Chapter 6 - University of Texas at Dallasgoeckner/emag2_class... · Web viewA specific value of θi can be obtained from equation 6.21 - Or (6.23) From equations 6.22 & 6.23 – This

Chapter 6

*Note: The information below can be referenced to: Iskander, M., Electromagnetic Fields and Waves, Waveland Press, Prospect Heights, IL, 1992, ISBN: 1-57766-115-X. Edminister, J., Electromagnetics (Schaum’s Outline), McGraw-Hill, New York, NY, 1993, ISBN: 0-07-018993-5. Hecht, E., Optics (Schaum’s Outlines), McGraw-Hill, New York, NY, ISBN: 0-07-027730-3. Serway, R., & Faughn, J., College Physics, ISBN: 0-03-022952-9.

Chapter 6

Oblique Incidence Plane Wave Reflection and Transmission

6.1 Plane Wave Propagation at Arbitrary Angle

Plane waves are not normally incident, so now we must consider the general problem of a plane wave propagating along a specified axis that is arbitrarily relative to a rectangular coordinate system. The most convenient way is in terms of the direction cosines of the uniform plane wave, the equiphase surfaces are planes perpendicular to the direction of propagation.

Definitions:

uniform planes – a free space plane wave at an infinite distance from the generator, having constant amplitude electric and magnetic field vectors over the equiphase surfaces.

equiphase surface – any surface in a wave over which the field vectors of a particular instant have either 0° or 180° phase difference.

For a plane wave propagating along the +z axis

(6.1)

Equation (6.1) states that each z equal to a constant plane will represent an equiphase surface with no spatial variation in the electric or magnetic fields. In other words,

¶

¶

¶

¶

x

y

=

=

0

Þ

for a uniform plane wave

It will be necessary to replace z for a plane wave traveling in an arbitrary direction with an expression when put equal to a constant (βz = constant), that will result in equiphase surfaces.

The equation of an equiphase plane is given by

b

b

b

×

=

×

r

n

r

The radial vector (r) from the origin to any point on the plane, and β is the vector normal to the plane is shown in Figure (6.1).

As you can see from figure 6.1, the plane perpendicular to the vector β is seen from its side appearing as a line P-W. The dot product nβ · r is the projection of the radial vector r along the normal to the plane and will have the constant value OM for all points on the plane. The equation β · r = constant is the characteristic property of a plane perpendicular to the direction of propagation β.

The equiphase equation is

β · r = βxx + βyy + βzz

= β (cos θxx + cos θyy + cos θzz)

= constant

r =

x

a

y

a

z

a

x

y

z

+

+

b

b

b

b

=

+

+

x

x

y

y

z

z

a

a

a

θx, θy, θz, are the angles the β vector makes with x, y, and z axes, respectively.

Definition:

transverse electromagnetic wave (TEM) – electromagnetic wave having electric field vectors and magnetic field vectors perpendicular to the direction of propagation.

H is perpendicular to E, and both E and H are perpendicular to the direction of propagation β. The expressions for

EMBED Equation.2

E

H

Ù

Ù

and

are

$

$

E

E

=

-

×

m

j

r

e

b

(6.2)

$

$

H

E

=

Ù

n

b

h

$

(

)

$

E

E

z

e

a

m

j

z

x

=

-

b

The unit vector nβ along β and η is the wave impedance in the propagation medium. See Figure 6.2 for the illustration of orthogonal relations between

E

H

Ù

Ù

and

and the direction of propagation.

q

x

EXAMPLE 6.1

The vector amplitude of an electric field associated with a plane wave that propagates in the negative z direction in free space is given by

$

E

m

x

y

a

a

V

m

=

+

2

3

Find the magnetic field strength.

Solution:

The direction of propagation nβ is –az. The vector amplitude of the magnetic field is then given by

$

$

H

E

m

x

y

z

x

y

n

a

a

a

a

a

A

m

=

=

-

=

-

æ

è

ç

ö

ø

÷

Ù

b

h

h

1

0

2

0

3

1

0

1

377

3

2

o

*note

o

o

o

h

m

e

=

120π~377Ω (Appendix D – Table D.1)

EXAMPLE 6.2

The phasor electric field expression in a phase is given by

(

)

[

]

$

$

(

.

.

)

E

E

=

+

+

+

-

-

+

x

y

y

z

j

x

y

a

a

j

a

e

2

5

2.

3

0

6

0

8

Find the following:

1.

$

E

y

.

2. Vector magnetic field, assuming

m

m

e

e

=

=

o

o

and

.

3. Frequency and wavelength of this wave.

4. Equation of surface of constant phase.

Solution:

1. The general expression for a uniform plane wave propagating in an arbitrary direction is given by

$

$

E

E

=

-

×

m

j

r

e

b

where the amplitude vector

$

E

m

, in general, has components in the x, y, and z directions. Comparing equation 6.3 with the general field equation for the plane wave propagating in an arbitrary direction, we obtain

β · r = βxx + βyy + βzz

= β (cos θxx + cos θyy + cos θzz)

= 2.3(-0.6x + 0.8y + 0)

Hence, a unit vector in the direction of propagation nβ is given by

nβ = -0.6ax + 0.8ay.

Because the electric field

$

E

must be perpendicular to the direction of propagation nβ, it must satisfy the following relations:

nβ ·

$

E

= 0

Therefore, (-0.6ax + 0.8ay) ·

(

)

[

]

x

y

y

z

a

a

j

a

+

+

+

=

$

E

2

5

0

Or

-0.6 + 0.8

$

E

y

= 0

Hence,

$

E

y

= 0.75. The electric field is given by

(

)

[

]

$

$

(

.

.

)

E

E

=

+

+

+

-

-

+

x

y

y

z

j

x

y

a

a

j

a

e

2

5

2.

3

0

6

0

8

2. The vector magnetic field

H

Ù

is given by

$

$

.

.

.

H

E

=

Ù

=

-

+

1

1

377

0

6

0

8

0

1

0

75

2

5

h

b

n

a

a

a

j

x

y

z

so that

(

)

$

.

(

)

.

.

H

x

j

j

=

+

=

+

*

-

0

8

2

5

377

4

24

10

6

10

3

$

H

y

EMBED Equation.2

(

)

=

+

=

-

*

-

0

6

2

5

377

3

18

7

95

10

3

.

(

)

.

.

j

j

$

H

z

EMBED Equation.2

(

)

=

+

=

-

*

-

0

6

0

75

0

8

377

3

31

10

3

.

.

.

.

The vector magnetic field is then given by

(

)

$

$

$

$

H

H

H

H

=

+

+

x

x

y

y

z

z

a

a

a

-

-

+

j

x

y

e

2.

3

0

6

0

8

(

.

.

)

3.The wavelength λ is given by

l

p

b

p

=

=

=

2

2

2

3

2

73

.

.

m

and the frequency

f

=

=

*

=

c

GHz

l

3

10

2

73

0

11

8

.

.

4.The equation of the surface of constant phase is

nβ · r = -0.6x + 0.8y = constant

The general expression of this equation in terms of the direction cosines is given by

nβ · r = (cos θxx + cos θyy + cos θzz) = constant

Comparison between equation 6.4 and the general expression shows that the plane given in equation 6.4 has no z dependence and, hence defines a plane parallel to the z axis. In other words, equation 6.4 can be obtained by substituting θx = π/2 in the general expression of the equiphase plane.

6.2 Reflection by Perfect Conductor – Arbitrary Angle of Incidence

By decomposing the general problem into two special cases we can simplify our analysis.

1. E field is polarized in the plane formed by the normal to the reflecting surface in the direction βi of the incident wave.

2. E field is perpendicular to the plane of incidence.

The plane formed by the normal to the reflecting surface and the direction of propagation β is known as the plane of incidence. The general case can be considered as a superposition of two cases –

· E is parallel to the plane of incidence

· E is perpendicular to the plane of incidence

6.2.1 E Field Parallel to Plane of Incidence

q

y

The figure shows an incident wave polarized with the E field in the plane of incidence and the power flow in the direction of

b

i

at angle

q

i

with respect to the normal to the surface of the perfect conductor.

The direction of propagation is given by the Poynting vector and the

b

i

, E, and H fields need to be arranged so that

b

i

is in the same direction as

E

H

i

i

Ù

at any time. The magnetic field is out of the plane of the paper,

H

H

=

$

y

y

a

for the direction of the electric field shown. There is no transmitted field within the perfect conductor; however there will be a reflected field with power flow at the angle

q

r

with respect to the normal to the interface. To maintain the power density flow

E

H

r

r

Ù

will be in the same direction

b

r

as. The expression for the total electric field in free space is

$

$

$

$

$

E

E

E

E

E

=

+

=

+

-

×

-

×

i

r

m

i

j

i

r

m

r

j

r

r

e

e

b

b

(6.5)

(

)

(

)

i

i

z

i

x

x

y

z

r

a

a

x

a

y

a

z

a

b

b

q

q

×

=

+

×

+

+

cos

sin

(

)

=

+

b

q

q

x

z

i

i

sin

cos

(6.6)

(

)

r

r

r

r

x

z

b

b

q

q

×

=

-

sin

cos

(6.7)

The total electric field has x and z components:

(

)

$

,

$

cos

$

cos

E

E

E

x

m

i

i

j

i

r

m

r

j

r

r

r

x

z

e

e

=

+

-

×

-

×

q

q

b

b

(

)

$

,

$

sin

$

sin

E

E

E

z

m

i

i

j

i

r

m

r

j

r

r

r

x

z

e

e

=

-

+

-

×

-

×

q

q

b

b

$

$

cos

$

cos

E

E

E

x

at

z

m

i

i

j

i

r

m

r

j

r

r

r

e

e

=

-

×

-

×

=

-

=

0

0

q

q

b

b

=

-

=

-

×

-

×

$

cos

$

cos

sin

sin

E

E

m

i

i

j

x

m

r

j

x

r

r

i

e

e

q

q

b

b

q

q

0

(6.8)

Equation 6.8 shows the relationship between the incident and reflected amplitudes for a perfect conductor the total tangential E field at the surface must be zero which satisfies the boundary condition. To be zero at all values of x along the surface of the conducting plane, the phase terms must be equal to each other –

q

q

i

r

=

(6.9)

Equation 6.9 is known as Snell’s law of reflection.

Definition:

Snell’s Law is a rule of Physics that applies to visible light passing from air (or vacuum) to some medium with an index of refraction different from air.

Substitute equation 6.9 into equation 6.8 –

$

$

E

E

m

i

m

r

=

(6.10)

Therefore, the total electric field in free space is

(

)

(

)

$

(

,

)

$

,

$

,

E

E

E

x

z

x

z

a

x

z

a

x

x

z

z

=

+

(

)

=

-

-

-

$

cos

sin

cos

cos

E

mi

i

j

x

i

j

z

i

j

z

i

x

e

e

e

a

q

b

q

b

q

b

q

(

)

-

+

-

-

$

sin

sin

cos

cos

E

m

i

i

j

x

i

j

z

i

j

z

i

e

e

e

z

a

q

b

q

b

q

b

q

=

-

2

j

z

e

a

m

i

i

j

x

i

i

x

$

cos

sin

cos

(

)

sin

E

q

b

q

b

q

(6.11)

(

)

-

-

2

j

z

e

a

m

i

i

i

j

x

i

z

$

sin

cos

cos

sin

E

q

b

q

b

q

[

=

-

2

$

cos

sin

cos

(

)

E

m

i

i

i

x

j

z

a

q

b

q

(

)

]

-

-

sin

cos

cos

sin

i

i

z

j

x

i

z

a

e

q

b

q

b

q

Take equation 6.11 and recover the time-domain form of the total electric field

(

)

(

)

(

)

E

E

r

t

r

e

j

t

,

Re

$

=

v

Observe the variation of the total field with the x variable indicating there is a traveling wave in the x direction with a phase constant

b

b

q

x

i

=

sin

And in the z direction the field forms a standing wave.

The total magnetic field is

(

)

(

)

(

)

(

)

$

,

$

,

$

,

$

,

H

H

H

H

x

z

x

z

a

x

z

a

x

z

a

y

y

y

i

y

y

r

y

=

=

+

Use the relation

H

E

=

Ù

n

b

h

for each of the incident and reflected fields to employ the expressions x and z components of the incident and reflected electric fields.

$

$

H

E

i

i

i

n

=

Ù

b

h

=

-

-

+

-

+

1

0

0

h

q

q

q

q

b

q

q

b

q

q

x

i

m

i

i

j

i

x

i

z

y

z

i

m

i

i

i

j

i

x

i

z

a

e

a

a

e

sin

$

cos

cos

$

sin

(sin

cos

)

(sin

cos

E

E

The solution of the determinant, the only nonzero component of

$

H

i

is the ay component given by

(

)

(

)

[

]

$

$

cos

$

sin

sin

cos

sin

cos

H

E

E

i

y

m

i

i

j

i

x

i

z

m

i

i

j

i

x

i

z

a

e

e

=

+

-

+

-

+

1

2

2

h

q

q

b

q

q

b

q

q

(

)

=

-

+

$

sin

cos

E

m

i

j

i

x

i

z

y

e

a

h

b

q

q

The reflected magnetic fields is given by

(

)

$

$

sin

cos

H

E

r

m

i

j

i

x

i

z

y

e

a

=

-

-

h

b

q

q

The total magnetic field

H

Ù

(x, z) is

(

)

$

(

,

)

$

cos

cos

sin

H

E

x

z

a

z

e

y

m

i

i

j

x

i

=

-

2

h

b

q

b

q

The average power flow parallel to the conducting surface is

(

)

[

]

ave

x

z

R

E

H

,

Re

$

$

=

Ù

1

2

=

*

1

2

0

0

0

Re

$

$

$

x

x

y

y

z

z

a

a

a

E

H

E

The cross product yields two components:

· One in the x direction

· One in the z direction

[

]

ave

z

y

x

x

y

z

a

a

R

E

H

E

H

=

-

+

*

*

1

2

Re

$

$

$

$

The expression of Pave will reduce to

(

)

[

]

ave

z

y

x

x

z

a

R

E

H

,

Re

$

$

=

-

*

1

2

[

]

2

2

$

sin

cos

cos

E

m

i

i

i

x

z

a

h

q

b

q

Glancing Incident:

(

)

(

)

i

ave

m

i

x

a

q

h

®

=

æ

è

ç

ö

ø

÷

o

90

2

2

,

$

R

E

, the power flow is at maximum.

Normal Incident:

i

x

ave

q

=

=

0

0

,

,

R

(Power flow in the x direction is zero)

Average power flow perpendicular to the conducting surface is zero, because the average Poynting Vector is zero in that direction

(

)

z

ave

x

y

P

,

Re

$

$

=

=

*

1

2

0

E

H

Why? Because

$

E

x

is multiplied by j, therefore

$

$

E

H

x

y

and

are out of phase by 90°. Therefore, a traveling-wave pattern occurs in the x direction, because the incident and reflected waves travel in the same direction, the standing-wave pattern will be observed in the z direction, because the incident and reflected waves travel in the opposite directions.

The location of zeros (nodes) of the

$

E

x

field can be found by letting sin

(

)

b

q

z

i

cos

= 0. At a distance z from the conducting plane given by

b

q

p

z

n

i

cos

=

Or

z = n

l

q

2

0

1

2

cos

,

,

,

.

.

.

i

n

=

The zeros will occur at distances larger than integer multiples of

l

2

. So, for normal incidence,

q

i

=

0

,

cos

q

i

=

1

, and the positions of the zeros will are the same as those discussed in chapter 5. For the oblique incidence, the locations of the standing-wave nodes are

l

2

apart along the direction of propagation. The wavelength measured along the z-axis is greater than the wavelength of the incident waves along the direction of propagation. As shown in Figure 6.4 the relation between these wavelengths is

z

i

l

l

q

=

cos

.

q

z

b

b

b

=

n

$

H

y

E

i

E

r

The plane of the zero

$

E

x

field occur at multiples of

l

2

along the direction of propagation, and they are located at integer multiples of

l

z

2

along the z-axis which appear separated by larger distances. Also note that the standing-wave pattern associated with the

$

E

z

component may appear as if there is no zero value of the electric field at

z = 0, but the

$

E

z

component is normal to the reflecting surface, therefore the boundary condition is not in violation.

6.2.2 Electric Field Normal to the Plane of Incidence

The entire electric field is (out of the paper) in the y direction and the magnetic field will have both x and z components. See Figure 6.5.

The incident electric and magnetic fields are

$

$

E

E

i

m

i

j

i

r

e

=

-

×

b

(

)

$

$

$

cos

sin

H

E

E

i

i

m

i

i

x

i

z

j

i

r

n

a

a

e

i

=

=

-

+

Ù

-

×

b

b

h

h

q

q

b

r

where

(

)

b

b

q

q

i

i

i

r

x

z

×

=

-

sin

cos

. Assume that the reflected field is also in the y direction so the magnetic field must be perpendicular to both E and the Poynting Vector P = E ^ H,

$

$

E

E

r

m

r

j

r

y

e

a

r

=

-

×

b

(

)

$

$

$

cos

sin

H

E

E

r

r

m

r

r

x

r

z

j

r

r

n

a

a

e

r

=

Ù

=

+

Ù

-

×

b

b

h

h

q

q

Where

(

)

b

b

q

q

r

r

r

r

x

z

×

=

-

sin

cos

.

Determine the angle of reflection

q

r

and the amplitude of the reflected electric field

$

E

m

r

by using the boundary conditions at z = 0. This also includes zero values of the tangential electrical field E and the normal component of the magnetic field H.

(

)

$

,

$

$

E

E

E

y

y

i

y

r

x

z

=

+

=

0

at z = 0

Therefore,

(

)

$

,

$

$

sin

sin

E

E

E

y

m

i

j

x

i

m

r

j

x

r

x

e

e

0

0

=

+

=

-

-

b

q

b

q

And

(

)

$

,

$

sin

$

sin

sin

sin

H

E

E

z

m

i

i

j

x

i

m

r

i

j

x

r

x

e

e

0

1

1

0

=

+

=

-

-

h

q

h

q

b

q

b

q

Note: These two conditions will provide the same results for the unknowns

r

m

r

and

q

$

E

, and be true for every value of x along z = 0 plane, so the phase factors must be equal.

q

q

r

i

=

And

$

$

E

E

m

r

m

i

=

-

Negative sign indicates the opposite direction of the reflected electric field (i.e. into the paper)

The total E field is

(

)

(

)

$

,

$

sin

cos

cos

E

E

y

m

i

j

x

i

j

z

i

j

z

i

x

z

e

e

e

=

-

-

-

b

q

b

q

b

q

(

)

[

]

=

-

-

2

j

z

e

m

i

i

j

x

i

$

sin

cos

sin

E

b

q

b

q

The total H field is

$

$

$

$

H

H

H

E

=

+

=

Ù

é

ë

ê

ù

û

ú

-

-

×

i

r

y

m

i

j

i

r

n

a

e

i

b

b

h

EMBED Equation.2

n

a

e

i

y

m

i

j

i

r

b

b

h

Ù

é

ë

ê

ê

ù

û

ú

ú

-

×

$

E

And the substitution of

$

$

E

E

m

r

m

i

=

has been made. The direction vectors of the incident and reflective wave are

b

q

q

i

r

i

x

i

z

n

a

a

,

sin

cos

=

±

And

b

q

q

i

r

y

i

z

i

x

n

a

a

a

,

sin

cos

Ù

=

m

The components of the total magnetic field are

(

)

(

)

$

,

$

cos

cos

cos

sin

H

E

x

m

i

i

i

j

x

i

x

z

z

e

=

-

-

2

h

q

b

q

b

q

(

)

(

)

&

&

,

$

sin

sin

cos

sin

H

E

z

m

i

i

i

j

x

i

x

z

j

z

e

=

-

-

2

h

q

b

q

b

q

There is a standing-wave in the z direction because the reflected and incident waves travel in the opposite direction along the z-axis. The fields traveling in the x direction and having the only nonzero power flow in the direction parallel to the interface.

The concept can be illustrated by considering the average density flow associated with the wave.

(

)

[

]

ave

x

z

R

E

H

,

Re

$

$

=

Ù

*

1

2

=

1

2

0

0

0

Re

a

a

az

T

S

W

x

y

(

)

=

2

2

$

sin

sin

cos

E

m

i

i

i

x

z

a

h

q

b

q

Þ

This indicates that the power flow is in the x direction.

(

)

S

z

e

m

i

i

i

j

x

i

=

-

+

2

$

cos

cos

cos

sin

E

h

q

b

q

b

q

,

(

)

T

z

e

m

i

i

j

x

i

=

-

-

2

$

sin

cos

sin

E

b

q

b

q

,

(

)

W

z

e

m

i

i

i

j

x

i

=

+

*

+

2

$

sin

sin

cos

sin

E

h

q

b

q

b

q

EXAMPLES:

Find the peak value of an induced surface current when a plane wave is incident at am angle on a large plane, perfectly conducting sheet. The surface of the sheet is located at z = 0 and

$

cos

E

i

y

t

x

z

a

V

m

=

-

-

æ

è

ç

ö

ø

÷

10

10

2

2

10

b

b

Solution

From the equation of the incident electric field, the propagation vector is given by

b

b

b

=

+

2

2

x

z

a

a

(

)

=

+

b

sin

cos

o

o

45

45

x

a

z

a

,

that is,

q

i

=

45

o

Because the electric field is along the y direction – that is, perpendicular to the plane of incidence, the equations given in the section above will be used.

The sheet current

$

J

(in ampere per meter) is determined by the total tangential magnetic field at the surface. From the boundary condition,

$

$

J

n

=

Ù

H

where the normal n to the surface for the geometry of Figure 6.5 is n = -az. The magnetic field in this case has two components:

(

)

$

$

cos

cos

cos

sin

H

E

x

m

i

i

i

j

x

i

z

e

=

-

-

2

h

q

b

q

b

q

(

)

$

$

sin

sin

cos

sin

H

E

z

m

i

i

i

j

x

i

j

z

e

=

-

-

2

h

q

b

q

b

q

The surface is then current is then

$

$

$

cos

sin

J

a

a

e

at

z

z

y

m

i

i

j

x

i

=

=

-

Ù

=

Ù

-

0

2

H

E

h

q

b

q

And the peak value of the surface current at z = 0 is given by

$

$

cos

(

)

cos

.

J

x

A

m

peak

value

m

i

i

=

=

=

-

2

2

10

45

377

3

75

10

2

E

h

q

o

EXAMPLE:

The electric field associated with a plane wave propagating in an arbitrary direction is given by

$

(

.

.

)

(

.

.

)

E

=

+

-

-

+

7

83

4

4

5

7

0

5

0

87

x

y

z

j

x

z

a

a

a

e

If this incident on a perfectly conducting plane oriented perpendicular to the z axis, find the following:

1. Reflected electric field.

2. Total electric field in region in front of the perfect conductor.

3. Total magnetic field.

Solution

Because a vector in the direction of propagation and a unit vector normal to the reflecting surface are contained in the x-z plane, we consider the x-z plane to be the plane of incidence as shown in Figure 6.6. The given electric field may, therefore, be decomposed into two components. The parallel polarization case in which the electric field is perpendicular to the plane of incidence

$

|

|

E

and the perpendicular polarization case in which the electric field is perpendicular to the plane of incidence

$

E

^

. From the given equation of the electric field,

$

(

.

.

)

|

|

(

.

.

)

E

=

-

-

+

7

83

4

5

7

0

5

0

87

x

z

j

x

z

a

a

e

Comparing this with the equation of the electric field in the parallel polarization case, where the incident electric field is given by

$

$

(cos

sin

)

(sin

cos

)

E

E

i

m

i

i

x

i

j

i

x

i

z

a

e

=

-

-

+

q

q

b

q

q

$

H

y

l

q

l

cos

i

z

=

Figure 6.6

Observe that:

cos

.

sin

.

i

i

q

q

=

=

ü

ý

þ

0

5

0

87

That is,

q

i

=

30

o

The magnitude of the incident electric field

$

E

m

i

is therefore = 7.83/0.87 = 9 or 4.5/0.5 = 9. Hence, the electric field associated with the parallel polarization case can be expressed in the form

$

(

.

.

)

|

|

(

.

.

)

E

i

x

z

j

x

z

a

a

e

=

-

-

+

9

0

87

0

5

0

5

0

87

Based on the analysis of section 6.2.1, we have

q

r

=

30

o

, and the amplitude of the reflected electric field

$

$

|

|

|

|

E

E

r

i

=

=

9

. Hence

$

(cos

sin

)

(sin

cos

)

E

r

x

z

j

x

z

a

a

e

=

-

-

-

9

30

30

7

30

30

o

o

o

o

We treat the perpendicular polarization case where

$

(

.

.

)

E

^

-

+

=

i

y

j

x

z

a

e

4

7

0

5

0

87

Based on the analysis of section 6.2.2, it can be shown that

$

(

.

.

)

E

^

-

+

=

-

r

y

j

x

z

a

e

4

7

0

5

0

87

The total reflected electric field is then

$

(

.

.

)

(

.

.

)

E

r

x

y

z

j

x

z

a

a

a

e

=

-

-

-

-

-

7

83

4

4

5

7

0

5

0

87

Parts 2 and 3 can easily be obtained by the following the analysis of section 6.2.

For example, the magnetic field associated with the electric field in the parallel polarization case is given by

$

|

|

(

.

.

)

H

i

j

x

z

y

e

a

=

-

+

9

7

0

5

0

87

h

The reflected magnetic field intensity for this polarization is

$

|

|

(

.

.

)

H

r

j

x

z

y

e

a

=

-

+

9

7

0

5

0

87

h

For the perpendicular polarization case, the magnetic field has two components,

$

(

cos

sin

)

(

.

.

)

H

^

-

+

=

-

+

i

i

x

i

z

j

x

z

a

a

e

4

7

0

5

0

87

h

q

q

=

-

+

æ

è

ç

ö

ø

÷

-

+

4

4

7

0

5

0

87

h

q

h

q

cos

sin

(

.

.

)

i

x

i

z

j

x

z

a

a

e

Because, for the case,

$

$

E

E

^

^

=

r

i

$

(cos

sin

)

(

.

.

)

H

^

-

+

=

-

+

r

x

z

j

x

z

a

a

e

4

30

30

7

0

5

0

87

h

o

o

The total reflected magnetic field is then

$

(

cos

sin

)

(

.

.

)

H

r

x

y

z

j

x

z

a

a

a

e

=

-

-

+

-

-

+

1

4

30

9

4

30

7

0

5

0

87

h

o

o

6.3 Reflection and Refraction at Plane Interface between Two Media: Oblique Incidence

Figure 6.7 shows two media with electrical properties

e

1

and

m

1

in medium 1, and

e

2

and

m

2

in medium 2. Here a plane wave incident angle

q

i

on a boundary between the two media will be partially transmitted into and partially reflected at the dielectric surface. The transmitted wave is reflected into the second medium, so its direction of propagation is different from the incidence wave. The figure also shows two rays for each the incident, reflected, and transmitted waves. A ray is a line drawn normal to the equiphase surfaces, and the line is along the direction of propagation.

q

r

q

i

Figure 6.7

The incident ray 2 travels the distance CB, while on the contrary the reflected ray 1 travels the distance AE. For both AC and BE to be the incident and reflected wave fronts or planes of equiphase, the incident wave should take the same time to cover the distance AE. The reason being that the incident and reflected wave rays are located in the same medium, therefore their velocities will be equal,

CB

V

AE

V

1

2

=

OR

AB

AB

i

r

sin

sin

q

q

=

With this being the case then it follows that

q

q

i

r

=

What is the relationship between the angles of incidence

q

i

and refraction

q

r

?

It takes the incident ray the equal amount of time to cover distance CB as it takes the refracted ray to cover distance AD –

CB

V

AD

V

1

2

=

And the magnitude of the velocity V1 in medium 1 is:

1

1

1

1

V

=

*

m

e

And in medium 2:

2

2

2

1

V

=

*

m

e

Also,

CB

AB

AD

AB

i

i

=

=

sin

sin

q

q

Therefore,

CB

AD

V

V

i

t

=

=

=

*

*

sin

sin

q

q

m

e

e

m

1

2

2

2

1

1

For most dielectrics

m

m

m

2

1

=

=

o

Therefore,

sin

sin

i

t

q

q

e

e

m

m

m

=

=

=

2

1

1

2

o

(6.12)

Equation 6.12 is known as Snell’s Law of Refraction.

6.3.1 Parallel Polarization Case – E is in Plane of Incidence

b

r

$

E

y

r

b

i

$

E

y

i

The unknown amplitudes of the reflected and transmitted electric fields

|

|

|

|

r

t

and

E

E

can be determined by simply applying the boundary conditions at the dielectric interface. The electric fields

|

|

|

|

r

t

and

E

E

will now be used in the analysis to emphasize the case of parallel polarization, instead of using the electric fields

m

m

t

r

and

E

E

.

The tangential component of H should be continuous across the boundary. Therefore,

$

|

|

H

i

j

i

r

y

e

a

-

×

+

b

EMBED Equation.2

$

|

|

H

r

j

i

r

y

e

a

-

×

=

b

EMBED Equation.2

$

|

|

H

t

j

i

r

y

e

a

-

×

b

There is no need to carry the ay vector, because the magnetic fields only have one component in the y direction. Recall that this relation is valid at z = 0,

$

|

|

(sin

)

H

i

j

i

i

x

e

-

+

b

q

EMBED Equation.2

$

|

|

(sin

)

H

r

j

i

i

r

x

e

-

=

b

q

EMBED Equation.2

$

|

|

(sin

)

H

t

j

i

t

x

e

-

b

q

(6.13)

b

1

&

b

2

1

are the magnitudes of

b

in regions 1 & 2, respectively. In order for this to be valid at any value of x at any point on the interface, and knowing

q

q

i

r

=

:

b

q

b

q

1

2

sin

sin

i

t

=

Or

sin

sin

i

t

V

V

V

V

q

q

b

b

w

w

=

=

=

2

1

2

1

1

2

* This is the same relation that was determined earlier from Snell’s Law. Substitute

sin

sin

i

t

V

V

q

q

=

1

2

into equation 6.13 to obtain

$

|

|

H

i

+

$

|

|

H

r

=

$

|

|

H

t

At z = 0

(6.14)

E and H are related by

h

, so equation 6.14 can be rewritten as

$

|

|

E

i

+

EMBED Equation.2

$

|

|

E

r

=

1

2

h

h

$

|

|

E

t

(6.15)

Tangential components of E must be continuous across the boundary, therefore

$

cos

|

|

E

i

i

q

-

EMBED Equation.2

$

cos

|

|

E

r

r

q

=

EMBED Equation.2

$

cos

|

|

E

t

t

q

EMBED Equation.2 At z = 0

(6.16)

*Remember the exponential terms cancel out z = 0, (Snell’s Law).

Equations 6.15 & 6.16 are solved by –

$

|

|

E

r

=

EMBED Equation.2

$

cos

cos

cos

cos

|

|

`

E

i

i

t

i

t

1

2

1

2

h

q

h

q

h

q

h

q

-

+

And

$

|

|

E

t

EMBED Equation.2

=

+

$

cos

cos

cos

|

|

`

E

i

i

i

t

2

1

2

2

h

q

h

q

h

q

(6.17)

*Making use of the fact that

q

q

i

r

=

. Define the reflection coefficient

$

|

|

G

and the transmission

$

|

|

T

:

$

$

|

|

|

|

|

|

G

E

E

=

r

i

EMBED Equation.2

=

-

+

=

-

+

=

=

2

1

2

1

2

1

2

1

1

2

h

q

h

q

h

q

h

q

q

e

e

q

q

e

e

q

m

m

m

cos

cos

cos

cos

cos

cos

cos

cos

t

i

t

i

t

i

t

i

o

And

$

$

|

|

|

|

|

|

T

E

E

=

=

t

i

2

2

2

1

2

1

2

1

1

2

h

q

h

q

h

q

h

q

q

q

e

e

q

m

m

m

cos

cos

cos

cos

cos

cos

cos

t

t

t

t

i

i

i

-

+

=

+

=

=

o

The total electric field in region 1 is

$

|

|

E

tot

=

EMBED Equation.2

$

|

|

E

i

+

EMBED Equation.2

$

|

|

E

r

=

$

(cos

sin

)

E

m

i

i

x

i

z

j

i

r

a

a

e

q

q

b

-

-

×

+

$

(

cos

sin

)

E

m

r

r

x

r

z

j

r

r

a

a

e

-

-

-

×

q

q

b

=

Ù

-

cos

$

sin

i

m

i

j

x

i

e

q

b

q

E

EMBED Equation.2

(

cos

-

j

z

i

e

b

q

EMBED Equation.2

+

Ù

|

|

)

cos

G

j

z

i

e

a

x

b

q

+

-

-

sin

$

sin

i

m

i

j

x

i

e

Traveling

wave

part

q

b

q

E

1

2

4

4

4

4

3

4

4

4

4

(

)

-

+

-

e

e

a

j

z

j

z

z

S

ding

plus

travelingw

aves

i

i

b

q

b

q

cos

cos

tan

$

|

|

G

1

2

4

4

4

4

4

4

3

4

4

4

4

4

4

(6.18)

Substituted

b

b

i

r

r

r

×

×

,

from expressions derived earlier, and

$

$

$

|

|

E

E

G

m

r

m

i

=

.

Equation 6.18 states that there is a traveling-wave field in the x direction, and a traveling and standing wave field in the z direction. The difference is that

$

|

|

G

¹

1

, but that

$

|

|

G

=

EMBED Equation.2

-

$

$

E

E

m

r

m

i

. By rearranging the second term in ax component of the total field –

(

)

(

)

[

]

1

2

-

+

-

$

$

cos

|

|

|

|

cos

G

G

j

z

i

i

e

z

b

q

b

q

This expression indicates that a wave of amplitude

(

)

1

-

$

|

|

G

is propagating in the z direction and another wave of amplitude

(

)

2

$

|

|

G

has the characteristics of a standing wave along the z axis. The characteristic of the wave along the z axis is a combination of a traveling and standing wave. If

$

|

|

G

=

1

the amplitude of the traveling wave will be zero, and the wave characteristic along the z axis will be a totally standing wave. If

$

|

|

G

=

0

, the amplitude of the standing wave will be zero and the wave characteristic in the z direction would be a totally traveling wave.

The magnetic field in region 1 is

$

|

|

H

tot

=

EMBED Equation.2

$

$

|

|

|

|

H

H

i

r

+

=

EMBED Equation.2

$

H

m

i

j

i

r

y

e

a

-

×

b

+

$

H

m

r

j

r

r

y

e

a

-

×

×

b

=

$

sin

E

m

i

j

x

i

e

h

b

q

1

-

-

+

j

z

i

e

b

q

cos

(

EMBED Equation.2

$

$

cos

)

E

E

m

i

m

r

j

x

i

y

e

a

-

b

q

=

Ù

-

$

sin

E

m

i

j

x

i

e

h

b

q

1

EMBED Equation.2

(

cos

-

j

z

i

e

b

q

EMBED Equation.2

-

$

)

|

|

cos

G

j

z

i

y

e

a

b

q

The transmitted fields in medium 2 are

(

)

$

$

cos

sin

|

|

E

E

i

m

t

t

x

t

z

j

r

a

a

e

t

=

-

-

×

q

q

b

=

(

)

$

$

cos

sin

|

|

T

E

m

i

t

x

t

z

j

r

a

a

e

t

q

q

b

-

-

×

And

$

$

$

$

|

|

|

|

H

H

T

E

t

m

t

y

j

r

m

i

j

r

y

a

e

e

a

t

t

t

=

=

-

×

Ù

-

×

×

b

b

h

2

Where

(

)

b

b

q

q

t

t

t

r

x

z

×

=

+

2

sin

cos

and

$

$

|

|

E

E

T

m

t

m

i

=

.

Definition:

Brewster Angle – (from Brewster’s Law), the polarizing angle of which (when light is incident) the reflected and refracted index is equal to the tangent of the polarizing angle. In other words, the angle of incidence of which there is no reflection.

From the reflection coefficient expression-

$

|

|

G

=

EMBED Equation.2

2

1

2

1

h

q

h

h

q

h

q

q

cos

cos

cos

cos

t

t

i

i

-

+

It can be seen that there is an angle of incidence at

$

|

|

G

=

0

. This angle can be obtained when

1

2

h

q

h

q

cos

cos

i

t

=

Or

cos

cos

i

t

q

h

h

q

=

2

1

(6.19)

The angle of incidence

q

i

, at which

$

|

|

G

=

0

, is known as the Brewster angle. The expression for this angle in terms of the dielectric properties of media 1 & 2, considering Snell’s Law for the special case

m

m

m

1

2

=

=

o

is

sin

sin

i

t

V

V

q

q

e

e

m

m

m

=

=

=

1

2

2

1

1

2

o

This condition is important, because it is usually satisfied by the materials often used in optical applications.

Equation 6.19 will take the form –

i

t

cos

cos

q

e

e

q

=

1

2

(6.20)

Square both sides of equation 6.20 and use Snell’s Law for the special case of

m

m

m

1

2

=

=

o

for the following result:

2

1

2

cos

i

q

e

e

EMBED Equation.2

(

)

=

=

-

2

1

2

2

1

cos

sin

t

t

q

e

e

q

(

)

=

-

1

2

1

2

e

e

q

sin

i

The last substitution was based on Snell’s Law of refraction. Therefore,

(

)

1

2

-

=

sin

i

q

1

2

e

e

-

1

2

2

2

2

e

e

q

sin

i

1

1

2

-

=

e

e

EMBED Equation.2

2

sin

i

q

EMBED Equation.2

1

1

2

2

2

-

æ

è

ç

ç

ö

ø

÷

÷

e

e

And

2

2

2

1

sin

i

q

e

e

e

=

+

(6.21)

The Brewster angle of incidence is

sin

i

q

e

e

e

=

+

2

2

1

(6.22)

A specific value of θi can be obtained from equation 6.21 -

1

2

2

2

1

-

=

+

cos

i

q

e

e

e

Or

2

2

2

1

1

cos

i

q

e

e

e

=

-

+

=

EMBED Equation.2

1

2

1

e

e

e

+

=

cos

i

q

e

e

e

=

+

1

2

1

(6.23)

From equations 6.22 & 6.23 –

tan

i

q

e

e

=

2

1

This specific angle of incidence

q

i

is called the Brewster angle

q

b

.

b

q

e

e

=

-

1

2

1

tan

6.3.2 Perpendicular Polarization case – E Normal to Plane of Incidence

As shown in figure 6.10 is a perpendicular polarized wave incident at angle

q

i

a dielectric medium 2. Snell’s Law states that a reflected wave will be at the same angle

q

q

r

i

=

, and the transmitted wave in medium 2 at angle

q

t

can be calculated using this law. The amplitude of the reflected and transmitted waves can be determined by applying the continuity of the tangential components of E & H at the boundary.

This is given by –

$

cos

H

^

i

i

q

EMBED Equation.2

-

^

$

cos

H

r

i

q

=

$

cos

H

^

t

t

q

i

H

e

m

2

2

,

e

m

1

1

,

Since E & H are related by

h

,

$

cos

E

^

i

i

1

h

q

EMBED Equation.2

-

=

^

$

cos

E

r

i

1

h

q

EMBED Equation.2

$

cos

E

^

t

t

2

h

q

(6.24)

$

E

^

i

EMBED Equation.2

+

^

$

E

r

EMBED Equation.2

=

^

$

E

t

At z = 0

(6.25)

*Note: The exponential factors were canceled after substituting z = 0 and using Snell’s Laws in the above two equations.

$

G

^

=

EMBED Equation.2

$

$

E

E

^

^

r

i

=

2

1

2

1

h

q

h

q

h

q

h

q

cos

cos

cos

cos

i

t

i

t

-

+

And for nonmagnetic materials,

m

m

m

1

2

=

=

o

,

$

G

^

=

cos

cos

cos

cos

t

t

t

t

q

e

e

q

q

e

e

q

-

+

2

1

2

1

at z = 0,

$

T

^

=

EMBED Equation.2

$

$

E

E

^

^

t

i

=

2

2

1

2

h

q

h

q

h

q

cos

cos

cos

i

i

t

+

For nonmagnetic material,

$

cos

cos

cos

T

^

=

2

2

1

i

i

t

q

q

e

e

q

6.4 Comparison between Reflection Coefficients

$

|

|

G

and

$

G

^

for Parallel and Perpendicular Polarizations

The significant differences between the two will be illustrated in the following example:

EXAMPLE

1. Define what is meant by the Brewster angle.

2. Calculate the polarization angle (Brewster angle) for an air water

(

)

e

r

=

81

interface at which plane waves pass from the following:

(a) Air into water.

(b) Water into air.

SOLUTION

1. Brewster angle is defined as the angle of incidence at which there will be no reflected wave. It occurs when the incident wave is polarized such that the E field is parallel to the plane of incidence.

2. (a) Air into water:

e

e

r

r

and

1

2

1

81

=

=

The Brewster angle is then given by

b

q

e

e

=

-

1

2

1

tan

= 6.34°

Therefore,

b

q

=

-

1

81

tan

= 83.7°

(b) Water into air:

e

e

r

r

and

1

2

8

1

1

=

=

Hence,

b

q

=

-

1

1

81

tan

= 6.34°

To relate the Brewster angles in both cases, let us calculate the angle of

refraction.

sin

sin

i

t

q

q

e

e

=

2

1

Therefore, in case a,

sin

sin

B

q

q

t

=

81

Therefore,

t

sin

sin

.

.

q

=

=

83

7

9

0

11

Or

q

t

=

6

34

.

o

, which is the same as the Brewster angle for case b. Also, the angle of refraction in case b is given by Snell’s Law as:

sin

sin

B

q

q

e

e

t

=

o

o

81

EMBED Equation.2

=

1

81

Therefore,

t

sin

sin

.

.

q

=

=

o

6

34

1

81

0

99

Or

t

q

=

o

83

7

.

, which is the Brewster angle for case a.

6.5 Total Reflection at Critical Angle of Incidence

In the previous section it was shown that for common dielectrics, the phenomenon of total transmission exists only where the electric field is parallel to the plane of incidence known as parallel polarization.

There is a second phenomenon existing for both polarizations:

· Total reflection occurring at the interface between two dielectric media

· A wave passing from a medium with a larger dielectric constant to a medium with smaller value of ε

Snell’s Law of refraction shows –

sin

sin

i

t

q

q

e

e

=

2

1

or

sin

sin

i

t

q

q

e

e

=

2

1

(6.26)

Therefore, if

e

e

q

q

1

2

>

>

,

and

t

i

then a wave incident at an angle

q

i

will pass into medium 2 at a larger angle

q

t

.

Definition:

q

c

, (critical angle of incidence) is the value of

q

i

that makes

q

t

= π/2, see Figure 6.13.

Substitute

q

t

= π/2 in equation 6.26 to get –

sin

c

q

e

e

=

2

1

, or

c

q

e

e

=

-

1

2

1

sin

q

i

Figure 6.13 illustrates the fact that

q

q

e

e

t

i

if

>

>

,

1

2

. The critical angle

q

c

is defined as the value of

q

i

at which

q

t

= π/2.

Envision a beam of light impinging on an interface between two transparent media where

n

n

i

t

<

. At normal incidence (

q

i

= 0) most of the incoming light is transmitted into the less dense medium. As

q

i

increases, more and more light is reflected back into the dense medium, while

q

t

increases. When

q

t

= 90°,

q

i

is defined to be

q

c

and the transmittance becomes zero. For

q

i

>

q

c

all of the light is totally internally reflected, remaining in the incident medium.

EXAMPLES:

· Use Snell’s Law to derive an expression for θc. Compute the value of θc for a water-air interface (

n

w

=1.33).

Rewrite

n

i

sin

q

i

=

n

ti

sin

q

t

As

sin

q

t

=

n

ti

sin

q

t

Where

n

ti

< 1. Requiring that

q

t

= 90° for

q

i

=

q

c

leads to

sin

q

c

=

n

ti

At water-air interface

q

c

EMBED Equation.2

sin

-

1

(

1

1

33

.

) =

sin

-

1

0.752 = 48.8°

· Imagine yourself lying on the floor of a pool filled with water, looking straight upwards. How larger a plane angle doe the field of view beyond the pool apparently subtend?

Rays striking the air-water interface from above at glancing incidence will enter the water at a transmission angle equal to

q

c

. The plane angle subtended at the observer is therefore 2

q

c

. Here,

sin

EMBED Equation.2

q

c

=

1

1

33

.

Whence

q

c

= 48.8° and 2

q

c

= 97.6°.

· Determine the critical angle for a water (

n

w

=1.33) –glass (

n

g

=1.50) interface. We have

sin

EMBED Equation.2

q

c

=

n

ti

Or

q

c

=

sin

-

1

1

33

1

50

.

.

=

sin

-

1

0.887 = 62.5°

6.6 Electromagnetic Spectrum

3 GHz

3 x 1012 Hz

3 x105 Hz

3 x1018 Hz

q

t

Frequency

6.7 Application to Optics

The figure above shows the spectrum of electromagnetic radiation extending from the long- wavelength radio waves to X rays and gamma rays the shortest wavelength.

Topics to be discussed will include control of polarization of incident waves, role of Brewster windows in light amplification, and use of the concept of angle of total reflection in optical fibers.

6.7.1 Polarization by Reflection

Definition

Unpolarized light – light in which the wave orientation is random around the axis of the beam.

Unpolarized light has both polarization cases

· Parallel polarization, where the electric field is the plane of incidence

· Perpendicular polarization where the electric field is perpendicular to the plane of incidence

In certain cases, there may be a need to separate the two polarizations. One method that can be used is the Brewster angle of incidence, also called the polarization angle, to separate the two orthogonal polarizations.

Example

Consider an Unpolarized light that is incident at the Brewster angle on a piece of glass with index of refraction

n

r

=

=

e

1

5

.

. The polarization with a electric field parallel to the plane of incidence will be entirely transmitted and the other polarization with a electric field perpendicular to the plane of incidence will be partially reflected and partially transmitted. Why is the electric field parallel to the plane of incident totally transmitted? *Because it is incident at the Brewster angle.

The second interface which is glass to air as illustrated in example 6.7 has an angle of incidence also known as the Brewster angle for light incident from the glass side to free space. So, again the polarization with E parallel to the plane of incident will be entirely transmitted, and E perpendicular will be partially reflected and partially transmitted.

In Figure 6.17:

· Reflected wave is entirely polarized, E perpendicular to the plane of incidence

· Transmitted wave possess both polarizations

· Larger amplitude is the E parallel to plane of incidence – entirely transmitted throughout the interfaces

· More glass elements and the transmitted light could be essentially completely polarized, E parallel to the plane of incidence

6.7.2 Brewster Windows or Brewster Cuts in LASER

In a normal situation there are more electrons in the ground state (level 1) than in the excited states (level 2 & 3). In other words, there are more electrons in level 1 ready to absorb photons that there are electrons in level 2 & 3 to emit photons. A net emission of photons could be the result if this situation could be inverted. Such a condition is called population inversion. This in fact is the fundamental principle involved in the operation of a laser. Figure 6.8 illustrates this principle.

Definition:

Laser (Light Amplification by Stimulated Emission) – A device that produces coherent radiation in the visible-light range, between 7500 and 3900 angstroms

Summarized steps leading to LASER action in three-level ruby laser material:

1. The laser material is in the shape of a long rod that is subjected to radiation from an extremely intense light source that causes interatomic transition from energy levels 1 to 3. (Figure 6.18b)

2. If the nonradiative transition between level 3 and level 2 is fast enough, then electrons in level 3 will transfer to level instead of returning to level 1.

3. As a result of direct transition the population of electrons in level 2 will increase from level 1. This is during the radiation from the light source, as well as the transfer from level 3. (Figure 6.18c)

4. If the pumping action is large and fast enough the electron population at level 2 can be made larger than level 1. Radiation of light quanta at frequency f21 occurs when the electrons can make the transition from level 2 to level 1.

5. By placing mirrors at the end of the laser and forcing the radiation to be reflected back and forth maintaining the high-photon density, stimulated emission will increase resulting in a large photon density build up or in other words an avalanche of photons.

6. An intense light beam will result emerging from the end of the laser rod.

q

r

|

|

i

E

|

|

r

H

e

m

2

2

,

q

r

q

i

b

r

|

|

r

E

b

i

q

t

Figure 6.17 Light polarizations by multiple reflections.

Figure 6.18 is a schematic diagram illustrating the sequence of events.

The role of the Brewster angle:

Known Factors

· The output of many lasers is linearly polarized

· The ratio of the light polarized in one direction exceeds the light polarized in the orthogonal direction by 1000:1

As in most cases, a high degree of linear polarization will be the result of a Brewster surface within the laser. A Brewster surface is usually used in the construction of a laser. The light must be transmitted out of the medium of the laser to avoid minimal loss.

b

t

|

|

t

E

Figure 6.18 Sequence of events occurring in laser action.

Figure 6.19 is a schematic illustrating the use of Brewster windows in a gas discharge laser. The Brewster angle makes sure that light in one polarization direction is transmitted out of the medium of the laser to the reflecting mirrors and back into the medium of the laser with no loss. Where the light is polarized perpendicular to the plane of incidence a large loss at the Brewster surface will take place due to the reflection out of the medium of the laser. The preferred polarization case (linear polarization) will lase (emit coherent light) that will account for the high degree of polarization taking place at the output.

The device in Figure 6.19 exhibits stimulated emission of radiation. For and example lets say the mixture of gases are helium and neon. These gases are confined to the glass tube sealed at both ends by mirrors. An oscillator is connected to the tube to that causes electrons to sweep through the tube, colliding with atoms of gas and raising them to exited states. Some neon atoms are excited to a higher state during this process that will also result in a collision with excited helium atoms. Stimulated emission occurs as the neon atoms make a transition to a lower state and neighboring excited atoms are stimulated to emit at the same frequency and phase. This will result in a production of coherent light.

6.7.3 Fiber Optics

Fiber optics deals with the transmission of light through small filamentary fibers called dielectric waveguides. This is based on the phenomenon of total internal reflection occurring at the point where the light is obliquely incident on an interface between two media with different refractive indexes at an angle greater than the critical angle. Light is incident at an angle θi as shown in Figure 6.20 and is required to determine the range of values of the index of refraction n so the internal reflections will occur for any value of θi.

Snell’s Law of refraction is the relationship between θi and θt as the wave enter the fiber is

sin

sin

i

t

n

q

q

e

e

=

=

2

1

e

e

1

=

o

(6.27)

If

q

2

is suppose to be larger than

q

c

, then

sin

EMBED Equation.2

q

2

= cos

q

t

≥ sin

q

c

(6.28)

Refraction from fiber to air sin

q

c

= 1/n, therefore, from equation 6.27 & 6.28 –

sin

cos

sin

sin

2

2

2

2

1

1

1

1

q

q

q

q

=

=

-

=

-

æ

è

ç

ö

ø

÷

³

t

t

i

n

n

(6.29)

Solve for n,

n

i

2

2

1

³

+

sin

q

(6.30)

For equation 6.30 to be true then = π/2, all incident light will be passed by the fiber requiring

n2 ≥ 2 or n ≥

2

Most types of glass have n ≈ 1.5; therefore, we have a valid equation.

E

^

r

^

t

H

r

� EMBED Equation.2 ��

W

P

z

y

x




O

M


p = P x H

power density flow

Y

X

Z

H

E

Point on the plane

Plane of constant phase

r

nβ



βi

� EMBED Equation.2 �� EMBED MSDraw.1.01 �� EMBED Equation.2 ��



θi

θr

Y

Z

X

Perfect Conductor


Incident wave fronts

λ

θi

Figure 6.4

Perfect Conductor

ax

az

Y









Figure 6.5

Y

Z

X

βi

1

2

Reflected rays

1

2



Incident

rays


A

Reflected rays

B



C

E



Region 2


X

Z

Y






(Out of paper)








βt

X



Z









Region 1


Figure 6.9








Radio

Microwave

Infrared

Vi

s

b

l

e

U

l

t

r

a

v

i

o

l

e

t

X rays

γ rays

1 m

10-3 m

10-6 m

10-9 m

1 mm1μm1nm

Wavelength

Figure 6.16 Electromagnetic spectrum from radio waves to X and γ rays.

n

n

n

n



Partially polarized light; mostly E parallel to plane of incidence

Polarized light with E perpendicular to plane of incidence

(a) Thermal equilibrium

(c)Nonradiative transfer to upper level

(b) Absorption of pump radiation

(d) Coherent radiative transition and emitting laser

Brewster windows

External mirror (totally reflecting)

External mirror (partially reflecting)

Output beam

Figure 6.19 Schematic illustrating the use of Brewster windows in a gas discharge LASER

Gas discharge tube (plasma)

θi

θt

θ2

Smallest critical angle

Reflected point

n ≥ � EMBED Equation.2 ��

Figure 6.20 Schematic illustrating the principle of light propagation in optical fibers.





PAGE

40

Notes by: Debbie Prestridge

^

t

E

^

r

H

e

m

2

2

,

q

r

q

i

b

r

b

i

e

m

1

1

,

^

i

E

e

m

1

1

,

q

p

t

=

2

q

c

e

1

e

2

e

e

1

2

>

q

i

q

t

q

b

1

q

b

2

2

H

r

H

|

|

i

H

|

|

t

H

^

i

_1120942590.unknown

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Documents

Chapter 6 - University of Texas at Dallasgoeckner/emag2_class... · Web viewA specific value of θi can be obtained from equation 6.21 - Or (6.23) From equations 6.22 & 6.23 – This