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Published by , 2016-09-14 08:26:12

12-pages-Stamnes-Refraction

12-pages-Stamnes-Refraction

FYS 263 32

k r local tangent plane

r kt
θ θt
i
θ

^n

ki
n1 n2

Point source

Figure 9.3: Generalisation of Snell’s law and the reflection law to include non-planar waves that are
incident upon a curved interface.

9.3 Reflection and refraction of plane electromagnetic waves

Note that the reflection law and the refraction law apply to all types of plane waves, i.e. to acoustic,
electromagnetic, and elastic waves. In the derivation we have only used that kq · r − ωt (q = i, r, t)
shall be the same for q = i, q = r, and q = t. Now we take a closer look at the reflection and
refraction of plane electromagnetic waves in order to determine how much of the energy in the
incident wave that is reflected and transmitted.

We know that a plane electromagnetic wave is transverse, i.e. that both E and B = µH are
normal to the propagation direction k = kˆs. In Fig. 9.1 we have chosen the z axis in the direction of
the interface normal. If E is normal to the plane of incidence, we have s polarisation (from German,
“Senkrecht”) or T E polarisation (“transverse electric” relative to the plane of incidence or the z
axis). And if E is parallel with the plane of incidence, we have p polarisation or T M polarisation,
since in this case B is normal to the plane of incidence or the z axis; hence the use of the term T M
or “transverse magnetic”.

A general time-harmonic, plane electromagnetic wave consists of both a T E and a T M compo-
nent. With the time dependence e−iωt suppressed, we have for the spatial part of the field

E = ET E + ET M ; B = BT E + BT M , (9.3.1)
(9.3.2)
ET E = ET E kt × eˆz eik·r, (9.3.3)
kt (9.3.4)
(9.3.5)
ET M = ET M k × (kt × eˆ)z eik·r,
kkt (9.3.6)

BT E = 1 k × ET E = ET E k × (kt × eˆz) eik·r,
k0 k0kt

BT M = 1 k × ET M = ETM 1 k× [k × (kt × eˆz)]eik·r.
k0 k0kkt

But since k × [k × (kt × eˆz)] = k[k · (kt × eˆz)] − (kt × eˆz)k · k = −k2kt × eˆz, we get

BT M = −k ET M kt × eˆz eik·r.
k0 kt


FYS 263 33

Note that the vectors eˆT E = kt × eˆz ; eˆT M = k × (kt × eˆz) ,
kt kkt
(9.3.7)

are unit vectors in the directions of ET E and ET M , respectively.

We represent each of the incident, reflected, and transmitted fields in the manner given above,

so that (q = i, r, t)

Eq = ET Eq + ET Mq ; Bq = BT Eq + BT Mq, (9.3.8)

ET Eq = ET Eq kt × eˆz eikq·r, (9.3.9)
kt (9.3.10)
(9.3.11)
ET Mq = ETMq kq × (kt × eˆz) eikq·r, (9.3.12)
kq kt
(9.3.13)
BT Eq = kq ET Eq kq × (kt × eˆz ) eikq ·r ,
k0 kq kt

BT Mq = −kq ET Mq kt × eˆz eikq·r,
k0 kt

where

ki = kt + kz1eˆz ; kt = kxeˆx + kyeˆy,

kr = kt − kz1eˆz ; kt = kt + kz2eˆz, (9.3.14)

kq = k1 = n1k0 for q = i, r (9.3.15)
k2 = n2k0 for q = t.

The continuity conditions that must be satisfied at the interface z = 0 are that the tangential

components of E and H = 1 B be continuous, i.e.
µ

eˆz × ET Ei + ET Er − ET Et + ET Mi + ET Mr − ET Mt = 0, (9.3.16)

eˆz × 1 BT Ei + BT Er − 1 BT Et + 1 BT Mi + BT Mr − 1 BT Mt = 0. (9.3.17)
µ1 µ2 µ1 µ2 (9.3.18)

Further, we have

eˆz × [kq × (kt × eˆz)] = (kq · eˆz) eˆz × kt,

eˆz × (kt × eˆz) = kt. (9.3.19)

By substituting from (9.3.9)-(9.3.12) into the boundary conditions (9.3.16)-(9.3.17) and using (9.1.2)
and (9.3.18)-(9.3.19), we get

kt ET Ei + ET Er − ET Et + eˆz × kt kz1 ET Mi − kz1 ET Mr − kz2 ET Mt = 0, (9.3.20)
k1 k1 k2

eˆz × kt 1 kz1 ET Ei − kz1 ET Er − 1 kz2 ET Et
µ1 k0 k0 µ2 k0

+kt 1 −k1 ET Mi − k1 ET Mr − −k2 ET Mt = 0. (9.3.21)
µ1 k0 k0 k0


FYS 263 34

Since kt and eˆz × kt are orthogonal vectors, the expression inside each of the {} parentheses in
(9.3.20) and (9.3.21) must vanish, i.e.

ETEi + ETEr = ETEt, (9.3.22)

kz1µ2 ET Ei − ET Er = kz2µ1ET Et, (9.3.23)

kz1k2 ET Mi − ET Mr = kz2k1ET Mt, (9.3.24)

k1µ2 ET Mi − ET Mr = k2µ1ET Mt. (9.3.25)
(9.3.26)
Now we define reflection and transmission coefficients as

RT E = ET Er ; TTE = ET Et ,
ET Ei ET Ei

RT M = ETMr ; TTM = ETMt (9.3.27)
ETMi ETMi ,

so that (9.3.22)-(9.3.25) give

1 + RT E = T T E ; 1 − RT E = kz2µ1 T T E , (9.3.28)
kz1µ2 (9.3.29)

1 − RT M = kz2k1 T T M ; 1 + RT M = k2µ1 T T M .
kz1k2 k1µ2

The two equations in (9.3.28) have the following solution

RT E = µ2kz1 − µ1kz2 ; T T E = 2µ2kz1 , (9.3.30)

µ2kz1 + µ1kz2 µ2kz1 + µ1kz2

whereas the two equations in (9.3.29) give

RT M = k22µ1kz1 − k12µ2kz2 ; TTM = 2k1k2µ2kz1 . (9.3.31)
k22µ1kz1 + k12µ2kz2 k22µ1kz1 + k12µ2kz2

The interpretation of the reflection and transmission coefficients follow from (9.3.26)-(9.3.27). Thus,

the reflection coefficient represents the amplitude ratio between the reflected and the incident E
field, whereas the transmission coefficient represents the amplitude ratio between the transmitted

and the incident E field.

Note that (9.3.22)-(9.3.23) and (9.3.28) contain only T E quantities, whereas equations (9.3.24)-
(9.3.25) and (9.3.29) contain only T M quantities. This implies that these two wave types are

independent or de-coupled upon reflection and refraction. Thus, an incident T E plane wave produces

a reflected T E plane wave and a transmitted T E plane wave, whereas an incident T M plane wave

produces a reflected T M plane wave and a transmitted T M plane wave. Upon reflection and
refraction there is no coupling between T E and T M waves.

From Fig. 9.1 it follows that

kz1 = ki · eˆz = k1 cos θi ; kz2 = kt · eˆz = k2 cos θt, (9.3.32)
so that if µ1 = µ2 = 1 the reflection and transmission coefficients become (9.3.33)
(9.3.34)
TTM = 2n1 cos θi ; RT M = n2 cos θi − n1 cos θt ,
n2 cos θi + n1 cos θt n2 cos θi + n1 cos θt

TTE = 2n1 cos θi ; RT E = n1 cos θi − n2 cos θt
n1 cos θi + n2 cos θt n1 cos θi + n2 cos θt ,


FYS 263 35
^ez
^e TMr kr kt e^ TMt
ri
ki kt
θ =θ θt

θi

^e TMi k i n1 n2

z=0 ^e TEi = e^ TEr = ^e TEt = ^e TE

Figure 9.4: Reflection and refraction of a plane electromagnetic wave at a plane interface between
two different media. Illustration of T E and T M components of the electric field.

These expressions are called the Fresnel formulas. By using Snell’s law (9.1.20), we can rewrite them
as (Exercise 9)

TTM = 2 sin θt cos θi ; RT M = tan(θi − θt) (9.3.35)
sin(θi + θt) cos(θi − θt) tan(θi + θt) ,

TTE = 2 sin θt cos θi ; RT E = − sin(θi − θt) (9.3.36)
sin(θi + θt) sin(θi + θt) .

At normal incidence where θi = θt = 0, we get from (9.3.33) and (9.3.34)

T T E = T T M = 2 ; RT M = −RT E = n − 1 ; n = n2 . (9.3.37)
n+1 n+1 n1

The fact that RT M = −RT E at normal incidence follows from the way in which ET E and ET E
are defined. From Fig. 9.4 we see that these two vectors point in opposite directions at normal

incidence.

9.3.1 Reflectance and transmittance

Fig. 9.4 shows the polarisation vectors eˆT Mq (q = i, r, t) and eˆT E for T M and T E polarisation.
These unit vectors are parallel with the electric field and follow from (9.3.9)-(9.3.12)

eˆT Ei = eˆT Er = eˆT Et = eˆT E = kt × eˆz ; |eˆT E| = 1, (9.3.38)
kt (9.3.39)

eˆT Mq = kq × (kt × eˆz ) ; |eˆT Mq| = 1.
kq kt

Let the angle between Eq and the plane of incidence spanned by kq and eˆT Mq, be αq [see Fig. 9.5],
so that

Eq = eˆT EEq sin αq + eˆT MqEq cos αq. (9.3.40)


FYS 263 36

q
E

e^ TMq

^e TEq

ETEq

αq

E TMq

q
k

Figure 9.5: Illustration of the angle αq between the electric vector Eq and the plane of incidence
spanned by kq and eˆT Mq.

Further, we let Ji, Jr, and Jt denote the energy flows of respectively the incident, reflected, and
transmitted fields per unit area of the interface. Then we have

J pq = Spq cos θq ; p = T E, T M ; q = i, r, t, (9.3.41)
where Spq is the absolute value of the Poynting vector, given by

Spq = c |Epq × Hpq| = c EpqHpq = c εq (E pq )2 . (9.3.42)
4π 4π 4π µq

Here we have used the relation à E pq = õqHpq. The reflectance Rp (p = T E, T M is the ratio
εq

between the reflected and incident energy flows. From (9.3.41)-(9.3.42) we have

RT M = JTMr = |ET Mr|2 = (RT M )2. (9.3.43)
JTMi |ET Mi|2

RT E = J T Er = |ET Er|2 = (RT E)2, (9.3.44)
J T Ei |ET Ei|2

Thus, the reflectance Rp is equal to the square of reflection coefficient Rp.
The transmittance T p (p = T E, T M ) is the ratio between the transmitted and incident energy

flows, and (9.3.41)-(9.3.42) give

T TM = JTMt = n2 µ1 cos θt (T T M )2, (9.3.45)
JTMi n1 µ2 cos θi

T TE = J T Et = n2 µ1 cos θt (T T E )2. (9.3.46)
J T Ei n1 µ2 cos θi

Thus, the transmittance T p is proportional to the square of the transmission coefficient T p (p =

T E, T M ). When µ2 = µ1 = 1, we find on substitution from (9.3.35)-(9.3.36) into (9.3.43)-(9.3.46)
the following expressions for the reflectance and the transmittance

RT M = tan2(θi − θt) (9.3.47)
tan2(θi + θt),


FYS 263 37

RT E = sin2(θi − θt) (9.3.48)
sin2(θi , (9.3.49)
(9.3.50)
+ θt)

T TM = sin 2θi sin 2θt ,
sin2(θi + θt) cos2(θi − θt)

T TE = sin 2θi sin 2θt
sin2(θi + θt) .

By use of these formulas one can show that

RT M + T T M = 1 ; RT E + T T E = 1, (9.3.51)

so that for each of the two polarisations the sum of the reflected energy and the transmitted energy
is equal to the incident energy.

From (9.3.41) and (9.3.42) we have

J pi = c ε1 |Epi|2 cos θi, (9.3.52)
4π µ1

which by the use of (9.3.40) gives

J T Ei = cos θi c ε1 |ET Ei|2 = cos θi c ε1 E2 sin2 αi, (9.3.53)
4π µ1 4π µ1 (9.3.54)

J T Mi = cos θi c ε1 |ET Mi|2 = cos θi c ε1 E2 cos2 αi. (9.3.55)
4π µ1 4π µ1

But since the total incident energy flow is given by

J i = cos θi c ε1 E2,
4π µ1

we find

J T Ei = J i sin2 αi ; J T Mi = J i cos2 αi. (9.3.56)

Thus, we have

R= Jr = JTMr + JTEr = JTMr cos2 αi + J T Er sin2 αi, (9.3.57)
Ji Ji JTMi J T Ei

which gives

R = RT M cos2 αi + RT E sin2 αi, (9.3.58)

and similarly we find

T = T T M cos2 αi + T T E sin2 αi. (9.3.59)

At normal incidence, θi = θt = 0, and the distinction between T E and T M polarisation disappears.
From (9.3.43)-(9.3.46) combined with (9.3.33)-(9.3.34), we find (when µ1 = µ2 = 1)

R = RT M = RT E = (RT E)2 = RT M 2 = n − 1 2 ; n = n2 , (9.3.60)
n+1 n1

T = T TM = T TE = TTE 2 = TTM 2 = 4n ; n = n2 . (9.3.61)
+ 1)2 n1
(n

When n → 1, we see that R → 0 and T → 1, as expected. Similarly, we find from (9.3.47)-(9.3.50)
that R → 0, R⊥ → 0, T → 1, T⊥ → 1 when n → 1.


FYS 263 38

1 kr
2
k i θ iB θ iB

θ tB
kt

Figure 9.6: Illustration of Brewster’s law.

9.3.2 Brewster’s law

From (9.3.47) it follows that RT M = 0 when θi + θt = π , since then tan(θi + θt) = ∞. We call this
2
particular angle of incidence θiB and the corresponding refraction or transmission angle θtB. By

using Snell’s law (9.1.20), we find

n2 sin θtB = n2 sin π − θiB = n2 cos θiB = n1 sin θiB, (9.3.62)
2

so that RT M = 0 when θi = θiB, where θiB is given by

tan θiB = n2 = n. (9.3.63)
n1

The angle θiB is called the polarisation angle or the Brewster angle. When the angle of incidence
is equal to θiB, the E vector of the reflected light has no component in the plane of incidence (Fig.

9.6). This fact is exploited in sunglasses with polarisation filter. The filter is oriented such that

only light that is polarised vertically (Fig. 9.6) is transmitted. Thus, one avoids to a certain degree

annoying reflections from e.g. a water surface.
Note that kr · kt = 0, i.e. kr and kt are normal to one another when θi = θiB, as shown in

Fig. 9.6.

9.3.3 Unpolarised light (natural light)

For natural light, e.g. light from an incandescent lamp, the direction of the E vector varies very

rapidly in an arbitrary or irregular manner, so that no particular direction is given preference. The

average reflectance R is obtained by averaging over all directions α. Since the average value of both

sin2 α and cos2 α is 1 , we find from (9.3.56) that
2

J T Mi = J icos2 αi = J T Ei = J isin2 αi = 1 J i. (9.3.64)
2

For the reflected components we find

JTMr = JTMr · JTMi = JTMr · 1Ji = 1 RT M J i, (9.3.65)
JTMi JTMi 2 2


FYS 263 39

J T Er = J T Er · 1Ji = 1 RT E J i, (9.3.66)
J T Ei 2 2

which shows that the degree of polarisation for the reflected light can be defined as

Pr = RT M − RT E |J T Mr − J T Er| (9.3.67)
RT M + RT E = JTMr + JTEr . (9.3.68)
(9.3.69)
The average reflectance is given by (9.3.70)

R= Jr = JTMr + JTEr = JTMr + J T Er = 1 RT M + RT E ,
Ji Ji 2J T Mi 2J T Ei 2

so that the degree of polarisation becomes

P r = 1 1 |RT M − RT E|,
R2

where |RT M − RT E| is called the polarised part of the reflected light.
Similarly, we find for the transmitted light

T = 1 (T T M + T T E) ; P t = 1 1 |T T M − T T E|.
2 T2

9.3.4 Rotation of the plane of polarisation upon reflection and refraction

Note that if the incident light is linearly polarised, then also the reflected and the transmitted light
will be linearly polarised, since the phases only change by 0 or π. This follows from the fact that the
reflection and transmission coefficients are real quantities [cf. (9.3.33)-(9.3.36)]. But the planes of
polarisation for the reflected and the transmitted light are rotated in opposite directions relative to
the polarisation plane of the incident light. The angles αi, αr, and αt that the planes of polarisation
of the incident, reflected, and transmitted light form with the plane of incidence, are given by [cf.
Fig. 9.5]

tan αi = ET Ei (9.3.71)
ETMi , (9.3.72)
(9.3.73)
tan αr = ET Er = ET Er ET Ei = RT E tan αi,
ETMr ET Ei ETMi RT M

ET Mr

ET Mi

tan αt = ET Et = ET Et ET Ei = TTE tan αi.
ETMt ET Ei ETMi TTM

ET Mt
ET Mi

By use of the Fresnel formulas (9.3.35)-(9.3.36) we can write

tan αr = − cos(θi − θt) tan αi, (9.3.74)
cos(θi + θt)

tan αt = cos(θi − θt) tan αi. (9.3.75)
(9.3.76)
Since 0 ≤ θi ≤ π and 0 ≤ θt ≤ π , we get
2 2

| tan αr| ≥ | tan αi|,

| tan αt| ≤ | tan αi|. (9.3.77)

In (9.3.76) the equality sign applies at normal incidence (θi = θt = 0) and at grazing incidence (θi =

π ), whereas in (9.3.77) the equality sign applies only at normal incidence. These two inequalities
2


FYS 263 40

imply that upon reflection the plane of polarisation is rotated away from the plane of incidence,

whereas upon transmission it is rotated towards the plane of incidence. Note that when θi = θiB,

so that θiB + θtB = π , then tan αr = ∞. Thus, we have αr = π in accordance with Brewster’s law.
2 2

9.3.5 Total reflection (9.3.78)

Snell’s law (9.1.20) can be written in the form

sin θt = sin θi ; n = n2 = ε2µ2 .
n n1 ε1µ1

Hence, it follows that if n < 1, then we get sin θt = 1 when θi = θic, where

sin θic = n. (9.3.79)

This implies that when θi = θic, we get θt = π , so that the transmitted light propagates along the
interface. If θi ≥ θic, we 2 i.e. no light will pass into the other medium. All
have total reflection,

light is then reflected. There exists a field in the other medium, but there is no energy transport

through the interface. When θi > θic, then sin θt > 1, which means that θt is complex. We have

from (9.3.78)

cos θt = ± 1 − sin2 θt = ±i sin2 θi −1 = ±i sin2 θi − n2 (9.3.80)
n2 .

n

The lower sign in (9.3.80) must be discarded. Otherwise the field in medium 2 would grow expo-
nentially with increasing distance from the interface. The electric field in medium 2 is

Ept = T pEpieˆptei(kt·r−ωt) (p = T E, T M ), (9.3.81)

where

kt · r = kxx + kyy + kz2z, (9.3.82)
with (cf. Fig. 9.1 and (9.3.80) with upper sign)

kz2 = k2 cos θt = ik2 1 sin2 θi − n2 ; n = n2 , (9.3.83)
n n1

so that

eikt·r = ei(kxx+kyy)e−|kz2|z ; |kz2| = k2 sin2 θi − n2 . (9.3.84)
n

We see that Ept represents a wave that propagates along the interface and is exponentially damped

with the distance z into medium 2.
From ∇ · Dt = ε2∇ · Et = 0 it follows that

kt · Et = 0, (9.3.85)

which gives

Ezt = − (kxExt + kyEyt ) . (9.3.86)
kz2

If we let the plane of incidence coincide with the xz plane, we have (cf. Fig. 9.7)

kx = −k1 sin θi ; ky = 0, (9.3.87)

Eyt = ET Etei(kxx−ωt)e−|kz2|z = T T E ET Eiei(kxx−ωt)e−|kz2|z , (9.3.88)


FYS 263 41

E TMr kr θt kt
θi E TMt ki

θt

z

θi

E TMi ki E TEi = E TEr = E TEt

x

Figure 9.7: Illustration of the refraction of a plane wave into an optically thinner medium, so that
θi < θt. When θi → θic, then θt → π/2, and we get total reflection.

Ext = −ET Mt cos θtei(kxx−ωt)e−|kz2| = −T T M ET Mi cos θtei(kxx−ωt)e−|kz2|z , (9.3.89)

Ezt = − kx Ext = kx T T M ET M iei(kxx−ωt)e−|kz2|z . (9.3.90)
kz2 k2

From these expressions for the components of Et and corresponding expressions for the components
of Ht one can show (Exercise 11) that the time average of the z component of the Poynting vector

is zero, which implies that there is no energy transport through the interface, as asserted earlier.

The reflection coefficients in (9.3.35)-(9.3.36) can be written as follows

RT M = sin θi cos θi − sin θt cos θt , (9.3.91)
sin θi cos θi + sin θt cos θt

RT E = − sin θi cos θt − sin θt cos θi . (9.3.92)
sin θi cos θt + sin θt cos θi

By combining Snell’s law (9.3.78) and (9.3.80) with the upper sign with (9.3.91)-(9.3.92), we get

RT M = n2 cos θi − i sin2 θi − n2 (9.3.93)
,
n2 cos θi + i sin2 θi − n2

RT E = cos θi − i sin2 θi − n2 (9.3.94)
.
cos θi + i sin2 θi − n2

Since both reflection coefficients are of the form z/z∗, where z is a complex number, it follows that

|RT M | = |RT E| = 1, (9.3.95)


FYS 263 42

which shows that for each polarisation the intensity of the totally reflected light is equal to the
intensity of the incident light.

But the phase is altered upon total reflection. Letting

Rp = E pr = eiδp = zp = e2iαp (p = T E, T M ), (9.3.96)
E pi zp∗

where [cf. (9.3.93)-(9.3.94)]

zT M = n2 cos θi − i sin2 θi − n2 = |zT M |eiαTM , (9.3.97)

zT E = cos θi − i sin2 θi − n2 = |zT E |eiαTE , (9.3.98)

we find

tan αT M = tan 1 δT M =− sin2 θi − n2 , (9.3.99)
2 n2 cos θi

tan αT E = tan 1 δT E =− sin2 θi − n2 (9.3.100)
2 cos θi .

The relative phase difference

δ = δTE − δTM, (9.3.101)
(9.3.102)
is determined by

1 tan 1 δT E − tan 1 δT M
2δ 2 2
tan = ,
1 + tan 1 δT E tan 1 δT M
2 2

which upon substitution from (9.3.99)-(9.3.100) gives

tan 1 δ = cos θi sin2 θi − n2 . (9.3.103)
2 sin2 θi

We see that δ = 0 for θi = π (grazing incidence) and θi = θic (critical angle of incidence). Between
2
these two values there is an angle of incidence θi = θim which gives a maximum phase difference

δ = δm, where θim is determined by

dδ (9.3.104)
= 0.

dθi θim

From (9.3.104) we find

sin2 θim = 2n2 (9.3.105)
1 + n2 ,

which upon substitution in (9.3.103) gives (Exercise 10)

tan 1 δm 1 − n2 (9.3.106)
=.
2 2n

If the phase difference δ is equal to ± π and in addition ET Mi = ET Ei, the totally reflected light
will be circularly polarised. 2 the angle αi between the polarisation plane and the plane
By choosing

of incidence equal to 45◦, we make ET Mi equal to ET Ei. In order to obtain δ = π we must have
that 1, which according to 2
δm π δm π
2 ≥ 4 . This means tan 2 ≥ tan 4 = (9.3.106) implies that

n2 + 2n − 1 ≤ 0. (9.3.107)
By completing the square on the left-hand side of (9.3.107), we find that


FYS 263 43

√ 1 = n1 ≥ √ = 2.41. (9.3.108)
n≤ 2−1 ; 2+1
n n2

Thus, n1 must exceed 2.41 in order that we shall obtain a phase difference of π in one single
n2 2
reflection.


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