Finally, a drastic simplification takes place if we make use of Snell’s law:
r⊥ =−sin(θi−θt)
sin(θi+θt), (3.30)
rk = +tan(θi−θt)
tan(θi+θt), (3.31)
t⊥ = +2 sinθtcosθi
sin(θi+θt) , (3.32)
tk= + 2 sinθtcosθi
sin(θi+θt) cos(θi−θt). (3.33)
Reflection and refraction 45
FIGURE 3.6: Amplitude reflection and transmission coefficients r⊥, rk, t⊥ and tk, respectively (from bottom to top), for ni = 1.0 and nt= 1.5.
with ⊥ polarization. In order to have a comprehensive idea of the behavior of all the amplitude reflection and transmission coefficients, we have plotted them in Fig. 3.6 forni = 1.0 and nt= 1.5. The Brewster angle for this case isθB= 56.3◦.
The situation changes drastically when ni > nt, resulting in θt > θi. In this caser⊥ is always positive while rk passes through zero at the Brewster angle (seeFig (3.7)). However, both the perpendicular and parallel amplitude reflection coefficients become complex beyond a critical angleθc given by
θc= sin−1(nt/ni). (3.35) It is easy to verify that θi = θc corresponds to an angle of refraction equal to π/2. For angles of incidence larger than the critical angle the magnitude of both the amplitude reflection coefficients equals one, implying thereby the return of all the energy to the medium of incidence. Hence the critical angle is also referred to as the angle of total internal reflection (TIR). The Brewster and the TIR angles for this case are given byθB = 33.7◦andθc= 41.8◦. Note the complementary nature of the Brewster angles for the cases of Figs. (3.6) and (3.7).
3.5.2 Phase shifts
We first look at the phase change in reflection for the case θi > θt, i.e., when light enters a denser medium from a lighter one (ni < nt). It is clear from Eq. (3.30) that, for any angle of incidence, r⊥ is negative. Thus the perpendicular (to plane of incidence) component of the electric field undergoes a phase shift of π under reflection when the incident medium has a lower refractive index than the transmitting medium.
0 20 40 60 80 0
0.5 1
Amplitude reflection
θi (o) r
r
θB θC
FIGURE 3.7: Amplitude reflection coefficientsrkandr⊥, respectively (from bottom to top), forni= 1.5 andnt= 1.0.
E
B E
B
B E (a)
E B
E B
E B (b)
FIGURE 3.8: Explanation of (a) out-of-phase and (b) in-phase components.
The situation is slightly more complicated for the components on the plane of incidence, since we need to define the terms in-phase and out-of-phase.
Fig. 3.8 is handy for understanding these notions. Two fields in the incidence plane arein-phase if their components along the unit normal to the surface are parallel and areout-of-phase if these components are antiparallel. If two E fields are antiparallel, then the same holds for the corresponding two B fields.
It is clear from Eq. (3.28) that rk is positive or the phase difference be- tween the reflected and incident components is zero (∆φk= 0) so long as the numerator is positive. The inequality that the numerator is positive can be rewritten in the form
sin(θi−θt) cos(θi+θt)>0. (3.36)
Reflection and refraction 47
0 20 40 60 80
0 0.5 1
phase in π units
0 50 100
0 0.5 1
θi (o) θi (o)
(a) (b)
FIGURE 3.9: Phase angles (a) ∆φ⊥ and (b) ∆φk components in units ofπ forni< nt.
Forni< nt, this translates into
θi+θt< π/2, (3.37)
while forni> nt, we have
θi+θt> π/2. (3.38)
Thus for ni < nt there will be zero phase lag between the reflected and incident components in the range (θi = 0−θB). Thereafter there will be a phase difference ofπ (see Fig. 3.9). In contrast, for a TIR case (forni > nt) rk is negative untilθreaches the Brewster angle, which implies that ∆φk=π.
FromθB toθc, ∆φk= 0. Beyond the critical angle the reflection coefficient is complex and ∆φkincreases gradually toπat 90◦. The results for this case are shown inFig. 3.10, where the last plot gives the difference between the parallel and perpendicular phases. These results will be used in order to understand the change of polarization state under total internal reflection and their use in various polarization devices.
3.5.3 Reflectance and transmittance
Let a light beam of circular cross-section be incident on the surface of a dielectric at an angle θi. Since we will be dealing with the reflected and transmitted intensities, it is better to recall the meaning and expressions of the Poynting vectorSand irradianceI. The Poynting vector gives the power per unit area crossing a normal surface and is given by
S=c2ε0E×B. (3.39)
The radiant flux density or the irradiance is the time-averaged Poynting vector and has the expression
< S >= cε0
2 E02. (3.40)
0 50 0
0.5 1
θi (o)
phase in units of π
0 50
0 0.5 1
0 50
0 0.5 1
θi (o) θi (o)
(a) (b) (c)
FIGURE 3.10: Phase angles (a) ∆φk and (b) ∆φ⊥ components in units of πfornt< ni. (c) gives the difference ∆φk−∆φ⊥.
The irradiance has the unitW/m2and it is the average energy crossing a unit area normal toS.
Let Ii, Ir and It be the incident, reflected and transmitted irradiances, respectively. The corresponding cross-sectional areas are Acosθi, Acosθr
andAcosθt, respectively. The incident, reflected and transmitted powers are IiAcosθi,IrAcosθr and ItAcosθt, respectively. The first quantity is the in- cident energy per unit time associated with the incident beam, and hence it is the incident power that falls onA. We define reflectance R as the ratio of reflected and incident power (flux) as
R= IrAcosθr
IiAcosθi
= Ir
Ii
. (3.41)
Similarly, transmittanceT can be defined as T= Itcosθt
Iicosθi, (3.42)
where
Ij =vjεjE20j, j =i, r, t. (3.43) Noting that both the medium of incidence and reflection are the same, Eq. (3.41) can be rewritten as
R= E0r
E0i
2
=r2. (3.44)
Likewise forT, we have
T =ntcosθt
nicosθi
E0t
E0i
2
=
ntcosθt
nicosθi
t2. (3.45)
Reflection and refraction 49
0 20 40 60 80
0 0.2 0.4 0.6 0.8 1
R, T
0 20 40 60 80
0 0.2 0.4 0.6 0.8 1
R, T
) b ( )
a (
θi (o) θi (o)
FIGURE 3.11: Reflectance (solid line) and transmittance (dashed line) for (a)⊥and (b)korientations, respectively, forni= 1.0 andnt= 1.5.
In Eq. (3.45) we made use of the relation µ0εj = µ0ε0n2j = n2j/c2 = 1/v2j (j=i, t). For normal incidence whenθi =θr=θt= 0, reflectance and trans- mittance are given by the ratios of the corresponding irradiances. Note thatT is not simply equal tot2for two reasons. First, the velocity of light in the two media is not the same. Hence the ratio of the refractive indices must appear inT. Second, the cross-sectional areas for incident and transmitted beams are not the same. The energy flow per unit area will be affected accordingly.
3.5.4 Energy conservation
The energy flowing into the areaAper unit time must be the same as that flowing out of it. Thus
IiAcosθi =IrAcosθr+ItAcosθt. (3.46) Multiplying both sides byc, Eq. (3.46) can be written as
niE0i2 cosθi=nrE0r2 cosθr+ntE0t2 cosθt, (3.47) expressed differently as
1 = E0r
E0i
2 +
ntcosθt
nicosθi
E0t
E0i
2
, or R+T = 1. (3.48) Eq. (3.48) just states that total reflectance and transmittance in passing through an interface add up to unity in absence of losses. The part that is not transmitted is bound to be reflected. In component form these relations read
as
R⊥=r2⊥, (3.49)
Rk=r2k, (3.50)
T⊥=
ntcosθt
nicosθi
t2⊥, (3.51)
Tk=
ntcosθt
nicosθi
t2k, (3.52)
R⊥+T⊥= 1, (3.53)
Rk+Tk= 1. (3.54)
The results for the reflectance and transmittance are shown in Fig. 3.11. It is easy to identify the Brewster angle at which there is null reflectance and unity transmittance for thep-polarized light. For normal incidence (θi = 0), the expressions for reflectance and transmittance reduce to
R=Rk=R⊥=
n1−n2
n1+n2
2
(3.55) and
T =Tk=T⊥= 4n1n2
(n1+n2)2. (3.56)
3.5.5 Evanescent waves
It can be easily verified that it is impossible to satisfy the boundary con- ditions if we assume that there is no transmitted wave in case of total internal reflection. In order to understand this, we rewrite Eqs. (3.24) and (3.28) in the form
r⊥ =nicosθi−ntcosθt
nicosθi+ntcosθt
=cosθi−(n2ti−sin2θi)1/2
cosθi+ (n2ti−sin2θi)1/2, (3.57) rk =ntcosθi−nicosθt
ntcosθi+nicosθt =n2ticosθi−(n2ti−sin2θi)1/2
n2ticosθi+ (n2ti−sin2θi)1/2, (3.58) wherenti=nt/ni<1 for a TIR situation. Forθi> θc, sin(θi)>sin(θc) =nti
and both r⊥ and rk are complex. Even then, r⊥r∗⊥ = rkrk∗ = 1 and the reflectance R equals unity. This means that though there is a transmitted wave, it cannot carry any energy across the boundary. In order to have a deeper understanding, let us write the expression for the transmitted wave as follows (we assume that the xz plane forms the plane of incidence with the rarer medium occupying the half-spacez >0):
Et=E0texpi(kt·r−ωt). (3.59)
Reflection and refraction 51 For the chosen geometry,
kt·r=ktxx+ktzz=ktsin(θt)x+ktcos(θt)z. (3.60) Using Snell’s law we have
ktz =ktcos(θt) =±kt
1−sin2θi
n2ti 1/2
=±i kt
sin2θi
n2ti −1 1/2
=±iβ.
(3.61) In writing Eq. (3.61), we used the fact that whenθi> θc, sinθi> nti. For the surface component we have
ktx= (kt/nti) sin(θi). (3.62) Thus for the transmitted wave we have the expression
Et=E0te∓βzei[(kt/nti)sin(θi)x−ωt]. (3.63) Here we neglect the positive exponential because it is unphysical. We thus have a wave whose amplitude decays as we see deeper in the rarer medium.
The wave advances along the surface as a surface or evanescent wave. The amplitude decays very fast (over a few wavelengths) as one moves away from the interface. Thus this represents an inhomogeneous wave. For this case the surfaces of constant phase (parallel to the yz plane) are perpendicular to surfaces of constant amplitude (parallel to thexyplane).
One of the areas where TIR is used extensively is to design beam-splitters with precise control of the transmission in two or more arms. Another impor- tant area involves the use of phase change beyond the critical angle to design various polarizers and for conversion of polarization from linear to circular and vice versa [8].
Chapter 4
Elements of polarization, anisotropy and birefringence
4.1 Basic types of polarization: Linear and elliptically polarized
waves . . . 54 4.1.1 Polarizers and analyzers . . . 56 4.1.2 Degree of polarization . . . 56 4.2 Stokes parameters and Jones vectors . . . 57
4.2.1 Representation of polarization states of a
monochromatic wave . . . 57 4.2.2 Measurement of Stokes parameters . . . 58 4.3 Anisotropy and birefringence . . . 59 4.3.1 Birefringence in crystals like calcite . . . 60 4.3.2 Polarizers based on birefringence: Nicol and Wollaston
prisms . . . 61 4.3.3 Retardation plates . . . 62 4.4 Artificial birefringence . . . 63 4.4.1 Stress-induced anisotropy . . . 63 4.4.2 Kerr effect . . . 64 4.5 Optical activity and rotation of plane of polarization . . . 64 Before Einstein’s theory of relativity and the famous Michelson-Morley ex- periment, scientists used to believe that light propagates through a ‘medium,’
the so-called ether, as longitudinal waves just like sound waves. The transverse character of light was first comprehended in the experiments carried out by Fresnel and Young with birefringent materials. The origin of the hypothesis was based on the interference of polarized light conducted by Fresnel. It was observed that light waves polarized in mutually orthogonal directions cannot interfere. In order to explain this phenomenon, Young put forth his theory of transverse nature of light waves. Despite the fact that this contradicted earlier theories of longitudinal character of light, Fresnel used this successfully to de- rive many useful results, including the Fresnel formulas, which were discussed in Chapter 3. In this chapter we define the basic states of polarization and also discuss how they can be changed as light propagates through different kinds of media.
54 Wave Optics: Basic Concepts and Contemporary Trends