A larger (more realistic) error.

Plot of normalized intensities Ī (0 to 1.0) against θ/degrees: red and blue data points for Ī_P and Ī_O with fitted curves crossing near 24 and 66 degrees. — Comparison of experimental data (normalized intensities $\bar{I}_P$ and $\bar{I}_O$ ) with fitting (equations (2.3) and (2.4)) using the calculated value of the phase difference $\Delta\phi$ .

Comparison of experimental data (normalized intensities $\bar{I}_P$ and $\bar{I}_O$ ) with fitting (equations (2.3) and (2.4)) using the calculated value of the phase difference $\Delta\phi$ .

GRAPH NOT REQUIRED!

Scatter plot of I/mV (0 to about 48) against θ̄/degrees (about -4 to 97): red points I_P fall from about 48 to a minimum near 45 degrees and rise again; blue points I_O rise from 0 to a maximum near 45 degrees and fall again. — Parallel $I_P$ and perpendicular $I_O$ intensities vs angle $\bar{\theta}$ .

Handwritten graph on IPhO 2009 graph paper: parabola fitted through I_O points against θ̄ from -4 to 8, minimum marked; note 'θ̄min ≈ 1.0 degrees', '4 points' and a small table θ̄ / (I_O/mV): -3.6 1.1; 0.0 0.2; 3.6 0.6; 7.2 2.0.

Handwritten graph on IPhO 2009 graph paper: Ī_O against sen²(2θ) with error bars, the main line and two extreme lines; notes 'Graphical analysis', 'm = 0.91 ± 0.08', 'b = -0.01 ± 0.04'.

Оригинал в Архива: exp2.pdf · официални решения: exp2_sol.pdf