light propagation (part 1) spatial behaviorcourses.egr.uh.edu/ece/ece5358/class notes/lectset 1-...

Light Propagation (part 1)Spatial BehaviorECE 5368/6358 han q le - copyrightedUse solely for students registered for UH ECE 6358/5368 during courses - DO NOT distributed (copyrighted materials)

Subset 2

5. Spatial aspects of light waves

5.1 General wave and harmonic wave

5.1.1 General waveLet’s simplify equations (4.1.10) (4.1.11):

2U

c22

t2U 0,

to just one-dimension to see what kind of solutions we get: 2

x2 U

c22

t2U 0, (5.1.1)

This has the general wave solution:f x v t where v c

. (5.1.2)

Thus, we can similarly reduce 2U if we can make coordinate reduction. One such reduction is x.s where s is a unit vector:

2 f x.s v t

c22

t2f x.s v t sx2 sy2 sz2

d2 Fd 2 v2

c2d2 Fd 2 0 (5.1.3)

d2 Fd 2 v2

c2

d2 Fd 2 1 v2

c2

d2 Fd 2 0

which is satisfied by: 1 v2

c2 0 or: v c

Looks like we have solved the wave equation! Does that mean we can solve all EM wave problems?

Well, not really. First, remember that the wave equation 2U

c22

t2U 0 is not strictly correct in a medium. It

is correct for vaccuum and it is OK to accept the above as a general wave equation. But a generic form f x v t is NOT very useful for particular problems, in which we need more details about the function f . But for now, it is useful just for the sake of illustration of the wave concept.

The above equation is called plane wave because every point on the plane x.s = constant has the same value at a

given time. The entire plane has the same value and it moves at the speed v. Here is an illustration:

f x_ : Sin x x ;Plot f x , x, 10, 10 , PlotRange All

10 5 5 100.2

0.2

0.4

0.60.8

1.0

Demonstration:

f x_ : Sin x x ;Manipulate k Cos , Sin ;AnimatePlot3D Sin k . x, y v t k . x, y v t , x, 10, 10 ,

y, 10, 10 , PlotRange 0.5, 1 , Mesh None, PlotPoints 80, 80 ,t, 10, 10, 0.05 , DisplayAllSteps True, AnimationRunning False ,, 4 , 0, 2 , v, 1 , 1, 1, 2

This is a faster demof x_ : Sin x x ;ManipulateAnimatePlot3D Sin Cos x Sin y v t Cos x Sin y v t , x, 10,

10 , y, 10, 10 , PlotRange 0.5, 1 , Mesh None, PlotPoints 80, 80 ,t, 10, 10, 0.05 , DisplayAllSteps True, AnimationRunning False ,, 4 , 0, 2 , v, 1 , 1, 1, 2

This is another f (waveshape) function

2 LectSet 1- Light propagation - spatial-ss2_p.nb

Plot Exp x^2 , x, 5, 5 , PlotRange All

4 2 2 4

0.2

0.4

0.6

0.8

1.0

ManipulateAnimatePlot3D Exp Cos x Sin y v t 2 , x, 5, 5 , y, 5, 5 ,PlotRange 0.5, 1 , Mesh None, PlotPoints 80, 80 ,t, 5, 5, 0.1 , DisplayAllSteps True, AnimationRunning False ,, 4 , 0, 2 , v, 1 , 1, 1, 2

Let’s look at a laser beam, how do you describe the beam 100 m from here, or a 1 M km from here? how about the beam from a flash light? How about the beam through a slit, a pin hole? a grating? How about the light coming out from a diode laser? why does one direction diverge than the other? how does that divergence depend on the laser structure? Afterall, all those beams are propagating in free space and satisfy the wave equation. Don’t we already have a solution?

Well, not all waves are plane wave! In fact, no real life wave is truly a plane wave! Even if we use plane wave in some problems, it is ONLY an approximation that is good enough for the problem of interest. Plane wave is an idealized model of waves. It is very useful, but clearly NOT enough to really describe so many other light waves that we deal around us and in many optical systems.We need specific solutions of waves for various problems of interest.

Is this a solution: g s f x.s v t s ? Which space does the integration extend?

5.1.2 Harmonic wave or monochromatic waveThe above general wave solution is not the only way to solve the wave equation. Let’s take another common approach: Solving by part.

2U

c22

t2U 0

We split U into 2 components:

LectSet 1- Light propagation - spatial-ss2_p.nb 3

U F x T t . (5.1.4a)

Then: T t 2F

c2 F x2 T t2

(5.1.4b)

or 1F x

2F

c21T t

2 T t2

constantC (5.1.4c)

DSolve1

T tt,t T t Cn, T t , t

T t Cn t C 1 Cn t C 2

What should Cn be?

Why don't we want solution where Cn is real? Does it make physical sense?

Let T t t. This is a fundamental harmonic wave: F x t. (See 3.2)

We are interested in this type of solutions because lightwaves are fundamentally harmonic waves. We will see later on that a photon has a unique frequency, and all lightwaves are simply made up of many photons (which all can have different frequencies).Thus, we are interested in learning specifically harmonic waves: waves that oscillate periodically.

If we substitute T t t in the equation above, then:

2F 2

c2 F 0: (5.1.5)

this is Helmholtz equation. As mentioned above, it is the frequency-domain approach to the Maxwell’s theory and it is more relevant to the real problem of optical media. It appears as if (5.1.1) is derived from the wave equations (4.1.10) and (4.1.11) and if the formers are not strictly correct, then how can (5.1.1) be correct? The answer is that (5.1.1) can be derived independently from (4.1.10) and (4.1.11) in frequency domain approach to the whole Maxwell’s equation shown in (3.2)

In the following, we will consider aspects of the spatial component F[x] of the wave, rather than the temporal component, which we will assume to be always t. The essence is to find solutions to the Helmholtz equation.We will study in later chapters waves with different values of frequency .

5.2 Harmonic plane waveWe see that a very easily recognized solution to the Helmholtz equation is the spatial harmonic functions: sin and cos. Or, we use the complex function which is more convenient than sin or cos:

F x x. k (5.2.1)(of course, this is NOT the only solution to the Helmholtz eq.).

Combine what we learn above from the general form of wave with harmonic wave, we have harmonic plane waves:

F x x. k and U F x T t x. k t (5.2.2)

Is there any relationship between k and ?Substitute in the Helmholtz equation:


k2 2

c2 0, or k c k0 (5.2.3)

Plane waves can actually be quite interesting.

5.2.1 Illustration and discussion of harmonic plane wavePlane wave demo link

So now we have harmonic plane waves. Is that enough for us to answer question above? (about laser beam etc...)

Let's look at the above solution again. We write: x. k kx x ky ykz z , we can think of this as the product of three waves: x, y, and z.

The requirement is: k2 kx2 ky2 kz2 2

c2 k02

What if we accidentally choose kx2 ky2 k02 ? Can we have a solution?

5.2.2 Evanescent waveLet's consider only x and z for simplicity (ky 0). We know this:

kx2 kz2 k02

c2 . (5.2.4)

Suppose kx2 k02 , what is the solution?

Clearly, kz2 k02 kx2 0. (5.2.5)

kz k02 kx2 . (5.2.6a)

where kx2 k02 (5.2.6b)

How about this solution: kx xkz z kx x z (5.2.7)

Animate 1; k0 2

; 2 ; k

2 1.2

;

k2 k02 ;Plot3D Re k x t z , x, 0, 5 , z, 0, 1

, BoxRatios 5, 1, 2 , PlotPoints 50, 10 ,ViewPoint 3, 1, 2 , Mesh False, PlotRange 1, 1

, ImageSize 300, 200 , t, 0, 1 , AnimationRunning False

Is it a physically meaningful wave? Where do you think they exist? What happens at z -> - The above wave is called evanescent wave in the z dimension. Discussion in class.

5.3 Wave in a medium: complex dielectric constant


In the above example, we see that plane wave can be made up of purely real and purely imaginary propagation constant kx x z . More generally, can we have wave x. k where k is complex? Why not? we see that the only requirement is that

kx2 ky2 kz2 2

c2 k02 (5.3.1)

- Case 1: we see that if k02 is real, but any component kx2, ky2, kz2 or the sum of 2 therof is greater than k0

2 , the remaining component(s) can be complex.- Case 2: if a medium has a complex , then k0

2 is complex and hence some or all of kx, ky, kz have to be complex. The 2 cases are different however, and we must make distinction.

Fundamentally, what is complex , ?Remember: classical optics wave theory is ONLY a description of lightwave behavior based on macroscopic average or approximation of the interaction of between photons and matters (electrons in atoms, molecules,...). As it turns out, many of these interactions result in a simplified description of lightwave in media with quantity , that are complex.A specifically and particularly important class of phenomena is absorption and/or stimulated emission. We are familiar with absorption all around us.We will learn about stimulated emission and optical gain later in discussing about lasers.In these phenomena, lightwave behavior can be well described with a complex dielectric constant.But... remember again: the real fundamental theory here is the quantum interaction between photons and atoms, molecules. There is no fundamental meaning in complex other than that.In other media, nanoscopic structures can also result in complex optical behaviors, e. g. metamaterials, negative index of refraction etc.

For Case 2, we write R I or 1 2.If I 0, k2 kR kI 2 k0

2 R I entails that kI 0; then the wave: x kR kI x kR kI x has both a spatially periodic term (wave) and a exponentially decaying term in the SAME direction. It means that the wave gets weaker as it propagates. The energy must be lost in the medium somehow (unless there is a reflection wave - but here, we don’t see any reflection wave, hence the energy is loss in the medium). On the reverse, the wave can gain energy from a medium.In general, we don't have to make any distinction for a lossy or amplifying medium. All wave parameters are assumed to have complex component whenever physically possible.

Illustration

Make distinction between these 2 cases:

Animate 1; k0 2

; 2 ; k

2 1.2

;

k2 k02 ;Plot3D Re k x t z , x, 0, 5 , z, 0, 1

, BoxRatios 5, 1, 2 , PlotPoints 50, 10 ,ViewPoint 3, 1, 2 , Mesh False, PlotRange 1, 1



2 ;Manipulate k 1 2 ;

Plot3D Re k z t , z, 0, 50 , x, 1, 1 , PlotRange 0, 50 , 1, 1 , 1, 1 ,Mesh False, BoxRatios 5, 1, 1 , PlotPoints 50, 3

, t, 0, 1 , , 0.05 , 0, 0.1

5.4 Wave in other geometries: spherical and cylindrical

5.4.1 Spherical wave in isotropic mediaLet's start from the Helmholtz equation:

2F k02 F 0.

Earlier we write: 2 x,x y,y z,z. What if we use a different coordinate?

Laplacian F r, , , r, , , "Spherical"

F 0,2,0 r,,r

F 1,0,0 r, ,

r F 2,0,0 r, ,

csc sin F 1,0,0 r, , cos F 0,1,0 r,,

r

csc F 0,0,2 r,,r

r

Let’s consider a wave that is purely spherical, it has no dependence on the angles:

Laplacian F r , r, , , "Spherical"

F r 2 F r

r


k0 .;

DSolve2F r

r F r k02 F r 0, F r , r

F r c1 r k02

rc2 r k02

2 r k02

The solution is k r

r c1 k r

r c2 (5.4.1)

where k k0

So, we have a spherical wave solution also: k r t

r. What do we do with this solution? What is the physical

meaning? what happens at r = 0?

1; k 2

; 2 ; n 20 ;

Animate Plot Re1

r k r t

, r, 1, 10, PlotStyle RGBColor 1., 0., 0 , Thickness 0.007 ,

RGBColor 0., 0., 1 , Thickness 0.007, PlotRange 0, 10 , 1, 1


2 4 6 8 10

1.0

0.5

0.0

0.5

1.0

Show Plot3D 0, x, 1, 2 , y, 1, 2 , Mesh False , SphericalPlot3D 1, , 0, Pi , , 0, 2 Pi ,PlotRange 1, 2 , 1, 2 , 1, 1 , BoxRatios 3, 3, 2

Let’s cut across the equator and plot the wave amplitude in density plot:


2 ; k 2 ;Manipulate

DensityPlot Rev k x2 y2 t

x2 y2, x, 1, 5 , y, 1, 5 ,

PlotPoints 40, 40 , PlotRange 1, 1 , ColorFunction "Rainbow", v, 1 , 1, 1 , t, 0, 1

2 ; k 2 ;Manipulate

Plot3D Rek v x2 y2 t

x2 y2, x, 1, 5 , y, 1, 5 ,

PlotPoints 40, 40 , Mesh False, PlotRange 1, 1, v, 1 , 1, 1 , t, 0, 1


Manipulate m ArcSin 1 r ;Show Graphics3D Blue, Tube r, 0, 0 , 0.2, 0, 0 ,

ParametricPlot3D r Cos Cos 1 , rCos Sin , r Sin ,, m, m , , m, m , Mesh False, PlotStyle Opacity 0.7 ,

PlotRange 1, 0.2 , 1, 1 , 1, 1 , BoxRatios 1.2, 2, 2,r,1,10

If the wave decays in amplitude, where does its energy go?

Where (in which phenomena) do you expect to see this type of wave?Remember: like plane wave, an ideal or perfect spherical wave doesn’t exist in nature. But they can be very useful approximation to numerous problems. For example, if we consider a very small source of light from a distance much larger than the extent of the source (far away) and many (a large number) wavelengths, it is as good as a spherical wave for any practical purpose calculation. The point source can be just a small aperture, or even a single atom or molecule emitting the photon. Of course each photon has a finite momentum k, hence a specific direction. But if k is purely random with equal probability in any direction, then the atom or the molecule is practically a source of spherical wave.We will see that in dealing with various problems in diffraction, spherical wave is a useful model.

Spherical wave in non-isotropic media can be quite more complicated (mathematically)

5.4.2 Cylindrical waveWe choose a different coord syst and we get an answer not expected from plane wave solutions. What if we choose another one like cylindrical?

Laplacian F , , , z , "Cylindrical"

F F

DSolve F F

k02 F 0, F ,

F c1 J0 k0 c2 Y0 k0

Again, it can be an approximated model for some some waves in some problem.

The solution now is Bessel J and Y function!. What are other coordinates? what are other solutions?


k 2 Pi ;Animate ParametricPlot3D Cos , Sin , Cos 2 t J0 k ,

, 0, 3 , , 0, 2 , PlotRange 3, 3 , 3, 3 , 1, 1 , Mesh False,PlotPoints 20, 20 , BoxRatios 1, 1, 1 , t, 0, 1 , AnimationRunning False

Notice that it behaves like a standing wave! which means that it must be a sum of incoming and outgoing wave. That's the only way to have finiteness at the center. Can we construct a radiating wave?

Where (in which phenomena) do you expect to see this type of waves?

5.5 Summary

There are many many (infinite) solutions to the Helmholtz wave equation. Many of them, if not all, have physical meaning one way or another. Our objective is to understand, gain insight, and have the ability to select the right solution (including most appropriate approximated solutions) to the physical problem of our interest. Use various wave models intelligently to do what we want with the given problems.

6. Some key light wave parameters

6.1 Review basic concepts, power and energy Light carries energy as we know. We learn from modern (quantum) physics that light consists of quantized package energy known as photons. (we will have a review of quantum physics). Each photon has an energy:

E h (6.1.1)where h is the Planck's constant.When we deal with a large number of photons, we don't have to worry about the quantum behavior of light, because we can treat them as an ensemble and we can use the ensemble statistics. Their ensemble statistical behavior is described by the Maxwell's equations, which is classical wave optics.

The energy density of an EM field is given by:u x 1

2 E2 x H2 x (6.1.2)

It is a function of space (position). For an EM wave, the energy is not stationary. As EM field forms a wave, the energy flows with the wave. If we measure the total energy passing a certain surface per unit time, we have power crossing that surface.

If we take the limit:


limS 0PS I (6.1.3)

we refer I as the light intensity (power density) at that point of the surface.

Excersise: What is the energy density of a plane wave

Excersise: What is the power density of a plane wave

6.2 Poynting vectorPoynting vector is defined as:

S c4

E H (6.2.1)

(note: use real representatives of the fields, i. e. S c4 Re E Re H ).

It can be shown that the Poynting vector represents the power density PA ) or intensity I of a light beam by taking

its time-averaged, which, for harmonic wave, is:

S c8 Re E

H

(6.2.2a)

(the brackets here is used to denote time-averaged quantity).

I S c8 Re E

H

(6.2.2b)

Plane wave in a medium: I S c8

Re EH

c8

E

2

(6.2.2c)

A best illustrative example is to apply the Poyting vector concept to a plane wave (HW).

Calculate the Poynting vector for a plane wave

6.3 A note on polarization (more in a different chapter)Remember that EM wave is vectorial in nature. It is not a scalar wave. Hence, although we generally care more about the absolute magnitude of the field rather than individual components such as Ex, Ey, Ez. The orientation of E and H vectors area also an important parameters of a light wave. Polarization is a concept used to describe this orientation. We will study this in another chapter.

7. Principle of linear superposition (spatial)

7.1 Discussion - conceptWe know that light waves all around us, in various technologies are certainly more complicated than those plane wave, spherical or cylindrical waves above. How do we describe these waves?Just like in differential equation, we know that there are infinite solutions, but all of them can be constructed from the basis set of solutions, the same is applicable to light waves. The solutions we discuss above are “basis” solution and using the linear superposition principle, we can construct any light wave.

Example:


DSolve y'' x a y x 0, y x , x

y x c2 sin a x c1 cos a x

Coefficients c1 and c2 are arbitrary and we can construct infinite solutions. Sin and Cos are basis functions. In many systems, the basis functions are orthogonal, and they can be normalized to become orthonormal.

A basis set may have an infinite number of orthonormal functions, such as Sin[n x], Cos[ n x], n= 0,1, 2,..Fourier series theorem allows us to construct any arbitrary-shape periodic function from the basis Sin and Cos function.

For discrete basis set, we write a general function as a sum: x, t n1

cn n x, t (7.1.1)For continuous basis set, we use integration:

x, t c k n k; x, t k (7.1.2)

We know that plane waves form a continuous set, so in principle, we can form a solution like this for a monochro-matic wave

c k x. k t k t

c k x. k k (7.1.3)So, this is the principle, let’s implement it in an example:

7.2 Example of linear superposition: 2D Gaussian beam Suppose we have a monochromatic 2D light beam in x and z. The beam is constant along y direction. For example, if we put a screen perpendicular to z (i. e. x, y plane), it looks like this:

Plot 2 x2 w2 . w 1 , x, 3, 3 , Filling Axis, FillingStyle Blue

3 2 1 1 2 3

0.2

0.4

0.6

0.8

1.0

DensityPlot 2 x2 w2 . w 1 , y, 10, 10 , x, 3, 3 ,ColorFunction GrayLevel, FrameLabel y, x , AspectRatio 0.3

This is what we call “Gaussian” profile (along x-direction). How do we describe the E field wave E[x,z,t] of this light beam? (the above give us ONLY a profile in x at z=0, and time-averaged.

Since it is monochromatic, we know that the time factor is simply t. It’s up to us to find out the z part.

We know that k2 kx2 kz2 2

c2 k02 . So only one of the 2 k's can be independent. Let's pick kx to

be independent and the integrating variable. Then, the formula (7.1.3) becomes:


t c kx x kx z kz kx (7.2.1a)

t c kx

x kx z k2 kx2

kx (7.2.1b)Now, what we really need is c kx . This is the “arbitrary” coefficient that allows us to construct any solution. How do we choose c kx in this case?

7.2.1 Integral expression for Gaussian profileFor the Gaussian profile above:

Plot x2 w2 . w 1 , x, 3, 3 , Filling Axis, FillingStyle Blue

3 2 1 1 2 3

0.2

0.4

0.6

0.8

1.0

This is what we have: (the solution at z=0) = x2 w2 .

We want: c kx

x kx z k2 kx2

kx at z = 0 = x2 w2

Or: c kx x kx kx x

2 w2 (7.2.2) How do we solve for c kx ? We need inverse Fourier transform:

c kx 12

x2 w2

x kx x (7.2.3)

k . ;1

2

x k x2 w2

x

ConditionalExpression

1

4k2 w2

2 1w2

, Re1

w2 0

So the solution is c kx = w

2

14kx2 w2

(7.2.4)

What is the meaning of c kx ?

c[k] describes the beam profile in terms of a "package" of several beams with different spatial frequency. Key concept: spatial frequency k. c[k] is the envelop or coefficient of the "package". (we'll study a time wave package later on). It can also be called the "angular spectrum".

Now we have our solution, complete with the time factor:

E x, z, t A w

2 t

14k2 w2

x k z k0

2 k2

k (7.2.5)

This cannot be integrated in a closed form, but we can numerically evaluate it, or later, do an approximation.


7.2.2 Graphic illustrationNumerical integration code:

Beam x_, z_, n_, _, w_ :Module kn, bR, bE , kn 2

n ;

bR w NIntegrate k x kn2k2 z 1

4k2 w2

, k, Min kn, 4. w , Min kn, 4. w ;

bE If kn 4. w,

2 w NIntegrate Re k x kn2k2 z 1

4k2 w2

, k, kn , 4. w ,

0 ; bR, bE

We can plot the E field:

0.6328 ; n 1; w 0.5 ;beamplot Plot3D Re Beam x, z, n , , w 1 , x, 5, 5 , z, 5, 5 ,

PlotPoints 25, 50 , Mesh False, FaceGrids All, BoxRatios 1, 1, 1 , PlotRange All

This is 1/2 of the beam along propagation axis

7.2.3 Paraxial approximation

Let’s find an approximation to do the ingration for the beam. Let's look at the integral:

1

4k2 w2

k x k02 k2 z k

If z is very large (far field) and the beam is paraxial (mostly in the propagation axis), we can think that kz is the dominant term and kx is small. Then, we can approxi-mate:


k02 k2 kn2 k2 kn 1 k2

kn2 kn 1 k2

2 kn2 kn k2

2 kn

14k2 w2

k x k02 k2 z k

14k2 w2

k x kn zk2

2 knz k

= kn z

1

4k2 w2

k x k2

2 knz k

= kn z

1

4k2 w2

2 zkn k x k

Now, we can perform integration. Define P w2 2 zkn

1

4k2 P k x k

ConditionalExpression2

x2

P

P, Re P 0

Hence:

w . ; n . ; z . ;

kn z2

x2

P

P. P w2

2 z

kn

2 z kn

x2

w22 z

kn

w2 2 zkn

Thus: E x, z, t A w

2

2

z knx2

w22 z

kn

w22 zkn

t A 1

1 2 z

w2 kn

z kn

x2

w22 z

kn t (7.2.6)


Show , BoxRatios 1, 1, 0.1

We can do some more mathematical manipulation to give us some insight. We can rewrite:

E x, z, t A 1

1 2 z

w2 kn

z kn

x2

w22 z

kn t (7.2.6)

E x, z, t A wW z

z kn t

x2

W z 2 (7.2.7a)

where W z w2 2 zkn

; W z 2 w2 2 zkn

(7.2.7b)

then, this is what we see: -The part: z kn t is just the plane wave in the z direction.

- The part x2

W z 2 is a Gaussian profile in the x-dimension that modifies the plane wave part z kn t . It has a Gaussian waist W z 2

- The only thing “unsual” is that the Gaussian waist is z-dependence and complex: W z 2 w2 2 zkn

We will learn more about the “meaning” of complex beam waist!


7.3 Illustration of some properties of 2D Gaussian beam

7.3.1 Phase front.The most obvious thing we see in the Gaussian beam calculation above is the circular “wave front”

The key insight to the wavefront is in this term: z kn

x2

w22 z

kn (7.3.1) Let’s do some reduction to separate the amplitude and the phase:

z kn x2

w22 zkn

kn z x2

kn w22 z kn z

x2 2 z kn w2

2 z kn w2 2 z kn w2

= kn z x2 2 z4 z2 kn2 w4 x2 kn 2 w2

4 z2 kn2 w4

= kn z x2

2 z kn2 w4

2 z

x2

w21

1 2 z

kn w2

2 kn z x2

2 z x2

w21

1 2 z

kn w2

2

But: kn z x2

2 z kn z2 x2 kn r

This is a circular (cylindrical) wavefront indeed with the term kn r

0.6328 ; n 1; w 0.5 ; kn 2 n ;ParametricPlot3D r Cos , r Sin , Re kn r , r, 0, 5 ,

, 0, 2 , Mesh False, PlotPoints 40, 15 , BoxRatios 1, 1, 0.1

Let’s compare:

A natural concept in describing wave is "phase front". We commonly think of "wave front" in layman term. It's the same. A phase front is a surface of all points that have the same continuous phase. Mathematically, if we express a wave as: x,y,z (where [x,y,z] is real), then, any surface defined by [x,y,z]=constant is a phase front. (other name: cophasal surfaces, wave surfaces)


Notice the phasefront, but what else do we need?

7.3.2 Beam divergenceFor the Gaussian beam, the amplitude is NOT circularly uniform, but it is rather distribute in the center lobe:

We need a term that make a “lobe” in the center along the propagation and the lobe should have a constant angular divergence.

We now can examine the second term (amplitude in the x-dimension) of the exponent:

z kn

x2

w22 z

kn kn r

x2

12 z

kn w2

2

kn r

x2

w2 2 z

kn w2

2 kn w2

2 z

21

kn r

x z 2

2

kn w

2 kn w2

2 z

21

kn r

Tan 2

2

kn w

2 kn w2

2 z

21

kn r

Tan 2

n

w

2 kn w2

2 z

21

What does the term

Tan 2

n

w

2 kn w2

2 z

21

tell us? It is a Gaussian profile with farfield divergence angle , and for z-> :

Tan 2

n

w

2

Tan 2

Tan 02 (7.3.2a)

where: Tan 0 nw

(7.3.2b)

In deed, this is the term that gives is the Gaussian profile along the propagation direction and with a divergence angle 0


0.6328 ; n 1; w 0.5 ; kn 2 n ;

ParametricPlot3D r Cos , r Sin ,Tan 2

n w2 kn w2

2 r Sin

21 , r, 0, 5 ,

, 0, 2 , Mesh False, PlotPoints 100, 15 , BoxRatios 1, 1, 0.1

Now, we see that:

x =>

Where (in which phenomena) do you expect to see this type of wave?(We’ll see that at r=0, the approximation for the second plot is not correct because we don’t have z-> with correct limit).

But, now we can plot and verify a key basic property: divergence vs. beam waist and wavelength.

7.3.3 Beam divergence.Recall Gaussian profile with farfield divergence angle for z-> :

Tan 2

n

w

2

Tan 2

Tan 02

where: Tan 0 nw


Manipulaten 1; kn 2 n ; T ArcTan n w ;

ParametricPlot3D r Cos , r Sin ,Tan 2

n w2

,r, 5, 20 , , 4 T, 4 T , Mesh False, PlotPoints 50, 20 ,

PlotRange 0, 20 , 20, 20 , 0, 1 , BoxRatios 1, 2, 0.2, , 0.5, 1 , w, 0.5, 3

What do we see of the divergence as function of and w?

Better illustration:

BeamApprox x_, z_, n_, _, w_ :Module kn , kn 2

n ;

1

w2 2 zkn

z kn

x2

w22 z

kn

We will vary the beam width and see what happens

n 1 ;ManipulatePlot3D Re BeamApprox x, z, n, , w , z, 0, 5 ,

x, 5, 5 , PlotPoints 50, 25 , Mesh False, FaceGrids None,BoxRatios 5, 10, 2 , PlotRange All, ImageSize 400, 400

, , 0.6328 , 0.5, 1 , w, 0.5, 3


Now we see how it moves vs. time:

n 1 ;ManipulateAnimatePlot3D Re Exp 2 t BeamApprox x, z, n, , w , z, 0, 2.5 ,

x, 2.5, 2.5 , PlotPoints 70, 30 , Mesh False, FaceGrids None,BoxRatios 5, 10, 2 , PlotRange All, ImageSize 400, 400

, t, 0, 1, 0.025 , DisplayAllSteps True, AnimationRunning False, , 0.6328 , 0.5, 1 , w, 0.5, 3


n 1 ; w 0.5 ; allbpl ;For i 0, i 39, i, dt 0.025 ;

0.65 ;AppendTo allbpl, Plot3D Re BeamApprox x, z, n, , w 2 idt , z, 0, 2.5 , x, 2.5, 2.5 ,

PlotPoints 70, 40 , Mesh False, FaceGrids None, BoxRatios 1, 2, 0.5 ,PlotRange All, ViewPoint 0.001, 0.093, 6.999 , ImageSize 400, 400

;

Manipulate allbpl i , i, 1, Length allbpl , 1 , DisplayAllSteps True

Another plotting technique is density field plot

n 1 ; w 0.5 ; allbplD ;For i 0, i 39, i, dt 0.025 ;

0.65 ;AppendTo allbplD, DensityPlot Re BeamApprox x, z, n, , w 2 idt ,

z, 0, 2.5 , x, 2.5, 2.5 , PlotPoints 70, 40 , Mesh False, PlotRange All, ColorFunction "Rainbow", AspectRatio 2

;

Manipulate allbplD i , i, 1, Length allbplD , 1 , DisplayAllSteps True


Summary

Summary:1. Any light packet (spatially as well as temporally) can be decomposed as a sum of many light components. (here, we will study spatial superposition first. Temporal superposition will be studied later).2. The components can be pure harmonic plane waves (this is Fourier optics) or harmonic waves in other systems3. The behavior of the beam is the net sum of those of the components4. Most important to develop an intuitive understand of the behavior of beam: "natural" description of the beam, not just any brute force solution

8. (Advanced topics 6358/7358) Introduction to the diffraction concept and far-field Fraunhofer approximation

8.1 Linear superposition and diffraction formulismThe general aspect of Gaussian beam properties we discussed above is NOT specific to just Gaussian beam. In fact, it comes from diffraction behavior, a topic we will study in depth later. We use Gaussian beam as a study object because it is mathematically convenient, and is also important in terms of practicality. But we can begin to gain some insight by consider the problem more generally.

In the above, we consider this expression:

E x, z; t t c kx

x kx z k02 kx2

kx (7.2.1b)

which is about constructing a solution with linear superposition of plane wave x kx z k0

2 kx2

weighed by the function c kx , which is called the spatial spectral function of the beam. We see that if we know the field at a particular plane, e. g. z=0, then


E x, z 0; t 0 c k x k k (8.1.1)

where we drop the subscript x for convenience.

Then:c k 1

2 E0 x x k x (8.1.2)

as the inverse Fourier transform of E0 x E x, z 0; t 0 .We can substitute (8.1.2) into (7.2.1b) above and eliminate c[k]:

E x, z; t t 12

E0 k

x k z k02 k2

k (8.1.3)

Note that we cannot use the same variable x because it is a dummy integration variable and hence we replace it with to distinguish it from the true variable x.

Now we can switch the integration order:

E x, z; t t E0

12

k x k z k0

2 k2

k (8.1.4)

The integral 12

k x k z k0

2 k2

k (8.1.5)

is a function that links the field in one plane E0 to the rest of field further away: this is the essence of the diffraction integral concept. But this integral is very difficult to calculate hence, we need approximation like we did


above.

8.2 Paraxial approximationFor paraxial approximation, we let:

k02 k2 kn2 k2 kn 1 k2

kn2 kn 1 k2

2 kn2 kn k2

2 kn (8.2.1)

Let’s take a detailed look at the paraxial approximation concept.

8.2.1 Paraxial phase front in r-space,The above approximation is made in k-space. But for illustration, we will look at the paraxial approximation in r-space, which is {x,y,z}. Of course the 2 spaces are different, but the math concept is equivalent. For r-space, let’s look at the phase front. This is the phase front of a spherical wave, which (obviously) is spherical!

y 0 ;

ContourPlot Arg 2 x^2y^2z^2 0, z, 10, 10 , x, 10, 10

10 5 0 5 1010

5

0

5

10

Now if we have very large z (or any other main propagation direction), and we look in the z x plane, we can compare 2 wavefronts: spherical and paraxial for a small viewing angle in the propagation direction:

y 0. ;

ContourPlot Arg 2 x^2y^2z^2 0, Arg 2 zx^2

2 z 0

, z, 40, 60 , x, 10, 10 , ContourStyle Thick, Red , Dashed, Thick, Blue

40 45 50 55 6010

5

0

5

10

We see that the paraxial approximation works very well along path nearly parallel to the axis: paraxial. Of course, we don't need contourplot to even plot it out. We know:

z x2

2 z m


z .; x . ;

Solve z x2

2 z m , x

x 2 m z z2 , x 2 m z z2

m 50;

Plot Table m2 z2 , m2 z2 , m, 40, 60, 2 ,

Table 2 m z z2 , 2 m z z2 , m, 40, 60, 2 , z, 30, 60 ,

PlotRange 0, 60 , 40, 40 , PlotStyle Red, Blue , AspectRatio 8. 6

10 20 30 40 50 60

40

20

20

40

We will see that even for angle where the concept of nearly “parallel” is a stretch, e. g. 30 degree, paraxial approx is still reasonable. At larger angle where the deviation is obvious, it is actually still fine if appropriate correction terms are added. Paraxial is not applicable only for large angle >45

ContinuedHere in k-space, the same paraxial approximation can be applied on the condition that the beam does not have much kx-component compared with kz. This means the beam has small divergence. However, we will still use the approximation with the understanding that the main beam in the narrow center is valid and usually what we care. (the beam portion with large divergence is usually very low in power).With the paraxial approximation:

12

k x k z k0

2 k2

k 12

kn z k

k x z k2

2 kn k (8.2.2)

which is can be integrated in closed form:

k . ;

k k x z

k2

2 q k

ConditionalExpression2

q x 2

2 z

zq

, Imz

q 0


Hence: 12

k k x z k2

2 kn k kn2 z

kn x 2

2 z (8.2.3)

Substitute (8.2.3) in (8.1.3):

E x, z; t kn z t kn2 z

E0 kn x 2

2 z (8.2.4)

We will see that for a finite aperture, this important expression (8.2.4) is equivalent with the Fresnel diffraction integral.We can expand the exponent term and obtain

E x, z; t kn z t kn2 z

E0 kn 2x22 x

2 z

E x, z; t kn z t knx2

2 zkn

2 z E0 kn

xz kn

2

2 z (8.2.5)

We can obtain an alternative approximation for (8.2.5) for very large z>>x:

E x, z; t kn r t

2 kn rkn

E0 kn xr kn

2

2 r (8.2.6)

8.3 IllustrationConsider we have a square beam:

Plot UnitBox x , x, 1, 1 , Filling 1 Axis, Blue

1.0 0.5 0.5 1.0

0.2

0.4

0.6

0.8

1.0

Such a beam is not realistic at the discontinuous edge, but we will consider that as an approximation for a sharp cut off beam, such as a plane wave with an hard aperture.

Applying (8.2.4): E0

kn x 2

2 z ww

kn x 2

2 z (8.3.1)where w is the beam half-width,

k . ;k

2 z w

w

k x 2

2 z

12

2z k

zerfi

1

2

2k xw

z erfi

1

2

2k wx

z

2 k


FullSimplify1

2

2

2

2

hence:

E x, z; t kn z t 2 Erfi 4 k xw

2 z Erfi 4 k xw

2 z (8.3.2)

Manipulate k 2 ;

Plot3D Re

2 k z Erfi

12

2k x w

z Erfi

12

2k x w

z, x, 5, 5 , z, 0, 5 ,

PlotRange All, BoxRatios 2, 1, 1 , Mesh False, PlotPoints 80, 40 , w, 0.5, 5

Manipulate k 2 ;

DensityPlot Re

2 k z Erfi

12

2k x w

z Erfi

12

2k x w

z,

z, 0, 5 , x, 5, 5 , PlotRange All, Mesh False, PlotPoints 80, 40 ,

ColorFunctionGrayLevel, AspectRatio 2 , w, 0.5, 5


compare with diffraction of water around an edge

Note that the solution at large angle is NOT accurate because it violates the assumption of paraxial approximation. The part with small divergence angle is more accurate. We will see that the farfield divergence vs. beam waist behaves the same as that of Gaussian beam.

8.4 Fraunhofer diffraction and farfield profile.In many cases, we are not interested the field near the beam waist or aperture as shown above, but care more about the beam divergence behavior at a large distance away, which referred to as “far field” (FF). In this case, we don’t care to describe the farfield in terms of cartesian coordinates x and y, but rather in terms of viewing angle , i. e. polar coordinate. It turns out that we can also make further approximation as well to make the integration a lot easier, yet the FF approximation is still very good, and in fact, much better than at near field.

Let’s look at the quadratic term in the exponent kn2

2 z of (8.2.5):

kn2

2 z 2

n z

n

z

n

z

(8.4.1)

In (8.4.1), is the integration variable over which E0 is appreciable. This approach is applicable only for E0

with finite spatial extent, i. e. we can set a rough value M for which E0 0 for > M . For Gaussian for

example, we can set M 2w. This is NOT an exact definition. Then, n

Mn

is roughly the size of the beam

(or aperture) relative to the medium wavelength n. Likewise, z is the ratio of the location where we observe the

beam relative to the beam size.

Now, we see that if zM

>> Mn

, then kn2

2 z 1 over the significant region of integration in

E0 kn

xz kn

2

2 z .

What it means is that we can drop kn2

2 z and our life gets easier (integration-wise) from (8.2.6):


E x, z; t kn r t

2 kn rkn

E0 kn xr kn

2

2 r

kn r t

2 kn rkn

E0 kn xr

(8.4.2)

Now, we notice that: xr Sin (8.4.3)

and we obtain the Fraunhofer diffraction expression:

E , z; t kn r t

2 kn rkn

E0 kn Sin (8.4.4)

what is this integral? E0 kn Sin (8.4.5)

It is the Fourier transform of E0 , the field at the aperture or in the plane of the field original measurement.

The far field (FF) beam profile (in the Fraunhofer FF condition) is approximately the Fourier transform of the nearfield beam profile. The Fraunhofer FF condition requires that the observation

distance from the source plane is much larger than the source Rayleigh length: z M2

n.

Note: if z M2

n or not much larger, it is said to be in the Fresnel zone to indicate that the quadratic

term of Fresnel diffraction is neccessary, as opposed to be in the Fraunhofer (farfield) zone of the above.

8.5 Fraunhofer diffraction illustrationLet’s consider a uniform field with an aperture to give us a top-hat shape field profile as in 8.3 above:

Plot UnitBox x , x, 1, 1 , Filling 1 Axis, Blue

1.0 0.5 0.5 1.0

0.2

0.4

0.6

0.8

1.0

We have seen that the field looks pretty complex at near field in 8.3. What about far field?

E0 kn Sin w

w kn Sin (8.5.1)

k . ; x .; z .; . ; a .;

w

w k Sin

2 csc sin k w sin

k

Hence: E , r; t kn r t

2 kn r2w kn

Sin kn w Sin kn w Sin (8.5.2)

What does it look like?

The part kn r t

2 kn r is a 2D circular wave as seen below:


Manipulate

DensityPlot Re 2 x2z2 t

x2 z2 1 4, z, 100, 110 , x, 20, 20 ,

ColorFunctionGrayLevel, AspectRatio 4, PlotPoints 40, 80 , Mesh False , t, 0, 1

The part Sin kn w Sin kn w Sin should be plotted in polar coordinate:

Manipulate

PolarPlot AbsSin 2 w Sin

2 w Sin , , 2, 2 ,

PlotStyle Thick, Red , PlotRange 0.1, 1 , 0.3, 0.3 , w, 0.5, 5

0.2 0.4 0.6 0.8 1.0

0.3

0.2

0.1

0.1

0.2

0.3

We see both part:


Manipulate

DensityPlot

Sin 2 w x

x2z2

w 2 Sin x

x2z2

Re 2 x2z2 t

x2 z2 1 4, z, 250, 270 ,

x, 50, 50 , ColorFunctionGrayLevel, AspectRatio 2.5, PlotPoints 80, 60 ,

Mesh False, PlotRange 1.

250,

1.

250, w, 1.5 , 0.5, 4 , t, 0, 1

Manipulate

ParametricPlot3D r Cos , r Sin ,Sin 2 w Sin

w 2 Sin Re

2 rt

r,

r, 250., 260 , , 0.2 2, 0.2 2 , BoxRatios 20, 150, 20 , PlotPoints 100, 80 ,

Mesh False, PlotRange 240, 260 , 75, 75 , 0.1, 0.1 , w, 2. , 1, 4 , t, 0, 1


Manipulate

ParametricPlot3D r Cos , r Sin ,Sin 2 w Sin

r w 2 Sin

2

, r, 250., 260 ,

, 0.5 2, 0.5 2 , BoxRatios 20, 150, 20 , PlotPoints 20, 70 ,

Mesh False, PlotRange 240, 260 , 75, 75 , 0, 0.005 , w, 2. , 1, 4

w

Manipulate z 250. ;

Off DensityPlot::exclul ;

Show DensityPlot If x 0., 1. ,

Sin 2 w x

x2z2

w 2 Sin x

x2z2

2

1

x2 z2, x, 75, 75 , y, 20, 20 ,

ColorFunctionGrayLevel, AspectRatio 0.4, PlotPoints 40, 2 , Mesh False, PlotRange All

, Graphics Hue 0.15, 1, 1, 0.3 , Rectangle w, 20 , w, 20

, w, 1.5 , 0.5, 4


HW: Divergence from a diode laser facet

8.6 Advanced assignment: Simulate Young’s double-slit experiment.


light propagation (part 1) spatial behaviorcourses.egr.uh.edu/ece/ece5358/class notes/lectset 1-...

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