• EM Waves at boundaries
• Fresnel Equations: Reflection and Transmission Coefficients
• Brewster’s Angle
• Total Internal Reflection (TIR) • Evanescent Waves
• The Complex Refractive Index • Reflection from Metals We will derive the Fresnel equations
rre: flection coefficient
22 E cosθθ−−n sin E Er r ==r θ TE 22 θr E cosθθ+−n sin 222 Er −+−nncosθ sin θ n1 rTM == E nn222cosθ +− sin θ n2
t : transmission coefficient θt Et 2cosθ E tTE == θθ t E cos+−n22 sin ntransmitted n2 Et 2cosn n ≡ = tTM == θ nincident n1 E n222cosθ +−n sin θ EM Waves at an Interface r r r Ei Eoi Er r r r r r Incident beam:exp E=⋅−E ⎡⎤ i(k rω t) knk= r r iiioi ⎣⎦i 10 k k r i r E r r r r or ⎡ r ⎤ knk= n1 Reflected beam:exp Err=⋅−Eor i()kr r t r 10 ⎣ ω ⎦ n r r 2 Eot r r r()knk= Transmitted beam:exp E=⋅−E ⎡ ik rr t ⎤ t 20 r t ot ⎣ t t ⎦ k ω t r E TE mode TM mode t
r n 2 n2
θi
n1 n1
Note the definition of the positive E-field directions in both cases. EM Waves at an Interface
r r r Incident beam: EE=⋅−exp ⎡⎤ ik rr ω t ioi⎣⎦( i i) rr r Reflected beam: EE=⋅−exp ⎡ ik rr t ⎤ ror⎣ ()r rω ⎦ rr r() Transmitted beam: E =⋅−Eirtexp ⎡ k r ⎤ tot⎣ t t⎦ ω At the boundary between the two media(), the x− y plane all waves must exist simultaneously, andthe tangential component must be equal on both sides of the interface . r r E nˆ r Therefore, for all time t and for all boundary points rr on thein terface, Ei oi Er r k r r i kr Eor n rrr 1 r rr nE) ×+ nE) ×=× n) E n irt 2 Eot rrrrr r r nE))×⋅−+×exp⎡ ikrrrrωωω t⎤⎡⎤⎡⎤ nE exp ikr⋅− t = nE ) × exp ik ⋅−rt k oi⎣ () i i⎦⎣⎦⎣⎦ or r r ot ( t t )t () r Et Assuming that the wave amplitudes are constant, the only way that this can be true over the entire interface and for all t is if : rrrrr r ⇒⋅−=⋅−=⋅− ()()()kriiωωω t kr rr t kr tt t : Phase matching at the boundary! EM Waves at an Interface
r r E E Phase matching condition: i r r knk= rr r r r i 10 rrr k k ()()()krii⋅−ω t = kr rr ⋅−ωω t = kr tt ⋅− t i r r n knk= 1 rr r 10 n2 r r knk= At r=0, this results in r t 20 kt ωωωirttt==t
⇒ it==r r nk ωωω Et 20 (Frequency does not change at the boundary!)
nk10 At t=0, this results in r rrrrrr kr ⇒ krirt⋅=kk⋅⋅rr= rr x (Phases on the boundary does not change!) r ki r ⇒⋅k rr = constant r irt,, kt r r → the equation for a plane perpendicular to kir,,t and r. Normal r rr ⇒ kir, k, and kt are coplanar in the plane of incidence. EM Waves at an Interface r r Ei Er θι θr At t =0, r r ki kr rrrrrr n1 kri ⋅= krr ⋅=krt ⋅=constant x n2 r Considering the relation for the incident and reflected beams, kt θt rrrr krir⋅= kr ⋅ ⇒ kr iirrsinθθ = kr sin r Et ω Since the incident and reflected beams are in the same medium,
ni kk== ⇒sin =sin ⇒ = : law of reflection nk20 ir c i θθr ir
nk10 θθ r Considering the relation for the incident and transmitted beams, k θ r rr r r kr⋅=rr kr ⋅ ⇒ krsinθθ = kr sin r it iit t r x k ω θi i θ But the incident and transmitted beams are in different media, t ω r nn k kk==⇒itnnsin= sin : law of refraction t i cct iitt θ Normal θ Development of the Fresnel Equations
From Maxwell', s EM field theory we have the boundary conditions at the interface TE-case for the TEcase:
EEir+= E t
BBiirrttcosθθ−= cosco B θs
The above conditions imply that the tangential rr components of both E and B are equal on both sides oftheinterface. We have also assumedμμ thatit≅≅ μ 0 , as is true for TM-case most dielectric materials.
For the TM mode :
EEiirrttcosθθ+= cos Ecos θ
−+BirBB =− t Development of the Fresnel Equations
⎛⎞c nE Recall that E==v B⎜⎟ B ⇒B = TE-case ⎝⎠n c n2
Let n1 = refractive index of incident medium n2 = refractive index of refracting medium n1 For the TE mode :
EEir+ = E t θθθ TM-case nE11i cosirrtt− nE cos = n2 E cos n2
For the TM mode :
n1 EEiircosθθ+ cosrt= E cos t
−nE1 i + n1 Ert=−θnE2 Development of the Fresnel Equations
Eliminating Et from each set of equations TE-case and solving for the reflection coefficient we obtain : n2
Er cosθit− n cos TE case: rTE == Eniicosθ + cosθ t θ n1
Er −+ncosit cos TM case: rTM == θθ Eiitncos+ cos n θθ where n = 2 n 1 TM-case
n2 We know that θθ sinit= n sin
2 θ sin θ n1 n cos =−nn1 sin222θ = 1 −i =− n sin θ ttn2 i Now we have derived the Fresnel Equations
Substituting we obtain the Fresnel equations for reflection coefficients r : cosθθ−−n22 sin Er ii TE-case TE case : rTE == E 22 n2 i cosiiθθ+−n sin 222 n2 Er −+−nncosii sin n ≡ TM case : rTM == E 222θθn1 i n cosii+−n sin θθ n1 For the transmission coefficiet n t : E 2cosθ TEcase: t ==tiθθ TE E 22 i cosii+−n sin En2cos TM case : t ==tiθ TM-case TM 222 Ei n cos+−θθn sin ii n2
TE: tTE= r TE + 1 These just mean the boundary conditions. TM: ntTM=− 1 r TM For the TE case: Eir+ E= E t n1 For the TM mode: − Bir+=− B B t Power : Reflectance (R) and Transmittance (T)
The quantities r and t are ratios of electric field amplitudes. The ratios R and T are the ratios of reflected and transmitted powers, respectively, to the incident power : P P RT==r t PPii
From conservation of energy : A
PPPirt=+ ⇒1 =+RT
We can express the power in each of the fields in terms of the product of an irradiance and area : θ PIAPIA== PIA = iiirrr ttt θ 2 ⇒ IAii= I rrA + IA tt ⎛⎞ Ioutcos out nEout outcosθ out IA cosθθθ= IA cos + IAcos Power_ ratio = =⎜2 ⎟ iirrtt I cos ⎜⎟ in in ⎝⎠nEin incos inθ Iiirrtcosθθθ= II cos+ cos t
1 1 11 ButInc= E 2 ⇒ n cE 2 cosθε=+ncE22 cos θ ε ncE θ cos 2 0 ε0 2 10 0i i 2210 0rr 20 0 tt EnE22cos E 2⎛⎞ cos E 2 ⇒= 1 020rttr +ε = 0 +nRT tt0 =+ 2 22θθ 2⎜⎟ 2 R = rr* =θr E010iiiinEcos E 0⎝⎠cos iiE 0 θθ 22 θ EE⎛⎞cos ⎛⎞ cos ⎛ cos ⎞ ⎛ cosθ ⎞ 2 ⇒ RrTn= 00rttt==2 θθ =n t 2 T = ⎜n t ⎟tt* = ⎜n t ⎟ t 22⎜⎟ ⎜⎟ ⎜ ⎟ ⎜ ⎟ E00iicos E i cos i cos cosθ ⎝⎝⎠⎠θ ⎝ i ⎠ ⎝ i ⎠
θ θθ23-2. External and Internal Reflection
θ t tTE, TM > 0 θ cos−−n 22 sin −+−nn22cosθ sin 2 θ r = i i r = ii TE 2 2 TM 222 cossinθ i +−n i n cosi +−n sin θ i rTE, TM > 0 rTM
External Reflection [nnn= 21/1> ]
⇒ nn> rTE 21 rTE, TM < 0 22 ⇒ nnn= 21 />⇒ 1() n − sinθ ≥ 0 n=1.50
⇒ rTE, TM are always real
⇒> If rTE, TM 0 then there are no phase changes after reflection.
o ⇒< If rTE, TM 0 then there are always π ( = 180 ) phase changes. iπ →= rrTE,, TM− TETM =e r TE , T M
Note for theTMcase: Brewster’s angle (or, polarizing angle) −1 ⇒== rwhenTM()θθ p 0p = tan n (No reflection of TM mode)
θ θθ
Internal Reflection [nnn= 21/1< ] θ rTE, TM > 0
22 22θ 2 cosi −−n sin i −+−nncosθ ii sin θ rTE = rTM = cossinθ +−n 2 2 n 222cos+−n sin θ i i i i TIR region
r < 0 nn1 > 2 ⇒=nnn2 /11 < TE, TM ⇒−()()no22 sinθθ > 0,rn ,22 − sin < 0
22 ⇒−> If (nr sinθ ) 0, TE, TM are always real
→> If rTE, TM 0 then there are no phase changes after reflection.
o →< If rTE, TM 0 then there are ( = 180 ) phase changes.
⇒−= If (nr22 sinθ ) 0, =1 TE, TM π θ critical angle →== sinc nnn (21 / ) −1 Note Brewster's angle(θ p = tan n)
22 for the TM case :0rTM = ⇒−< If (nr sinθ ) 0, TE, TM =1, BUT rTE, TM are complex! → r =1 TE, TM Total internal reflection (TIR) when θ > θc iiφφ →= rreeTE,, TM TE TM = →+ (-φπ ~ ) π phase change may occur after reflection Derivation of Brewster’s Angle
Brewster's angleθ p (): for polarizing angle θc
θ 222 −+−nnθθcospp sin r ()==0 θp θ TM p 222 p nncosθ pp+ − sin R 42 2 2 θ ⇒=−nncosθθpp sin nn42 cosθθ−+ 2 sin 2 pp external 2222⎡⎤ internal reflection =− (nnc 1)⎣ ospp − sin0⎦ = TE TM reflection θ θ −1 θ ⇒ p = tan n
For n == 1.50,p 56.31° θ
Brewster ‘s angle : tanθ p = n : n >1 or n <1 Æ External & Internal reflections, but TM-polarization only
Critical angle : sinθc = n : n < 1 Æ TE & TM polarizations, but Internal reflection only Total Internal Reflection (TIR)
n Internal reflection : n =<2 1 R n1 R = 1 θc
−1 Forθθ≥ c = sin n,called total internal reflection( TIR ), ⇒ rRr===1 and r*1 for both (TE and TM) cases. ⇒ risacomplexnumber internal reflection
E cosθθ−−i sin22n r = r = ii TE E 22 i cosiiθθ+−i sin n
E −+nn222cosi sin − r ==r ii TM E 222θθ r i n cosii+−i sin n θθ Complex value 23-3. Phase changes on reflection
External reflection tTE, TM > 0 Phase shift after External Reflection
rTE, TM > 0 rTM rTE, TM is always a real number for external reflection ,
then the phase shift is0°> for rTE, TM 0,
rTE, TM < 0 and the phase shift is 180(°=π )0for rTE, TM < . rTE n=1.50
External Reflection External Reflection
TE TM
For TE case, π phase shift for all incident angles For TM case, π phase shift for θ < θp
No phase shift for θ > θp Phase shift after Internal Reflection Internal reflection
−1 ⇒> rnTE0 for θθ <= c sin Complex value ⇒> rTE is complex in TIR region where θ θ c
iiφφTE TE →= rreeTE TE = In TIR region
−1 ⇒> rnTM0 for θθ <= p tan :θ > θc ⇒< rTM0 for θθθ p << c
iφTM →=−= rrerTM TM TM →=φ TM π
⇒> rTM is complex in TIR region where θ θ c
iiφφTM TM →= rreeTM TM =
TIR TIR
For TM case, no phase shift for θ < θp For TE case, no phase shift for θ < θc π phase shift for θp < θ < θc φTE(θ) phase shift for θ > θc φTM(θ) phase shift for θ > θc Phase shifts on total Internal Reflection for both TE- and TM-cases θθ
When≥ c (TIRcase ) then riscomplex and for both the TE and TM cases has the form :
−iα aib−−cosαα αi sin e−ii2 sin b ree== ==+i =⇒tan == = − 2 aib++cosα i sin eaα αφ cos αα φ is the phase shift on total internal reflection ( TIR ). θθ
22 E cosii−−θθin sin TE case: r ==r α TE θθE 22 i cosii+−in sin φα abn==−cos sin 22 Internal reflection iiα TIR φ 2 2 (Complex r ) ⎛⎞TE sin i − n ⇒=−=tanφ tan ⎜ ⎟ θ ⎝⎠2cos θ ⎛⎞sin22− n θ =−2tan−1 ⎜⎟i :θ > θ TE ⎜⎟cos icθ ⎝⎠i
Asimiφπlar analysis for the TM case gives :
⎛⎞sin22θ − n =−2tan−1 ⎜⎟i :θ > θ TM ⎜⎟n2 cos ic ⎝⎠θi Therefore, r TE , TM after TIR is ……….. Internal reflection
rTE, TM
For TIR case () θincident > θc
E cosθθ−−i sin22n r ==r ii Complex value TE E 22 i cosiiθθ+−i sin n E −+nn222cosi sin − r ==r ii TM E 222θθ i nncosii+−i sin θθ φπ
⎛⎞sin22θ − n φTE, TM =−2tan−1 ⎜⎟i TM ⎜⎟n2 cos φ ⎝⎠θ ⎛⎞sin22−θn =−2tan−1 ⎜⎟i TE ⎜⎟cos ⎝⎠θ Summary of Phase Shifts on Internal Reflection
⎧ Internal reflection o ' TIR ⎪ 0 θθ < p (Complex r ) ⎪ ⎪ ⎪ o ' φTM ==⎨πθθθ( 180 ) p < < c ⎪ ⎪ ⎛⎞sin22− n π − 2tan−1 ⎜⎟i θ θθ < ⎪ ⎜⎟n2 cos c φ ⎩⎪ ⎝⎠ θ ⎧ o 0 θ θ <θc ⎪ ⎪ ⎛⎞22 TE = ⎨ sin θ− n −2tan−1 ⎜⎟i θ >θ ⎪ ⎜⎟cos c ⎩⎪ ⎝⎠ φφ φ
o ⎧= 0 θ <θp Δφ ⎪ φ Δ= − TM TM TE ⎨=<π θθθpc < ⎪ o ⎩ ><0 θc θ φTE Fresnel Rhomb
3π Note φφ−= θ near =53o when n =1.5 TM TE4 i → After two consequentive TIRs, Δφ φTM 3 →−= TMφφ TE 2 π φ →Δ = − = TE TMφφ TE φ2 →− Quarter wave retarderπ
Linearly polarized light (45o)
Circularly Polarized light Quarter-wave retardation after TIR
φ Δφ π TM Note φφ−= θ near =69o when n =??? TM TE2 i →Δ = − = φφTM φ TE 2 φ →− Quarter wave retarderπ TE
Linearly Circularly polarized light Polarized (45o) light
n 23-5. Evanescent Waves at an Interface
rr r Incident beam:exp E=⋅− E⎡⎤ i k rω t ioii⎣⎦( i) rr()r Reflected beam:exp E=⋅− E⎡⎤ i k r t rorr()⎣⎦ rω rr r Transmitted beam:exp E=⋅− E ⎡⎤ik rtr tott⎣⎦ t () ω
For the transmitted beam : r ω EE=⋅− exp ⎡⎤ ikrr t tott⎣⎦ t θ r r ) ))) kttt⋅=rk()() sin xkθ+⋅+ttcosθ zxxzz () =+kxtt sin z cos tθ θ
2 2 θ sin But, cos= 1−− sin= 1 i ttn θ θ
Whensini > n ()total internal reflection ,:then θ sin2 θ cos =−ii 1 ⇒ a purely imaginary number t n Evanescent Waves at an Interface
For the transmitted beam with an TIR condition (sin i > n), we can write the phase factor as :
⎡⎤⎛⎞kxttsinθ 2 E =−−Eiexpω t exp α z r ⎛⎞sinθ sin θ tt0 ⎢⎥⎜⎟() kr⋅=r k⎜⎟ x t +iz i −1 ⎣⎦⎝⎠n tt⎜⎟n n ⎝⎠θ
Defining the coefficient : α z 22α sini θθ2 sin i = kt −−11= n2 nnt π n > n h λ 1 2 x We can write the transmitted wavea s :
⎡⎤⎛⎞kxttsinθ Et = Ei0t exp ⎢⎥⎜⎟− texp()− z n ⎣⎦⎝⎠n ω 1 α The evanescent wave amplitude will decay rapidly as it penetrates into the lower refractive index medium. Note that the incident and reflection waves form a standing wave in x direction ⎛⎞1 1α λ Penetration depth: EEtot=⇒⎜ ⎟ h == ⎝⎠e sin2 21π iθ− n2 Frustrated TIR
d
Tp = fraction of intensity transmitted across gap
n1=n2=1.517 1.65
Zhu et al., “Variable Transmission Output Coupler and Tuner for Ring Laser Systems,” Appl. Opt. 24, 3610-3614 (1985). d/λ Frustrated Total Internal Reflectance
Pellin-Broca prism
d
Zhu et al., “Variable Transmission Output Coupler and Tuner for Ring Laser Systems,” Appl. Opt. 24, 3610-3614 (1985). d= 1 ~ λ: changing the reflectance Rotation: changing the wavelength resonant at θB 23-6. Complex Refractive Index σ
σ ⎛ σ ⎞ For a material with conductivity ( ) : ni% =+1 ⎜⎟ = ninRI + εω ⎝⎠εω0
222⎛⎞ ni% =+ 1 ⎜⎟ = nninRI − + 2 RIn ⎝⎠0 Solving for the real and imaginary components we obtain :
2 2 σ nnRI−=12σσ nn RI = ⇒nR = εω0 2nI 0 ωεω εω 22 ⎛⎞ ⎛⎞σ ⇒−=⇒−−=nnn24210 ⎜⎟III ⎜⎟ε ⎝⎠22nI 00 ⎝⎠ω
From the quadratic solution we obtain :
2 2 ⎛⎞σ ⎛⎞ 114 114 ±+⎜ ⎟ ε ++⎜⎟ 2 ⎝ 2 0 ⎠ ω 2 ⎝ 2 0 ⎠ n = ⇒ n = I 2 I 2
We need to take the positive root because nI is a real number. σ εω Complex Refractive Index
Substituting our expression for the complex refractive index back into our expression for the electric field we obtain r r r EE=⋅−exp ⎡⎤ ikrtr 0 ⎣ ()ω ⎦ r ⎧⎫⎡⎤ =+⋅−Eininurtexp ˆ r 0 ⎨⎬⎢⎥()RkωI () ⎩ ⎣⎦c ⎭ r ⎧⎫⎡⎤n ⎡⎤n R ˆ r ω I ˆ r = E0 exp ⎨⎬iurtω ⎢⎥()k ⋅− exp ⎢⎥− ()urk ⋅ ⎩ ⎣⎦c ⎭ ⎣⎦c
The first exponential term is oscillatoryω.
The EM wave propagates with a velocity of ncR / .
The second exponential has a real argument (absorbed) . Complex Refractive Index
rr ⎧⎫⎡⎤n ⎡ n ⎤ R ˆ r I ˆ r EE= 0 exp ⎨⎬iurω ⎢⎥()k ⋅−t exp ⎢− ()uk ⋅r ⎥ ⎩ ⎣⎦c ⎭ ⎣ c ⎦
The second term leads to absorption of ωthe beam in metals due to inducing a current in the medium. This causes the irradiance to decrease as the wave propagates through the medium.
r rr** r r ⎡⎤2nurIkω ()ˆ ⋅ IEEEE≡=00exp ⎢⎥ − ⎣⎦c ω α ⎡⎤2nu()ˆ ⋅rr Ik ˆ r II=−0 exp ⎢⎥=−⋅Iur0 exp ⎡⎣⎦()k ⎤ ⎣⎦c
α 24nnωπ The absorption coefficient is defined : ==II c λ 23-7. Reflection from Metals
Reflection from metals is analyzed by substituting the complex refractive index n% intheFresnelequatiosn : E cosθθ−−n22 sin TE case: r ==r ii% TE E 22 i cosiiθθ+−n% sin
Reflectance E −+−n222cosn sin TM caser: ==r % ii% TM E 222θθ i n% cosii+−n% sin θθ
Substitutingnni% = RI+ n we obtain :
22 2 E cosθθiRIi−−−+()nn sin inn() 2 RI TE case : r r == θ E 22 2 i i cosiRIiθθ+−−+()nn sin inn() 2 RI
⎡⎤22 22 2 E −−+()nnRI inn()2cos RI i +() nn RI −− sin2 i + inn() RI TM case : r ==r ⎣⎦ 22 22 2 Ei ⎡⎤nn−+ inn2cos + nn −− sin2 +inn ⎣()RI() RI⎦ i() RIiRI() θθ
θθ Reflection from Metals at normal incidence (θi=0)
At normal incidence,0:θ = ° At normal incidence i (from Hecht, page 113) cosθθ−−n22 sin 1− n r ==ii% % TE 221+ n cosiiθθ+−n% sin % −+−n222cosn sin 1− n r ==% ii% % TM 222θθ1+ n n% cosii+−n% sin % θθ
1 −−()ninRI ∴ r = () visible 1 +−niRIn λ Thepower reflectance R is given by
Rrr= * ⎡⎤⎡⎤11−−()nin −+() nin ⎛⎞12−++nnn22 ==RI RI RRI ⎢⎥⎢⎥()()⎜⎟22 ⎣⎣⎦11+−ninRI⎦ ++ nin RI⎝⎠1 +2 n RR ++ n nI 2 ()nn−+1 2 R = RI ()2 2 nnRI+1 +