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Formation of birefringence patterns under everyday conditions

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2014 Eur. J. Phys. 35 055008

(http://iopscience.iop.org/0143-0807/35/5/055008)

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Formation of birefringence patterns undereveryday conditions

Youtian Zhang1, Lan Chen1, Sihui Wang and Huijun Zhou

School of Physics, Nanjing University, Jiangsu 210093, People’s Republic of ChinaE-mail: wangsihui@nju.edu.cn

Received 12 April 2014, revised 11 May 2014Accepted for publication 30 May 2014Published 1 July 2014

AbstractThe interference of two polarized lights is always discussed in general opticscourses, and birefringence is considered to be a way to cause polarized light.In this article, we investigate the formation of birefringence patterns undereveryday conditions. We firstly present a simple method to demonstratecoloured patterns in a plastic plate with everyday utensils. Then we investigatethe possibility of observing colours in plastics without Polaroids under naturalincident light. We prove by theoretical calculation that polarization of theincident light is unnecessary to produce birefringence patterns except for in thenormal incidence condition.

Keywords: birefringence, photoelasticity without polaroids, coloured patterns

(Some figures may appear in colour only in the online journal)

1. Introduction

Colourful fringes can be often observed when light encounters a transparent plastic object.For example, when looking at a blank CD case under sunlight, we can observe light coloursfrom a certain angle (see figure 1).

This phenomenon is common in everyday life and is usually associated with the bire-fringence of the objects originating from residual stress or an applied load [1]. The methodcalled photoelasticity has been widely applied in industry or laboratories to determine thestress distribution in transparent objects [2]. The standard setup in laboratories to examine thebirefringence patterns is a plane polariscope, which consists of two crossing Polaroids and alight source [3]. The Polaroids are important to meet the interference conditions. The firstPolaroid (polarizer) makes the incident light linearly polarized. The birefringent specimen(the plastic plate) decomposes the polarized light along its two principal axes with phase

European Journal of Physics

Eur. J. Phys. 35 (2014) 055008 (9pp) doi:10.1088/0143-0807/35/5/055008

1 Contributing equally to this article.

0143-0807/14/055008+09$33.00 © 2014 IOP Publishing Ltd Printed in the UK 1

retardation. The second Polaroid (analyzer) recombines the two light components so thatfringes are produced. However, the colourful patterns in plastics seen in everyday life areusually observed without Polaroids, and the phenomenon is more obvious under reflectionconditions. As we have comprehended the significance of Polaroids so far, it is natural toconsider some mechanism acting as hidden Polaroids.

In this article, we will investigate the formation of birefringence patterns under daylightconditions. In section 2, a simple method is introduced to demonstrate coloured patterns in aplastic plate with everyday utensils. In section 3, we investigate the possibility of observingcolours in plastics without Polaroids under natural incident light. One common explanationfor the polarization mechanism is attributed to the scattering of sunlight that partiallypolarizes the sunlight [4, 5]. Such polarization will enhance the brightness of the resultingpatterns, of course. However, we will prove that polarization of the incident light is unne-cessary to produce birefringence patterns except for in the normal incidence condition. Anexample of a uniaxial crystal is given to calculate the coloured patterns of the transmis-sion light.

2. A simple demonstration

The standard method demonstrating birefringence patterns requires two crossing Polaroids,with the sample placed in-between. First of all, we will present a simple method without thepolariscope. The experimental setup includes: a computer LCD as the light source, a plasticplate and a pair of polarization sunglasses. The LCD is a linearly polarized light source, so thefirst Polaroid of the polariscope can be removed. The sunglasses are used as the sec-ond Polaroid.

Turn on the computer LCD and make the screen display white. We can see that thebrightness of the LCD changes when we rotate the orientation of the sunglasses. Fix thesunglasses at the orientation at which the brightness of the LCD is minimized (it can be totallydark for ‘good’ sunglasses). Now, a homemade polariscope with crossed Polaroids is made.Place the plastic sample (the cover of a candy box) between the LCD and the sunglasses, anda colourful pattern is formed (see figure 2).

The birefringence in the sample can be described by the propagation of ordinary (o) lightand extraordinary (e) light in the sample. The refraction indexes of the o-light and e-light ofthe wave plate are no and ne, respectively. If a light with wavelength λ and intensity3 [6] I

Figure 1. Coloured patterns without Polaroids.

3 The intensity here is defined as the time-averaged magnitude of the Poynting vector. I Enc 02

2= ε , where n is therefractive index, c is the speed of light in a vacuum and ε0 is the vacuum permittivity.

Eur. J. Phys. 35 (2014) 055008 Y Zhang et al

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penetrates through the setup, the intensity of the transmitted [7] light is:

I In n d

cos ( ) sin 2 sin 2 sin2 ( )

(1)2 2 e oα β α βπ

λ′ = − −

−⎡⎣⎢

⎤⎦⎥

where α or β are the angles between the fast axis of the specimen and the polarizer oranalyzer, and d is the thickness of the specimen. Here α− β = π/2, so that equation (1) can besimplified as

I In n d

sin 2 sin2 ( )

. (2)2 2 e oαπ

λ′ =

−

In figure 3(a), the dark strips are called ‘isoclinic fringes’. Isoclinic fringes appear whenthe fast axis (or slow axis) of the sample coincides with either of the polarizer’s axes.According to equation (2), when α= 0 or π/2, extinction occurs irrespective of the light’s

Figure 2. Coloured fringes of an anisotropic plane with Polaroids.

Figure 3. Experimental (a) and theoretical (b) results using a plastic plate. The darkcrossing lines are isoclinic fringes.

Eur. J. Phys. 35 (2014) 055008 Y Zhang et al

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wavelength. The other coloured fringes are called isochromatic fringes. According toequation (2), extinction occurs in the sequence of wavelengths for an identical optical pathdifference.

The centro-symmetrical pattern in figure 3(a) indicates that the specimen’s internal stressis also centro-symmetrical. We can ignore the inhomogeneity of the plastic plate along the zdirection (defined as shown in figure 6) as the thickness is small. For the non-uniform samplewe used, equation (2) can be applied locally in the x–y plane. Assume that the optical pathdifference decreases linearly from the centre to the rim (d= 5 mm, ne–no decreases from0.00011 to 0.000052). The interference pattern with a white light source can be simulated byMATLAB using equation (2), as shown in figure 3(b). There are slight differences betweenthe simulation and the experimental results in colour, because of the deviation in reproducingthe colours.

3. Formation of birefringence patterns without Polaroids

Consider a uniaxial crystal with its optic axis on the interface between the air and thecrystal. ( )ε is the dielectric tensor. The angle between the optics axis and the incidence planeis α. Its direction is denoted by (cos , sin , 0)α α . The incident angle of a homogenous light is

iθ , while the refraction angles are oθ and eθ for ordinary light and extraordinary light,

respectively. We have the wave vectors of refraction light ( )k sin , 0, coso o oθ θ = −and ( )k sin , 0, cos .e e eθ θ = −

Those colourful patterns caused by birefringence without Polaroids are due to the dif-ferent observed transmitted intensities of different wavelengths. To calculate the transmittedintensity of a certain wavelength, there is a need to study the propagation directions andamplitudes of two refraction lights within the birefringent crystal. Their optical path differ-ence upon leaving the crystal is then used when superimposing the two waves.

Suppose that the crystal’s principal dielectric constants are given as oε , eε and oε , then inthe orthogonal coordinate system shown in figure 4, we have

sin cos 0cos sin 0

0 0 1

0 00 00 0

sin cos 0cos sin 0

0 0 1

o

e

o

1

εα αα α

εε

ε

α αα α =

−⋅ ⋅

− −⎛⎝⎜⎜

⎞⎠⎟⎟

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟

⎛⎝⎜⎜

⎞⎠⎟⎟

Figure 4. Light propagation in birefringence materials.

Eur. J. Phys. 35 (2014) 055008 Y Zhang et al

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A wave normal ellipsoid is often used to investigate light propagation in birefringentmaterials (see figure 5). The section perpendicular to the propagation direction is an ellipse,the length of whose principal axes are inversely proportional to the phase velocities. Thedirections of the principle axes are the same as dielectric displacements [8].

To determine the direction of refracted light at the boundary surface, the relation betweenthe propagation direction and propagation velocity is determined according to the wavenormal ellipsoid. The vectorial difference between the incident wave vector and therefraction wave vector is perpendicular to the boundary surface caused by the tangential

continuity of the electric field [8]. In other words, ( )k k xOyirefraction − ⊥ , i.e.

k k xOyirefraction refraction airε ε − ⊥⎛⎝⎜

⎞⎠⎟ . In addition, for ordinary light, refraction oε ε= , which

corresponds to the square of the semi-minor axis of the section. For extraordinary light,

refractionε equals the length of the semi-major axis of the section, which changes with the

refraction angle.The angle between the refracted light (sin , 0, cos )θ θ− and the optical axis

(cos , sin , 0)α α is arccos (sin cos )γ θ α= . The semi-minor axis ( )oε of the ellipse

section perpendicular to the propagation direction does not change, so the relation of oεand θ is a semi-circle. However, the length of the semi-major axis is ( )sin cos

1/22

e

2

o+γ

εγ

ε

−and the

relation is not a semi-circle any more (see figure 6).

Because k k xOyirefraction refraction airε ε − ⊥⎛⎝⎜

⎞⎠⎟ , then

sin sin sin ,io o e e airε θ ε θ ε θ= =

Figure 5. Normal ellipsoid. The red line is the direction of the wave normal vector andthe light propagating in this direction will decompose into two rays of light. Therefraction index of either light is determined by the principle axes of the section.

Figure 6. Different refraction angles for e- and o-light. The arcs in the figure are themagnitude of oε and eε in different directions. The x components of the refractedand incident light must be equal.

Eur. J. Phys. 35 (2014) 055008 Y Zhang et al

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which means the x components of the three vectors in figure 6 are the same.

sinsin

(3)io

air2

o

θε θ

ε=

sinsin

( ) sin cos. (4)i

ie

o air2

o e e o air2 2

θε ε θ

ε ε ε ε ε θ α=

− −

The dielectric displacement of the o-light is perpendicular to the principal plane, and thatof the e-light is in the principal plane [8],

( )( )

D(cos , sin , 0) sin , 0, cos

(cos , sin , 0) sin , 0, cos(5)o

o o

o o

α α θ θα α θ θ

ˆ =× −× −

( ) ( )( ) ( )

Dsin , 0, cos (cos , sin , 0) sin , 0, cos

sin , 0, cos (cos , sin , 0) sin , 0, cos. (6)e

e e e e

e e e e

θ θ α α θ θ

θ θ α α θ θˆ =

− × × −

− × × −

⎡⎣ ⎤⎦⎡⎣ ⎤⎦

Then, E E Do o

1

oε = −, E E De e

1

eε = −, where L0 and Le are the magnitudes of Eo and Ee. Notice

that ( )H k D1 = × εμ . According to the electromagnetic boundary conditions at the medium

interface:

E E E E E E E E, (7)x x x x y y y yi r o e i r o e+ = + + = +

H H H H H H H H, , (8)x x x x y y y yi r o e i r o e+ = + + = +

where the subscripts i, r, o and e indicate the incident, reflected, ordinary and extraordinaryrays of light, we can get Eo and Ee. So the intensities of the two refracted rays of light aredetermined.

The analysis procedure of the lower surface is similar to that of the upper surface, and thedouble reflection phenomenon should also be considered.

The optical path difference between two transmitted rays of light is 2Δ π= σλ . Taking

equations (3) and (4) into the expression for the optical path difference, we have

( )n n d

d

cos cos

1 sin cos

1sin

( ) sin cos1

. (9)i

i

i

e e o o

e

0

e

o

air

0

2 2

e air2

o e e o air2 2

o

0

air

o

σ θ θ

εε

εε

εε

θ α

ε ε θε ε ε ε ε θ α

εε

εε

= − ⋅

=+ −

× −− −

− −

⋅

⎡

⎣

⎢⎢⎢⎢⎢

⎛⎝⎜

⎞⎠⎟

⎤

⎦

⎥⎥⎥⎥⎥

When 0α ≠ , the polarization of the two transmitted rays of light can be non-perpendicular.

We combine equations (7)–(9) and a numerical calculation is adopted to demonstrate thephenomenon. The electric amplitudes of incident light are taken as unity. We calculated theintensity of the transmitted light for P and S polarization and unpolarized light respectivelywith respect to different incident angles. The wavelengths for the seven colours are taken as660 nm, 610 nm, 570 nm, 550 nm, 460 nm, 440 nm and 410 nm. We consider the weakdispersion condition, and the refraction indexes are taken as no = 1.5 and ne = 1.51, d= 2 mm.

The results for P and S polarization and unpolarized light are shown in figure 7. We seethat the transmitted light follows the tendency given by the Fresnel equations and the intensity

Eur. J. Phys. 35 (2014) 055008 Y Zhang et al

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of the light oscillates with smaller periods as the incident angle increases. In addition, theresults coincide with those given by the Fresnel equations when 0.α =

The intensity with a shorter wavelength has a smaller period as well. The oscillation fordifferent colours at a large incident angle results in the colour observed on the plastic sample.

Figure 7. The oscillation for different colours at a large incident angle results in colouron the sample. I and I0 are the intensities of the transmitted and incident light.

Eur. J. Phys. 35 (2014) 055008 Y Zhang et al

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Comparing figures 7(c) with 7(a) and 7(b), we see that fluctuation is less evident for unpo-larized light.

Based on figures 7(a)–(c), we can demonstrate how the colour changes with the incidentangle, as shown in figures 8(a)–(c). For a normal incidence condition or a small incidentangle, figure 8 shows that no colour appears for either polarized or unpolarized light. Thismeans that two Polaroids are necessary to produce birefringence patterns at small angles.Comparing figures 8(c) with 8(a) and 8(b), we see that polarization may enhance thebrightness of the resulting colours. However, figure 8(c) shows that polarization is unne-cessary to produce colours at a large incident angle.

The formation of birefringence patterns under reflection is similar but far more com-plicated. But a simple argument may explain why coloured fringes are more apparent underreflection conditions. As is well known, reflection near the Brewster angle can make thereflected light partially polarized. This is also true for birefringent materials, so the brightnessof the colours can be enhanced. It is also effective when reflection occurs outside the bire-fringent samples.

4. Conclusion

The standard method of producing birefringence patterns in a laboratory requires a polari-scope to meet the interference conditions. In this article, we have investigated the formation ofbirefringence patterns under everyday conditions.

First of all, we made a simple demonstration setup with a computer LCD (as thepolarized white light source), a candy box (as the birefringence sample) and a pair ofpolarizing sunglasses. Isoclinic fringes and isochromatic fringe patterns were observed andexplained. The interference pattern was also calculated using MATLAB, supposing that thebirefringence of the sample originating from residual stress decreases linearly from the centreto the rim.

Figure 8. Colour changes for different polarization of the light source. The colour in (c)is faint in contrast with (a) and (b). The polarization of the light source is unnecessaryfor the formation of colour patterns, although it can enhance this phenomenon. It tendsto be dark when the incident angle is large because the intensity is quite weak.

Eur. J. Phys. 35 (2014) 055008 Y Zhang et al

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Secondly, we tried to explain the common phenomenon that plastic objects revealcolours in transmission without the help of Polaroids under natural incident light. Wederived the equation of light propagation in birefringent materials in the case of uniaxialcrystals. From the constraints of the boundary conditions for electromagnetic waves, wenotice that light can be polarized due to refraction upon the interface. We calculated theintensities of the transmitted light through a uniaxial crystal plate for different wavelengthswith respect to incident angles. The result shows that at large incident angles, polarization ofthe incident light is unnecessary to produce birefringence patterns. Reflection and diffractionor other forms of polarization may enhance the brightness of the resulting patterns though.

Acknowledgments

The authors are grateful to Professor Jin Zhou from the Fundamental Physics Laboratory ofNanjing University for valuable advice.

References

[1] Bond M M and Hadley D W 1974 Photoelasticity without polaroids Phys. Educ. 9 411–3[2] Khan A S and Wang X 2001 Strain Measurement and Stress Analysis (New York: VCH)[3] Phillips J W 1998 Experimental Stress Analysis (Urbana Champaign: Univ. of Illinois)[4] Smith G S 2007 The polarization of skylight: an example from nature Am. J. Phys. 75 25–35[5] Können P 1985 Polarized Light in Nature (Cambridge: Cambridge University Press)[6] Paschotta R 2008 Optical intensity Encyclopedia of Laser Physics and Technology (Berlin: Wiley)[7] Gao W 2000 Optics (Nanjing: Nanjing University Press) pp 188–9[8] Bloss F D 1961 An Introduction to the Methods of Optical Crystallography (New York: Holt,

Rinehart and Winston)

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