ece 318 f fundamentals of optics lecture 2qianli/publications/lecture_on_tir.pdfece 318 fundamentals...

ECE 318 Fundamentals of Optics

Lecture 16 & 17

ECE 318 Fundamentals of Optics

Lecture 16 & 17

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.

© Li Qian2 Lecture 16 &17

Total Internal ReflectionTotal Internal Reflection

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.


Critical AngleCritical AngleCritical 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:

nt

ni

θcki

kr

kt

ttii nn θθ sinsin = (1)

Therefore:

=

2sinsin πθ tci nn (2)

or

≡ −

i

tc n

n1sinθ (3)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.


What happens to Snell’s Law beyond the critical angle?What happens to Snell’s Law beyond the critical 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.

1sinsin >=t

iit n

n θθ (4)

1sincos2

−

=

t

iit n

ni θθ(5)

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


“Transmitted” wave vector is complex “Transmitted” wave vector is complex

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:

nt

niθi θrki kr

kt=?x

y

ttytxt nc

kk ω=+= 22k

And in order to satisfy the boundary phase-matching condition, we must have

( ) ( )

β

θωθωθ

i

nnn

cinn

ckkkkk

t

iitiitiittxtty

±≡

−

±=−=−=−= 1sinsinsin

2222222

And in order to satisfy both (6) and (7), and noting that nisinθi > nt, kty must become imaginary!

iiiiixtx nc

kkk θωθ sinsin ===

(6)

(7)

Above: Boundary conditions cannot be satisfied with a real kt .

(8)


“Transmitted” wave is evanescent“Transmitted” wave is evanescentWhat 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:

( )[ ]

+−=+−+= φωθωφω β txn

cietykxki ii

ytotytxtot sinexpexp EEE

Amplitude exponentially

decays with –y.A wave propagating in the x direction with vφ=c/(nisinθi)

(9)

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?


Penetration DepthsPenetration DepthsFrom Eq (9), we can draw three conclusions regarding the amplitude of the transmitted Efield:

(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

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

(10)

(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?


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

( )[ ]zE ˆexp φωβ +−= txkieE ixy

tot (12)

Now if we calculate the transmitted H field from the E field using one of the Maxwell’s Equations:

tt ∂∂

−=∂∂

−=×∇HBE µ

We obtain:(13)

( )[ ]φωµωβµωβµ β +−−=⇒=⇒

∂∂

−=∂∂ tkieEiHHiE

tH

yE

txy

toxxtxt exp

( )[ ]φωµω

µωµ β +−−=⇒=−⇒∂

∂−=

∂∂

− tkieEkHHiEikt

HxE

txy

totx

yyttxyt exp

(14)

(15)

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.)


“Transmitted” wave is not transmitting energy across the boundary* “Transmitted” wave is not transmitting energy across the 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:

( )zE ˆcos φωβ +−= txkeE txy

tot

( ) ( )

+−−+−= yxH ˆcosˆsin φω

µωφω

µωββ txkktxkeE tx

txtx

ytot

( ) ( ) ( )xyHE ˆcosˆsincos 22222 φωµω

φωφωµωβ ββ +−++−+−=× txkeEktxktxkeE tx

yto

txtxtx

ytott

( ) ( )

( ) 0222sin2

sincosˆ

22

22

=+−=

+−+−=⋅×

φωωµβ

φωφωωµβ

β

β

txkeE

txktxkeE

txy

to

txtxy

tott yHE

(16)

Poynting vector of the “transmitted” wave becomes:

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.

(18)

(17)

(19)

Q16-7: How do you reconcile the fact that there is finite field penetration into the ntmedium, but no net energy transport?


Complex Fresnel Reflection CoefficientsComplex Fresnel Reflection CoefficientsAs 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:

( )( ) 1sincos

1sincos2

2

−+

−−=

tiitii

tiitiiTE

nninn

nninnr

θθ

θθ(23)

( )( ) 1sincos

1sincos2

2

−+

−+−=

tiiiit

tiiiitTM

nninn

nninnr

θθ

θθ(20)

TEiTE er φ=∴ TMi

TM er φ=∴(21) (24)

( )

−−= −

ii

tiitTE n

nnnθ

θφ

cos1sin

tan22

1 ( )π

θθ

φ +

−−= −

it

tiiiTM n

nnncos

1sintan2

21(22) (25)

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.


Phase Shifts upon Total Internal ReflectionPhase Shifts upon Total Internal ReflectionThe figure below shows φTE, φTM, and the relative phase difference (φTM – φTE) as a function of incident angle for a glass-to-air interface.

0θiP

hase

θc π/2

φΤΕ

φΤΜ

∆φ=φΤΜ−φΤΕ

−π

0

π

−π/2

π/2

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


Reflectance and TransmittanceReflectance and Transmittance

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).

-1

-0.5

0

0.5

1

0

ni=1.5, nt=1

Fre

snel

Ref

lect

ion

Coe

ffic

ient

s

Incident Angle (θi)0 π/2

rTM

θp θc

rTERegion of

Total

Internal

Reflection

0

0.2

0.4

0.6

0.8

1

0

ni=1.5, nt=1

0 π/2

Incident Angle (θi)

Tra

nsm

itta

nce

or R

efle

ctan

ceRTM

RTE

TTM

TTE

θp θc

Region of

Total

Internal

Reflection

RTE=RTM=1

TTE=TTM=0


Frustrated Total Internal ReflectionFrustrated Total Internal ReflectionEven 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.

FTIR has been used in numerous applications, including:

- Prism coupling: coupling light into and out of waveguides;

- Optical directional couplers with variable ratios;- Coupling light into the “Whisper-Gallery” mode of

a high Q resonator

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

nt

ni

θi θrki kr

x

y

nt2

kt2


Example: Fresnel RhombExample: Fresnel RhombThe 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.

θ

θ

θ

Eout

Ei

cross-sectionview

Ei

Eout

inputfacet output

facet

If the linearly polarized input has equal TE and TM components, say, with an amplitude of Eo, then the input can be expressed by:

( ) ( )[ ]φω +−⋅+= tiE TMTEoi rkuuE expˆˆ (26)

( ) ( )[ ]

( ) ( )[ ]( )( ) ( )[ ]TETM

iTEo

TMi

TEi

o

TMTMTMTETETEoout

tieE

tieeE

tirrrrE

TETM

TMTE

φφω

φω

φω

φφ

φφ

2expˆˆ

expˆˆ

expˆˆ

2

22

++−⋅+=

+−⋅+=

+−⋅+=

− rkuu

rkuu

rkuuE

After two total internal reflections, and ignoring the facet reflections, the output wave becomes:

(27)


Example: Fresnel Rhomb (Cont’d)Example: Fresnel Rhomb (Cont’d)In order to convert to circular polarization, one requires:

( ) ππφφ mTETM +=− 22 (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:

(29)oi 6.54≈=θθ

Q16-9: Given a Fresnel rhomb with a proper apex angle, would linearly polarized inputalways 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?


Example: Right-angle PrismExample: Right-angle Prism

The right-angle glass prism is often used in optical experiments as a retro-reflector / delay stage (top figure). A circularly polarized collimated laser beam (bottom graph) enters the prism through the front facet at normal incidence. The refractive index of the prism is 1.5.

(a) Find the reflectance of the prism.(b) Calculate the phase difference between the TE

and TM polarizations of the output beam. (c) Plot the time-evolution of the output E field vector

using the co-ordinate system specified in the top figure.

(d) Can the TE-TM phase difference due to the two reflections be compensated exactly by another identical prism. Why or why not?

45o

zy

xincid

ent

z’

y’

x’ref

lected

frontfacet

x

yEin


Example: Right-angle Prism (Cont’d)Example: Right-angle Prism (Cont’d)Solution:

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

As shown on the left, the laser beam is refracted and reflected at 4 interfaces:

1) Normal incidence (air-to-glass) at the front facet. Let t1 and T1 denote the Fresnel transmission coefficient and transmittance, respectively.

2) Internal reflection (glass-to-air) at 45o angle. Let r2and 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.

n=1.5

45o

45o

45o

frontfacet


Example: Right-angle Prism (Cont’d)Example: Right-angle Prism (Cont’d)Using Fresnel Equations, one can easily find

96.0111

2

41 =

+−

−==nnTT (30)

For R2 and R3, the incident angle is 45o, and since

oo

n8.411sin45 1 =

> − (31)

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

132 == RR (32)

The reflectance of the prism becomes:

%2.924321 ==≡ TRRTpowerinputpoweroutputRprism (33)


Example: Right-angle Prism (Cont’d)Example: Right-angle Prism (Cont’d)

(b) Calculate the phase difference between the TE and TM polarizations of the output beam.

(34)

Using the co-ordinate system specified, we can write the expression for the incident polarization:

+= yxE ˆˆ 2

πi

oin eE

Note we have omitted the 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:

( )[ ]φω +− tkziexp

+= 'ˆ'ˆ 2

_4_3_2_1_4_3_2_1 yxEπi

TETETETETMTMTMTMoout etrrttrrtE (35)

For normal incidence, there is no polarization dependence, and therefore,

12

1_1_1 +=≡=

nttt TMTE

12

4_4_4 +=≡=

nnttt TMTE

(36)

(37)

Note that both t1 and t4 are real and positive, which means no phase change after transmission.



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

TEiTETE err φ== _3_2 (38)

(39)TMiTMTM err φ== _3_2

where( ) o

o

o

TE radn

n9.366435.0

45cos145sin

tan22

1 −=−=

−−= −φ (40)

( ) oo

o

TM radnn

3.10685.145cos

145sintan2

21 ==+

−−= − πφ (41)

Substitute (36)-(39) into (35), we get the output polarization expression:

+=

'ˆ'ˆ 22

41 yxE φ TETMi

oout eettE (42) +2 πφi



Therefore, the phase difference between TE and TM polarizations at the output is:

oTMTE rad 3.19643.32

22 −=−=−+=∆ φπφφ (43)

(c) Plot the time-evolution of the output E field vector.

x’

y’Eout

First, we can define the “rectangular box” that confines the time-evolution trace. From (42), we notice that the x’ and y’ components of Eout have equal magnitude, and therefore the “box” is a square. Second, we need to determine qualitatively the orientation of the ellipse and the handedness of the rotation. To do so, we use the real representations to express the variation of the two components with time:

( )φω ∆+−Φ= tEE oyout 0'_ cos'

( )tEE oxout ω−Φ= 0'_ cos'

( ) ωπ 2+Φ= ot

ωot Φ=where Φo is not time varying, and ∆φ is shown in (43).


Example: Right-angle Prism (Cont’d)Example: Right-angle Prism (Cont’d)(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:


More QuestionsMore Questions

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?


HomeworkHomework

1. *From Maxwell’s Equations and boundary conditions, derive the Fresnelreflection 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.

ece 318 f fundamentals of optics lecture 2qianli/publications/lecture_on_tir.pdfece 318 fundamentals...

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