14. Total Internal Reflection and Evanescent Waves...Total Internal Reflection n i n t k i k t θ i...
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14. Total Internal Reflection and Evanescent Waves
Phase shifts in reflection
Total internal reflection and applications
Evanescent waves

Reminder: the Fresnel equations

cos( ) cos( )
cos( ) cos( )
−=+
i t t i
i t t i
n nr
n n
θ θθ θ

2 cos( )
cos( ) cos( )=
+i i
i t t i
nt
n n
θθ θ
2 cos( )
cos( ) cos( )⊥ = +
i i
i i t t
nt
n n
θθ θ
cos( ) cos( )
cos( ) cos( )⊥
−=+
i i t t
i i t t
n nr
n n
θ θθ θ
spolarized light: ppolarized light:
And, for both polarizations: sin( ) sin( )=i i t tn nθ θ
incident wave
transmitted wave
interface
incident wave
transmitted wave
interface

[ ][ ]⊥
−= =
+i t
i t
n nr r
n n
So there will be destructiveinterference between the incident and reflected beams near the surface, where they overlap in space.
At normal incidence, θi = 0, we find:
Phase Shift in Reflection
Note: for ppolarized light, the sign of r changes
for angles above Brewster’s angle.
Analogously, if ni > nt (glass to air), r > 0, and there will be constructive
interference.
Incidence angle, θi
Reflection c
oeffic
ien
t, r
1.0
.5
0
.5
1.0
r
r
0° 30° 60° 90°
Incidence angle, θi
Reflection c
oeffic
ien
t, r
1.0
.5
0
.5
1.0
r
r
0° 30° 60° 90°
from air to glass
Brewster’s angle
0.2= −1
=1.5i
t
n
n
=
for
If ni < nt, then r < 0 at normal incidence.

The obvious answer is the front of the object, which sees the higher intensity first.
But constructive interference happens at the back surface between the incident light and the reflected wave.
If you slowly turn up a laser intensity, where does damage happen first, the front or the back?
2(1 0.2) 1.44+ =
This yields an irradiance that is 44% higher just inside the back surface (for nglass = 1.5)!

Phase shifts in reflection (air to glass)
ni < nt
180° phase shift
for all angles
180° phase shift for angles below
Brewster's angle;
0° for larger angles
0° 30° 60° 90°Incidence angle
0° 30° 60° 90°Incidence angle
π
0
π
0
┴

In this case, the reflection coefficients are real numbers, so the phase of the reflected wave is either 0 or π(relative to the incident wave).

Total Internal Reflection
ni
nt
ik�
tk�
θi
θt
Ei
Et
Interface
Snell’s Law:
sin sin=i i t tn nθ θ
Solve for θt :
1 sin sin−
=
it i
t
n
nθ θ
But is impossible to take the arcsin of a number larger than one!
sinθi is always ≤ 1, so if ni < nt then this never becomes an issue.
BUT: if ni > nt then the argument of the sin1 can exceed one!
When does this occur? As θi increases, θt also increases. When θt reaches π/2, the transmitted wave is grazing the interface. This occurs at a value of θi given by: ( )1 sin−=critical t in nθ
In this illustration, the light
wave bends away from the normal because nt < ni.

Total Internal Reflection occurs just as thetransmitted beam grazes the surface.
0º
0º
15º
20º
30º
42º
45º
70º
60º
??
Total internal reflection is 100% efficient.
As the angle of incidence increases from 0º…
… the refracted ray becomes dimmer
… the reflected ray becomes brighter
… the angle of the refracted ray approaches 90º

Transmitted intensity T (glass to air)
Incidence angle
0º 30º 60º 90º
0%
100%
Tra
nsm
itte
d inte
nsity
┴

Critical angle
Transmitted power goes smoothly to zero as the critical angle is approached.

An application of total internal reflection
Charles Kao (1965): first proposed that fiber could be used as a practical communication technology if
the attenuation could be reduced below 20 dB/km.
He showed that the loss was dominated by chemical impurities in the glass core.
Corning (1970): first commercial fiber for
telecommunications. Attenuation = 17 dB/km
State of the art today (commercial fibers):
Attenuation = 0.2 dB/km
data rate = 40 Gb/sec
“Wavelength division multiplexing”
(WDM) – using multiple wavelengths,
each carrying an independent data stream, on a single fiber
Current record (2011): 370 WDM channels, 273 Gb/sec in each channel

Beam steerersused to compressthe path insidebinoculars
Beam steerers
Another application of total internal reflection

A thought experiment
Suppose I have a glass prism, oriented as shown, with a laser beam undergoing total internal reflection from the internal surface.
nn
Total internal reflection
Now I bring a second identical prism close to the first one.
Question: at what point do I see a transmitted beam?
Answer: when the prisms are close together, but not yet touching!
Conclusion: something interesting must be happening on the low index (air) side of the interface. There must be a wave there, but it doesn’t propagate away (until the 2nd prism is very close).

Nonpropagating waves
( ) ( )0 0exp expE j kx t E j j x t− → − ω β ω
We know that E0ejkxejωt is a solution to the wave equation.
What if we allow the wave vector k to be a purely imaginary
number? Let k be replaced by jβ, where β is a real number.
( ) ( )0 exp expE x j t= − −β ωThis no longer oscillates as a function of position  it exponentially decays! But it still oscillates as a function of time.
x = 0
x
E(x)at t = 0
x
E(x)at t = π/ω
x
E(x)at t = 2π/ω
Real part: ( ) ( )0 exp cosE x t−β ω

Is this still a solution to the wave equation?
( ) ( ) ( )0, exp exp= − −E x t E x j tβ ω
Nonpropagating waves
( ) ( ) ( )
( ) ( ) ( )
22 2
02
22 2
02
exp exp ,
exp exp ,
= − − =
= − − − = −
d EE x j t E x t
dx
d EE x j t E x t
dt
β β ω β
ω β ω ω
So this is a solution to the wave equation! But it is a very different kind of solution from the ones we’re used to seeing.
It does not propagate in space. It is localized. It is known as an evanescent wave.
Such a wave can be found in a number of situations. In particular, evanescent waves are always present in the case of total internal reflection.

The Evanescent Wave
The evanescent wave is the
"transmitted wave" when total
internal reflection occurs. A very
mystical quantity! So we'll do a
mystical derivation. Assume that
θi is greater than θcritical.
2
2 2cos( ) 1 sin ( ) 1 sin ( ) a negative number
= − = − =
it t i
t
n
nθ θ θ
[ ][ ]
0
0
cos( ) cos( )
cos( ) cos( )
i i t tr
i i i t t
n nEr
E n n
θ θθ θ⊥
−= =
+
Since sin(θt) > 1, θt doesn’t exist. So how can we compute r⊥?
Use Snell’s Law to eliminate θt from the equation:
Substitute this imaginary number into the formula for r⊥. We find that
r⊥ is a complex number with magnitude of one. We redefine the
reflectance R as: R = r⊥ r⊥* = r⊥2. Thus R = 1 for all angles above the
critical angle.
So, all of the power is reflected; the evanescent wave contains no power.

Phase shifts in reflection (glass to air)
nt < ni
0° phase shift for
angles below the critical angle
0° phase shift for angles below
Brewster's angle;
180° for larger angles up to the
critical angle0° 30° 60° 90°
Incidence angle
0° 30° 60° 90°Incidence angle
π
0
π
π
0
π
┴

In the case of total internal reflection, the phase shift for the reflected wave is a complicated function of θi.
r is a real
numberr is complex
and r = 1.
r is a real
numberr is complex
and r = 1.

The EvanescentWave kvector
Using Snell's Law, sin(θt) = (ni /nt) sin(θi), so ktx is a real quantity.
ˆ ˆ= +�
txk k x j yβUsing this complex kvector , we find:
Et(x,y,t) = E0 exp[–β y] exp[ j ( kt (ni /nt) sin(θi) x – ω t )]
The evanescent wave kvector must have x and y components:
Along surface: ktx = kt sin(θt)
Perpendicular to it: kty = kt cos(θt)
As before: kt cos(θt) = kt [1 – sin2(θt)]1/2 = kt [1 – (ni /nt)2 sin2(θi)]1/2
= a pure imaginary quantity; let’s call it jβ (β = real)
The evanescent wave decays exponentially away from the interface.
Here, kt = 2πnt/λ as usual  a positive real number
The quantity 1/β sets the length scale for the decay of the wave along y.

http://www.andrew.cmu.edu/user/dcprieve/Evanescent%20waves.htm
TIR and the evanescent wave
θi
evanescent wavereflected wave incident wave
total wave:
sum of incident and reflected waves
evanescent wave: decay length = 1/β

The evanescent wave: an example
So how far does the evanescent wave extend away from the interface? Let’s work an example.
n
nair = 1
nglass = 1.5
θi = 45°
θcrit = sin1(2/3) = 41.8°
For our example, let’s assume that the light is green: λ = 532 nm
Then we find: β = 0.0042 nm1 1/β = 240 nm
Typical result: somewhat less than the freespace wavelength
We saw earlier that:
where kt is the wave vector in the transmission medium (air):
kt = 2π nair / λ
2
2sin ( ) 1it it
nk
nβ θ
= −

Frustrated Total Internal Reflection (TIR)
By placing another surface in contact with a totally internallyreflecting one, total internal reflection can be “frustrated.”
nn
Total internal reflection
nn
Frustrated total internal reflection
We can now calculate how close the prisms have to be before TIR is
frustrated.
The quantity 1/β tells us how far the evanescent wave extends beyond the surface of the first prism, which tells us how close the second prism needs to be in order to frustrate the TIR.

An application of frustrated TIR
The ridges on a finger act as locations where TIR is frustrated, so less light comes from there. But between the ridges, there is still TIR so more light is reflected.

Total internal reflection fluorescence (TIRF) microscopy
Only objects within ~100 nm of the interface are illuminated.
conventional fluorescence microscope image is blurred due to fluorescence from outoffocus sources
TIRF image is sharper

Fiber optic sensors
rays propagating in a largecore fiber
The value of the critical angle depends on the ratio nt/ni.
evanescent wave
1 1.1 1.2 1.3 1.4 1.540
50
60
70
80
90
critica
l a
ngle
index of transmitted medium
ni = 1.5 If the value of nt increases,
then TIR is less likely to occur (a smaller range of angles experience TIR).
This effect can be used to sense small refractive index changes.