PHYS20312 Wave Optics - Section 2: Polarization 2.1 Polarization States. Consider two orthogonally polarised plane waves travelling in the �-direction: �! = ��!��� �� − �� �! = ��!��� �� − �� + � The amplitudes of the two wave (�! and �!) and their relative phase, �, determine the resultant polarization state: Polarization State Relative amplitudes Phase, � Linear Can take any value 0 or � � Left-hand circular �! = �! + 2 � Right-hand circular �! = �! − 2 Elliptical �! ≠ �! arBitrary N.B. For an oBserver looking toward the source of light, � rotates clockwise for right-hand circularly polarized light, and anti-clockwise for left-hand circularly polarized light. Unpolarised & Partially Polarized Light The emission from atoms and molecules in natural light sources (e.g. the sun, a light bulb, a candle flame) is unsynchronised and so produces a resultant polarization that changes randomly and too rapidly to measure. This ‘smears out’ the polarization direction over 360°. We describe this as ‘unpolarized light’. If the polarization changes rapidly but tends to spend more time in a particular direction then this is described as partially polarized light. 2.2 Polarization by Reflection and Scattering. Both are the consequence of the anisotropic emission of dipoles (see animation in Powerpoint presentation); in particular, dipoles do not emit in the direction parallel to the original �. 2.2.1 Polarization by reflection. Reflectivity is polarization dependent, i.e.: �! �! ≠ �! !"#!$%"& �! !"#$"%&'() 1 | Page PHYS20312 Wave Optics - Section 2: Polarization where for an ‘s-polarized’ wave � is perpendicular to the plane of incidence (as defined by the incident and reflected rays), and for a ‘p-polarized’wave it is parallel to the plane of incidence – see Fig 1 Below. Figure 1 The electric field orientation of the incident, transmitted and reflected rays for (left) an s-polarised wave and (right) a p-polarised wave. 2.2.3 Brewster’s Angle. For p-polarized light, at a certain angle of incidence, �!, the axis of the induced dipole, which is perpendicular to the transmitted ray, will be parallel to the reflected ray and thus the reflected intensity will be zero–see Fig 2. Figure 2 A p-polarized ray incident at Brewster's angle, �� 2 | Page PHYS20312 Wave Optics - Section 2: Polarization Remembering that �! is also the angle of reflection, we see that this is equivalent to saying that zero amplitude will occur when the angles of incidence and transmission sum to 90°: �! + �! = 90° This special angle of incidence is called ‘Brewster’s angle’ or the ‘polarization angle’. Using Snell’s law: �! sin �! = �! sin �! = �� sin 90° − �� = �� cos �� !! �! �! = tan �! (2.1). 2.2.4 Polarization by Scattering. Optical E-field induces dipole in an atom or molecule (e.g. air) For polarized incident light, the polarization is maintained But the amplitude of scattered light changes with direction, as determined by the dipole emission – see Figs 3 and 4. In particular, horizontally polarized light is not scattered in the horizontal plane, and vertically polarized light not in the vertical plane Figure 3 Scattering of horizontally polarized light. Figure 4 Scattering of vertically polarized light 3 | Page PHYS20312 Wave Optics - Section 2: Polarization For unpolarized light, the polarization state changes with direction, see fig 5, the effects of which can be seen in nature (see Powerpoint). Figure 5 The scattering of unpolarized light. 2.3 Dichroism. Some materials selectively absorB one polarization. For instance, polaroid sheets consist of aligned, long and thin molecules (e.g. polyvinylalcohol). The E-field parallel to the long axis does work in moving charge and is aBsorBed. In contrast, the E-field perpendicular to this axis doe slittle work and passes through – see Fig 6. Figure 6 Polarization-dependent absorption in a polaroid sheet. The transmitted field component is thus � cos � and the transmitted intensity is ! � � = �! cos � (2.2) which is known as “Malus’ Law”. 4 | Page PHYS20312 Wave Optics - Section 2: Polarization 2.4 Optical Anisotropy. 2.4.1 Anisotropic materials. For isotropic dielectric materials, � is a scalar and hence can be factored out of Gauss’ equation: ∇. � = 0 �∇. � = 0 ∇. � = 0 which is a result used in the derivation of the wave equation. However, the structure of some crystals (e.g. quartz, ice, calcite) is anisotropic, which means that it is easier to polarize them in one direction compared to the other – see fig 7. Figure 7 Anisotropic polarization in calcite. Now the dielectric constant is descriBed instead By a tensor: � = �� where we can choose axes so that this tensor is diagonalised �! 0 0 � = 0 �! 0 �! 0 0 �! If �! ≠ �! ≠ �! then the crystal is “biaxial” If �! = �! ≠ �! then the crystal is “uniaxial”, and we will restrict ourselves to this case for simplicity. ! ! Let �! = �! = �! and �! = �! 5 | Page PHYS20312 Wave Optics - Section 2: Polarization Now as Before ∇. � = 0 i.e. ��! ��! ��! + + = 0 �� �� �� ��! ��! ��! � + � + � = 0 ! �� ! �� ! �� but since now �! = �! ≠ �! ∇. � ≠ 0 which, as we will see in the next section, changes the wave equation. 2.4.2 The wave equation in anisotropic materials. We will now re-derive the wave equation taking into account that in anisotropic materials ∇. � ≠ 0: Faraday’s law: ∇×� = −� apply the curl operator: ∇× ∇×� = −∇×� but Ampere’s law is ∇×� = ��� and using the identity ∇× ∇×� = ∇ ∇. � − ∇!� yields a wave equation: ∇!� − ∇ ∇. � = ��� (2.3) In isotropic materials, the second term on the LHS is zero Because ∇. � = 0 but for anisotropic materials it is not and so this wave equation is different from the one we’ve met previously. This suggests that light will propagate in a different way through anisotropic materials (and it does). We now examine what happens differently by considering the case of a plane wave which, Because any optical field can be described by a linear combination of them, is quite general. Consider a plane wave of the following form propagating through thecrystal: � = ����� � �. � − �� Hence, ∇!� = −�!�; ∇. � = ��. �; � = −�!�; and ∇ ��. � = −� �. � SuBstituting these into equation 2.3 gives 6 | Page PHYS20312 Wave Optics - Section 2: Polarization −�!� + � �. � = −���!� writing this out explicitly for each component: ! ! −� �! + �! �. � = −��!�!� �! ! ! −� �! + �! �. � = −��!�!� �! (2.4) ! ! −� �! + �! �. � = −��!�!� �! We can also write this set of the 3 equations in a matrix form: �. � = 0 where �!�! ! − �! + �! � � � � �! ! � � � � ! ! � �! � = � � − �! + �! � � � � �! ! � � �!�! � � � � ! − �! + �! � � � � �! ! !! N.B. �� � = ! etc ! ! !! The solution to this set of equations is given By ���� = 0 which after a large amount of algeBra yields: ! ! ! ! ! ! ! ! �! �! �! � �! �! �! � ! + ! + ! − ! ! + ! + ! − ! = 0 �! �! �! � �! �! �! � (2.5) Equation 2.5 thus has two solutions: solution 1 corresponds to when the first Bracket is zero and solution 2 to when the second Bracket is zero. As we will see in the next sections, these distinct solutions to the wave equation correspond to different rays, which Behave in different ways. 3.4.3. Birefringence. What is the refractive index for the rays corresponding to the two solutions of equation 2.5? The refractive index for a ray, � �� � = = v! � 1 �! ∴ = �! �!�! 7 | Page PHYS20312 Wave Optics - Section 2: Polarization First consider solution 1: ! ! ! ! �! �! �! � ! + ! + ! = ! �! �! �! � but ! ! ! ! �! + �! + �! = � 1 �! 1 1 ∴ ! = ! ! = ! �! � � � � = ±�! i.e. a constant.(Here the negative solution corresponds to – v!) However, we get something different for solution 2: ! ! ! ! �! �! �! � ! + ! + ! = ! �! �! �! � Now we can’t find a common factor for the LHS as for solution so instead we will suBstitute in expressions for �!, �! and �!, which we can write down by considering Fig 8: Figure 8 The orientation of the k, Eo and Ee relative to the crystal axes. Hence, �! = � sin � cos � �! = � sin � sin � �! = � cos � 8 | Page PHYS20312 Wave Optics - Section 2: Polarization SuBstitute these into solution 2: �! cos!� + sin!� sin!� �! cos!� �! ! + ! = ! �! �! � 1 sin!� cos!� ! = ! + ! � �! �! (2.6) Note that in this case the refractive index is angle-dependent, i.e. � = � � (we write � � from now on to reflect this. Hence, a Beam travelling through an anisotropic crystal will experience two different refractive indices, �! and � � , which implies different velocities and different phases. In other words, the two solutions of equation 3.5 correspond to two distinct rays. A Beam of light entering an anisotropic crystal will thus split into separate rays: 1. The ordinary ray or “o-ray”, corresponding to solution 1, for which � = �! 2. The extraordinary or “e-ray”, corresponding to solution 1, for which � = � � Note however that � � = 0° = �! and hence at this angle the distinction is lost. This special direction of � = 0° is called the “Optics Axis”. If �! = �! ≠ �! then there will Be one optic axis and then hence the crystal is descriBed as uniaxial. However, if �! ≠ �! ≠ �! then there will Be two optic axes and the crystal is descriBed as biaxial. 2.4.4 E vectors in anisotropic crystals We are now going to find out what we can aBout the � vectors associated with Both the o-ray and the e-ray: for the o-ray (��): From equation 2.4 we have for the �-component (and a similar expression for the �-component): ! �! −�!� + � �. � = −�� � �!� = − �!� !,! ! � ! ! !,! �! !,! but we have found in the previous section that for an o-ray (i.e. solution 1 of equation 2.5) ! �! �! = �! �! Hence ! ! −� �!,! + �! �. �� = −� �!,! and thus �. ��