9 12 Fresnel Equations 8

8/13/2019 9 12 Fresnel Equations 8

1/17

6.

1

silen

ts


2/17

6.

Boundary Conditions for EM waves

Derivation of Fresnel Equations

Consequences of Fresnel Equations

Amplitude of reflection coefficients

Phase shifts on reflection

Brewsters angle

Conservation of energy

2


3/17

6.

B dA = 0

E

ds =

d

dt

B

dA

E dA =

q

B

ds =

J dA+

d

dt

E dA

When an EM wave propagates across aninterface, Maxwells equations must be satisfiedat the interface as well as in the bulkmaterials. The constraints necessary for this tooccur are called the boundary conditions

3

1, 1 2, 2


4/17

6.

B dA = 0

E

ds =

d

dt

B

dA

E dA =

q

Gauss law can be used to find the boundaryconditions on the component of the electric fieldthat is perpendicular to the interface.

If the materials are dielectrics there will be nofree charge on the surface (q=0)

41 1 = 2 2

01E1

2E2 =

q

B

ds =

J dA+

d

dt

E dA

1, 1 2, 2


5/17

6.

B dA = 0

E

ds =

d

dt

B

dA

E dA =

q

5

B

ds =

J dA+

d

dt

E dA

1 = 2E2E1 =

dt

B

dA

0

1, 1 2, 2

Faradays law can be applied at the interface. Ifthe loop around which the electric field iscomputed is made to have an infintesimal areathe right side will go to zero giving arelationship between the parallel components ofthe electric field


6/17

6.

B

ds =

J dA+

d

dt

E dA

B dA = 0

E

ds =

d

dt

B

dA

E dA =

q

Gauss law for magnetism gives a relationshipbetween the perpendicular components of themagnetic field at the interface

61AB2A = 0 1 = 2

1, 1 2, 2


7/176.

B dA = 0

E

ds =

d

dt

B

dA

E dA =

q

Amperes law applied to a loop at the interfacethat has an infintesimal area gives a relationshipbetween the parallel components of themagnetic field. (Note that in most commonmaterials !=!o)

7

B

ds =

J dA+

d

dt

E dA

B1

1 L

B2

2 L=

J

dA+

d

dt

EdA

B1

1=

B2

2

0

0

1, 1 2, 2


8/17

6. 8

Plane of the interface(here theyz plane) (perpendicular to page)

ni

nt

!i !r

!t

Ei Er

Et

Interface

x

y

z

s polarization (senkrecht, aka TE orhorizontal) has an E field that is

perpendicular to the plane of incidence

p polarization (parallel aka TM orvertical) has an E field that is parallel

to the plane of incidence

The reflection and transmissioncoefficients at an interface can befound using the boundary conditions,but they depend on the polarization of

the incident light

B1

1=

B2

2

B1 = B2

1 1 = 2 2

1 = 2


9/17

6.

The tangential electric field is continuous

9


ni

nt

!i !r

!t

Ei Er

Et

Interface

x

y

z

B1

1=

B2

2

B1 = B2

1 = 2

Ei(y = 0, t) + Er(y= 0, t) = Et(y= 0, t)

Using "i="rand B=nE/c and considering only theam litude of the waves at the boundar

ni(E0r E0i)cos i = nt(E0r+E0i)cos t

*It's actually the tangential B/!, but we're assuming !=!0

Bi Br

BtThe tangential magnetic field is continuous*

Bi(y= 0, t)cos i+Br(y= 0, t)cos r

= Bt(y= 0, t)cos t


10/17

6.

rearranging to find r=Eor/Eoigives

10


ni

nt

!i !r

!t

Ei Er

Et

Interface

x

y

z

Bi Br

Bt

ni(E0r

E0i)cos i =nt(E0r+E0i)cos t

and similarly t=Eot/Eoiis

T

T

r =E0r

E0i=

nicos i ntcos t

nicos i + ntcos t

t =E0t

E0i=

2nicos i

nicos i + ntcos t


11/17

6.

The tangential electric field is continuous

11


ni

nt

!i !r

!t

EiEr

Et

Interface

x

y

z

B1

1=

B2

2

B1 = B2

1 1 = 2 2

1 = 2

Ei(y = 0, t)cos i+Er(y= 0, t)cos r

= Et(y= 0, t)cos t

*It's actually the tangential B/!, but we're assuming !=!0

The tangential magnetic field is continuous*

Bi(y= 0, t) +Br(y= 0, t) =Bt(y= 0, t)

Bi Br

Bt

Using "i="rand E=cB/n and considering only theamplitude of the waves at the boundary

nt(E0r E0i)cos i= ni(E0r+E0i)cos t


12/17

6.

rearranging to find r=Eor/Eoigives

12


ni

nt

!i !r

!t

Ei Er

Et

Interface

x

y

z

Bi Br

Bt

and similarly t|| =Eot/Eoiis

||

nt(E0r

E0i)cos i = ni(E0r+E0i)cos t

r=

E0r

E0i=

nt cos i nicos t

ni cos t + ntcos i

t =E0t

E0i=

2nicos i

nicos t + ntcos i


13/17

6.

At normal incidence

At Brewsters angle

At grazing incidence

13

r =E0r

E0i=

ni cos i nt cos t

ni cos i + nt cos t

t =E0t

E0i

=2nicos i

nicos

i +

ntcos

t

r =

E0r

E0i=

ntcos i ni cos t

ni cos t + ntcos i

t =E0t

E0i=

2nicos i

nicos t + ntcos i

reflection and transmission at an air-glass interface

r =nt ni

nt + ni

r = 0

limi90

o

r =1

How can r||differ from r at "=0 where sand p-polarization are degenerate?

T

Why isnt t||=1 when r||=1? Ifnone of the field is reflected,shouldnt it all be transmitted?


14/17

6.14

r =nt ni

nt + ni Plane of the interface(here theyz plane) (perpendicular to page)

ni

nt

!i !r

!t

EiEr

Et

Interface

x

y

z

Bi Br

Bt


ni

nt

!i !r

!t

Ei Er

Et

Interface

x

y

z

Bi Br

Bt

Considering our definition forwhat we consider positive Ernotice that as "#0 we havepositive values for Erpointing indifferent directions for s and p-

polarization, hence the reflectioncoefficients need to have oppositesign for them to converge to thesame physical solution

Note that r2+t2=1 indicating

energy is conserved at theboundary

t =2nt

nt + ni


15/17

6.

When the incident electric field oscillationsexcite dipole oscillation in the material in adirection parallel to the reflected beam thedipoles cannot radiate along the direction of

the reflected beam

At this angle, called Brewsters angle r||=0.There are many practical applications of this

polarize the reflected light

minimize reflection off the surface oflaser mirrors

15


ni

nt

!i !r

!t Et

Interface

x

y

z

Ei

Bi

Er

Br

Bt


16/17

R =Ir cos r

Ii cos i= r

2

T =It cos t

Ii cos i= t

2cos t

cos i

6.

Irradiance is proportional to the square ofthe field so if we are interested in thereflected and transmitted irradiance we use

the square of the field reflectivity rand

transmissivity t(i.e. r2and t2)

The power is irradiance times area, and thecross sectional area of the beam is differentfor the incident and transmitted beams

The power reflection and transmissioncoefficients for a beam are R and T and are

called the Reflectanceand Transmittance

R+T=1 so energy is conserved

16

Acos"i

A

Acos"r

Acos"t


17/17

A full electromagnetic treatment of the fieldsat the boundary of two dielectrics leads to theFresnel equations for transmissivity andreflectivity

At normal incidence

At Brewsters angle the reflectivity of the

P-polarized field goes to zeroThe power reflectivity and transmissivity of abeam are

6.17

r =nt ni

nt + nit =

2nt

nt + ni

T = t2cos t

cos iR = r

2

9 12 Fresnel Equations 8

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