13. Fresnel's Equations for Reflection and Transmission

Incident, transmitted, and reflected beams

Boundary conditions: tangential fields are continuous

Reflection and transmission coefficients

The "Fresnel Equations"

Brewster's Angle

Total internal reflection

Power reflectance and transmittance Augustin Fresnel

1788-1827

Posing the problem

What happens when light, propagating in a

uniform medium, encounters a smooth interface

which is the boundary of another medium (with a

different refractive index)?

k-vector of the

incident light

boundarynincident

ntransmitted

First we need to

define some

terminology.

Definitions: Plane of Incidence and plane of the interface

Plane of incidence (in this

illustration, the yz plane) is the

plane that contains the incident

and reflected k-vectors.x

y

z

Plane of the interface (y=0, the xz plane) is the plane that

defines the interface between the two materials

Definitions: “S” and “P” polarizations

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

1. “S” polarization is the perpendicular polarization, and it sticks up out of the plane of incidence

The plane of the interface (y=0)

is perpendicular to this page.

Here, the plane of

incidence (z=0) is the

plane of the diagram.x

y

z

I R

T

A key question: which way is the E-field pointing?

There are two distinct possibilities.

reflected light

reflecting medium

Definitions: “S” and “P” polarizations

The amount of reflected (and transmitted) light is

different for the two different incident polarizations.

Note that this is a different use of the word “polarization”

from the way we’ve used it earlier in this class.

ni

nt

ik�

rk�

tk�

θi θr

θt

EiBi

Er

Br

Et

Bt

Interface x

y

zBeam geometry for

light with its electric

field sticking up out of

the plane of incidence

(i.e., out of the page)

We treat the case of s-polarization first:

This red dashed

line represents the

xz plane (y = 0),

which is the plane

of the interface.

Augustin Fresnel was the first to do this calculation (1820’s).

Fresnel Equations—Perpendicular E field

ni

nt

ik�

rk�

tk�

θi θr

θt

EiBi

Er

Br

Et

Bt

Interface

Boundary Condition for the ElectricField at an Interface: s polarization

x

y

z

In other words,

The Tangential Electric Field is Continuous

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

The component of the E-field that lies in the xz plane is continuous as you move across the plane of the interface.

Here, all E-fields are in the z-direction, which is in the plane of the interface.

(We’re not explicitly writing

the x, z, and t dependence,

but it is still there.)

Boundary Condition for the MagneticField at an Interface: s polarization

ni

nt

ik�

rk�

tk�

θi θr

θt

Ei

Bi

Er

Br

Et

Bt

Interface

x

y

z

θi

θi

*It's really the tangential B/µ, but we're assuming µi = µt = µ0

–Bi(y = 0) cosθi + Br(y = 0) cosθr = –Bt(y = 0) cosθt

The Tangential Magnetic Field* is Continuous

In other words,

The total B-field in the plane of the interface is continuous.

Here, all B-fields are in the xy-plane, so we take the x-components:

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 Bθ θ θ

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 Eθ θ

0 0 0( ) cos( ) cos( )− = −i r i i t t tn E E n Eθ θ

0 0 /( / ) / .But and= = =i rB E c n nE c θ θSubstituting into the second equation:

Reflection & Transmission Coefficientsfor 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

θ θθ θ θ θ

− = − ++ = −

Rearranging yields

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

0 0/ , istransmission coefficientAnalogously, the , t iE E

[ ] [ ]0 0/ cos( ) cos( ) / cos( ) cos( )r i i i t t i i t tr E E n n n nθ θ θ θ⊥ = = − +0 0/ Solving for yields t reflection coefficienthe :r iE E

These equations are called the Fresnel Equations for

perpendicularly polarized (s-polarized) light.

ni

nt

ik�

rk�

tk�

θi θr

θt

Ei

Bi Er

Br

EtBt

Interface

×

Fresnel Equations—Parallel electric field

x

y

z

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

Note that the reflected magnetic field must point into the screen to achieve for the reflected wave. The x with a circle around it means “into the screen.”

E B k× ∝�� �

Note that Hecht uses a different notation for the reflected field, which is confusing!

Ours is better!

This leads to a difference in the signs of some equations...

Now, the case of P polarization:

Reflection & Transmission Coefficientsfor 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 nθ θ θ θ= = − +

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

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(θi) + E0rcos(θr) = E0tcos(θt)

To summarize…

||

cos( ) cos( )

cos( ) cos( )

−=+

i t t i

i t t i

n nr

n n

θ θθ θ

||

2 cos( )

cos( ) cos( )=

+i i

i t t i

nt

n n

θθ θ

2 cos( )

cos( ) cos( )⊥ =

+i i

i i t t

nt

n n

θθ θ

cos( ) cos( )

cos( ) cos( )⊥

−=+

i i t t

i i t t

n nr

n n

θ θθ θ

s-polarized light: p-polarized light:

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

incident wave

transmitted wave

interface

incident wave

transmitted wave

interface

E-field vectors are red.

k vectors are black.

Reflection Coefficients for an Air-to-Glass Interface

Incidence angle, θi

Re

fle

ctio

n c

oe

ffic

ien

t, r

1.0

.5

0

-.5

-1.0

r||

r┴

0° 30° 60° 90°

The two polarizations are indistinguishable at θ = 0°

Total reflection at θ = 90°for both polarizations.

nair ≈ 1 < nglass ≈ 1.5

Brewster’s angle

r||=0!Zero reflection for parallel polarization at:

“Brewster's angle”

The value of this angle depends on the value of the ratio ni/nt:

θBrewster = tan-1(nt/ni)

Sir David Brewster

1781 - 1868

For air to glass (nglass = 1.5), this is 56.3°.

Incidence angle, θi

Re

fle

ctio

n c

oe

ffic

ien

t, r

1.0

.5

0

-.5

-1.0

r||

r┴

0° 30° 60° 90°

Brewster’s

angle

Total internal

reflection

Critical

angle

Critical

angle

Total internal reflection

above the "critical angle"

θcrit ≡ 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

http://www.ub.edu/javaoptics/index_en.html

The obligatory java applet.

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’s area doesn’t change on reflection.

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

A = Area

20 00

2

cI n E

ε =

θiwi ni

nt

θr wi

2R r=So: since

2

0 2

2

0

=r

i

Er

E

Transmittance (T)

t t

i i

I A

I A= A = Area

20 00

2

cI n E

ε =

cos( )

cos( )t t t

i i i

A w

A w

θθ

= =

θt

θiwi

wt

ni

nt

If the beam has 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 n wT t

cI A w n wn E wn E

ε

ε

= = = =

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

θθ

=

⇒

T ≡ Transmitted Power / Incident Power

Reflectance and Transmittance for anAir-to-Glass Interface

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

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

Perpendicular polarization

Incidence angle, θi

1.0

.5

00° 30° 60° 90°

R

T

Parallel polarization

Incidence angle, θi

1.0

.5

00° 30° 60° 90°

R

T

Brewster’s

angle

Perpendicular polarization

Incidence angle, θi

1.0

.5

00° 30° 60° 90°

R

T

Reflectance and Transmittance for aGlass-to-Air Interface

Parallel polarization

Incidence angle, θi

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

Reflection at normal incidence, θi = 02

t i

t i

n nR

n n

−= + ( )2

4 t i

t i

n nT

n n=

+

When θi = 0, the Fresnel equations reduce to:

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

R = 4% and T = 96%

The values are the same, whichever direction the light travels, from air to glass or from glass to air.

This 4% value has big implications for photography.

If it weren’t for “lens flare,” there

would be no J. J. Abrams.

Windows look like mirrors at night (when you’re in a 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 partial reflectors (also aluminum-coated one-way mirrors).

Smooth surfaces can produce pretty good mirror-like reflections, even though they are not made of metal.

Where you’ve seen Fresnel’s Equations in action

Optical fibers only

work because of total

internal reflection.

Fresnel’s Equations in optics

R = 100%R = 90%Laser medium

0% reflection!

0% reflection!

Many lasers use Brewster’s

angle components to avoid

reflective losses: