sco 1617 1 - w3.ualg.ptw3.ualg.pt/~jlongras/aulas/sco_1617_03_090217.pdf · from: s.o. kasap,...

From: S.O. Kasap, Optoelectronics and Photonics: Principles and Practices, Second Edition, © 2013 Pearson Education, USA 46

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A light wave traveling in a medium with a greater refractive index (n1 >

n2) suffers reflection and refraction at the boundary – see figure below.

(Notice that λt

is slightly longer than λ.)We can use constructive interference to show that there can only be one reflected wave which occurs at an angle equal to the incidence angle. The two waves along Ai and Bi are in phase.

When these waves are reflected to become waves Ar and Br then they must still be in phase, otherwise they will interfere destructively and destroy each other. The only way the two waves can stay in phase is if θr = θi. All other angles lead to the waves Ar and Brbeing out of phase and interfering destructively.

Reflection and refraction of waves9-Feb-2017


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Snell's Law

1

2

sin

sin

n

n

t

i =θθ

This is Snell's law which relates the angles of incidence and

refraction to the refractive indices of the media.


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When n1 > n2 then obviously the

transmitted angle is greater than

the incidence angle as apparent in

the figure.

When the refraction angle θt

reaches 90°, the incidence angle is

called the critical angle θc

which is

given by

ti nn θθ sinsin 21 =

1

2sinn

nc =θ


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Light wave traveling in a more dense medium strikes a less dense medium. Depending on the incidence angle with respect to θc, which is determined by the

ratio of the refractive indices, the wave may be transmitted (refracted) or reflected. (a) θi < θc (b) θi = θc (c) θi > θc and total internal reflection (TIR).

Total Internal Reflection


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When the incidence angleθi exceedsθc then there is no transmittedwave but only a reflected wave. The latter phenomenon is called totalinternal reflection (TIR). TIR phenomenon that leads to thepropagation of waves in a dielectric mediumsurrounded by amediumof smaller refractive index as inoptical waveguides, e.g.optical fibers.

Although Snell's law forθi > θc shows that sinθt > 1 and henceθt isan "imaginary" angle of refraction, there is however an attenuatedwave called theevanescent wave.

1

2sinn

nc =θ

Snell’s Law


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Prisms


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Light travels by total internal reflection in optical fibers

An optical fiber link for transmitting digital information in communications. The fiber corehas a higher refractive index so that the light travels along the fiber inside the fiber coreby total internal reflection at the core-cladding interface.


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A small hole is made in a plastic bottle full of water to generate a water jet. When the hole is illuminated with a laserbeam (from a green laser pointer), the light is guided by total internal reflections along the jet to the tray. The lightguiding by a water jet was first demonstrated by Jean-Daniel Colladan, a Swiss scientist (Water with air bubbles wasused to increase the visibility of light. Air bubbles scatter light.) [Left: Copyright: S.O. Kasap, 2005] [Right: ComptesRendes, 15, 800–802, October 24, 1842; Cnum, Conservatoire Numérique des Arts et Métiers, France


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Fresnel's Equations

Light wave traveling in a more dense medium strikes a less dense medium. The plane of incidence is the plane of the paper and is perpendicular to the flat interface between the two media. The electric field is normal to the direction of

propagation. It can be resolved into perpendicular and parallel components.


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Describe the incident, reflected and refracted waves by the exponential representation of a traveling plane wave, i.e.

Ei = Eioexpj(ωt − ki⋅r ) Incident wave

Er = Eroexpj(ωt − kr⋅r ) Reflected wave

Et = Etoexpj(ωt − kt⋅r ) Transmitted wave

Fresnel's Equations

These are traveling plane waves


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wherer is the position vector, the wave vectorski, kr

and kt describe the directions of the incident, reflected and transmitted waves and Eio, Ero and Eto are the respective amplitudes.

Any phase changessuch as φr and φt in the reflected and transmitted waves with respect to the phase of the incident wave are incorporated into the complex amplitudes, Ero and Eto. Our objective is to find Ero and

Eto with respect to Eio.

Fresnel's Equations


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Fresnel's Equations

Incident wave Ei = Eioexpj(ωt − ki⋅r )

Reflected wave Er = Eroexpj(ωt − kr⋅r )

Transmitted waveEt = Etoexpj(ωt − kt⋅r )


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Applying the boundary conditions to the EMwave goingfrom medium 1 to 2, the amplitudes of the reflected andtransmitted waves can be readily obtained in terms ofn1, n2

and the incidence angleθi alone. These relationships arecalled Fresnel's equations. If we define n = n2/n1, as therelative refractive index of medium2 to that of 1, then thereflection and transmission coefficientsfor E⊥ are,

[ ][ ] 2/122

2/122

,0

,0

sincos

sincos

ii

ii

i

r

n

n

E

E

θθθθ

−+−−==

⊥

⊥⊥r

Fresnel's Equations


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[ ] 2/122,0

,0

sincos

cos2

ii

i

i

t

nE

E

θθθ

−+==

⊥

⊥⊥t

There are corresponding coefficients for the E// fields with corresponding

reflection and transmission coefficients, r// and t//,

[ ][ ] ii

ii

i

r

nn

nn

E

E

θθθθ

cossin

cossin22/122

22/122

//,0

//,0//

+−−−==r

[ ] 2/1222//,0

//,0//

sincos

cos2

ii

i

i

t

nn

n

E

E

θθθ

−+==t

Fresnel's Equations


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Further, the above coefficients are related by

r// + nt// = 1 and r⊥ + 1 = t⊥

For convenience we take Eio to be a real number so that phase angles ofr⊥ and t⊥ correspond to the phase changesmeasured with respect to the incident wave.

For normal incidence (θi = 0) into Fresnel's equations we find,

21

21// nn

nn

+−== ⊥rr

Fresnel's Equations


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We find a special incidence angle, labeled asθp, by solvingthe Fresnel equation forr// = 0. The field in the reflectedwave is then always perpendicular to the plane of incidenceand hence well-defined. This special angle is called thepolarization angleor Brewster's angle,

1

2tann

np =θ

Reflection and Polarization Angle

[ ][ ] 0

cossin

cossin22/122

22/122

//,0

//,0// =

+−−−==

ii

ii

i

r

nn

nn

E

E

θθθθ

r

For both n1 > n2

or n1 < n2.


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Polarized Light

A linearly polarized wave has its electric field oscillations defined along a line perpendicular to the direction of propagation, z. The field vector E and z define a plane of polarization.

x

y

z

E

Plane of polarization


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Brewster's angle

Reflected light at θi = θp has only E⊥

for both n1 > n2 or n1 < n2.

E⊥


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In internal reflection (n1 > n2), the amplitude of the reflected wave from TIR is equal to the amplitude of the incident wave but its phase has shifted.

What happens to the transmitted wave when θi > θc?

According to the boundary conditions, there must still be an electric field in medium 2, otherwise, the boundary conditions cannot be satisfied. When θi > θc, the field in medium 2 is attenuated (decreases with y, and is called theevanescent wave.

Evanescent Wave


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When �i > �c, for a plane wave that is reflected, there is an evanescent wave at the boundarypropagating along z.


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)(exp),,( 2, zktjtzyE iz

yt −∝ −

⊥ ωαe

where kiz

= kisinθ

iis the wavevector of the incident wave along the z-axis, and α2

is an attenuation coefficient for the electric field penetrating into medium 2

2/1

2

2

2

122 1sin

2

−

= in

nn θλπα

Evanescent wave when plane waves are incident and reflected


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Penetration depth of evanescent wave

α2 = Attenuation coefficient for the electric field penetrating into medium 2

α2 =2πn2

λn1

n2

2

sin2θ i −1

1 / 2

The field of the evanescent wave is e−1 in medium 2 when

y = 1/α2 = δ = Penetration depth


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Goos-Hänchen Shift

∆z = 2δtanθi

θi

n2

n1 > n2

Incidentlight

Reflectedlight

θr

∆z

Virtual reflecting plane

dz

y

A

B


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Beam SplittersFrustrated Total Internal Reflection (FTIR)

(a)A light incident at the long face of a glass prism suffers TIR; the prism deflects the light.

(b) Two prisms separated by a thin low refractive index film forming a beam-splitter cube. The incident beam is split into two beams by FTIR.


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Two prisms separated by a thin low refractive index film forming a beam-splitter cube. The incident beam is split

into two beams by FTIR.

Beam splitter cubes (Courtesy of CVI Melles Griot)


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When medium B is thin (thicknessd is small), the field penetrates from the ABinterface into medium B and reaches BC interface, and gives rise to a transmittedwave in medium C. The effect is the tunneling of the incident beam in A through Bto C. The maximum fieldEmax of the evanescent wave in B decays in B alongy and

but is finite at the BC boundary and excites the transmitted wave.

Optical Tunneling


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Optical Tunneling

Light propagation along an

optical guide by total

internal reflections

Coupling of laser light into a thin layer

- optical guide - using a prism. The

light propagates along the thin layer.


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Light approaches the boundary from the lower index side, n1 < n2

This is external reflection. Light becomes reflected by the surface of an optically denser (higher refractive index) medium.

r⊥ and r// depend on the incidence angle θi. At normal incidence, r⊥ and r// are negative.In external reflection at normal incidence there is a phase shift of 180°. r// goes through zero at the Brewster angle, θp. At θp, the reflected wave is polarized in the E⊥ component only.

Transmitted light in both internal reflection (when θi < θc) and external reflection does not experience a phase shift.

External Reflection


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Intensity, Reflectance and Transmittance

Reflectance R measures the intensity of the reflected light with respect to that of

the incident light and can be defined separately for electric field components

parallel and perpendicular to the plane of incidence. The reflectances R⊥ and R//are defined by

2

2

2

⊥

⊥

⊥⊥ == rR

,

,

io

ro

E

Eand

2

//2

//,

2

//,

// rR ==io

ro

E

E


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At normal incidence

2

21

21//

+−=== ⊥ nn

nnRRR

Since a glass medium has a refractive index of around 1.5 this means that typically 4% of the incident radiation on an air-glass surface will be reflected back.


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Example: Reflection at normal incidence. Internal and external reflection

Consider the reflection of light at normal incidence on a boundary between a glass medium of refractive index 1.5 and air of refractive index 1.

(a) If light is traveling from air to glass, what is the reflection coefficient and the intensity of the reflected light with respect to that of the incident light?

(b) If light is traveling from glass to air, what is the reflection coefficient and the intensity of the reflected light with respect to that of the incident light?


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Solution(a) The light travels in air and becomes partially reflected at the surface of the glass which corresponds to external reflection. Thus n1 = 1 and n2 = 1.5. Then,

r// = r⊥ =n1 − n2

n1 + n2

=1−1.5

1+1.5= −0.2

This is negative which means that there is a 180° phase shift. The reflectance (R), which gives the fractional reflected power, is

R = r//2 = 0.04 or 4%.


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(b) The light travels in glass and becomes partially reflected at the glass-air interface which corresponds to internal reflection. n1 = 1.5 and n2 = 1. Then,

r// = r⊥ =n1 − n2

n1 + n2

=1.5−1

1.5+1= 0.2

There is no phase shift. The reflectance is again 0.04 or 4%.

In both cases (a) and (b) the amount of reflected light is the same.


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Example 2: solar cells

When light is incident on the surface of a semiconductor it becomes partially reflected. Partial reflection is an important energy loss in solar cells.

The refractive index of Si is about 3.5 at wavelengths around 700 - 800 nm. Reflectance with n1(air) = 1 and n2(Si) ≈ 3.5 is

R =n1 − n2

n1 + n2

2

=1− 3.5

1+ 3.5

2

= 0.309

30% of the light is reflected and is not available for conversion to electrical

energy; a considerable reduction in the efficiency of the solar cell.


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Snell’s Law or Descartes’s Law?9-Feb-2017

sco 1617 1 - w3.ualg.ptw3.ualg.pt/~jlongras/aulas/sco_1617_03_090217.pdf · from: s.o. kasap,...

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