Lec 2 :1

Outline1. Ray Optics Review

1. Image formation (slide 2)

2. Special cases of plane wave or point source (slide 3)

3. Pinhole Camera (slide 4)

2. Wave Optics Review1. Definitions and plane wave example (slides 5,6)

2. Point sources and spherical waves (slide 7,8)

3. Introduction to Huygen’s wavelets1. Plane wave example (slide 9)

4. Transmission of a plane wave through a single slit1. Narrow slits and Huygen’s wavelets (slides 10,11)

2. Wider slits up to the geometric optics limit (slide 12)

3. Time average intensity patterns from single slits (slide 13)

5. Transmission of a plane wave through two slits1. Narrow slits and Huygen’s wavelets (slides 14-17)

2. Calculating the time averaged intensity pattern observed on a screen (slides 20-22)

6. Wave view of lenses (slides 24,25)

7. 4f optical system and Fourier Transforms1. Introduction (slide 26) and reminder that the laser produces a plane

wave in the z direction, so the kx and ky components of the propagation vector are 0

2. Demonstration that the field amplitude pattern g(x,y) produced by the LCD results in diffraction that creates non-zero k vector components in the x and y directions such that the amplitude of the efield in the fourier transform plane corresponds to the kx kycomponents of the fourier transform of g(x,y)

Ray Optics IntroAssumption: Light travels in straight lines

Ray optics diagrams show the straight lines along which the light propagates

Rules for a converging lens with focal length f, with the optic axis  defined as the line that goes through the center of the lens perpendicular to the plane containing the lens: 

1. Rays passing through the center of the lens are not deflected

2. Rays parallel to the optic axis are deflected so they pass through the focal point, which is the point on the optic axis  one focal length in front of the lens 

Rules for finding  an image using ray optics diagrams

1. Draw a line from an edge of the object that passes undeflectedthrough the center of the lens  

2. Draw a line from an edge of the object that travels parallel to the optic axis, this ray will deflect and pass through the focal point 

3. The image is the point where the two rays intersect

Object

FocalLength

FocalPoint Image

PositionOpticAxis

Ray Optics Examples

PointImage

FocalLength

Object at infinity=Plane wave, focuses at the focal point

PointSource

FocalLength

A point source at the focal point, produces a propagating plane wave

Lec 2 :4

By en:User:DrBob (original); en:User:Pbroks13 (redraw) -

Pinhole Camera

Wave Optics

Linearly polarized plane wave propagating in the z direction is characterized by single k vector in the z direction with magnitude 2A wave crest moves a distance during T,one oscillation period of the wave.  Since the speed of light is c, T=c. T=2 and k= so 2=2/(kc) ‐> w=kcE(x,y,z)= Eo Cos(kz – t)  ŷ)= Eo Cos(k( z – c t ))  ŷComplex Notation

E(x,y,z)= Eo Exp[i(kz – t)]  ŷ)= Eo Exp[ik (z –c t )]  ŷ

Light is a wave whose propagation is governed by a wave equation.   At a given time, at each position in space the wave can be described by specifying the amplitude and the phase.

Wavelength=2/k

The wave moves at 

velocity v. It moves a 

distance in one oscillation period T=

k

Lec 2 :6

Propagating Plane Wave

Out[172]=

Out[173]=1

4

Out[174]=

Out[175]=12

Out[176]=

Out[177]=3

4

Out[178]=

Out[179]= 1

Out[180]=

WavesRe[Exp[ i (kz-wt)]

Cos[kz-wt]

Out[285]=

Out[286]=1

4

Out[287]=

Out[288]=12

Out[289]=

Out[290]=3

4

Out[291]=

Out[292]= 1

Out[293]=

Instantaneous IntensityCos2[kz-wt]

Time Average Intensity

1/T 0 ∫ TCos2[kz-wt] dt =1/2 E E*= 1/2

Out[421]=-2 -1 1 2

-1.0

-0.5

0.5

1.0

Out[420]=-2 -1 1 2

-1.0

-0.5

0.5

1.0

Out[419]=-2 -1 1 2

-1.0

-0.5

0.5

1.0

Out[749]=

Out[750]=1

4

Out[751]=

Out[752]=12

Out[753]=

Out[754]=3

4

Out[755]=

Out[756]= 1

Out[757]=

Lec 2 :7

Wave propagation from a point source

Efield amplitude=Re[Exp[ i (kr-wt)]/r

=Cos[kr-wt)/rIn spherical coordinates, for a point source at the origin, the wave vector always points in the r direction

Lec 2 :8

Two In Phase Sources Very Close(gives resolution limit of microscopes)

Looks like 1 SourceKey Point: Can’t resolve objectsmuch closer than a wavelength

Field dueto 2 Sources

Intensity dueto 2 Sources

Lec 2 :9

Huygen’s Wavelets Example• He proposed that the forward propagation of a wave

can be reproduced by choose a wavefront (surface of constant phase) and in phase point sources at every position on that wavefront. In practice Using sources separated by the wavelength is good enough.

• Below a plane of sources creates plane waves

Lec 2 :10

Plane Wave Transmission through a Small Slit

http://www.fas.harvard.edu/~scidemos/OscillationsWaves/RippleTank/RippleTank004.jpg

Real Picture of a Water Wave Going

Through a Slit

slit < almost no wave gets through

slit ~ wave reaches everywhere behind the slit

slit > wave reaches everywhere behind the slit, but is less intense at large angles

Lec 2 :11

Huygen’s Wavelets Example• For a plane wave going through a slit, choose a

phase front at the slit

• Replace the plane wave by spherical waves all along the open slit.

• For a narrow slit all of the point sources for Huygen’s wavelets are very close to each other -> propagating wave looks like a spherical wave

Lec 2 :12

Transmission through slits with increasing d/

Ray Optics are Limit of Wave Optics when  << system sizeParticles follow ray optics

Plane wave traveling through a single slit

Wave Picture

Time Average Intensity Picture

Lec 2 :14

Fields due to 2 separate sources

Lec 2 :15

Increasing the Spacing Between the Sources Increases the Number

of Nodes

http://webphysics.davidson.edu/Applets/ripple4/

Lec 2 :16

What you will measure is the Resulting Time Averaged

Intensity at a Screen

http://www.youtube.com/watch?v=DfPeprQ7oGc

Lec 2 :17

Electron Double Slit( both wave and particle clearly visible)

http://www.physics.brocku.ca/courses/1p22/images/electron_two_slit.jpg

100 Electrons

3000 Electrons

70000 Electrons

stc: 18

Calculating Intensity at Different Positions on the Detection Screen

At the top of the screen, the phase difference is pi (180 degrees).The waves interfere destructively and the intensity is zero

At the center of the screen (half way up), the phase difference is 0.The waves interfere constructively and the intensity 4.

3/4 of the way up the screen, the phase difference is pi/2 (90 degrees). The intensity is 2.

¼ of the way up the screen, the phase difference is -pi/2 (-90 degrees).The intensity 4.

At the bottom of the screen, the phase difference is - pi (-180 degrees).The waves interfere destructively and the intensity is zero

Lec 2 :19

Top view of the total wave and Graph of the Intensity at the

screen as a function of position

• With both sources the probability of reaching a given point on the screen is NOT the sum of the probabilities for the individual sources. At a given point, the probability can be higher or lower than the sum of the individual probabilities

Lec 2 :20

How can you find the pattern without a Computer?

• Draw Pictures showing the wave fronts– each red line represents a peak in the displacement

– peaks are separated by lambda

• For two sources draw pictures for each source separately

– the TOTAL intensity is a maximum where the wave fronts coincide indicating peaks for BOTH waves

Lec 2 :21

Line intersection -> Constructive interference

Half separation -> Destructive interference

Lec 2 :22

2 Slits using Huygen’s Wavelets

Just After Sources Startspherical waves can be seen coming from each source

At a later timethe two waves are interfering everywhereNote the maxima, where the wave fronts from both waves coincide and minima where the wavef ronts from the two sources are as far separated as possible

http://www.phy.ntnu.edu.tw

Lec 2 :23

Atoms are Waves

NIST

Prentiss Group

Lec 2 :24

Linking Lensing in Geometric Optics with

Lensing in Wave Optics

Point source at the focal point creates a plane wave

PointSource

FocalLength

Plane Wave

PointSource

Lec 2 :25

Wave Picture of a Lens Converting a diverging Spherical Wave from the

focus point to a Plane wave( convert green sphere phase front to green plane phase front)

R=f

(x) + k L = k R -> (x) = k (R –L) ~ k R ( 1- (1+1/2 x2/R2)) = ½ k x2/R. If the phase shift at the center of the lens is k nglass do, then the phase shift everywhere else should be smaller by ½ k x2/R. so (x) = k nglass do( 1 - ½ x2/R), so it decreases quadratically with x.

Lens thickness d(x) given by do( 1 - ½ x2/f),

Phase Lag=

L=Sqrt(R2+x2)

L

Phase Lag=

Lens thickness d(x,y)= do( 1 - ½ (x2+y2)/f)

x

K is in the z direction

Lec 2 :26

Huygen’s Wavelets and Fourier Optics

(x,y,z=0) (x’,y’,z’=f) (x’’,y’’,z’=2f)

G(kx,ky)= 1/(2)2-∞ ∫ ∞ -∞ ∫ ∞ g(x,y) Exp[i(kx x +ky y) dx dy

Goal: Use a “4f” optical system to explore two dimensional Fourier transforms. In particular, we will be studying a plane wave traveling in the z direction that interacts with an optical mask at position z=0. Let g(x,y) is amplitude of the electric field amplitude in the z=0 plane. The spatial dependence g(x,y) results in diffractions that gives the light k vector components along the x and y directions. The vector components along the x and y directions will cause represent the k vector Fourier transforms due to the optical mask pattern g(x,y). In the Fourier transform plane (z=2f) we have positioned an aperture. A bright spot at a position (x’’,y’’,z=2f) corresponds to the square of the amplitude of a k vector. A bright spot at x’=y’’=0 corresponds to a k vector with no component along the x and y directions. All other points represent k vectors with non-xero components.

K from the laser is in the z direction only

For z< 0 light is a plane wave in the z direction

Propagating through the lens changes the phase of the light, where the phase change is give by= - ½ k x2/f.

f f

Computer controlled optical mask at z=0 creates a spatial distribution in the electric field g(x,y) that produces diffraction creating k vectors in the x and y direction, though |kz|>>|kx|and |kz|>>|ky|

Diffracted light propagates freely for a distance f= focal length

Light propagates freely for a distance f= focal

length

The electric field at (x’’,y’’) corresponds to the Fourier transform G(kx,ky)= of g(x,y), where the amplitude at each position (x’’,y’’) corresponds to the amplitude of the k vector component in the x and y directions represented by that position.

Computer controlled optical

mask

Lens with focal length f Aperture

Lec 2 :27

Huygen’s Wavelets and Fourier Optics

(x,y,z=0) (x’,y’,z’=f) (x’’,y’’,z’=2f)

K from the laser is in the z direction only

For z< 0 light is a plane wave in the z direction

To get the total field at any p position (x’,y’) on the lens you must add up the Huygens wavelets from every single point source point in the optical mask.

f f

Computer controlled optical mask at z=0 creates a spatial distribution in the electric field g(x,y) that produces diffraction creating k vectors in the x and y direction, though |kz|>>|kx|and |kz|>>|ky|

Diffracted light propagates freely for a distance f= focal length

Light propagates freely for a distance f= focal

length

To get the total field E(x’’,y’’) at the aperture in the Fourier transform plan, add up spherical waves starting at each point (x’,y’) in the lens with the phase and amplitude to the field at (x’,y’) after the lens. This is a generalization of Huygen’s idea that does not require placing the sources on a wavefront.

Computer controlled optical

mask

Lens with focal length f Aperture

Lec 2 :28

Huygen’s Wavelets and Fourier Optics

(x,y,z=0) (x’,y’,z’=f) (x’’,y’’,z’=2f)

Overall Outline of Wave Optics Calculation1. The optical electrical field after the optical mask has an amplitude as a function of position

give by g(x,y) that is created by the spatial pattern that you put on the mask2. The total electric field at an position (x’,y’) just before the lens is given by the sum of all of the

Huygen’s wavelets starting at positoin (x,y) in the z=0 plane and freely propagating to (x’,y’,z=f) along a vector r1=( x’-x,y’-y,z’-z) where the computer controlled mask at z=0 is approximated by an infinite plane. Example r1 vectors from the z=0 plane to one single chosen point (x’,y’,z=f) are shown below. Geometric optics are NOT assumed. These are not ray optics paths, they are propagation distances for Huygen’s wavelets

E(x’,y’)before lens = -∞ ∫ ∞ -∞ ∫ ∞ g(x,y) Exp[i k r1] dx dy

3. The total electric field at an position (x’,y’) just after the lens is the electric field before the lens with an additional phase shift due to the lens that depends on x’ and y’. The x,y, x’,y’, and x’’,y’’ independent component of the phase shift = kglass do can be left out since it is a constant phase factor common to every wavelet so it can be left out of the calculation, but the x’,y’ component must be included, so

E (x’,y’) after lens =Exp [- i ½ (x’2+y’2)/f)] -∞ ∫ ∞

-∞ ∫ ∞ g(x,y) Exp[i k r1] dx dy

1. 4. The total electric field at an position (x’’,y’)’ just before the lens is given by the sum of all of spherical waves starting at position(x’,y’)z=0 ) after the lens and freely propagating to (x’’,y’’,z=2f) along a vector r2=( x’’-x’,y’’-y,’z’’-z’=f)

E(x’’,y’’)at aperture= -∞ ∫ ∞ -∞ ∫ ∞ E after lens (x’,y’) Exp[i k r2] dx’ dy’

K from the laser is in the z direction only

f f

Computer controlled optical

mask

Lens with focal length f Aperture

r1 vectors from different (x,y,z=0) points to a single (x’,y’,z=f) point

r2 vectors from different (x’,y,’z=f) points to a single (x’’,y’’,z=2f) point

Lec 2 :29

Huygen’s Wavelets and Fourier OpticsG(kx,ky)= 1/(2)2

-∞ ∫ ∞ -∞ ∫ ∞ g(x,y) Exp[i(kx x +ky y) dx dy

K is in the z direction only

Example r1

f

r1=((x’-x)2+(y’-y)2+z2)1/2 =z ((x’-x)2/z2+(y’-y)2/z2+1)2)1/2

Taylor Series expand~z(1+1/2 (x’-x)2/z2 +1/2 (y’-y)2/z2) =z+1/2 (x’-x)2/z +1/2 (y’-y)2/z

Diffracted light propagates freely for a distance f= focal length

Step 1: Calculate the Electric field before the lens given a plane wave before the mask, and an electric field after the mask whose amplitude is given by g(x,y). This requires adding up the Huygen’s wavelet contributions from every point in the optical mask.

Step A: Consider one particular propagation vector from the mask to the lens. Thus start at the lens at a position (x,y) and consider the propagation to a point (x’,y’) on the lens. This represents one optical path. This is NOT a ray optics picture. This is one particular wave optics path.

Step 2: Substitute into the expression for the electric field after the lens.Exp [i kglass do] is a constant phase factor independent

E (x’,y’) after lens =Exp [-i k ½ (x’2+y’2)/f)] -∞ ∫ ∞ -∞ ∫ ∞ g(x,y) Exp[i k r1] dx dy=Exp [-i k ½ (x’2+y’2)/f) Exp[i k z]

-∞ ∫ ∞ -∞ ∫ ∞ g(x,y) Exp[i ½(x’-x)2/z+ ½ (y’-y)2/z] dx dy

Example r2

f

E (x’,y’) after lens = Exp [-i k ½ (x’2+y’2)/f)

-∞ ∫ ∞ -∞ ∫ ∞ g(x,y) Exp[i ½(x’-x)2/z+ ½ (y’-y)2/z] dx dy

Exp[i k z] is independent of x,y, x’,y’, and x’’,y’’ so it is a global phase shift that can be left out giving

Lec 2 :30

Huygen’s Wavelets and Fourier OpticsG(kx,ky)= 1/(2)2

-∞ ∫ ∞ -∞ ∫ ∞ g(x,y) Exp[i(kx x +ky y) dx dy

K is in the z direction only

Example r1

fDiffracted light

propagates freely for a distance f= focal length

Step 3: Calculate the Electric field at one single point(x’’,y’’, z’’’=2f) in the Fourier transform plane by adding up spherical waves starting from each point (x’,y’,z=f) just after the lens. This requires adding up the Huygen’s wavelet contributions from every point in the optical mask.

E (x’’,y’’) at aperture=

-∞ ∫∞

-∞ ∫∞

(Exp [-i k ½ (x’2+y’2)/f) -∞ ∫ ∞ -∞ ∫ ∞ g(x,y)Exp[i/(2f) ((x’-x)2 +(y’-y)2] dx dy )Exp[i ½(x’’-x’)2/z+ ½ (y’’-y’)2/f] dx’ dy’

Example r2

f

Step 3: Do the integral over dx’ and dy’. To do this, reverse the order of the integrals over x’,y’ and x,y. The combine terms containing x’, and y’.

E (x’’,y’’) at aperture=

-∞ ∫ ∞ -∞ ∫ ∞ g(x,y)

(-∞ ∫ ∞ -∞ ∫ ∞ Exp[i(k/2f) (- (x’2+y’2)+((x’-x)2+(y’-y)2+(x’’-x’)2+(y’’-y’)2]dx’dy’ ) dx dy

Step 4: Since the x’ and y’ terms are perfectly symmetrical, do just just the x’ first. Begin by simplify the terms in the exponent for the x,x’, and x’’ coordinates only since there are no cross terms between x and y

- x’2+x’2-2xx’ +x2+x’’2-2 x’’x’ + x’2= -2x’(x+ x’’) +x’2+ (x’’2 +x2)

Lec 2 :31

Huygen’s Wavelets and Fourier OpticsG(kx,ky)= 1/(2)2

-∞ ∫ ∞ -∞ ∫ ∞ g(x,y) Exp[i(kx x +ky y) dx dy

K is in the z direction only

Example r1

fDiffracted light

propagates freely for a distance f= focal length

Example r2

f

Step 5: The integral over dx’ is a definite integral with value Exp[ i k x x’’/f] Sqrt[f /(2i k)]. The integral over y’ is analogous, so the total result of the two integrals is a multiplicative factor Exp[ i k /f (x x’’ + y y’’) ] f /(2i k). If one defines kx’’= k/f x’’ and ky’’= k/f y’’ the electric field in the fourier transform plane becomes

=-∞ ∫∞-∞ ∫∞ g(x,y) (-∞ ∫∞ -∞ ∫∞ Exp[i(k/2f) (-2x’(x+ x’’)+x’2+x’’2+x2]dx’ dy’ ) dx dy

E(x’’,y’’)at aperture = (f /(2i k) -∞ ∫∞ -∞ ∫ ∞ g(x,y)Exp[i(x kx’’ + y ky’’) ]dx dy

Step 5: Notice that the integral is the same as the integral in the definition of the Fourier transform. Thus, E(x’’,y’’)at aperture is proportional to the Fourier transform of g(x,y) the spatial distribution of the electric field at the light modulator.

Lec 2 :32

The End