(a) a light incident at the long face of a glass prism...

ELEC425-summer, 2012 Assignment 5

1

Assignment 5.

1.13, 7.1, 7.2, 7.5, 7.11, 7.12, 7.15, 7.21

1.13. TIR and FTIR

a) By considering the electric field component in medium B in Figure 1.20 (b),

explain how you can adjust the amount of transmitted light.

Figure 1.20. from Optoelectronics, S. O. Kasap

b) What is the critical angle at the hypotenuse face of a beam splitter cube

made of glass with n₁ = 1.6 and having a thin film of liquid with n₂ = 1.3.

Can you use 45° prism with normal incidence?

c) Explain how a light beam can propagate along a layer of material between

two different media as shown in Figure 1.34 (a). Explain what the

requirements are for the indices n₁, n₂, n₃. Will there be any losses at the

reflections?

d) Consider the prism coupler arrangement in Figure 1.34 (b). Explain how this

arrangement works for coupling an external light beam from a laser into a

thin layer on the surface of a glass substrate. Light is then propagated inside

the thin layer along the surface of the substrate. What is the purpose of the

adjustable coupling gap?

Incident

light

Reflected

light

i > c

TIR

(a)

Glass prism

i > c

FTIR

(b)

n1

n1n

2 n1

B = Low refractive index

transparent film ( n2)

ACA

Reflected

Transm itted

(a) A light incident at t he long face of a glass prism suffers T IR; t he prism deflects thelight.(b) T wo prisms separated by a thin low refractive index film forming a beam-splitt er cube.T he incident beam is split into two beams by FTIR.

Incident

light

© 1999 S.O. Kasap, Optoelectronics (Prentice Hall)


2

Figure 1.34. from Optoelectronics, S. O. Kasap

Solution.

a) Consider the prism A when the neighboring prism C in Figure 1.20 (b) in far

away. When the light beam in prism A is incident on the A/B interface,

hypotenuse face, it suffers TIR as θi > θc. There is however an evanescent

wave whose field decays exponentially with distance in medium B. When

we bring prism C close to A, the field in B will reach C and consequently

penetrates C. (The tangential field must be continuous from B to C). One

cannot just use the field expression for the evanescent wave because this was

derived for a light beam incident at an interface between two media only; no

third medium. The transmitted light intensity from A to C depends on the

thickness of B.

b) For the prism A in Figure 1.20 (b), n₁ = 1.6 and n₂ = 1.3 so that the critical

angle for TIR at the hypotenuse face is

3.54

6.1

3.1sinsin 1

1

21

n

nc

in this case 45° prism cannot be employed, Figure 1.20 (a).

c) If the angle of incidence θi at the n₁/n₂ layer is more than the critical angle

θc12 and if angle of incidence θi at the n₁/n₃ layer is more than the critical

angle θ₁₃ then the light ray will travel by TIR zigzagging between the

boundaries as sketched in Figure 1.20 (a). For example, suppose that n₁ = 2

(thin layer), n₂ = 1 (air), n₃ = 1.6 (glass),

n1

n3

n2Air

Glass substrate

Thin layer

(a)

Glass substrate

Thin layer

PrismLaser light

d = Adjustable coupling gap

(b)

(a) Light propagation along an opt ical guide. (b) Coupling of laser light int o a t hin layer -optical guide - using a prism. The light prop agates along t he t hin layer.

© 1999 S.O. Kasap, Optoelectronics (Prentice Hall)


3

1.532

6.1sinsin

8.382

1sinsin

1

1

31

13

1

1

21

12

n

n

n

n

c

c

So that θi > 53.1° will satisfy TIR. There is no loss in TIR as the magnitude

of the amplitude of the reflected way is the same as that of the incident

wave.

Note: There is an additional requirement that the waves entering the thin

film interfere constructively, otherwise the waves will interfere destructively

to cancel each other. Thus there will be an additional requirement, called the

waveguide condition (Chapter 2).

d) The light ray entering the prism is deflected towards the base of the prism.

There is a small gap between the prism and the thin layer. Although the light

arriving at the prism base/gap interface is reflected, because of the close

proximity of the thin layer, some light is coupled into the thin layer as it was

discussed in part (a) due to frustrated TIR. This arrangement is a much more

efficient way to couple the light into the thin layer because the incident light

is received by the large hypotenuse face compared with coupling the light

directly into the thin layer.

7.1. Polarization.

Suppose that we write the Ex and Ey components of a light wave generally as:

kztEE xx cos0 and kztEE yy cos0

Show that at any instant Ex and Ey satisfy the ellipse equation on the Ey vs. Ex

coordinate system:

2

00

2

0

2

0

sincos2

y

y

x

x

y

y

x

x

E

E

E

E

E

E

E

E

Sketch schematics what this ellipse look like assuming Ex0 = 2Ey0. When would

this ellipse form an (a) ellipse with its major axis on the x-axis, (b) a linearly

polarized light at 45° to the x-axis, (c) right and left circularly polarized light?


4

Solution.

Consider the LHS of equation

22222222

2222

22222

22

22222

22

22

0

0

0

0

2

0

0

2

0

0

00

2

0

2

0

sinsin]sin[cossinsinsincos

sinsincos1cos

sinsincoscoscos

sincossincos2coscos2

sincossincos2

sinsincoscoscos

cossinsincoscoscos2

sinsincoscoscos

coscoscos2coscos

coscoscos

2coscos

cos2

kztkztkztkzt

kztkzt

kztkztkzt

kztkztkzt

kztkzt

kztkztkzt

kztkztkzt

kztkztkzt

kztkztkztkzt

E

kztE

E

kztE

E

kztE

E

kztE

E

E

E

E

E

E

E

E

y

y

x

x

y

y

x

x

y

y

x

x

y

y

x

x

The equation is proved.

Consider the general expression

2

00

2

0

2

0

sincos2

y

y

x

x

y

y

x

x

E

E

E

E

E

E

E

E

This is a quadratic equation in Ey (or Ex),

2

2

000

2

2

sin ;cos1

2 ;1

where

0

x

x

yx

x

yo

yy

E

Ec

EE

Eb

Ea

cbEaE

Thus, for a given ϕ and a given Ex0 and Ey0, the graph Ey vs. Ex can be sketched

where Ex determines b and c and Ey is given by


5

a

acbbEy

2

42

If Ex0 = 2Ey0 then

a) for ϕ = π/2, Ey0 = 1 and Ex0 = 2 we will obtain an ellipse with its major axis

on the x-axis


6

b) For ϕ = 0, Ey0 = 1 and Ex0 = 1 we obtain the line (linear) polarization at an

angle π/4 to the x-axis

c) For ϕ = π/2, Ey0 = 1 and Ex0 = 1we obtain a circle, if ϕ= π/2 right circular

polarization; shift ϕ = -π/2 results in left circular polarization


7

7.2. Linear and circular polarization

Show that a linearly polarized light wave can be represented by two circularly

polarized light waves with opposite rotations. Consider the simplest case of a wave

linearly polarized along the y-axis. What is your conclusion?

Solution.

if the wave is polarized along the y-direction then circularly polarized light has

retarded electric field in x-direction and E₀ = Ex0 = Ey0 = 1

for right circularly polarized light:

2cos0

kztEERx , kztEERy cos0

for left circularly polarized light:

2cos0

kztEELx , kztEELx cos0

Total x and y components of two polarizations are


8

02

cos2

cos 00

kztEkztEEx

kztEEy cos2 0

There is only one component which proves the linear polarization along y-axis

In case of the linear polarization in x-axis, the retarded component will be in y-

direction (

2cos0

kztEERy , kztEEx cos0 )

+ =

7.5. Jones Matrices

When we represent the state of polarization of a light wave using a matrix, called a

Jones matrix (or vector) then various operations on the polarization state

correspond to multiplying this matrix with another matrix that represents the

optical operation. Consider a light wave travelling along z with field components

Ex and Ey along x and have a phase difference ϕ between them. If we use the

exponential notation then

xxx kztjEE exp0 and yyy kztjEE exp0

Jones matrix is a column matrix whose elements are Ex and Ey without the

common exp[j(ωt-kz)] factor


9

yy

xx

y

x

jE

jE

E

E

exp

exp

0

0E (1)

Usually Eq. (1) is normalized by dividing by the total amplitude 2/12

0

2

00 yx EEE .

We can also factor out exp(jϕx) to further simplify to obtain the Jones matrix:

jE

E

E y

x

exp

1

0

0

0

J (2)

where ϕ=ϕy – ϕx.

a) Table 7.3 shows Jones vectors for various polarizations. Identify the state of

polarization for each matrix.

b) Passing a wave of given Jones vector Jin through an optical device is

represented by multiplying Jin by the transmission matrix T of the device.

If Jout is the Jones vector for the output light through the device, then Jout = T

Jin. Given

jT

0

01 (3)

Determine the polarization state of the output wave given the Jones vectors

in Table 7.3, and the optical operation represented by T. Hint: Use

1

1as

input for determining T.

Table 7.3 Jones vectors

Jones vector

Jin

0

1

1

1

2

1

sin

cos

j

1

2

1

j

1

2

1

Polarization ? ? ? ? ?

Transmission

matrix T

00

01

j

j

e

e

0

0

j0

01

10

01

je0

01

Optical

operation

? ? ? ? ?

Solution.

a)

Polarization state for each matrix


10

Jones vectors

Jones vector

Jin

0

1

1

1

2

1

sin

cos

j

1

2

1

j

1

2

1

Polarization Linear;

horizontal E

Linear; E at

45° to x-axis

Linear; E at

θ to x-axis

Right

circularly

polarized

Left

circularly

polarized

b) Optical operation represented by T is

Jones vectors

Jones vector

Jin

0

1

1

1

2

1

sin

cos

j

1

2

1

j

1

2

1

Polarization Linear,

horizontal E

Linear; E at

45° to x-axis

Linear; E at

θ to x-axis

Right

circularly

polarized

Left

circularly

polarized

Transmission

matrix T

00

01

j

j

e

e

0

0

j0

01

10

01

je0

01

Optical

operation

Linear

polarizer;

horizontal

transmission

axis

Isotropic

phase

changer or

phase

retarder

Quarter-

wave plate

Half-wave

plate

Wave

retarder; fast

axis along x

The polarization state of the output waves from quarter-wave plate are

Jones vectors

Jones vector

Jin

0

1

1

1

2

1

sin

cos

j

1

2

1

j

1

2

1

Polarization Linear

horizontal

polarization,

horizontal E

Linear; E at

45° to x-axis

Linear; E at

θ to x-axis

Right

circularly

polarized

Left

circularly

polarized

Transmission

matrix T

j0

01

j0

01

j0

01

j0

01

j0

01

Jones vector

Jout

0

1

Linear

horizontal

polarization,

horizontal E

j

1

2

1

Right

circularly

polarized

sin

cos

j

Right

circularly

polarized

1

1

2

1

Linear; E at

(2π-π/4) to

x-axis

1

1

2

1

Linear; E at

π/4 to x-axis


11

The polarization state of the output waves are

Jones vectors

Jones vector

Jin

0

1

1

1

2

1

sin

cos

j

1

2

1

j

1

2

1

Polarization Linear,

horizontal E

Linear; E at

45° to x-axis

Linear; E at

θ to x-axis

Right

circularly

polarized

Left

circularly

polarized

Transmission

matrix T

00

01

j

j

e

e

0

0

j0

01

10

01

je0

01

Optical

operation

Linear

polarizer;

horizontal

transmission

axis

Isotropic

phase

changer or

phase

retarder

Quarter-

wave plate

Half-wave

plate

Wave

retarder; fast

axis along x

Jones vector

Jout

0

1

Linear,

horizontal

polarization

1

1

2

je

Linear, E at

45° to x-axis

with Ex

leading Ey

sin

cos

j

Right

elliptical

polarization

j

1

2

1

Left

circular

polarized

je

1

2

1

Linear,

retarded y-

component,

fast axis

along x

7.11. Glan-Foucault prism

Figure 7.37 shows the cross section of a Glan-Foucault prism which is made of two

right angle calcite prisms with a prism angle of 38.5°. Both have their optic axes

parallel to each other and to the block faces as in the figure. Explain the operation

of the prisms and show that the o-wave does indeed experience total internal

reflection.


12

Solution.

Calcite is optically anisotropic material, birefringent. This is negative uniaxial

crystal since two of their principle indices the same (n₁ = n₂) and n₃ < n₁.

The light has two orthogonally polarized components in uniaxial crystal, (o)

ordinary and (e) extraordinary waves. The refractive indices (from Table 7.1) are

no = 1.658 and ne = 1.486.

The critical angles for TIR in e- and o-waves are

3.42/1arcsin

09.37/1arcsin

ec

oc

nwavee

nwaveo

If the angle of incidence is θ at calcite/air interface then from Figure 7.37

90°+(90°-θ)+38.5° = 180° → θ = 38.5° > θc (o-wave)

θ = 38.5° < θc (e-wave)

Thus, the o-wave suffers TIR while the e-wave does not. Hence the beam that

emerges is the e-wave, with a field Ee along optic axis.


13

O ptic axis

38.5

Absorbe r

Air-ga p

Calcite

e -ray

o -ray

The Glan-Foucau lt p rism prov ides linearly po larized light

© 1999 S .O. K asap, O ptoelectronics (P rentice H al l)

θ

7.12. Faraday Effect.

Application of a magnetic field along the direction of propagation of a linearly

polarized light wave through a medium results in the rotation of the plane of

polarization. The amount of rotation θ is given by

BL

where B is the magnetic field (flux density), L is the length of the medium, and ϑ is

the so-called Verdet constant. It depends on the material and the wavelength. In

contrast to optical activity, sense of rotation of the plane of polarization is

independent of the direction of light propagation. Given that glass and ZnS have

Verdet constants of about 3 and 22 minutes of arc Gauss⁻¹meter⁻¹ at 589 nm

respectively, calculate the necessary magnetic field for a rotation of 1° over a

length 10 mm. What is the rotation per unit magnetic field for a medium of length

1 m? (Note: 60 minutes of arc = 1° and 10⁴ Gauss = 1Tesla).

Solution.

Rotation per unit magnetic field

For glass: Tesla 0.2or Gauss 20001010'3

'60311

mmeterGaussL

B

For ZnS: Tesla 0.27or Gauss 2731010'22

'60311

mmeterGaussL

B


14

7.15 Transverse Pockels cell with LiNbO₃

Suppose that instead of the configuration in Figure 7.20, the field is applied along

the z-axis of the crystal, the light propagates along the y-axis. The x-axis is the

polarization of the ordinary wave and z-axis that of the extraordinary wave. Light

propagates through as o- and e-waves. Given that Ea = V/d, where d is the crystal

length along z, the indices are

aeee

aooo

Ernnn

Ernnn

33

3,

13

3,

2

1

2

1

Show that the phase difference between the o- and e-waves emerging from the

crystal is,

d

Vrnrn

Lnn

Loeoeoe 13

3

33

3

2

122

where L is the crystal length along the y-axis.

Explain the first and second terms. How would you use two such Pockels cells to

cancel the first terms in the total phase shift for the two cells?

If the light beam entering the crystal is linearly polarized in the z-direction, show

that

d

VrnLLn ee

2

22 33

3

Consider a nearly monochromatic light beam of the free-space wavelength λ = 500

nm and polarization along z-axis. Calculate the voltage Vπ needed to change the

output phase ∆ϕ by π given a LiNbO₃ crystal with d/L = 0.01 (see Table 7.2).

Solution.

Consider the phase change between the two electric field components


15

d

Vrnrn

Lnn

Loeoeoe 13

3

33

3

2

122

The first term is the natural birefringence of the crystal and occurs all the time,

even without applied electric field. The second term is the Pockels effect, applied

field inducing a charge in the refractive indices.

x

x

y

light

d

L

Ea

z

y

∆ϕ dEa

L

Light

z

Two transverse Pockels cell phase modulators together cancel the natural

birefringence in each crystal.

If the light is linearly polarized with its field along z, we only need to consider the

extraordinary ray (ϕ₀ = 0).

d

VrnLLn oee

2

22 13

3

The first term does not depend on the voltage. The voltage V that changes the

output phase by π is

d

VrnL o

2

2 13

3

Or

V 5.15108.30187.2

1050001.0

123

9

13

3

rnL

dV

o

7.21. Optical Kerr effect

Consider a material in which the polarization does not have the second order term:

3

3010 EEP or 2

310/ EEP


16

The first term with the electric susceptibility χ₁ corresponds to the relative

permittivity εr and hence to the refractive index no of the medium in the absence of

the third order term, i.e. under low fields. The E² term represents the irradiance I of

the beam. Thus, the refractive index depends on the intensity of the light beam, a

phenomenon called the optical Kerr effect:

2

322

4

3 and

o

on

nInnn

And η=(μ₀/ε₀)1/2 = 120π = 377 Ω, is the impedance of the free space.

a) Typically, for many glasses, χ₃≈ 10⁻²¹ m²/W; for many doped glasses, χ₃ ≈

10⁻¹⁸ m²/W; for many organic substances, χ₃ ≈ 10⁻¹⁷ m²/W; for

semiconductors, χ₃ ≈ 10⁻¹⁴ m²/W. Calculate n₂ and the intensity of light

needed to change n by 10⁻³ for each case.

b) The phase ϕ at a point z is given by

z

Inntz

nt

20

00

22

It is clear that the phase depends on the light intensity I and the change in the

phase along ∆z due to light intensity alone is

zIn

t

2

0

2

As the light intensity modulates the phase, this is called self-phase

modulation. Obviously light is controlling light.

When the light intensity is small n₂I ≪ n₀, obviously the instantaneous

frequency

0/ t

Suppose we have an intense beam and the intensity I is time dependent

I=I(t). Consider a pulse of light traveling along the z-direction and the light

intensity vs. t shape is a “Gaussian” (this is approximately so when a light

pulse propagate in an optical fiber, for example). Find the instantaneous

frequency ω. Is this still ω₀? How does the frequency change with “time”, or

across the light pulse? The change in the frequency over the pulse is called

chirping. Self-phase modulation therefore changes the frequency spectrum

of the light pulse during propagation. What is the significance of this result?


17

c) Consider a Gaussian beam in which intensity across the beam cross section

falls with radial distance in a Gaussian fashion. Suppose that the beam is

made to pass through a plate of nonlinear medium. Explain how the beam

can become self-focused? Can you envisage a situation where diffraction

effects trying to impose divergence are just balanced by self-focusing

effects?

Solution.

a)

The fractional change in the refractive index is

3

0

3

0

2

0

0

4

3

n

I

n

In

n

nnn

Thus 3

3

0

3

4

nI n

If 0.1%or 10 3n

For glasses: n₀ ≈ 1.5, χ₃ = 10⁻²¹ m²/W, I = 1.2·10¹⁶ W/m²

for doped glasses: n₀ ≈ 1.5, χ₃ = 10⁻¹⁸ m²/W, I = 1.2·10¹³ W/m²

for organics: n₀ ≈ 1.5, χ₃ = 10⁻¹⁷ m²/W, I = 1.2·10¹² W/m²

for semiconductors: n₀ ≈ 2.5, χ₃ = 10⁻¹⁴ m²/W, I = 5.5·10⁹ W/m²

large intensities.

b)

the phase ϕ at a point z is given by

z

Inntz

nt

20

00

22

So that the instantaneous frequency is

t

Izn

tt

2

0

2


18

When the light intensity is small n₂I≪n₀, ω = ∂ϕ/∂t=ω₀.

We have an intense beam with time dependent intensity I(t). Consider a pulse of

light traveling along the z-direction and the light intensity vs. t shape is a

“Gaussian”. Then the instantaneous frequency is,

t

Izn

tt

2

0

2

So that the frequency changes with “time”, or across the light pulse. Since ∂I/∂t is

rising at the leading edge and falling in the trailing edge, the two ends of the pulse

contain different frequencies. The change in the frequency over the pulse is called

chirping. Self-phase modulation therefore changes the frequency spectrum of the

light pulse during propagation. The second term above results in the broadening of

the frequency spectrum and hence leads to more dispersion for a Gaussian pulse

propagating in an optical fiber.

c)

If an intense Gaussian optical beam is transmitted through a plate of nonlinear

medium, it will change the refractive index in the medium with the maximum

change of the refractive index in the center of the beam. Therefore, the plate will

act as a graded-index medium and change the curvature of the wavefront. Under

certain conditions the plate can act as a lens with a power-dependent focal length,

producing a co-called self-focusing of the beam. Similarly, if an intense Gaussian

beam propagates through a nonlinear medium, the medium can act as a graded-

index waveguide. In this case, under certain conditions, the self-focusing effect can

compensate the divergence of the beam due to diffraction, and the beam will be

confined to its self-created waveguide. Such self-guided beams are called spatial

solutions.


19

Figure 1. propagation of the light through nonlinear medium.

Ref. http://www.absoluteastronomy.com/topics/Gradient_index_optics

The intensity variation across the beam cross section leads to a similar refractive

index variation in the nonlinear medium. Thus, the medium resembles a graded

index guide or a GRIN rod and can focus the beam.

(a) a light incident at the long face of a glass prism...

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