Download PHYS20312 Wave Optics -‐ Section 2: Polarization

PHYS20312 Wave Optics -­‐ Section 2: Polarization 2.1 Polarization States. Consider two orthogonally polarised plane waves travelling in the 𝑧-­‐direction: 𝑬! = 𝒙𝐸! 𝑒𝑥𝑝 𝑘𝑧 − 𝜔𝑡 𝑬! = 𝒚𝐸! 𝑒𝑥𝑝 𝑘𝑧 − 𝜔𝑡 + 𝜙 The amplitudes of the two wave (𝐸! and 𝐸! ) and their relative phase, 𝜙 , determine the resultant polarization state: Polarization State Relative amplitudes Phase, 𝝓 Linear Left-­‐hand circular Can take any value 𝐸! = 𝐸! Right-­‐hand circular 𝐸! = 𝐸! Elliptical 𝐸! ≠ 𝐸! 0 or 𝜋 𝜋
+ 2
𝜋
− 2
arbitrary N.B. For an observer looking toward the source of light, 𝑬 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 𝑬. 2.2.1 Polarization by reflection. Reflectivity is polarization dependent, i.e.: 𝐸!
𝐸!
≠
!"#!$%"&
𝐸!
𝐸!
!"#$"%&'()
1 | P a g e PHYS20312 Wave Optics -­‐ Section 2: Polarization where for an ‘s-­‐polarized’ wave 𝑬 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, 𝜃! , 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, 𝜽𝑩 2 | P a g e PHYS20312 Wave Optics -­‐ Section 2: Polarization Remembering that 𝜃! 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°: 𝜃! + 𝜃! = 90° This special angle of incidence is called ‘Brewster’s angle’ or the ‘polarization angle’. Using Snell’s law: 𝑛! sin 𝜃! = 𝑛! sin 𝜃! = 𝑛𝑡 sin 90° − 𝜃𝐵 = 𝑛𝑡 cos 𝜃𝐵
𝜃! = tan!!
𝑛!
𝑛!
(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 | P a g e 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 𝐸 cos 𝜃 and the transmitted intensity is 𝐼 𝜃 = 𝐼! cos ! 𝜃 (2.2) which is known as “Malus’ Law”. 4 | P a g e PHYS20312 Wave Optics -­‐ Section 2: Polarization 2.4 Optical Anisotropy. 2.4.1 Anisotropic materials. For isotropic dielectric materials, 𝜀 is a scalar and hence can be factored out of Gauss’ equation: ∇. 𝑫 = 0 𝜀∇. 𝑬 = 0 ∇. 𝑬 = 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: 𝑫 = 𝜺𝑬 where we can choose axes so that this tensor is diagonalised 𝜀!
𝜀= 0
0
0
𝜀!
0
0
0 𝜀! 𝜀!
If 𝜀! ≠ 𝜀! ≠ 𝜀! then the crystal is “biaxial” If 𝜀! = 𝜀! ≠ 𝜀! then the crystal is “uniaxial”, and we will restrict ourselves to this case for simplicity. Let 𝜀! = 𝜀! = 𝑛!! and 𝜀! = 𝑛!! 5 | P a g e PHYS20312 Wave Optics -­‐ Section 2: Polarization Now as before ∇. 𝑫 = 0 i.e. 𝜕𝐷! 𝜕𝐷! 𝜕𝐷!
+
+
= 0 𝜕𝑥
𝜕𝑦
𝜕𝑧
𝜀!
𝜕𝐸!
𝜕𝐸!
𝜕𝐸!
+ 𝜀!
+ 𝜀!
= 0 𝜕𝑥
𝜕𝑦
𝜕𝑧
but since now 𝜀! = 𝜀! ≠ 𝜀! ∇. 𝑬 ≠ 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 ∇. 𝑬 ≠ 0: Faraday’s law: apply the curl operator: ∇× ∇×𝑬 = −∇×𝑩 ∇×𝑬 = −𝑩 but Ampere’s law is ∇×𝑩 = 𝜀𝜇𝑬 and using the identity ∇× ∇×𝑬 = ∇ ∇. 𝑬 − ∇! 𝑬 yields a wave equation: ∇! 𝑬 − ∇ ∇. 𝑬 = 𝜀𝜇𝑬 (2.3) In isotropic materials, the second term on the LHS is zero because ∇. 𝑬 = 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: 𝑬 = 𝑬𝟎 𝑒𝑥𝑝 𝑖 𝒌. 𝒓 − 𝜔𝑡 Hence, ∇! 𝑬 = −𝑘 ! 𝑬; ∇. 𝑬 = 𝑖𝒌. 𝑬; 𝑬 = −𝜔 ! 𝑬; and ∇ 𝑖𝒌. 𝑬 = −𝒌 𝒌. 𝑬 Substituting these into equation 2.3 gives 6 | P a g e PHYS20312 Wave Optics -­‐ Section 2: Polarization −𝑘 ! 𝑬 + 𝒌 𝒌. 𝑬 = −𝜇𝜀𝜔 ! 𝑬 writing this out explicitly for each component: −𝑘 ! 𝐸! + 𝑘! 𝒌. 𝑬 = −𝜇𝜀! 𝜀! 𝜔 ! 𝐸!
−𝑘 ! 𝐸! + 𝑘! 𝒌. 𝑬 = −𝜇𝜀! 𝜀! 𝜔 ! 𝐸! −𝑘 ! 𝐸! + 𝑘! 𝒌. 𝑬 = −𝜇𝜀! 𝜀! 𝜔 ! 𝐸!
(2.4) We can also write this set of the 3 equations in a matrix form: 𝑴. 𝑬 = 0 where 𝜔 ! 𝑛!!
− 𝑘 ! + 𝑘!!
𝑐!
𝑴=
N.B. 𝜇 𝜀! 𝜀! =
!!!
!!
𝑘𝒙 𝑘𝒚
𝑘𝒙 𝑘𝒛
𝑘𝒙 𝑘𝒚
𝜔 ! 𝑛!!
− 𝑘 ! + 𝑘!!
𝑐!
𝑘𝒚 𝑘𝒛
𝑘𝒙 𝑘𝒛
𝑘𝒚 𝑘𝒛
𝜔 ! 𝑛!!
− 𝑘 ! + 𝑘!!
𝑐!
etc The solution to this set of equations is given by 𝑑𝑒𝑡𝑴 = 0 which after a large amount of algebra yields: 𝑘!! 𝑘!! 𝑘!! 𝜔 !
+
+
−
𝑛!! 𝑛!! 𝑛!! 𝑐 !
𝑘!! 𝑘!! 𝑘!! 𝜔 !
+
+
−
= 0 𝑛!! 𝑛!! 𝑛!! 𝑐 !
(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, 𝑛=
∴ 𝑐
𝑐𝑘
= v!
𝜔
1
𝜔!
=
𝑛! 𝑐 ! 𝑘 !
7 | P a g e PHYS20312 Wave Optics -­‐ Section 2: Polarization First consider solution 1: 𝑘!! 𝑘!! 𝑘!! 𝜔 !
+
+
= ! 𝑐
𝑛!! 𝑛!! 𝑛!!
but 𝑘!! + 𝑘!! + 𝑘!! = 𝑘 ! ∴
1
𝜔! 1
1
=
= ! !
!
!
𝑐 𝑘
𝑛
𝑛!
𝑛 = ±𝑛! i.e. a constant.(Here the negative solution corresponds to – v! ) However, we get something different for solution 2: 𝑘!! 𝑘!! 𝑘!! 𝜔 !
+
+
= ! 𝑐
𝑛!! 𝑛!! 𝑛!!
Now we can’t find a common factor for the LHS as for solution so instead we will substitute in expressions for 𝑘! , 𝑘! and 𝑘! , 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, 𝑘! = 𝑘 sin 𝜃 cos 𝜙 𝑘! = 𝑘 sin 𝜃 sin 𝜙 𝑘! = 𝑘 cos 𝜃 8 | P a g e PHYS20312 Wave Optics -­‐ Section 2: Polarization Substitute these into solution 2: 𝑘 ! cos ! 𝜙 + sin! 𝜙 sin! 𝜃 𝑘 ! cos ! 𝜃 𝜔 !
+
= ! 𝑐
𝑛!!
𝑛!!
1
sin! 𝜃 cos ! 𝜃
=
+
𝑛!
𝑛!!
𝑛!!
(2.6) Note that in this case the refractive index is angle-­‐dependent, i.e. 𝑛 = 𝑓 𝜃 (we write 𝑛 𝜃 from now on to reflect this. Hence, a beam travelling through an anisotropic crystal will experience two different refractive indices, 𝑛! and 𝑛 𝜃 , 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 𝑛 = 𝑛! 2. The extraordinary or “e-­‐ray”, corresponding to solution 1, for which 𝑛 = 𝑛 𝜃 Note however that 𝑛 𝜃 = 0° = 𝑛! and hence at this angle the distinction is lost. This special direction of 𝜃 = 0° is called the “Optics Axis”. If 𝜀! = 𝜀! ≠ 𝜀! then there will be one optic axis and then hence the crystal is described as uniaxial. However, if 𝜀! ≠ 𝜀! ≠ 𝜀! 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 𝑬 vectors associated with both the o-­‐ray and the e-­‐ray: for the o-­‐ray (𝑬𝒐 ): From equation 2.4 we have for the 𝑥 -­‐component (and a similar expression for the 𝑦-­‐component): −𝑘 ! 𝐸!,! + 𝑘! 𝒌. 𝑬𝒐 = −𝜇𝜀! 𝜀! 𝜔 ! 𝐸!,! = −
𝑛!! !
𝜔 𝐸!,! 𝑐!
but we have found in the previous section that for an o-­‐ray (i.e. solution 1 of equation 2.5) 𝑘! =
𝑛!! !
𝜔 𝑐!
Hence −𝑘 ! 𝐸!,! + 𝑘! 𝒌. 𝑬𝒐 = −𝑘 ! 𝐸!,! and thus 𝒌. 𝑬𝒐 = 0 9 | P a g e PHYS20312 Wave Optics -­‐ Section 2: Polarization Substituting this into the expression for the 𝑧-­‐component from equation 2.4: −𝑘 ! 𝐸!,! + 𝑘! 𝒌. 𝑬𝒐 = −
𝑛!! !
𝜔 𝐸!,! 𝑐!
yields 𝑛!! 𝐸!,! = 𝑛!! 𝐸!,! , which since 𝑛!! ≠ 𝑛!! implies 𝐸!,! = 0. Using this result back in 𝒌. 𝑬𝒐 = 0, we find that 𝑘! 𝐸!,! + 𝑘! 𝐸!,! + 0 = 0 𝐸!,! = −
so 𝑘!
𝐸 𝑘! !,!
1
1
!!
−
𝑬! =
𝐸!,! = − cot 𝜙 𝐸!,! !!
0
0
(2.7) So since 𝒌. 𝑬𝒐 = 0 and 𝐸!,! = 0 we conclude the 𝑬𝒐 is perpendicular to both 𝒌 and the 𝑧-­‐axis – as shown in Fig 8. for the e-­‐ray: Re-­‐arranging solution 2 of equation 2.5, we get: 𝑛!! 𝜔 !
𝑘!! 𝑘!!
!
= 𝑛! ! + ! + 𝑘!! 𝑐!
𝑛! 𝑛!
Substituting this into the 𝑥 and 𝑦 parts of equation 2.4, we get: 𝑘!! + 𝑘!!
𝑛!!
− 1 𝐸!,! + 𝑘! 𝒌. 𝑬𝒆 = 0 𝑛!!
𝑘!! + 𝑘!!
𝑛!!
− 1 𝐸!,! + 𝑘! 𝒌. 𝑬𝒆 = 0 𝑛!!
and dividing the first of these by the second yields: 𝐸!,! 𝑘!
=
= cot 𝜙 𝐸!,! 𝑘!
Solution 2 can also be re-­‐arranged to give: 𝑛!! 𝜔 !
𝑛!! !
!
!
=
𝑘
+
𝑘
+
𝑘! !
!
𝑐!
𝑛!!
and substituting this in to the 𝑧 part of equation 2.4 𝑘!!
𝑛!!
− 1 𝐸!,! + 𝑘! 𝒌. 𝑬𝒆 = 0 𝑛!!
10 | P a g e PHYS20312 Wave Optics -­‐ Section 2: Polarization Comparing this expression with the one for 𝑥 : 𝐸!,!
𝑛!! 𝑘!! + 𝑘!!
𝑛!! tan 𝜃
=− !
=− !
𝐸!,!
𝑛! 𝑘! 𝑘!
𝑛! cos 𝜙
We can now write each of the components of 𝑬𝒆 in terms of 𝐸!,! : 𝑬𝒆 =
𝑛!!
− !
𝑛!
1
𝑘!
𝑘!
𝑘!! + 𝑘!!
𝑘! 𝑘!
𝐸!,! =
1
tan 𝜙
𝑛!! tan 𝜃 𝐸!,! − !
𝑛! cos 𝜙
(2.8) Now note that: i.
ii.
𝑬𝒐 . 𝑬𝒆 = 0 𝒌. 𝑬𝒆 ≠ 0 i.e. the o-­‐ray and e-­‐ray are polarized orthogonally i.e. 𝑬𝒆 is not perpendicular to the wave vector 2.4.5 Double Refraction. The direction of a ray corresponds to the Poynting vector, 𝑺 = 𝑬×𝑩 Taking the example of a plane wave for which 𝑬 = 𝑬𝟎 𝑒𝑥𝑝 𝑖 𝒌. 𝒓 − 𝜔𝑡 𝑩 = 𝑩𝟎 𝑒𝑥𝑝 𝑖 𝒌. 𝒓 − 𝜔𝑡 Faraday’s Law gives us in this case: ∇×𝑬 = −𝑩 𝑖𝒌×𝑬 = 𝑖𝜔𝑩 Hence, 𝑺 = 𝑬×𝑩 =
1
1
𝑬× 𝒌×𝑬 =
𝒌 𝑬. 𝑬 − 𝑬 𝑬. 𝒌 𝜔
𝜔
which implies that 𝑺 is only parallel to 𝒌 if 𝑬. 𝒌 = 0 i.e. when cos 𝛼 = 0 where 𝛼 is the angle between 𝑬 and 𝒌. Let’s now calculate 𝛼 for both the o-­‐ray and e-­‐ray using cos 𝛼 =
𝑬. 𝒌
𝑬 𝒌
11 | P a g e PHYS20312 Wave Optics -­‐ Section 2: Polarization for the o-­‐ray: We showed in the previous section 𝒌. 𝑬𝒐 = 0 and so cos 𝛼 = 0 for the e-­‐ray: Using the expressions found in the previous section: 𝑬𝒆 . 𝒌 = 𝐸!,! 𝑘! +
𝑘!! 𝑛!! 𝑘!! + 𝑘!!
−
𝑘! 𝑛!!
𝑘!
= 𝑘𝐸!,!
sin 𝜃
𝑛!!
1 − ! cos 𝜙
𝑛!
and 𝑬𝒆 𝒌 = 𝑘𝐸!,! 1 +
where the identity 1 + tan! 𝜙 =
!
!"#! !
tan! 𝜙
𝑛!! tan! 𝜃 𝑘𝐸!,!
𝑛!!
+ !
=
1 + ! tan! 𝜃 𝑛! cos ! 𝜙 cos 𝜙
𝑛!
has been used to simplify the expression. Hence cos 𝛼 =
𝑬. 𝒌
=
𝑬 𝒌
sin 𝜃 1 −
1+
𝑛!!
𝑛!!
𝑛!!
tan! 𝜃
𝑛!!
(2.9) which ≠ 0 except when: i.
ii.
𝜃 = 0° i.e. for propagation along the optics axis 𝜃 = 90° i.e. for propagation perpendicular to the optics axis Hence, the o-­‐ray and e-­‐ray propagate in different directions except when 𝒌 is parallel to or perpendicular to the optic axis. The propgagtion of the o-­‐ray and e-­‐ray in difference directions is called “double refraction” and is illustrated in Fig 9 below. 12 | P a g e PHYS20312 Wave Optics -­‐ Section 2: Polarization Figure 9 Double refraction. Let’s now re-­‐cap what we’ve learnt: •
•
•
•
A beam propagating though an anisotropic crystal splits into two components of orthogonal polarization – the o-­‐ray and the e-­‐ray Each ray experiences a different refractive index, which is called birefringence, and hence different v! and phase, where for the o-­‐ray 𝑛! =constant and for e-­‐ray 𝑛 = 𝑓 𝜃 The ‘optic axis’ is that direction such that refractive indices for the two rays are equal. The o-­‐ and e-­‐rays travel in different directions, which is called double refraction, except when 𝒌 is either parallel or perpendicular to the optic axis. 13 | P a g e PHYS20312 Wave Optics -­‐ Section 2: Polarization 2.4.6 The Polarising Beam-­‐splitter. For total internal reflection, the angle of incidence, 𝜃! , must be greater than the critical angle, 𝜃! , where 𝜃! = sin!!
𝑛!
𝑛!
(2.10) and where 𝑛! > 𝑛! . Hence, in a birefringent crystal, 𝜃! , will be different for the o-­‐ray and e-­‐ray. This can be exploited to split an unpolarized beam into two orthogonally polarized components – as shown in Fig 10 below. Figure 10 A polarizing beam-­‐splitter. •
•
•
The optic axis is oriented so that the rays propagate in the same direction until they reach the interface i.e. 𝜃 = 90° The second piece of calcite ensures that the e-­‐eay is parallel to the input beam. This device is also known as a ‘Glan-­‐Thompson’ polarizer or a ‘Glan-­‐Foucalt’ polarizer if no cement is used and there is just an air gap between the calcite pieces. 14 | P a g e PHYS20312 Wave Optics -­‐ Section 2: Polarization 2.4.7 Waveplates. Consider the phases of the o-­‐ and e-­‐rays: 𝜙! = 𝑘! 𝑧 − 𝜔𝑡 and 𝜙! = 𝑘! 𝑧 − 𝜔𝑡 but 𝑘! =
!!
!
𝑛! and 𝑘! =
!!
!
𝑛 𝜃 Hence, the phase difference between the rays is Δ𝜙 = 𝜙! − 𝜙! = 𝑘! − 𝑘! 𝑧 So, for a crystal of thickness, 𝑑 Δ𝜙 𝜃 =
2𝜋
𝑑 𝑛! − 𝑛 𝜃 𝜆
If we orient the crystal such that 𝜃 = 90° then the rays will not separate and 𝑛 𝜃 = 𝑛! hence, Δ𝜙 𝜃 = 90° =
2𝜋
𝑑 𝑛! − 𝑛! 𝜆
This is called a waveplate and is shown below in Fig 11: Figure 11 A waveplate. Terminology: •
The direction for which is 𝑛 is a minimum and hence v! is a maximum is called the ‘fast axis’ •
The direction for which is 𝑛 is a maximum and hence v! is a minimum is called the ‘slow axis’ 15 | P a g e PHYS20312 Wave Optics -­‐ Section 2: Polarization Now consider two special cases: 1. Set 𝑑 so that Δ𝜙 = 2𝑚 + 1 𝜋 where 𝑚 is an integer. This produces a 𝜆 2 phase difference between the o-­‐ and e-­‐rays and is hence called a “half-­‐wave plate”. In the example sheet, you will see how it can be used to flip the polarization direction about the optic axis. 2. Set 𝑑 so that Δ𝜙 = 4𝑚 + 1 𝜋 2 where 𝑚 is an integer. This produces a 𝜆 4 phase difference between the o-­‐ and e-­‐rays and is hence called a “quarter-­‐wave plate”. In the example sheet, you will see how it can be used to change circular polarization to linear and vice versa. 2.5 The Faraday Effect. The plane of polarization is rotated as light passes through a length of material, 𝑑 , in the presence of a magnetic field,𝐵, as shown below: Figure 12 The Faraday Effect (Image from http://en.wikipedia.org/wiki/Faraday_effect). This rotation angle, 𝛽 , is given by: 𝛽 = 𝑉𝐵𝑑 where 𝑉 is the “Verdet” constant – some example values are given in the Table below: 16 | P a g e PHYS20312 Wave Optics -­‐ Section 2: Polarization -1
-1
Material
V (rad.T .m )
Glass
9.2×10
Water
3.8 ×10
Air
1.8 ×10
Quartz
4.8 ×10
Terbium Gallium Garnet
-40
-4
-4
-7
-4
Note that the direction of rotation depends on the sign of both 𝑉 and 𝐵 but not on the direction of propagation ( is defined as positive for anticlockwise rotation for 𝐵 pointing towards the viewer). This enables an “Optical Diode” to be produced (otherwise known as a Faraday Isolator) i.e. a device that allows light to pass in one direction but not the other. Consider Fig 13: Figure 13 Light passing through (left) and being blocked by (right) an optical diode. For the left-­‐hand diagram: •
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light passes through polariser P1 (becoming vertically polarized if not already) and then through the material in the presence of a B-­‐field and is rotated by 45° clockwise this allows it to pass without loss through the second polarizer P2 oriented at 45° to the vertical For the right-­‐hand diagram: •
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light passes through P2 (and thus becomes polarized at 45° to the vertical) and then through the material in the presence of a B-­‐field and is rotated by 45° clockwise it is now polarised horizontally and so is blocked by P1 17 | P a g e