On Total Internal Reflection?

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ECEECE 318318 FundamentalsFundamentals ofof OpticsOptics LectureLecture 1616 && 1717

Light as Waves VI

• Total Internal Reflection

* Topics, questions, or homework problems with an asterisk (*) are at a difficulty level or having a scope beyond the expectation of this course. Only qualitative understanding is required. TotalTotal InternalInternal ReflectionReflection

So far we have considered the cases where there always exists a propagating transmitted wave. Snell’s law is satisfied with a transmitted

angle θt between 0 and π/2.

Now we are going to consider a special case called Total Internal Reflection (TIR). Two conditions must be satisfied for TIR to occur:

(1) ni>nt, that is, the incident light wave is in the denser medium (2) θi >θc , that is, the incident angle is greater than the critical angle

In what follows, we’ll show that, upon TIR, both TE and TM polarizations are 100% reflected, however, they will experience different phase shifts.

Q16-1: Next time you go for a swim, think about how you could verify first hand the zero transmission phenomenon when total internal reflection occurs at the water-air boundary.

2 © Li Qian Lecture 16 &17 CriticalCritical AngleAngle

Critical angle refers to the incident angle when the refracted angle is exactly π/2. This is a special case that occurs only for internal reflections.

Snell’s Law: (1) ni sinθi = nt sinθt k θc r ki Therefore: n i  π  k n t ni sinθc = nt sin  (2) t  2  or  n  −1 t  θc ≡ sin   (3)  ni  Above: Solving for θc graphically using k-matching at the boundary.

Example: For a glass-air interface, the critical angle occurs at 41.8o, for a water-air interface, this occurs at 48.8o.

3 © Li Qian Lecture 16 &17 WhatWhat happenshappens toto Snell’sSnell’s LawLaw beyondbeyond thethe criticalcritical angle?angle?

It can be shown that (see Homework #1), even in the case of total internal reflection, Snell’s law still holds, except that the sine of the transmitted angle becomes greater than 1, and the cosine of the transmitted angle becomes imaginary. Though there is no physical interpretation of the “transmitted angle”, it is still useful as a mathematical tool. The Fresnel reflection

formulae (Lecture 14&15, Eq (24) & (26)) still apply if we continue to use the cosθt term, despite its imaginary value.

ni sinθi sinθt = >1 (4) nt

2  n sinθ   i i  cosθt = i   −1  nt  (5)

Q16-2: What’s the origin of Snell’s Law from the Wave Optics point of view?

4 © Li Qian Lecture 16 &17 “Transmitted”“Transmitted” wavewave vectorvector isis complexcomplex

Let’s analyze the wave vector kt of the “transmitted wave”, assuming it exists. In order to satisfy the wave equations (Lecture 9, Eq (27) & (28)), we must have: y ω k = k 2 + k 2 = n (6) t tx ty c t And in order to satisfy the boundary phase- ki θi θr kr ni matching condition, we must have ω x (7) n ktx = kix = ki sinθi = ni sinθi t kt=? c And in order to satisfy both (6) and (7), and

noting that nisinθi > nt, kty must become Above: Boundary conditions cannot be imaginary! satisfied with a real kt . 2 ω ω  n sinθ  2 2 2 2 2 2  i i  kty = kt − ktx = kt − ()ki sinθi = nt − ()ni sinθi = ±i nt   −1 c c  nt  (8) ≡ ±iβ

5 © Li Qian Lecture 16 &17 “Transmitted”“Transmitted” wavewave isis evanescentevanescent

What does an imaginary kty signify and what is the consequence? If we substitute (7) and (8) into our usual wave expression for the transmitted wave, we will have:

βy   ω  Et = Eto exp[]i()ktx x + kty y −ωt +φ = Etoe expi ni sinθi x −ωt +φ  (9)   c  Amplitude exponentially A wave propagating in the x decays with –y. direction with vφ=c/(nisinθi)

Note 1. In deriving (9), only kty=–iβ is taken, since kty=+iβ leads to an unphysical result. Note 2. For finite incident beam size, the x-direction propagation of the evanescent wave is sustained only within the extent of the incident beam. The wave expressed by (9) is a peculiar wave: it differs from the regular plane wave in that its amplitude exponentially decays in the directions orthogonal to its propagation direction. It is called an evanescent wave. Q16-3: Does this evanescent wave have planar wavefronts? Q16-4: In an absorbing medium, the E field also decays exponentially (see Lecture 12&13, Eq(15)). How is the wave propagating in an absorbing medium different from the evanescent wave described here? 6 © Li Qian Lecture 16 &17 PenetrationPenetration DepthsDepths

From Eq (9), we can draw three conclusions regarding the amplitude of the transmitted E field:

(1) E field is finite in the nt medium, that is, there is field penetration into the transmitted medium. The field does not stop abruptly at the interface. (2) E field is constant in the planes parallel to the interface, i.e., for y = constant. (3) E field decays exponentially with the distance away from the interface.

We can define two penetration depths:

(1) The distance from the interface at which the E field decays to its 1/e value Field penetration depth = 1/β (10)

(2) The distance from the interface at which the intensity decays to its 1/e value Intensity penetration depth = 1/2β (11)

Q16-5: Penetration depth is a function of incident angle. At what angle is penetration depth the greatest?

Q16-6: For optical waves, what’s the approximate length scale of the penetration depths for large incident angles? 7 © Li Qian Lecture 16 &17 “Transmitted”“Transmitted” wavewave isis non-transverse*non-transverse*

It can be shown that the evanescent wave is non-transverse, that is, it has an E or an H component in the direction of propagation (x-direction). For example, if the incident polarization is TE, then the transmitted wave can be written as:

βy (12) Et = Etoe exp[i(kix x −ωt +φ)]zˆ

Now if we calculate the transmitted H field from the E field using one of the Maxwell’s Equations: ∂B ∂H ∇ ×E = − = −µ ∂t ∂t (13) We obtain: ∂E ∂H β t = −µ x ⇒ βE = iµωH ⇒ H = −i E eβy exp[]i()k −ωt +φ (14) ∂y ∂t t x x µω to tx ∂E ∂H k − t = −µ y ⇒ − ik E = iµωH ⇒ H = − tx E eβy exp[]i()k −ωt +φ (15) ∂x ∂t tx t y y µω to tx

Eq (14) indicates that the H field has a finite x-component. Note that Hx is π/2 out-of-phase with Ez . This is another peculiar feature of the evanescent wave. (Recall that, for a “normal” propagating wave, E and H are in phase.)

8 © Li Qian Lecture 16 &17 “Transmitted”“Transmitted” wavewave isis notnot transmittingtransmitting energyenergy acrossacross thethe boundary*boundary* We can also show that there is no net energy transport across the boundary. Again, take TE polarization as an example, from (12), (14) & (15), we can write the E and H fields using real representations: βy Et = Etoe cos(ktx x −ωt +φ)zˆ (16)

βy  β ktx  Ht = Etoe  sin()ktx x −ωt +φ xˆ − cos()ktx x −ωt +φ yˆ  (17)  µω µω  Poynting vector of the “transmitted” wave becomes: β k E ×H = E 2 e2βy cos()k x −ωt +φ sin()k x −ωt +φ yˆ + tx E 2 e2βy cos2 ()k x −ωt +φ xˆ (18) t t µω to tx tx µω to tx The time-average of the y-component of the Poynting vector is zero, as shown below. This signifies that, on average, no net energy flow in the y-direction. β E × H ⋅yˆ = E 2 e2βy cos()k x −ωt +φ sin()k x −ωt +φ t t ωµ to tx tx

β = E 2 e2βy sin()2k x − 2ωt + 2φ = 0 2ωµ to tx (19)

Q16-7: How do you reconcile the fact that there is finite field penetration into the nt medium, but no net energy transport? 9 © Li Qian Lecture 16 &17 ComplexComplex FresnelFresnel ReflectionReflection CoefficientsCoefficients

As mentioned before, the Fresnel reflection coefficients can still be applied to the TIR case,

as long as we incorporate the imaginary cosθt term, expressed by Eq (5).

TE-case: TM-case:

n cosθ − in ()n sinθ n 2 −1 − n cosθ + in ()n sinθ n 2 −1 r = i i t i i t (20) r = t i i i i t (23) TE 2 TM 2 ni cosθi + int ()ni sinθi nt −1 nt cosθi + ini ()ni sinθi nt −1

iφTE iφTM ∴ rTE = e (21) ∴ rTM = e (24)

 n ()n sinθ n 2 −1   n ()n sinθ n 2 −1  φ = −2tan −1 t i i t  (22) φ = −2 tan−1 i i i t  +π (25) TE  n cosθ  TM  n cosθ   i i   t i 

Important conclusions: (1) The wave is 100% reflected; (2) The reflected wave is not in phase with the incident wave; (3) The additional phase incurred in the reflected wave is different from TE and TM polarized waves.

10 © Li Qian Lecture 16 &17 PhasePhase ShiftsShifts uponupon TotalTotal InternalInternal ReflectionReflection

The figure below shows φTE, φTM, and the relative phase difference (φTM – φTE) as a function of incident angle for a glass-to-air interface. π

∆φ=φΤΜ−φΤΕ π/2

φΤΜ

0 θi Phase 0 θc π/2

−π/2 φΤΕ

−π

Q16-8: Do the wavefronts of the incident, reflected and transmitted wave match at the boundary upon total internal reflection?

11 © Li Qian Lecture 16 &17 ReflectanceReflectance andand TransmittanceTransmittance

From (21) and (24), rTE and rTM both have a magnitude of unity in the region of Total Internal Reflection (left plot). Therefore, the reflectances of both TE and TM polarizations are unity as well, and the transmittances are zero (right plot).

ni=1.5, nt=1 ni=1.5, nt=1 1 1 RTE=RTM=1 TTM 0.8 0.5 r TE TTE Region of Region of 0.6 Total Total 0 0 θp θc Internal Internal 0.4 Reflection rTM Reflection -0.5 R 0.2 TE RTM Fresnel Reflection Coefficients Fresnel Reflection Transmittance or Reflectance TTE=TTM=0 0 -1 0 π/2 0 π/2 θp θc Incident Angle (θ ) i Incident Angle (θi)

12 © Li Qian Lecture 16 &17 FrustratedFrustrated TotalTotal InternalInternal ReflectionReflection

Even though the evanescent wave does not transport any energy to the nt medium, the finite E field in the nt medium can still polarize the atoms in the medium. If a higher index medium with index nt2 is placed near the ni/nt interface, and as long as the boundary conditions can be satisfied by a propagating wave in the nt2 medium, a propagating transmitted wave will result (see graph below). Consequently, a finite amount of energy will be transmitted to the nt2 medium, and we say that the initial total internal reflection is frustrated. y FTIR has been used in numerous applications, including:

- Prism coupling: coupling light into and out of waveguides; k θ i i θr kr - Optical directional couplers with variable ratios; ni - Coupling light into the “Whisper-Gallery” mode of a high Q resonator nt x

nt2 With multiple dielectric layers separated by air gaps, kt2 one can create a resonant tunneling condition and make highly selective wavelength filters.

13 © Li Qian Lecture 16 &17 Example:Example: FresnelFresnel RhombRhomb

The Fresnel rhomb can convert linear polarization into circular polarization, or vice versa. It is made of a transparent, homogenous, isotropic material, such as glass. Its cross-section is a parallelogram with an apex angle θ, as shown. Incoming light enters the rhomb normal to its input facet, experiences two total internal reflections, and exits normal to its output facet. You are asked to calculate the apex angle. The refractive index of the rhomb is 1.5. If the linearly polarized input has equal TE and TM E components, say, with an amplitude of Eo, then the input i can be expressed by: input output facet E E (uˆ uˆ )exp[]i(k r t ) facet i = o TE + TM ⋅ −ω +φ (26) θ After two total internal reflections, and ignoring the facet Ei reflections, the output wave becomes: θ Eout = Eo (rTE rTEuˆ TE + rTM rTM uˆ TM )exp[i(k ⋅r −ωt +φ)] Eout

i2φTE i2φTM = Eo ()e uˆ TE + e uˆ TM exp[]i()k ⋅r −ωt +φ cross-section θ

view i2()φTM −φTE = Eo ()uˆ TE + e uˆ TM exp[]i()k ⋅r −ωt +φ + 2φTE (27) Eout

14 © Li Qian Lecture 16 &17 Example:Example: FresnelFresnel RRhombhomb (Cont’d)(Cont’d)

In order to convert to circular polarization, one requires:

2(φTM −φTE )= π 2 + mπ (28)

where m is an integer. From the plot shown on Page 11, the most appropriate m value is 1.

Substitute (22) and (25) into (28), and using ni=1.5, nt=1.0, we can solve for θi:

o (29) θ = θi ≈ 54.6

Q16-9: Given a Fresnel rhomb with a proper apex angle, would linearly polarized input always be converted to circularly polarized light at the output? What is the other condition that must be satisfied?

Q16-10: Can one use the Fresnel rhomb to convert circularly polarized light into linearly polarized light? Into elliptically polarized light?

15 © Li Qian Lecture 16 &17 Example:Example: Right-angleRight-angle PrismPrism

front The right-angle glass prism is often used in optical 5o facet 4 experiments as a retro-reflector / delay stage (top figure). A circularly polarized collimated laser beam d y’ cte (bottom graph) enters the prism through the front facet fle at normal incidence. The refractive index of the prism re nt is 1.5. x’ de y ci in z’ z (a) Find the reflectance of the prism. x (b) Calculate the phase difference between the TE and TM polarizations of the output beam. y (c) Plot the time-evolution of the output E field vector Ein using the co-ordinate system specified in the top figure. (d) Can the TE-TM phase difference due to the two x reflections be compensated exactly by another identical prism. Why or why not?

16 © Li Qian Lecture 16 &17 Example:Example: Right-angleRight-angle PrismPrism (Cont’d)(Cont’d)

Solution:

(a) Find the reflectance of the prism (i.e., the ratio of output power to input power)

n=1.5 As shown on the left, the laser beam is refracted and reflected at 4 interfaces: front 45o 1) Normal incidence (air-to-glass) at the front facet. facet 45o Let t1 and T1 denote the Fresnel transmission coefficient and transmittance, respectively. 45o o 2) Internal reflection (glass-to-air) at 45 angle. Let r2 and R2 denote the Fresnel reflection coefficient and reflectance, respectively. 3) Another internal reflection (glass-to-air) at 45o

angle. Use r3 and R3 for notation. 4) Normal incidence (glass-to-air) at the front facet.

Use t4 and T4 for notation.

Using Fresnel Equations, one can easily find

2  n −1 T1 = T4 =1−   = 0.96 (30)  n +1

o For R2 and R3, the incident angle is 45 , and since  1  45o > sin −1  = 41.8o (31)  n 

The incidence angle is greater than the critical angle. Therefore, total internal reflection occurs at these two interfaces. Hence,

R2 = R3 =1 (32) The reflectance of the prism becomes: output power Rprism ≡ = T1R2 R3T4 = 92.2% (33) input power

(b) Calculate the phase difference between the TE and TM polarizations of the output beam. Using the co-ordinate system specified, we can write the expression for the incident polarization: π  i  E = E xˆ + e 2 yˆ  (34) in o     Note we have omitted the exp [ i ( kz − ω t + φ ) ] component in (34), as it is not relevant to our polarization analysis. In fact, (34) is simply the Jones vector for Ein. The output polarization can be expressed as: π  i  E = E t r r t xˆ'+t r r t e 2 yˆ' out o  1_TM 2_TM 3_TM 4_TM 1_TE 2_TE 3_TE 4_TE  (35)   For normal incidence, there is no polarization dependence, and therefore, 2 t = t ≡ t = (36) 1_TE 1_TM 1 n +1 2n t = t ≡ t = 4_TE 4_TM 4 n +1 (37)

Note that both t1 and t4 are real and positive, which means no phase change after transmission. 19 © Li Qian Lecture 16 &17 Example:Example: Right-angleRight-angle PrismPrism (Cont’d)(Cont’d)

For the two total internal reflections, however, there are phase changes, and furthermore, the TE and TM components experience different phase changes.

iφTE r2_TE = r3_TE = e (38)

iφTM r2_TM = r3_TM = e (39)

where  o 2  −1 (nsin 45 ) −1  o φTE = −2 tan = −0.6435rad = −36.9 (40)  ncos 45o   

 o 2  −1 n (nsin 45 ) −1  o φ = −2 tan +π =1.85rad =106.3 (41) TM  cos 45o    Substitute (36)-(39) into (35), we get the output polarization expression:

 π   i 2φTE +   E = E t t ei2φTM xˆ'+e  2 yˆ' out o 1 4   (42)  

Therefore, the phase difference between TE and TM polarizations at the output is: π ∆φ = 2φ + − 2φ = −3.43rad = −196.3o (43) TE 2 TM (c) Plot the time-evolution of the output E field vector. y’ First, we can define the “rectangular box” that confines Eout the time-evolution trace. From (42), we notice that the x’ and y’ components of Eout have equal magnitude,

t = ()Φo +π 2 ω and therefore the “box” is a square. Second, we need to determine qualitatively the orientation of the ellipse x’ and the handedness of the rotation. To do so, we use the real representations to express the variation of the two components with time:

Eout _ x' = E'o cos()Φ0 −ωt

Eout _ y' = E'o cos()Φ0 −ωt + ∆φ t = Φo ω where Φo is not time varying, and ∆φ is shown in (43).

(d) Can the TE-TM phase difference due to the two reflections be compensated exactly by another identical prism. Why or why not?. Yes. It can be compensated by another identical prism. One needs to switch the role of TE and TM after the first prism, so that the original x-polarized and y-polarized

components experience the same total phase shift (2φTE+2φTM) after going through two prisms. This is achieved by rotating the second prism by 90o, so that the plane of incidence is perpendicular to each other, as shown:

Q16-11: Consider the polarization state change upon reflection at an interface between

two media of refractive indices ni and nt. List all the conditions under which each of the following scenario occurs: (a) Polarization state unchanged (b) Linear polarization (incident) changes to an elliptical polarization (reflected) (c) Linear polarization (incident) changes to a different linear polarization (reflected) (d) Random polarization (incident) changes to linear polarization (reflected)

Q16-12: Can we apply Fresnel’s Equations in the case of frustrated total internal reflection? Why or why not?

1. *From Maxwell’s Equations and boundary conditions, derive the Fresnel reflection coefficients in the case of total internal reflection (Eq (20) and

(23)), and therefore show that one can use Eq (5) for cosθt, even though an imaginary cosθt is not physical.