Optoelectronics EE 430.423.001 2016. 2nd Semester

Lecture 02 : Chapter 2. The vectorial nature of light

2016. 9. 8~9.29

Changhee Lee School of Electrical and Computer Engineering Seoul National Univ. [email protected]

1/34 Changhee Lee, SNU, Korea Optoelectronics Maxwell equations EE 430.423.001 2016. 2nd Semester 1 ∂ 2U Wave equation ∇2U = u 2 ∂t 2 ∂ ∂ ∂      ∇ = ˆi + ˆj + kˆ ; ∇ expi(k ⋅r −ωt)= ik expi(k ⋅r −ωt) ∂x ∂y ∂z ∂     expi(k ⋅r −ωt)= −iω expi(k ⋅r −ωt) ∂t   ∂H    ∇× E = −µ k × E = µωH ∂t      ∂E k × H = −εωE ∇× H = ε ∂t    ⋅ = ∇ ⋅ E = 0 k E 0    ∇ ⋅ H = 0 k ⋅ H = 0

2/34 Changhee Lee, SNU, Korea Optoelectronics Energy flow. The Poynting vector EE 430.423.001 2016. 2nd Semester

εω ω c 1 2 H = E = εuE, u = = , c = , ε ≈ ε on , k k n ε oµo nE µ ∴H = , Z = o Z o ε o o ∂     expi k ⋅r −ωt = −iω expi k ⋅r −ωt µo ( ) ( ) Impedance of free space Zo = ≈ 377Ω ∂t ε o    The Poynting vector S = E × H

        Harmonic waves E = Eo cos(k ⋅r −ωt), H = H o cos(k ⋅r −ωt)

      1   S = E × H cos2 (k ⋅r −ωt), S = E × H o o 2 o o   1 k 1 n 2 S = Eo H o = Inˆ, I = Eo H o = Eo 2 k 2 2Zo

3/34 Changhee Lee, SNU, Korea Optoelectronics Linear Polarization EE 430.423.001 2016. 2nd Semester         E = Eo cos(k ⋅r −ωt), H = H o cos(k ⋅r −ωt)

If the amplitudes Eo and Ho are constant real vectors, the wave is said to be linearly polarized (or plane polarized).

In optics, it is traditional to designate the direction of the E field as the direction of polarization.

linear polarization elliptical polarization circular polarization

4/34 Changhee Lee, SNU, Korea Optoelectronics Partial Polarization EE 430.423.001 2016. 2nd Semester • linear polarizer: A device that produces linearly polarized light from unpolarized light (e.g. Polaroid film – dichroic sheet which absorbs one component of polarization more strongly than the other). • The transmission axis of a polarizer is the direction of the E field for a light wave that is transmitted with little or no loss. A light wave whose E field is at right angles to the transmission axis is absorbed or attenuated.

E1 = E cosθ 2 2 I1 = I cos θ For unpolarized light, =1/2.

I pol Degree of polarization: P = I pol + Iunpol I − I For partial linearly polarization, P = max min Imax + Imin 5/34 Changhee Lee, SNU, Korea Optoelectronics Scattering EE 430.423.001 2016. 2nd Semester Rayleigh scattering, named after the British physicist Lord Rayleigh (John William Strutt): elastic scattering of light. Electromagnetic radiation by particles much smaller than the wavelength of the radiation. d 6 Rayleigh scattering cross-section σ ∝ https://en.wikipedia.org/wiki/Rayleigh_scattering s λ4 Why is the sky blue? Molecules and atoms in the atmosphere scatter the sunlight according to Rayleigh scattering, inversely proportional to the fourth power of wavelength. Sky appears blue since the shorter wavelengths are scattered more than the longer wavelengths (yellow and especially red light).

Why is the sky blue and not violet? The most strongly scattered violet wavelength is absorbed in the upper atmosphere, so there is less violet in the light. Our eyes are also less sensitive to violet. Violet light stimulate the red cones slightly as well as the blue, and therefore violet appears blue with an added red tinge. Eye response curves Why does the sunset appear yellow/orange colors? The light from the sun has passed a long distance through air and shorter wavelengths have been scattered away from the direct line-of-sight.

http://math.ucr.edu/home/baez/physics/General/BlueSky/blue_sky.html 6/34 Changhee Lee, SNU, Korea Optoelectronics Polarization of scattered light EE 430.423.001 2016. 2nd Semester

The effects of a polarizing filter (right image) on the sky in a photograph. A properly oriented polarizing filter in front of the camera can block out the scattered light, increasing the contrast between the sky and the white clouds. https://en.wikipedia.org/wiki/Polarization_(waves)

Can we determine the location of the sun by measuring the polarization of the light from the blue sky? Yes, if we have the help of polarizing devices. Many insects whose eyes are sensitive to polarization can locate the sun even when it is behind a cloud.

http://www.science.oregonstate.edu/~giebultt/COURSES/ph332/Reading/Ch13.1.pdf 7/34 Changhee Lee, SNU, Korea Optoelectronics Circular Polarization EE 430.423.001 2016. 2nd Semester Consider two linearly polarized waves of the same amplitudes E polarized orthogonally to each other. o  iE cos(kz −ωt)  o jEo sin(kz −ωt)

Right-circularly polarized wave    E = Eo[i cos(kz −ωt) + jsin(kz −ωt)]

At a given point in space, E rotates clockwise when viewed against the direction of propagation. At a given instant in time, E describes right-handed spirals.

Left-circularly polarized wave    E = Eo[i cos(kz −ωt) − jsin(kz −ωt)]

   E = iEo expi(kz −ωt) + jEo expi(kz −ωt ±π / 2)] Take the real part of this expression:   - for right-circularly polarized wave = ± −ω Eo (i ij)expi(kz t) + for left-circularly polarized wave 8/34 Changhee Lee, SNU, Korea Optoelectronics Elliptic Polarization EE 430.423.001 2016. 2nd Semester If the component (real) fields are not of the same amplitude, the resultant wave is said to be elliptically polarized.

complex vector amplitude    Eo = iEo + jEo '   E = Eo expi(kz −ωt)

9/34 Changhee Lee, SNU, Korea Optoelectronics Quarter-Wave Plate EE 430.423.001 2016. 2nd Semester The thickness of the quarter-wave plate is given by λ d = o 4(n1 − n2 )

The effect of the quarter-wave plate is to introduce a iπ/2 phase shift term e = i between the slow (n1) and doubly refracting fast (n2) components of the wave. transparent crystal, such as calcite or mica    E = Eo (i + j)expi(kz −ωt)

↓ passing through the quarter - wave plate

   E = Eo (i + ij)expi(kz −ωt)

10/34 Changhee Lee, SNU, Korea Optoelectronics Matrix Representation of Polarization: Jones Calculus EE 430.423.001 2016. 2nd Semester    iφx iφy E = iEox + jEox , Eox = Eox e , Eoy = Eoy e

A convenient notation for the above pair of complex amplitudes is the Jones vector:

iφx E   E e  1 Eox  ox = ox normalized      iφy  2 2 E E e Eoy  oy   oy  Eox + Eox  

A 1   = A  Linearly polarized wave in the x direction 0 0 0 0   = A  Linearly polarized wave in the y direction A 1

A 1 o   = A  A wave that is linearly polarized at 45 relative to the x axis A 1 1  1    Left circularly polarization   right circularly polarization i − i 11/34 Changhee Lee, SNU, Korea Optoelectronics Applications of Jones Notation EE 430.423.001 2016. 2nd Semester

Addition of two waves of given polarizations = addition of Jones vectors:  1  1  1+1  2 1   +   =   =   = 2  Linearly polarized wave in the x direction having its − i i − i + i 0 0 amplitude twice that of the circular components

The optical elements are represented by 2x2 Jones matrices. a bA A'    =   c dB B'

If light is sent through a train of optical elements, the result is given by matrix multiplications:

an bn  a2 b2 a1 b1 A A'   ⋅⋅⋅     =   cn dn  c2 d2 c1 d1 B B'

12/34 Changhee Lee, SNU, Korea Optoelectronics Jones matrices for some optical elements EE 430.423.001 2016. 2nd Semester

Example: Suppose a quarter-wave plate is inserted into a beam of linearly polarized light and the incoming beam is polarized at 45o with respect to the horizontal (x axis). What is the polarization of the emerging beam? :

1 01 1    =   0 i 1 i

The emerging beam is left circularly polarized.

13/34 Changhee Lee, SNU, Korea Optoelectronics Orthogonal polarization EE 430.423.001 2016. 2nd Semester Two waves whose states of polarization are represented by the complex vector amplitudes E and E are said to   1 2 * be orthogonally polarized if E1 ⋅E2 = 0

Two Jones vectors are orthogonal if A*  2 = * + * = [A1 B1 ] *  A1 A2 B1B2 0 B2  Examples: • Two waves linearly polarized at right angles to one another • right and left circularly polarized waves

Light of arbitrary polarization can always be resolved into two orthogonal components.

A 1 0 1  1  1 1   = A  + B  = (A + iB)  + (A − iB)  B 0 1 2 − i 2 i

14/34 Changhee Lee, SNU, Korea Optoelectronics Eigenvectors of Jones matrices EE 430.423.001 2016. 2nd Semester

a bA A iψ Eigenvalue equation    = λ  , λ = λ e eigenvalue c dB B

Physically, an eigenvector of a given Jones matrix represents a particular polarization of a wave which, upon passing through the optical element in question, emerges with the same polarization as when it entered.

However, depending on the value of λ, the amplitude and the phase may change.

a − λ b A    = 0  c d − λB

a − λ b = 0 secular equation c d − λ

(a − λ)(d − λ) − bc = 0 → two roots λ1, λ2

To each root there is a corresponding eigenvector. (a − λ)A − bB = 0 cA + (d − λ)B = 0 15/34 Changhee Lee, SNU, Korea Optoelectronics Example: Jones matrix for a λ/4 plate with fast-axis horizontal EE 430.423.001 2016. 2nd Semester 1 0A A Eigenvalue equation    = λ  0 i B B

(1− λ)(i − λ) = 0 → two roots λ1 =1, λ2 = i

Normalized eigenvectors (1− λ)A = 0 (i − λ)B = 0 1 For λ1 =1, A ≠ 0, B = 0. ∴ eigenvector =   0 0 For λ2 = i, A = 0, B ≠ 0. ∴ eigenvector =   1 Physically, light that is linearly polarized in the direction of either the fast axis or the slow axis is transmitted without change of polarization. There is no change in amplitude since | λ |=1 for both cases, but a relative phase iπ/2 change of π/2 occurs since λ2/λ1=i= e .

16/34 Changhee Lee, SNU, Korea Optoelectronics Reflection and refraction at a plane boundary EE 430.423.001 nd    2016. 2 Semester  ⋅ −ω ei(k ⋅r −ωt) ei(k ' r t)

  ei(k "⋅r −ωt)

      k ⋅r −ωt = k '⋅r −ωt = k"⋅r −ωt (at boundary)

17/34 Changhee Lee, SNU, Korea Optoelectronics Reflection and refraction at a plane boundary EE 430.423.001 2016. 2nd Semester

k sinθ = k'sinθ '= k"sinφ law of reflection θ = θ ' ( k = k', two waves traveling in the same medium)

18/34 Changhee Lee, SNU, Korea Optoelectronics Reflection and refraction at a plane boundary EE 430.423.001 2016. 2nd Semester

n n = 2 : relative index of refraction n1

k" ω / u" c / u" n2 sinθ = = = = n = Snell’s law of refraction k ω / u c / u n1 sinφ 19/34 Changhee Lee, SNU, Korea Optoelectronics Amplitudes of reflected and refracted waves EE 430.423.001 2016. 2nd Semester  1   H = k × E (incident) µω  1   H'= k' × E' (reflected) µω  1   H"= k" × E" (transmitted) µω TE (transverse electric) polarization (s-polarization): incident E // boundary plane, i.e. E ⊥ plane of incidence TM (transverse magnetic) polarization (p-polarization): incident H // boundary plane, i.e. H ⊥ plane of incidence s from senkrecht (German for perpendicular).

20/34 Changhee Lee, SNU, Korea Optoelectronics Amplitudes of reflected and refracted waves EE 430.423.001 nd TE polarization 2016. 2 Semester E + E'= E" − H cosθ + H 'cosθ = −H"cosφ  1   H = k × E µω − kE cosθ + k' E'cosθ = −k"E"cosφ

TM polarization H − H '= H" kE − k' E'= k"E" E cosθ + E'cosθ = E"cosφ

 E'  E' = = Coefficients of reflection rs   rp    E TE  E TM  E"   E"  = = Coefficients of transmission ts   t p    E TE  E TM 21/34 Changhee Lee, SNU, Korea Optoelectronics Amplitudes of reflected and refracted waves EE 430.423.001 2016. 2nd Semester TE polarization TM polarization (1) E + E'= E" (1) H − H '= H" (2) − H cosθ + H 'cosθ = −H"cosφ  1   kE − k' E'= k"E" ( H = k × E)  1   µω H = k × E µω k = k', k"/k = n − kE cosθ + k' E'cosθ = −k"E"cosφ (1)' E − E' = nE" k = k', k"/k = n (2) E cosθ + E'cosθ = E"cosφ cosφ cosφ (2)' − E + E'= −E"n (2)' E + E'= E" cosθ cosθ cosφ cosφ (1) − (2)' : 2E = E"(1+ n ) (1)'+(2)' : 2E = E"(n + ) cosθ cosθ  E"  2cosθ  E"  2cosθ = = = = ts   t p    E TE cosθ + ncosφ  E TM ncosθ + cosφ  E'  E"  cosθ − ncosφ  E'  E"  − ncosθ + cosφ = = − = = = − = rs     1 rp   1 n   E TE  E TE cosθ + ncosφ  E TM  E TM ncosθ + cosφ 22/34 Changhee Lee, SNU, Korea Optoelectronics Amplitudes of reflected and refracted waves EE 430.423.001 2016. 2nd Semester For normal incidence θ = φ = 0. 1− n ∴ r = r = s p 1+ n

If n>1, r=E’/E<0: The phase of reflected wave is changed by180o relative to the incident wave. Thus, 180o phase change occurs when light is partially reflected upon entering a dense medium

(n2>n1), for example, from air to glass.

23/34 Changhee Lee, SNU, Korea Optoelectronics Fresnel’s equations EE 430.423.001 2016. 2nd Semester TE polarization Snell’s law of refraction cosθ − ncosφ 2cosθ sinθ sin(θ −φ) rs = ts = n = r = − cosθ + ncosφ cosθ + ncosφ sinφ s sin(θ +φ) θ φ θ − φ θ φ − θ φ θ −φ 2cos sin cos ncos cos sin sin cos sin( ) ts = rs = = = − θ +φ cosθ + ncosφ cosθ sinφ + sinθ cosφ sin(θ +φ) sin( ) tan(θ −φ) 2cosθ 2cosθ sinφ 2cosθ sinφ rp = − ts = = = tan(θ +φ) cosθ + ncosφ cosθ sinφ + sinθ cosφ sin(θ +φ) 2cosθ sinφ t = TM polarization p sin(θ +φ)cos(θ −φ)

tan(θ −φ) cos(θ +φ)sin(θ −φ) (cosθ cosφ − sinθ sinφ)(sinθ cosφ − cosθ sinφ) − ncosθ + cosφ 2cosθ = = tan(θ +φ) cos(θ −φ)sin(θ +φ) (cosθ cosφ + sinθ sinφ)(sinθ cosφ + cosθ sinφ) rp = ts = sinθ cosθ (sin 2 φ + cos2 φ) − (sin 2 θ + cos2 θ )sinφ cosφ sinθ cosθ − sinφ cosφ ncosθ + cosφ ncosθ + cosφ = = sinθ cosθ (sin 2 φ + cos2 φ) + (sin 2 θ + cos2 θ )sinφ cosφ sinθ cosθ + sinφ cosφ − ncosθ + cosφ − sinθ cosθ + sinφ cosφ tan(θ −φ) r = = = − p ncosθ + cosφ sinθ cosθ + sinφ cosφ tan(θ +φ) 2cosθ 2cosθ sinφ 2cosθ sinφ t = = = p ncosθ + cosφ sinθ cosθ + sinφ cosφ sin(θ +φ)cos(θ −φ)

24/34 Changhee Lee, SNU, Korea Optoelectronics Fresnel’s equations EE 430.423.001 2016. 2nd Semester TE polarization cosθ − ncosφ cosθ − n2 − sin 2 θ rs = = cosθ + ncosφ cosθ + n2 − sin 2 θ

sinθ 2 1 2 Snell’s law of refraction n = cosφ = 1− sin φ = 1− sin θ sinφ n2

TM polarization 1 − ncosθ + 1− sin 2 θ − ncosθ + cosφ 2 − n2 cosθ + n2 − sin 2 θ r = = n = p 2 2 2 ncosθ + cosφ 1 2 θ + − θ ncosθ + 1− sin θ n cos n sin n2

2 2 2 E' 2 E' Reflectance Rs = rs = Rp = rp = E TE E TM

25/34 Changhee Lee, SNU, Korea Optoelectronics Fresnel’s equations EE 430.423.001 2016. 2nd Semester

For normal incidence (θ = 0) 2 1− n R = R = s p 1+ n

For grazing incidence (θ ~ 90o )

Rs ≈ Rp ≈1, independent of n

26/34 Changhee Lee, SNU, Korea Optoelectronics External and internal reflection EE 430.423.001 2016. 2nd Semester • External reflection if n>1 (n2>n1): The incident wave approaches the boundary from the side with the smaller index of refraction. The amplitude ratios are real for all values of θ.

• Internal reflection if n<1 (n2 n. 1 critical angle θ = sin −1( ) critical n

https://en.wikipedia.org/wiki/Total_internal_reflection

When the internal angle of incidence θ ≥ θ critical, the amplitude ratio E’/E becomes complex and R=1 (total internal reflection). cosθ − i sin 2 θ − n2 − n2 cosθ + i sin 2 θ − n2 rs = rp = cosθ + i sin 2 θ − n2 n2 cosθ + i sin 2 θ − n2

27/34 Changhee Lee, SNU, Korea Optoelectronics Fiber optics and optical waveguides EE 430.423.001 2016. 2nd Semester

Acceptance angle = maximum semiangle of the vertex of a cone of rays entering the fiber from one end

−1 2 2 α = sin n1 − n2 (Prob. 2.20)

28/34 Changhee Lee, SNU, Korea Optoelectronics Brewster angle EE 430.423.001 2016. 2nd Semester 2 2 2 − n cosθ + n − sin θ −1 rp = = 0 if θBrewster = tan n n2 cosθ + n2 − sin 2 θ

2 2 2 4 2 2 2 n cosθB = n − sin θB , n (1− sin θB ) = n − sin θB n4 − n2 n2 (n2 −1) n2 tan 2 θ 2 θ = = = = B sin B 4 2 2 2 2 n −1 (n −1)(n +1) n +1 1+ tan θB −1 ∴ tanθB = n, θB = tan n

An arbitrary beam incident at Brewster's angle yields a reflected beam that is totally polarized parallel to the interface. That's why Polaroid glasses, with the transmission axis vertical, help to reduce glare off a horizontal surface. https://en.wikipedia.org/wiki/Brewster%27s_angle

For glass of n=1.5

−1 o θBrewster = tan 1.5 ≈ 57 for external reflection 1 θ = tan −1 ≈ 33o for internal reflection Brewster 1.5 Light is entirely transmitted (perfect window) 29/34 Changhee Lee, SNU, Korea Optoelectronics Evanescent wave in total reflection EE 430.423.001 2016. 2nd Semester

In spite of the fact that the incident energy is totally reflected when the angle of incidence θ ≥ θ critical, there is still an electromagnetic wave field (evanescent wave) in the region beyond the boundary.     i(k"⋅r-ωt) Etrans = E"e   sin 2 θ k" ⋅r = k"xsinφ − k"y cosφ = k"xsinφ − ik"y −1 n2

  2 −α y ω sin θ k"sinθ E = E"e ei(k1x- t) , α = k" −1, k = trans n2 1 n

30/34 Changhee Lee, SNU, Korea Optoelectronics Phase change in total internal reflection EE 430.423.001 2016. 2nd Semester

For total internal reflection, θ ≥ θ critical, the amplitude ratio E’/E becomes complex and R=1. 2 2 −iα cosθ − i sin θ − n − δ ae + α = = i s = δ = α i = θ + 2 θ − 2 rs e +iα , s 2 , ae cos i sin n , cosθ + i sin 2 θ − n2 ae δ sin 2 θ − n2 tan s = tanα = 2 cosθ

2 2 2 −iβ − n cosθ + i sin θ − n −iδ be = = − p = − δ = β +iβ = 2 θ + 2 θ − 2 rp e +iβ , p 2 , be n cos i sin n n2 cosθ + i sin 2 θ − n2 be 2 2 δ p sin θ − n tan = tan β = 2 n2 cosθ ∆ tan β − tanα ∆ = δ −δ , tan = tan(β −α) = p s 2 1+ tan β tanα sin 2 θ − n2 sin 2 θ − n2 2 − = n cosθ cosθ sin 2 θ − n2 sin 2 θ − n2 1+ n2 cosθ cosθ

∆ cosθ sin 2 θ − n2 tan = 2 sin 2 θ (Prob. 2.21) 31/34 Changhee Lee, SNU, Korea Optoelectronics Fresnel’s rhomb EE 430.423.001 2016. 2nd Semester A Fresnel rhomb is designed in 1817 by Augustin-Jean Fresnel for converting linearly polarized light into circularly polarized light. While the device performs the same function as a quarter wave plate, the Fresnel rhomb does not depend on a material's birefringence. Rather, the relative phase shift between the s and p polarizations is a result of total internal reflection, two of which each contribute a nominal retardance of π/4=45°. https://en.wikipedia.org/wiki/Fresnel_rhomb

32/34 Changhee Lee, SNU, Korea Optoelectronics Reflection matrix EE 430.423.001 2016. 2nd Semester

− rp 0 − rp 0 A A' reflection matrix   Jones vector for the reflected light     =    0 rs   0 rs  B B'

t p 0 t p 0 A A" transmission matrix   Jones vector for the transmitted light     =    0 rs   0 ts  B B"

− (1− n) /(1+ n) 0  1− n −1 0 reflection matrix for normal incidence   = ( )   0 (1− n) /(1+ n) 1+ n  0 1

Jones vector of the reflected light if the incident light is right circularly polarized 1− n −1 0 1  1− n −1 n −1 1    =   =   right circularly polarized ↔ left circularly polarized 1+ n  0 1− i 1+ n − i n +1 i

− δ − δ i p = i s Consider the case of total internal reflection. rp = −e rs e

−iδ p −iδ p A' e 0 A Ae  −iδ  A  = = = p    −iδ    −iδ  e  i∆  The reflected light is elliptically polarized. B'  0 e s B  Be s  Be  33/34 Changhee Lee, SNU, Korea Optoelectronics EE 430.423.001 2016. 2nd Semester

Homework set #2.

Solve Problems 2.4, 2.8, 2.9, 2.11, 2.12, 2.13, 2.17, 2.20, 2.22, 2.23.

Due date: 2016. 10. 6 (목)

34/34 Changhee Lee, SNU, Korea