Chapter 23: Fresnel equationsChapter 23: Fresnel equations

Recall basic laws of optics

Law of reflection: ri

qi

normal

n1

n2

qr

qt

Law of refraction“Snell’s Law”:

1

2

sin

sin

n

n

t

i

Easy to derive on the basis of:Huygens’ principle: every point on a wavefront may be regarded as a

secondary source of wavelets

Fermat’s principle: the path a beam of light takes between two points is the one which is traversed in the least time

Incident, reflected, refracted, and normal in same plane

Today, we’ll show how they can be derived when we consider light to be an

electromagnetic wave

E and B are harmonic

)sin(

)sin(

0

0

t

t

rkBB

rkEE

Also, at any specified point in time and space,

cBE where c is the velocity of the propagating wave,

m/s 10998.21 8

00

c

We’ll also determine the fraction of the light reflected vs. transmitted

and the change in the phase upon reflection

external reflection,

Incidence angle, qi

1.0

.5

0

0° 30° 60° 90°

R

TR

T

qi

Let’s start with polarization…light is a 3-D vector field

linear polarization circular polarization

y

x

z

Plane of incidence: formed by and k and the normal of the interface plane

…and consider it relative to a plane interface

k

normal

kBE

TE: transverse electrics: senkrecht polarized

(E-field sticks in and out of the plane)

Polarization modes (= confusing nomenclature!)

TM: transverse magneticp: plane polarized

(E-field in the plane)E

M

M

E

E Eperpendicular ( ), horizontal parallel ( || ), vertical

always relative to plane of incidence

y

x x

y

Plane waves with k along z directionoscillating electric field

Any polarization state can be described as linear combination of these two:

“complex amplitude” contains all polarization info

yeExeE yx

tkziy

tkzix ˆˆ 00

E

tkziiy

ix eyeExeE yx ˆˆ 00E

Derivation of laws of reflection and refraction

boundary point

using diagram from Pedrotti3

At the boundary point:

phases of the three waves must be equal:

true for any boundary point and time, so let’s take 0r

ttt tri

tri or

hence, the frequencies are equal

and if we now consider 0t

)()()( ttt ttrrii rkrkrk

rkrkrk tri

which means all three propagation vectors lie in the same plane

rkrkrk tri

focus on first two terms:

rkrk ri

rrii rkrk sinsin

incident and reflected beams travel in same medium; same l

ri kk

hence we arrive at the law of reflection:

ri

Reflection

rkrkrk tri

now the last two terms:

rkrk tr

ttrr rkrk sinsin

reflected and transmitted beams travel in different media (same frequencies; different wavelengths!):

cnvk rr // 1

which leads to the law of refraction:

tr nn sinsin 21

cnvk tt // 2

Refraction

Boundary conditions from Maxwell’s eqns

yE

yE

yE

0

0

0

ˆ

ˆ

ˆ

tt

rr

ii

E

E

E

for both electric and magnetic fields, components parallel to boundary plane must be continuous as boundary is passed

complex field amplitudes

electric fields:

TE waves

continuity requires:

tri EEE

parallel to boundary plane

Boundary conditions from Maxwell’s eqnsfor both electric and magnetic fields, components parallel to boundary plane must be continuous as boundary is passed

)(

)(

)(

)ˆsinˆcos(

)ˆsinˆcos(

)ˆsinˆcos(

tittt

tirrr

tiiii

t

r

i

eBB

eBB

eBB

rk

rk

rk

zxB

zxB

zxB

magnetic fields:

coscoscos tri BBB continuity requires:

same analysis can be performed for TM waves

TE waves

TE waves TM waves

n2

tri EEE

ttrrii BBB coscoscos

tri BBB

ttrrii EEE coscoscos

BvBE nc

Summary of boundary conditions

n1

iE

iE

rE

rE

tE

tE

tB

tB

iB

iB

rB

rB

amplitudes are related:

TE waves TM waves

tri EEE

tttiriiii EnEnEn coscoscos

ttriii EnEnEn

ttirii EEE coscoscos

tnn

i

tnn

i

i

rTE

i

t

i

t

E

Er

coscos

coscos

tinn

tinn

i

rTM

i

t

i

t

E

Er

coscos

coscos

For reflection: eliminate Et , separate Ei and Er , and take ratio:

Get all in terms of E and apply law of reflection (qi = qr):

Apply law of refraction and let :ttii nn sinsin

Fresnel equations

ii

iiTE

n

nr

22

22

sincos

sincos

ii

iiTM

nn

nnr

222

222

sincos

sincos

i

t

n

nn

ii

i

i

tTE

nE

Et

22 sincos

cos2

For transmission: eliminate Er , separate Ei and Et , take ratio…

And together:

ii

i

i

tTM

nn

n

E

Et

222 sincos

cos2

TE waves TM waves

Fresnel equations

TETE rt 1 TETM rtn 1

Fresnel equations, graphically

External and internal reflections

internal reflection:

External and internal reflections

external reflection: 21 nn

21 nn

11

2 n

nn

11

2 n

nn

occur when

n = n2/n1 = 1.5

External reflections (i.e. air-glass)

normal grazing

- at normal and grazing incidence, coefficients have same magnitude- negative values of r indicate phase change- fraction of power in reflected wave = reflectance =

- fraction of power transmitted wave = transmittance =

2

2

i

r

i

r

E

Er

P

PR

2

cos

costn

P

PT

i

t

i

t

RTM = 0(here, reflected light TE polarized;RTE = 15%)

22

1

1

n

nrR

at normal : 4%

Not

e: R

+T =

1

http://www.ray-ban.com/clarity/index.html?lang=uk

Glare

- incident angle where RTM = 0 is:

- both and reach values of unity before q=90°

total internal reflection

p

2TETE rR 2

TMTM rR

1

211 sin)(sinn

nnc

Internal reflections (i.e. glass-air)

n = n2/n1 = 1.5

total internal reflection

Internal reflections (i.e. glass-air)

Conservation of energy

tri PPP

1TR

it’s always true that

and

in terms of irradiance (I, W/m2)

ttrrii AIAIAI using laws of reflection and refraction, you can deduce

2

2

0

0 rE

E

I

I

P

PR

i

r

i

r

i

r

and2

cos

costnT

i

t

Summary:Reflectance and Transmittance for anAir-to-Glass Interface

Perpendicular polarization

Incidence angle, qi

1.0

.5

00° 30° 60° 90°

R

T

Parallel polarization

Incidence angle, qi

1.0

.5

00° 30° 60° 90°

R

T

Summary:Reflectance and Transmittance for aGlass-to-Air Interface

Perpendicular polarization

Incidence angle, qi

1.0

.5

00° 30° 60° 90°

R

T

Parallel polarization

Incidence angle, qi

1.0

.5

00° 30° 60° 90°

R

T

Back to reflections

Brewster’s angleor the polarizing angle

is the angle qp, at which RTM = 0: 1

211 tantann

nnp

at qp, TM is perfectly transmitted with no reflection

Brewster’s angle for internal and external reflections

at Brewster’s angle, “s skips and p plunges”s-polarized light (TE) skips off the surface; p-polarized light (TM) plunges in

Brewster’s angle

Punky Brewster Sir David Brewster

(1781-1868)(1984-1986)

http://www.brewstersociety.com/cbs_sundaymorning_09.html

Brewster’s other angles: the kaleidoscope

Phase changes upon reflection

-recall the negative reflection coefficients

-indicates that sometimes electric field vector reverses direction upon reflection:

-p phase shiftexternal reflection: all angles for TE and at for TM internal reflection: more complex…

ErEr )(

0)(

0 titii eEreEreEr rkrk

p

Phase changes upon reflection: internal

in the region , r is complexc

222

222

22

22

sincos

sincos

sincos

sincos

nin

ninr

ni

nir

TM

TE

reflection coefficients in polar form: ierr

f phase shift on reflection

Phase changes upon reflection: internaldepending on angle of incidence, - < < p f p

cos

sin

2tan

cos

sin

2tan

2

2222

n

nn TMTE

Exploiting the phase differencecircular polarization

-consists of equal amplitude components of TE and TM linear polarized light, with phases that differ by ±p/2

-can be created by internal reflections in a Fresnel rhomb

each reflection produces a π/4 phase delay

http://www.halbo.com/fr_rhmb.htm

Summary of phase shifts on reflection

TE mode TM mode

airglass

external reflection

TE mode TM mode

airglass

internal reflection

A lovely example

How do we quantify beauty?

Case study for reflection and refraction

You are encouraged to solve all problems in the textbook (Pedrotti3).

The following may be covered in the werkcollege on 21 September 2011:

Chapter 23:1, 2, 3, 5, 12, 16, 20

Exercises