9-14 total internal reflection

8/13/2019 9-14 Total Internal Reflection

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7.

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7.

Conditions for total internal reflection

The evanescent wave

Uses for total internal reflection

PrismsBeamsplitters

Fiber Optics

Laser slabs

Phase shift on total internal reflection

Reflection from metals

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7.

Snells law tells us light bends towards the normalwhen going from low-index to high-index materials

Going from high-index to low-index light must bendaway from the normal

At some critical angle, the transmitted beam in thelow index material will be at 90

As the incident beam angle increases the transmittedbeam angle cannot increase!

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!i

!tnt

ni !i

!tni

nt


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Snells law allows us to calculate the angle of thebeam transmitted through an interface. Are thereconditions that prevent there from being a real

mathematical solution?

What happens when there is no real mathematicalsolution?

7. 4

ni sin i = ntsin t

sin t = i

nt

sin i 1

i sin1nt

ni


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7.

Consider the Fresnelreflection coefficients

at the critical angle,!c=sin-1(nt/ni)

Beyond the critical angle what do we get forthe transmitted angle?

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r =E0r

E0i=

nicos i ntcos t

nicos i + ntcos t

r = E0r

E0i= ntcos i

nicos t

nicos t + ntcos i

r = 1 r = 1


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7.

Beyond the critical angle what do we get forthe transmitted field?

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sin =ei/2

e e

i/2e

2i=

e + e

2 = cosh

Et = tE0ieik0(sin tx+cos ty)

cos = e

i/2e + ei/2e

2 =

ie ie

2 = i sinh

Et = tE0ieik0cosh(t)x+k0sinh(t)y

sin = e

i

e

i

2i> 1 let = /2 +i

The transmitted field is a traveling wave

in the direction along the interface

The transmitted field exponentially decays

as it gets further from the interface

Plane of the interface(here theyz plane) (perpendicular to page)

ni

nt

!i !r

!t

Ei Er

Et

Interface

x

y

z


7/207.

Beyond the critical angle the reflectioncoefficients are complex

imaginary part of coefficient implies a phase

shift

Magnitude of reflection coefficient is 1,indicating 100% reflection

Power reflectivity coefficient must begeneralized to allow for complex reflectioncoefficients

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R = rr

Er = rE0ei(t+)


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7.

Because the transmitted field is an evanescent

wavethat decays exponentially to zero, it doesnot carry energy away from the interface

The evanescent wave is still necessary tosatisfy the boundary conditions at the interface

100% of the power is contained in the reflected

field, i.e. there is total internal reflection

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7.

Incident and reflected fields on reflection froma high-index to low-index material are in-phase

Without a transmitted field the E field would be

discontinuous across the boundary

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E


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7.10

By placing a high-index material in the presence of theevanescent wave power can be coupled through thelow-index gap, frustrating the total internal reflection

nn

nn

n=1 n=1total internal reflection frustrated total

internal reflection

The prisms must be within a few wavelengths (wherethe evanescent field is non-zero) for this to work

This is the principle of operation for cube beamsplitters


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7.11

zig-zag laser slabs

fiber optics

prisms

fingerprinting


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7.

The circulating beam in many high-power lasersis made to zig-zag through the laser crystal toaverage over the thermal gradient in the

crystal. Having many reflections requires thereflectivity at each interface be high

12

0 2.5 5 7.5 10 12.5 15 17.5 2

0.25

0.5

0.75

1

Teff = RN

N


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7.

Prisms are used for reflecting beams with unitefficiency via TIR. Various configurations allowmany interesting properties

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7.

Glass fibers are used aswaveguides to transmit lightover great distance

High index core guides the light

A low index cladding protects the interfaceof the core

The acceptance angle of a fiber determineswhat light will be guided through the fiber

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7.15

fingertip valleys reflect light via TIR, whilefinger tip ridges in contact with prism frustratethe reflection


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7.16

nt< ni

!

||

above the critical angle,TIR field shows an

interesting phase shift

A !phase shift occurs atBrewsters angle indicating

a change in the reflectioncoefficient sign as itpasses through zero

nt< ni


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7.

For a perfect conductor, there can be nointernal electric fields, hence the boundarycondition requires E||=0, so for the parallelcomponent of the field Er=-EiEt=0

Reflection coefficient is r=1, R=1

Transmission coefficient is t=0, T=0

Does a real metal behave like this?

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7.

The free electrons in a metal can be thought ofas a gas or plasma with a plasma frequency(natural frequency of oscillation) of

The refractive index of metals is given by

When ""p, the metal is transparent

typical metals have a value for "pin the UV 18

p =

Ne2

0me

n2

= 1

p

2


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7.19


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7.

When light passes from a dense material to a less dense material itbends away from the normal

When the incident angle is large enough the transmitted angle if90 and cannot increase

Beyond the critical angle 100% of the power is reflected

An evanescent wave is present in the transmitted material thatmatches the boundary conditions at the interface, but carries nopower away from the interface

A high index material in the presence of the evanescent wave cancouple light through the low index gap causing frustrated total

internal reflection

The reflected field acquires a phase shift upon totally internallyreflecting

Metals reflect light efficiently below their plasma frequency

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9-14 total internal reflection

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