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

s-polarized light: p-polarized 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 p-polarized 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 sin-1 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).

Non-propagating waves

( ) ( )0 0exp expE j kx t E j j x t− → − ω β ω

We know that E0ejkxe-jωt 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 ω

This 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

Non-propagating 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 Evanescent-Wave k-vector

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 k-vector , we find:

Et(x,y,t) = E0 exp[–β y] exp[ j ( kt (ni /nt) sin(θi) x – ω t )]

The evanescent wave k-vector 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 = sin-1(2/3) = 41.8°

For our example, let’s assume that the light is green: λ = 532 nm

Then we find: β = 0.0042 nm-1 1/β = 240 nm

Typical result: somewhat less than the free-space wavelength

We saw earlier that:

where kt is the wave vector in the transmission medium (air):

kt = 2π nair / λ

2

2sin ( ) 1it i

t

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 out-of-focus sources

TIRF image is sharper

Fiber optic sensors

rays propagating in a large-core 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.