Chapter 6

Light at Interfaces – Law of Reflection and Refraction Part I

Introduction Law of Reflection and Refraction Law of Reflection and Refraction

normal Law of reflection:

i  r

i r

n1 n2 Law of refraction “Snell’s Law”:

 sin  n t i  2 sint  n1 Incident, reflected, refracted, and surface normal in same plane Huygens’ Principle Every point on a wavefront may be regarded as a secondary source of wavelets Huygens’ Principle

Law of refraction Fermat’s Principle The path a beam of light takes between two points is the one which is traversed in the least time

Law of reflection Fermat’s Principle

Law of refraction SO OP t   vi vt

h2  x2 b2  (a  x)2 t   vi vt To minimize t(x) with respect to x, dt x  (a  x)   dx 2 2 2 2 vi h  x vt b  (a  x)  0.

sin sin Thus, i  t , vi vt so,

ni sini  nt sint Fermat’s Principle For a ray passing through a multilayer (m layers) material s s s t  1  2   m v1 v2 vm Then 1 m t  ni si , c i1 The summation is known as the optical path length (OPL) traversed by the ray. P (OPL)  S n(s)ds. The route light travels having the smallest optical path length EM Wave Point of View Light is an electromagnetic wave, obeys the Maxwell’s equations

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

normal

   EB  k EM Wave Point of View     i(tkr) E  E0e     i(tkr) B  B0e E  cB 1. What happens to and at the interface? 2. Can one derive law of reflection and refraction? 3. Or obtain more information? Part II

Fresnel Equations Plane of Incidence Polarization of Light Always relative to plane of incidence

TE: Transverse electric TM: Transverse magnetic s: senkrecht polarized p: plane polarized (E-field sticks in and out of the plane) (E-field in the plane)

perpendicular, horizontal parallel, vertical Polarization of Light Linear polarized light decomposition y

x

Any polarization state can be described as y linear combination of these two:  ikzt  ikzty  E  E e x xˆ  E e yˆ 0x 0 y x  ix iy ikzt  E  E0xe xˆ  E0 ye yˆ e

“complex amplitude” contains all polarization info Waves at Interface

boundary point Waves at Interface At the boundary point, phases of the three waves must be equal        (k i r it)  (k r r rt)  (k t r tt) r is at the boundary  This is true for any boundary point and time, so let’s take r  0

it  rt  tt

i  r  t Thus, the frequencies are equal If we now consider t  0       k i r  k r r  k t r which means all three propagation vectors lie in the same plane Law of Reflection     Since k i r  k r r

kir sini  kr r sinr

Both incident and reflected waves travel in same medium,

ki  kr

Thus one obtains the law of reflection:

 i   r Law of Refraction     Since k r r  k t r

kr r sinr  kt r sint Both reflected and transmitted waves travel in different media (same frequencies; different wavelengths!):

kr   / vr  n1 / c

kt   / vt  n2 / c which leads to the law of refraction:

n1 sinr  n2 sint Part III

Fresnel Equations Maxwell’s Equations

   D – Displacement field D  dS   dV   D     H – Magnetic Intensity   B   B E  dr     dS   E   t t P – Polarization       D     D M – Magnetization    H  dr   j    dS   H  j   t  t  - Magnetic permeability     B  dS  0   B  0  - Permittivity

e - Dielectric Susceptibility

     m - Magnetic Susc. D   E  P P   E o e  – Conductivity  1     H  B  M M  m H j – Current density o       j  E D  E B  H Boundary Conditions I    B  0 A small cylindrical Gauss surface cross the interface gives

The normal components of magnetic field are continuous across the interface. Boundary Conditions II    D  

A small cylindrical Gauss surface cross the interface gives

is the surface free charge density. If , then the normal component of D is continuous across the interface. But E may not unless  will be the same. Boundary Conditions III   B   E   t A small rectangular loop cross the interface gives

The tangential component of the electric field is continuous across the boundary. Boundary Conditions IV    D   H  j  t A small rectangular loop cross the interface gives

is the surface free current density. If , then the normal component of H is continuous across the interface. E-fields at the Boundary For both electric and magnetic fields, the tangetial components must be continuous as boundary is passed

All waves only have tangential electric fields: TE waves  E  E yˆ  0i i E  E yˆ  0r r E0t  Et yˆ complex field amplitudes

Continuity requires:

E i  E r  E t B-fields at the Boundary

TE waves The magnetic fields:

   i(ki r t) Bi  (Bi cosi xˆ  Bi sini zˆ)e    i(kr r t) Br  (Br cosr xˆ  Br sinr zˆ)e    i(kt r t) Bt  (Bt cost xˆ  Bt sint zˆ)e

Continuity requires:

Bi cosi  Br cosr  Bt cost

Same analysis can be performed for TM waves Boundary Conditions at the Interface

 TE waves  TM waves E E i  i  Er Er n1   Bt B n2   t   B B i Br i Br   Et Et

Ei  Er  Et Bi  Br  Bt

Bi cosi  Br cosr  Bt cost  Ei cosi  Er cosr  Et cost Fresnel Equations TE waves (s-polarization) TM waves (p-polarization)

Get all in terms of E and apply law of reflection (i = r):

Ei  Er  Et ni Ei  ni Er  nt Et ni Ei cosi  ni Er cosi  nt Et cost  Ei cosi  Er cosi  Et cost

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

Er ni cosi  nt cost Er nt cosi  ni cost rTE  r   rTM  r||   Ei ni cosi  nt cost Ei nt cosi  ni cost

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

Et 2ni cosi Et 2ni cosi tTE  t   tTM  t||   Ei ni cosi  nt cost Ei nt cosi  ni cost Fresnel Equations

TE waves (s-polarization) TM waves (p-polarization) And together:

nt t 1 r t|| 1 r|| ni

Using Snell’s law, , one has tan(  ) sin(i t ) i t r   r||    tan(  ) sin(i t ) i t

2sint cosi 2sint cosi t   t||   sin(i t ) sin(i t )cos(i t ) Part IV

Understand Fresnel Equations Perpendicular Incident When , then n  n n  n r  i t r  t i  || n  n ni  nt t i 2n i 2ni t  t||  ni  nt ni  nt Thus, At the air/glass interface, r  r|| therefore t  t|| The plane of incidence is no longer specified when i  t  0 Grazing Incident When , but

r  1 r||  1

t  t||  0 Reflection and Transmission Coefficients

nt Calculate r and t versus i, let: n  ni

cos  n2  sin 2  2 2 2 i i n cosi  n  sin i r  r  cos  n2  sin 2  || 2 2 2 i i n cosi  n  sin i

2cos t  i 2ncosi  2 2 t  || 2 2 2 cosi  n  sin i n cosi  n  sin i

Mathematica plot of transmitteded/reflected light http://demonstrations.wolfram.com/FresnelEquations/ Reflection and Transmission Coefficients

1.0

t 0.5 || t r||

0.0

Coefficients -0.5 r Brewster's angle 56.3o -1.0 0 20 40 60 80 Incident Angle  i Reflection and Transmission Coefficients

1.0

0.5 r

0.0

41.8o r|| Coefficients -0.5 Critical angle Brewster's angle 33.7o -1.0 0 10 20 30 40 50 60 70 80 90 Incident Angle  i Phase Shift    k r t  ˆ Ei  Eie j    k r t  ˆ Er  Ere j At the Air  Glass interface

One expect a reversal of sign in the electric fields, compared to

ei  1 for    .

i.e., after reflection, the phase of the E-field of the reflected light changes with respect to the incident E-field Phase Shift

We need to define phase shift for two cases: A. When two fields E or B are  to the plane of incidence, they are said to be (i) in-phase ( = 0) if the two E or B fields are parallel and (ii) out-of-phase ( = ) if the fields are anti-parallel. B. When two fields E or B are parallel to the plane of incidence, the fields are (i) in-phase if the y-components of the field are parallel and (ii) out-of-phase if the y-components of the field are anti-parallel. Phase Shift

  (a) Bi , Bt  (b)    in  phase,   0     Ei , Et  E , E , E     i r t all in  phase   0  Ei , Er  Bi , Br , Bt     out  of  phase,     Bi , Br  Phase Shift

Analogy between a wave on a string and an E-M wave traversing the air-glass interface. Phase Shift Analogy between a wave on a string and an E-M wave traversing the air-glass interface.

Glass (n = 1.5) Air (n = 1)

r 0  = 0  = 0

Air (n = 1) Glass (n = 1.5)

 =   = 0 r 0 Phase Shift

1.0

0.5 t t|| r||

0.0

Coefficients -0.5 r

-1.0 0 20 40 60 80 Incident Angle  i Phase Shift

1.0

0.5 r

0.0

Coefficients -0.5 r||

-1.0 0 10 20 30 40 50 60 70 80 90 Incident Angle  i Reflectance and Transmittance

From the geometry and total area A of the beam at the interface, the power (P) for the (i) incident, (ii) reflected and (ii) transmitted beams are:

(i) Pi  Ii Acosi

(ii) Pr  Ir Acosr

(iii) Pt  It Acost Cross-sectional Areas Reflectance and Transmittance

P I Acos R  r  r r  r 2 Pi Ii Acosi

Pt It Acost  nt cost  2 T     t Pi Ii Acosi  ni cos i 

2 2  E   n cos  E  Therefore, or t t ot 1         R T  Eoi   ni cosi  Eoi  We can write this expression in the form of components  and ||:

 n cos   n cos  2 2  t t  2  t t  2 R  r and R||  r|| , T   t and T||   t||  ni cosi   ni cosi 

Therefore, R  T 1 and R||  T|| 1 Reflectance and Transmittance

TM TE (p) (s)

Reflectance and Transmittance for an Air-to-Glass Interface Reflectance and Transmittance

Perpendicular polarization Parallel polarization 1.0 1.0 R R .5 .5 T T 0 0 0° 30° 60° 90° 0° 30° 60° 90°

Incidence angle, i Incidence angle, i

Reflectance and Transmittance for an Glass-to-Air Interface Reflectance and Transmittance

2  n  n  When   0, R  R  R   t i  i ||     nt  ni 

4nt ni and T  T||  T  2 nt  ni 

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. Reflectance and Transmittance When does a window become a mirror? at night (when you’re in a brightly lit room)

Iin >> Iout Indoors Outdoors Brewster’s Angle

2 2 2 n cosi  n  sin i When r||   0 2 2 2 n cosi  n  sin i

1.0 2 2 2 n cosi  n  sin i  0

0.5 t t  || 1 1 nt r||  p  tan n   tan ni 0.0  p t  90 Coefficients -0.5 r

-1.0 0 20 40 60 80 Incident Angle  i Brewster’s Angle Brewster’s Angle Polarized Sunglasses • Glare consists almost entirely of horizontally polarized light • Vertically polarized light is useful for human vision

• Polarized sunglasses can selectively block certain polarizations of light Brewster’s Angle

Polarized Sunglasses

Regular Sunglasses Polarized Sunglasses Brewster’s Angle Polarized Camera Lenses

Unpolarized Polarized Brewster’s Angle Brewster Windows • Whenever a wave passes through a transparent window, some degree of optical loss will occur • Even with an anti-reflective coating, a normal glass pane will have a reflectivity of around 0.2% on each side

• Brewster windows are tilted to create an angle of incidence close to Brewster's angle • Result in at least 10 times lower losses Brewster’s Angle Brewster Windows

R = 100% 0% reflection! Laser medium R = 90%

0% reflection!

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

When and at , TIR happens

1.0

0.5 r

0.0 Brewster’s angle

Coefficients -0.5 r||

-1.0 0 10 20 30 40 50 60 70 80 90 Incident Angle  i

Total Internal Reflection Total Internal Reflection

Beam steerers used to compress the path inside binoculars ncore > ncladding The Evanescent Wave When and if , TIR happens, what happens to the boundary conditions?

For both TE and TM wave, the reflected E-field and incident E-field

are in-phase. Thus there must be transmitted wave in nt.

1.0 r 0.5

0.0 r|| Coefficients -0.5

-1.0 0 10 20 30 40 50 60 70 80 90 Incident Angle  i The Evanescent Wave When and if , TIR happens, and the reflection coefficients can be calculated as

2 2 2 2 2 cosi  n  sin i n cosi  n  sin i r  r||  2 2 2 2 2 cosi  n  sin i n cosi  n  sin i

Since , both and are complex quantities, but one can find that . Thus T = 0, i.e. the transmitted wave does not carry energy across the boundary. The Evanescent Wave The transmitted wave will decay into the nt media, and is called the evanescent wave The evanescent wave decays exponentially in the transverse direction. The Evanescent Wave Sensors The Evanescent Wave Sensors