Fresnel's Equations for Reflection and Refraction

Incident, transmitted, and reflected beams at interfaces

Reflection and transmission coefficients

The Fresnel Equations

Brewster's Angle

Total internal reflection

Power reflectance and transmittance

Phase shifts in reflection

The mysterious evanescent wave

Definitions: Planes of Incidence and the Interface and the polarizationsPerpendicular (“S”) polarization sticks out of or into the plane of incidence.

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

Plane of incidence (here the xy plane) is the plane that contains the incident and reflected k-vectors.

ni

nt

ik

rk

tk

i r

t

Ei Er

Et

Interface

x

y

zParallel (“P”) polarization lies parallel to the plane of incidence.

Incident medium

Transmitting medium

Shorthand notation for the polarizations

Perpendicular (S) polarization sticks up out of the plane of incidence.

Parallel (P) polarization lies parallel to the plane of incidence.

Fresnel Equations

We would like to compute the fraction of a light wave reflected and transmitted by a flat interface between two media with different refractive indices.

Ei Er

Et

where E0i, E0r, and E0t are the field complex amplitudes.

We consider the boundary conditions at the interface for the electric and magnetic fields of the light waves.

We’ll do the perpendicular polarization first.

0 0/r ir E E

0 0/t it E E 0 0/r ir E E

0 0/t it E E

for the perpendicular polarization

for the parallel polarization

The Tangential Electric Field is Continuous

In other words:

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

Here, all E-fields arein the z-direction, which is in the plane of the interface (xz), so:

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

ni

nt

ik

rk

tk

i r

t

EiBi

Er

Br

Et

Bt

Interface

Boundary Condition for the ElectricField at an Interface

x

y

z

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 arein the xy-plane, so we take the x-components:

–Bi(x, y=0, z, t) cos(i) + Br(x, y=0, z, t) cos(r) = –Bt(x, y=0, z, t) cos(t)

*It's really the tangential B/, but we're using 0

Boundary Condition for the MagneticField at an Interface

ni

nt

ik

rk

tk

i r

t

Ei

Bi

Er

Br

Et

Bt

Interface x

y

z

ii

Reflection and Transmission for Perpendicularly (S) Polarized Light

Canceling 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 EB B B

0 0 0 0( ) cos( ) ( ) cos( )i r i i t r i tn E E n E E

0 0 0( ) cos( ) cos( )i r i i t t tn E E n E

0 0/( / ) / :r iB E c n nE c But and

0 0 0 0 :t i r tE E E E Substituting for using

Reflection & Transmission Coefficientsfor Perpendicularly Polarized Light

Fresnel EquaThese equations are called the perpendicular

for polarized

tlly i

ionsght.

0 0 0 0( )cos( ) ( ) cos( ) :i r i i t r i tn E E n E E Rearranging yields

0 0/ cos( ) cos( ) / cos( ) cos( )r i i i t t i i t tr E E n n n n

0 0/ 2 cos( ) / cos( ) cos( )t i i i i i t tt E E n n n

0 0/t iE EAnalogously, the transmission coefficient, , is

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

0 0cos( ) cos( ) cos( ) cos( )r i i t t i i i t tE n n E n n

Simpler expressions for r┴ and t┴

cos( )cos( )

t t

i i

wm

w

t

iwi

wt

ni

nt

Recall the magnification at an interface, m:

Also let be the ratio of the refractive indices, nt / ni.

cos( ) cos( ) / cos( ) cos( ) 1 / 1i i t t i i t tr n n n n m m

2 cos( ) / cos( ) cos( ) 2 / 1i i i i t tt n n n m

Dividing numerator and denominator of r and t by ni cos(i:

11

mrm

2

1t

m

Fresnel Equations—Parallel electric field

x

y

z

Note that the reflected magnetic field must point into the screen to achieve . The x means “into the screen.”E B k

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

ni

nt

ik

rk

tk

i r

t

Ei

Bi Er

Br

EtBt

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

This B-field points into the page.

Reflection & Transmission Coefficientsfor Parallel (P) Polarized Light

For parallel polarized light, B0i - B0r = B0t

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

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

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

These equations are called the Fresnel Equations for parallel 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

Simpler expressions for r║ and t║

/r m m

2 /t m

Again, use the magnification, m, and the refractive-index ratio, .

And again dividing numerator and denominator of r and t by ni cos(i:

mrm

2t

m

|| 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

Reflection Coefficients for an Air-to-Glass Interface nair 1 < nglass 1.5

Note that:

Total reflection at = 90°for both polarizations

Zero reflection for parallel polarization at Brewster's angle (56.3° for these values of ni and nt).

(We’ll delay a derivation of a formula for Brewster’s angle until we do dipole emission and polarization.) Incidence angle, i

Ref

lect

ion

coef

ficie

nt, r

1.0

.5

0

-.5

-1.0

r||

r┴

0° 30° 60° 90°

Brewster’s angler||=0!

Reflection Coefficients for a Glass-to-Air Interface

nglass 1.5 > nair 1

Note that:

Total internal reflectionabove the critical angle

crit arcsin(nt /ni)

(The sine in Snell's Lawcan't be > 1!):

sin(crit) nt /ni sin(90) Incidence angle, i

Ref

lect

ion

coef

ficie

nt, r

1.0

.5

0

-.5

-1.0

r||

r┴

0° 30° 60° 90°

Total internal reflection

Brewster’s angle

Criticalangle

Criticalangle

Transmittance (T)T Transmitted Power / Incident Power

20 002

cI n E

2 2coscos

t t

i i

nT t mt

n

cos( )cos( )

t t t

i i i

A wm

A w

t

iwi

wt

ni

nt

t t

i i

I AI A

A = Area

20 020

0 22

20 0 00

cos( )2cos( )

2

t tt t tt t t t t

i i i i ii i ii i

cn E n E wI A w nT tcI A w nn E wn E

Compute the ratio of the beam areas:

The beam expands in one dimension on refraction. 20 2

20

t

i

Et

E

The Transmittance is also called the Transmissivity.

1D beam expansion

Reflectance (R)

R Reflected Power / Incident Power

2R r

r r

i i

I AI A

Because the angle of incidence = the angle of reflection, the beam area doesn’t change on reflection.

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

So:

20 002

cI n E

A = Area

iwi ni

nt

r wi

The Reflectance is also called the Reflectivity.

Reflectance and Transmittance for anAir-to-Glass Interface

Note 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

Reflectance and Transmittance for aGlass-to-Air Interface

Note 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

Reflection at normal incidenceWhen i = 0, 

and 

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. The 4% has big implications for photography lenses.

2

t i

t i

n nRn n

2

4 t i

t i

n nT

n n

Windows look like mirrors at night (when you’re in the brightly lit room)

One-way mirrors (used by police to interrogate bad guys) are just partial reflectors (actually, aluminum-coated).

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

Lasers use Brewster’s angle components to avoid reflective losses:

Practical Applications of Fresnel’s Equations

Optical fibers use total internal reflection. Hollow fibers use high-incidence-angle near-unity reflections.

R = 100%R = 90%Laser medium

0% reflection!

0% reflection!

Phase shifts in reflection (air to glass)

180° phase shiftfor all angles

180° phase shiftfor angles belowBrewster's angle;0° for larger angles

0° 30° 60° 90°Incidence angle

0° 30° 60° 90°Incidence angle

0

0

┴

||

Incidence angle, i

Ref

lect

ion

coef

ficie

nt, r

1.0

0

-1.0

r||

r┴

0° 30° 60° 90°

Brewster’s angle

Phase shifts in reflection (glass to air)

0° 30° 60° 90°Incidence angle

0° 30° 60° 90°Incidence angle

0

0

┴

||

Incidence angle, i

Ref

lect

ion

coef

ficie

nt, r

1.0

0

-1.0

r||

r┴

0° 30° 60° 90°

Total internal

reflectionBrewster’s

angle

Criticalangle

Criticalangle

Interesting phase above the critical angle

180° phase shiftfor angles belowBrewster's angle;0° for larger angles

Phase shifts vs. incidence angle and ni /nt

Li Li, OPN, vol. 14, #9,pp. 24-30, Sept. 2003

ni /nt

ni /nt

Note the general behavior above and below the various interesting angles…

i

i

If you slowly turn up a laser intensity incident on a piece of glass, where does damage happen first, the front or the back?

The obvious answer is the front of the object, which sees the higher intensity first.

But constructive interference happens at the back surface between the incident light and the reflected wave.

This yields an irradiance that is 44% higher just inside the back surface!

2(1 0.2) 1.44

Phase shifts with coated opticsReflections with different magnitudes can be generated using partial metallization or coatings. We’ll see these later.

But the phase shifts on reflection are the same! For near-normal incidence:

180° if low-index-to-high and 0 if high-index-to-low.

Example:

Laser Mirror

Highly reflecting coating on this surface

Phase shift of 180°

Total Internal Reflection occurs when sin(t) > 1, and no transmitted beam can occur.

Note that the irradiance of the transmitted beam goes to zero (i.e., TIR occurs) as it grazes the surface.

Total internal reflection is 100% efficient, that is, all the light is reflected.

Brewster’s angle

Total Internal Reflection

Applications of Total Internal Reflection

Beam steerers

Beam steerersused to compressthe path insidebinoculars

Three bounces: The Corner Cube

Corner cubes involve three reflections and also displace the return beam in space. Even better, they always yield a parallel return beam:

Hollow corner cubes avoid propagation through glass and don’t use TIR.

If the beam propagates in the z direction, it emerges in the –z direction, with each point in the beam (x,y) reflected to the (-x,-y) position.

Optical fibers use TIR to transmit light long distances.

Fiber Optics

They play an ever-increasing role in our lives!

Core: Thin glass center of the fiber that carries the light

Cladding: Surrounds the core and reflects the light back into the core

Buffer coating: Plastic protective coating

Design of optical fibers

ncore > ncladding

Propagation of light in an optical fiber

Some signal degradation occurs due to imperfectly constructed glass used in the cable. The best optical fibers show very little light loss -- less than 10%/km at 1,550 nm.

Maximum light loss occurs at the points of maximum curvature.

Light travels through the core bouncing from the reflective walls. The walls absorb very little light from the core allowing the light wave to travel large distances.

Microstructure fiber

Such fiber has many applications, from medical imaging to optical clocks.

Photographs courtesy of Jinendra Ranka, Lucent

Air holes

Core

In microstructure fiber, air holes act as the cladding surrounding a glass core. Such fibers have different dispersion properties.

Frustrated Total Internal ReflectionBy placing another surface in contact with a totally internallyreflecting one, total internal reflection can be frustrated.

How close do the prisms have to be before TIR is frustrated?

This effect provides evidence for evanescent fields—fields that leak through the TIR surface–and is the basis for a variety of spectroscopic techniques.

nn

nn

Total internal reflection Frustrated total internal reflectionn=1 n=1

FTIR and fingerprinting

See TIR from a fingerprint valley and FTIR from a ridge.

The Evanescent WaveThe evanescent wave is the "transmitted wave" when total internal reflection occurs. A mystical quantity! So we'll do a mystical derivation:

2

2 2

) 1,:

cos( ) 1 sin ( ) 1 sin ( )

t t

t

it t i

t

rR

nn

Since sin( doesn't exist, so computing is impossible.Let's check the reflectivity, , anyway. Use Snell's Law to eliminate

Neg. Number

Substitutin

* 1

ra bi a biR R r ra bi a bi

g this expression into the above one for and

redefining yields:

0

0

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

i i t tr

i i i t t

n nErE n n

ni

nt

ik

rki

t

xy

tk

So all power is reflected; the evanescent wave contains no power.

The Evanescent-Wave k-vectorThe evanescent wave k-vector must have x and y components:

Along surface: ktx = kt sin(t)

Perpendicular to it: kty = kt cos(t)

Using Snell's Law, sin(t) = (ni /nt) sin(i), so ktx is meaningful.

And again: cos(t) = [1 – sin2(t)]1/2 = [1 – (ni /nt)2 sin2(i)]1/2

= ± i

Neglecting the unphysical -i solution, we have:

Et(x,y,t) = E0 exp[–ky] exp i [ k (ni /nt) sin(i) x – t ]

The evanescent wave decays exponentially in the transverse direction.

ni

nt

ik

rki

t

xy

tk

Optical Properties of MetalsA simple model of a metal is a gas of free electrons (the Drude model).

These free electrons and their accompanying positive nuclei can undergo "plasma oscillations" at frequency, p.

where:

22

0

22

2

1

0,

pe

p

p p

N em

n

n n

The refractive index for a metal is :

When is imaginary, and absorption is strong.So for metals absorb strongly. For meta .ls are transparent

Reflection from metals

At normal incidence in air:

Generalizing to complex refractive indices:

2

2

*

*

( 1)( 1)( 1)( 1)( 1)( 1)

nRn

n nRn n