optics lecture-fresnel equations

7/28/2019 Optics Lecture-fresnel Equations

1/19

13. Fresnel's Equations for Reflection

and Refraction

Incident, transmitted, and reflected beams

Boundary conditions: tangential fields are continuous

Reflection andtransmission

coefficients

The "Fresnel Equations"

Brewster's Angle

Total internal reflection

Power reflectanceand transmittance

Augustin Fresnel1788-1827


2/19

Posing the problem

What happens when light, propagating in auniform medium, encounters a smooth interfacewhich is the boundary of another medium with adifferent refractive index?

k-vector of the

incident light

boundarynincident

ntransmitted

This is really two problems, not just one.


3/19

Definitions: Planes of Incidence, the

interface and S and P polarizationsS polarization is the perpendicular polarization, and it sticks upout of the plane of incidence

Plane of the interface (y=0, the xzplane) (perpendicular to screen)

P polarization is the parallel polarization, and it lies parallelto the plane of incidence.

Plane of incidence (z=0)is the plane thatcontains the incident

and reflected k-vectors.

x

y

z

I R

T


4/19

ni

nt

ik

rk

tk

i r

t

EiBi

Er Br

Et

Bt

Interface

Fresnel EquationsWe would like to compute the fraction of a light wave reflected andtransmitted by a flat interface between two media with different refrac-tive indices. Fresnel was the first to do this calculation (early 1800s).

x

y

zBeam geometry for

light with its electric

field sticking up out ofthe plane of incidence

(i.e., out of the page)

We treat the case of S polarization first:

the xz plane (y = 0)


5/19

ni

nt

ik

rk

tk

i r

t

EiBi

ErBr

Et

Bt

Interface

Boundary Condition for the Electric

Field at an Interface: s polarizationx

y

z

In other words,

The Tangential Electric Field is Continuous

So: Ei(y= 0) +Er(y= 0) = Et(y= 0)

The total E-field inthe plane of theinterface is

continuous.

Here, all E-fields arein the z-direction,

which is in the planeof the interface.

(Were not explicitly writingthe x, z, and t dependence,

but it is still there.)


6/19

Boundary Condition for the Magnetic

Field at an Interface: s polarization

ni

nt

ik

rk

tk

i r

t

Ei

Bi

ErBr

Et

Bt

Interface

x

y

z

ii

*It's really the tangential B/, but we're usingi =t =0Bi(y= 0) cosi +Br(y= 0) cosr = Bt(y= 0) cost

The Tangential Magnetic Field* is Continuous

In other words,

The total B-field in theplane of the interface iscontinuous.

Here, all B-fields are inthe xy-plane, so we takethe x-components:


7/19

Reflection and Transmission for

Perpendicularly Polarized Light

Ignoring the rapidly varying parts of the light wave and keeping

only the complex amplitudes:

0 0 0

0 0 0

cos( ) cos( ) cos( )

+ =

+ =

i r t

i i r r t t

E E E

B B B

0 0 0 0

0 0 0 0

:

( )cos( ) ( )cos( )

Substituting for using + =

= +

t i r t

i r i i t r i t

E E E E

n E E n E E

0 0 0( )cos( ) cos( ) = i r i i t t tn E E nE

0 0/( / ) / .But and= = =

i rB E c n nE c

Substituting into the second equation:


8/19

Reflection & Transmission Coefficients

for Perpendicularly Polarized Light

[ ] [ ]

0 0 0 0

0 0

( )cos( ) ( )cos( ) :

cos( ) cos( ) cos( ) cos( )

i r i i t r i t

r i i t t i i i t t

n E E n E E

E n n E n n

= +

+ =

Rearranging yields

[ ]0 0/ 2 cos( )/ cos( ) cos( )t i i i i i t tt E E n n n = = +

0 0/ , ist iE EAnalogously, the transmission coefficient,

[ ] [ ]0 0/ cos( ) cos( ) / cos( ) cos( )r i i i t t i i t tr E E n n n n = = +

0 0/r iE ESolving for yields the reflection coefficient:

These equations are called the Fresnel Equations for

perpendicularly polarized (s-polarized) light.


9/19

ni

nt

ik

rk

tk

i r

t

EiBi Er

Br

EtBt

Interface

Fresnel EquationsParallel electric field

x

y

z

Beam geometryfor light with itselectric fieldparallel to theplane of incidence(i.e., in the page)

Note that the reflected magnetic field must point into the screen toachieve for the reflected wave. The x with a circle

around it means into the screen.

E B k

Note that Hechtuses a differentnotation for the

reflected field,which is confusing!

Ours is better!

This leads to adifference in thesigns of someequations...

Now, the case of P polarization:


10/19

Reflection & Transmission Coefficients

for Parallel Polarized Light

These equations are called the Fresnel Equations for

parallel polarized (p-polarized) light.

[ ] [ ]|| 0 0/ cos( ) cos( ) / cos( ) cos( )r i i t t i i t t ir E E n n n n = = +

[ ]|| 0 0/ 2 cos( )/ cos( ) cos( )t i i i i t t it E E n n n = = +

Solving for E0r / E0i yields the reflection coefficient, r||:

Analogously, the transmission coefficient, t|| = E0t / E0i, is

For parallel polarized light, B0i B0r = B0t

and E0icos(i) +E0rcos(r) = E0tcos(t)


11/19

To summarize

||

cos( ) cos( )

cos( ) cos( )

=

+

i t t i

i t t i

n nr

n n

||

2 cos( )

cos( ) cos( )=

+

i i

i t t i

nt

n n

2 cos( )cos( ) cos( )

=

+

i i

i i t t

ntn n

cos( ) cos( )

cos( ) cos( )

=

+

i i t t

i i t t

n nr

n n

s-polarized light: p-polarized light:

And, for both polarizations: sin( ) sin( )=i i t tn n

planeofincidenceincidentwave

transmitted wave

interface

planeofincidenceincidentwave

transmitted wave

interface

E-field vectors are red .k vectors are black.


12/19

Reflection Coefficients for an

Air-to-Glass Interface

Incidence angle, i

Reflection

coefficient,

r

1.0

.5

0

-.5

-1.0

r||

r

0 30 60 90

The two polarizations areindistinguishable at = 0

Total reflection at = 90for both polarizations.

nair 1 < nglass 1.5

Brewsters angle

r||=0!Zero reflection for parallelpolarization at 56.3

Brewster's angle

The value of this angledepends on the value ofthe ratio ni/nt:

Brewster = tan-1(nt/ni)

Sir David Brewster1781 - 1868


13/19

Incidence angle, i

Reflection

coefficient,

r

1.0

.5

0

-.5

-1.0

r||

r

0 30 60 90

Brewstersangle

Total internalreflection

Critical

angle

Criticalangle

Total internal reflection

above the "critical angle"

crit

sin-1

(nt /ni) 41.8 for glass-to-air

nglass > nair

(The sine in Snell's Law

can't be greater than one!)

Reflection Coefficients for a

Glass-to-Air Interface


14/19

Reflectance (R)

R Reflected Power / Incident Power r r

i i

I A

I A=

Because the angle of incidence = the angle of reflection,the beam area doesnt change on reflection.

Also, n is the same for both incident and reflected beams.

A = Area

20 0

02

c

I n E

=

iwi nint

r wi

2R r=So: since

2

0 2

2

0

=r

i

Er

E


15/19

Transmittance (T)t t

i i

I A

I A=

A = Area

20 00

2

cI n E

=

cos( )cos( )

t t t

i i i

A wA w

= =

t

iwi

wt

nint

If the beamhas width wi:

20 020

0 2

220 0 0

0

2

2

t tt t tt t t t t

i i i i ii i ii i

cn E

n E wI A w nwT t

cI A w nwn E w

n E

= = = =

The beam expands (or contracts) in one dimension on refraction.

since

2

0 2

2

0

t

i

Et

E=

( )( )( )

( )

2cos

cos

t t

i i

nT t

n

=

T Transmitted Power / Incident Power


16/19

Reflectance and Transmittance for an

Air-to-Glass Interface

Note that it is NOT true that: r +t= 1.

But, it is ALWAYS true that: R + T = 1

Perpendicular polarization

Incidence angle, i

1.0

.5

00 30 60 90

R

T

Parallel polarization

Incidence angle, i

1.0

.5

00 30 60 90

R

T

Brewstersangle


17/19

Perpendicular polarization

Incidence angle, i

1.0

.5

00 30 60 90

R

T

Reflectance and Transmittance for a

Glass-to-Air Interface

Parallel polarization

Incidence angle, i

1.0

.5

00 30 60 90

R

T

Note that the critical angle is the same for both polarizations.

And still, R + T = 1


18/19

Reflection at normal incidence, i = 02

t i

t i

n nR

n n

=

+ ( )2

4 t i

t i

n nT

n n=

+

When i = 0, the Fresnelequations reduce to:

For an air-glass interface (ni = 1and nt = 1.5),

R = 4% andT = 96%

The values are the same, whicheverdirection the light travels, from air toglass or from glass to air.

This 4% value has big implicationsfor photography.

lens flare


19/19

Windows look like mirrors at night (when youre in the brightly lit room)

One-way mirrors (used by police to interrogate bad guys) are just

partial reflectors (actually, with a very thin aluminum coating).

Disneyland puts ghouls next to you in the haunted house using partialreflectors (also aluminum-coated).

Optical fibers use total internal reflection.

Where youve seen Fresnels Equations in action

R = 100% R = 90%Laser medium

0% reflection!

0% reflection!

Some lasers use Brewsters angle components to avoid reflective losses:

optics lecture-fresnel equations

Documents