Outline 1. 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 Review 1. Definitions and plane wave example (slides 5,6) 2. Point sources and spherical waves (slide 7,8) 3. Introduction to Huygen’s wavelets 1. Plane wave example (slide 9) 4. Transmission of a plane wave through a single slit 1. 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 slits 1. 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 Transforms 1. 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 ky components of the fourier transform of g(x,y) Lec 2 :1 Ray Optics Intro

Assumption: 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 undeflected through 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

Focal Point Image Optic Position Axis Object

Focal Length Ray Optics Examples

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

Point Image

Focal Length

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

Point Source

Focal Length Pinhole Camera

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

Lec 2 :4 Wave Optics 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. Linearly polarized plane wave propagating in the z direction is characterized by single k vector in the z direction with magnitude 2 A 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=kc

E(x,y,z)= Eo Cos(kz – t) ŷ)= Eo Cos(k( z –c t )) ŷ Complex Notation

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

k

Wavelength= 2/k

The wave moves at velocity v. It moves a distance in one oscillation period T= Propagating Plane Wave Waves Instantaneous Intensity Time Average Intensity Re[Exp[ i (kz-wt)] 2 Cos [kz-wt] T 2 Cos[kz-wt] 1/T 0 ∫ Cos [kz-wt] dt =1/2 E E*= 1/2

1.0 1.0 1.0

0.5 0.5 0.5

Out[419]= Out[420]= Out[421]= -2 -1 1 2 -2 -1 1 2 -2 -1 1 2

-0.5 -0.5 -0.5

-1.0 -1.0 -1.0

Out[749]= Out[172]= Out[285]=

1 1 1 Out[750]= Out[173]= Out[286]= 4 4 4

Out[751]= Out[174]= Out[287]=

1 1 1 Out[752]= Out[175]= Out[288]= 2 2 2

Out[753]= Out[176]= Out[289]=

3 3 3 Out[754]= Out[177]= Out[290]= 4 4 4

Out[755]= Out[178]= Out[291]=

Out[756]= 1 Out[179]= 1 Out[292]= 1

Out[757]= Out[180]= Out[293]=

Lec 2 :6 Wave propagation from a point source

Efield amplitude= In spherical Re[Exp[ i (kr-wt)]/r coordinates, for =Cos[kr-wt)/r a point source at the origin, the wave vector always points in the r direction

Lec 2 :7 Two In Phase Sources Very Close (gives resolution limit of microscopes)

Looks like 1 Source Key Point: Can’t resolve objects much closer than a wavelength

Field due to 2 Sources

Intensity due to 2 Sources

Lec 2 :8 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 :9 Plane Wave Transmission through a Small Slit slit <  almost no wave gets through Real Picture of a Water Wave Going Through a Slit

slit ~  wave reaches everywhere behind the slit

slit >  wave reaches everywhere behind the slit, but is less intense at large http://www.fas.harvard.edu/~scidemos/Oscillati angles onsWaves/RippleTank/RippleTank004.jpg Lec 2 :10 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 :11 Transmission through slits with increasing d/

Lec 2 :12 Ray Optics are Limit of Wave Optics when  << system size Particles follow ray optics

Plane wave traveling through a single slit Wave Picture

Time Average Intensity Picture Fields due to 2 separate sources

Lec 2 :14 Increasing the Spacing Between the Sources Increases the Number of Nodes

http://webphysics.davidson.edu/Applets/ripple4Lec 2 :15 / What you will measure is the Resulting Time Averaged Intensity at a Screen

http://www.youtube.com/watch?v=DfPeprQ7oGcLec 2 :16 Electron Double Slit ( both wave and particle clearly visible)

100 Electrons

3000 Electrons

70000 Electrons

http://www.physics.brocku.ca/courses/1p22/images/electron_two_slit.jpgLec 2 :17 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

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

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

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 stc: 18 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 :19 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 :20 Line intersection -> Constructive interference Half separation -> Destructive interference

Lec 2 :21 2 Slits using Huygen’s Wavelets

Just After Sources Start spherical waves can be seen coming from each source

At a later time the two waves are interfering everywhere Note 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.twLec 2 :22 Atoms are Waves

NIST

Prentiss GroupLec 2 :23 Linking Lensing in Geometric Optics with Lensing in Wave Optics

Point source at the focal point creates a plane wave

Point Source Focal Plane Length Wave

Point Source

Lec 2 :24 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) Lens thickness d(x,y)= do( 1 - ½ (x2+y2)/f) Phase Lag = 

L  x R=f K is in the z direction

L=Sqrt(R2+x2) Phase Lag =  (x) + k L = k R -> (x) = k (R –L) ~ k R ( 1- (1+1/2 x2/R2)) 2 = ½ k x /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. 2 so (x) = k nglass do( 1 - ½ x /R), so it decreases quadratically with x. Lens thickness d(x) given by do( 1 - ½ x2/f),

Lec 2 :25 Huygen’s Wavelets and Fourier Optics 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. 2 ∞ ∞ G(kx,ky)= 1/(2) -∞ ∫ -∞ ∫ g(x,y) Exp[i(kx x +ky y) dx dy Computer Lens with focal controlled optical length f Aperture mask

f K from the f laser is in Diffracted light Light propagates freely the z propagates freely for a for a distance f= focal direction distance f= focal length length only (x,y,z=0) (x’,y’,z’=f) (x’’,y’’,z’=2f) For z< 0 Computer The electric field at Propagating light is a controlled optical (x’’,y’’) corresponds to through the lens plane mask at z=0 the Fourier transform changes the phase wave in creates a spatial G(k ,k )= of g(x,y), of the light, where x y the z distribution in the where the amplitude the phase change direction electric field g(x,y) at each position that produces is give by 2 (x’’,y’’) corresponds to diffraction creating = - ½ k x /f. the amplitude of the k k vectors in the x vector component in and y direction, the x and y directions though |kz|>>|kx| represented by that and |kz|>>|ky| position. Lec 2 :26 Huygen’s Wavelets and Fourier Optics

Computer Lens with focal controlled optical length f Aperture mask

f K from the f laser is in Diffracted light Light propagates freely the z propagates freely for a for a distance f= focal direction distance f= focal length length only (x,y,z=0) (x’,y’,z’=f) (x’’,y’’,z’=2f) For z< 0 Computer light is a controlled optical To get the total field To get the total plane mask at z=0 E(x’’,y’’) at the field at any p wave in creates a spatial aperture in the position (x’,y’) on the z distribution in the Fourier transform the lens you must direction electric field g(x,y) plan, add up add up the that produces spherical waves Huygens wavelets diffraction creating starting at each point from every single k vectors in the x (x’,y’) in the lens with point source point and y direction, the phase and in the optical mask. though |kz|>>|kx| amplitude to the field and |kz|>>|ky| at (x’,y’) after the lens. This is a generalization of Huygen’s idea that does not require placing the sources on a wavefront.

Lec 2 :27 Huygen’s Wavelets and Fourier Optics Overall Outline of Wave Optics Calculation 1. 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 mask 2. 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 = 2 2 ∞ ∞ Exp [- i ½ (x’ +y’ )/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’ Computer Lens with focal controlled optical length f Aperture mask r2 vectors from different (x’,y,’z=f) points to a single (x’’,y’’,z=2f) point

K from the laser is in r1 vectors from the z different (x,y,z=0) direction points to a single (x’,y’,z=f) point only (x,y,z=0) f (x’,y’,z’=f) f (x’’,y’’,z’=2f)Lec 2 :28 Huygen’s Wavelets and Fourier Optics 2 ∞ ∞ G(kx,ky)= 1/(2) -∞ ∫ -∞ ∫ g(x,y) Exp[i(kx x +ky y) dx dy 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.

Example r1

K is in the z direction f f only Diffracted light propagates freely for a Example r distance f= focal length 2 2 2 2 1/2 2 2 2 2 2 1/2 r1=((x’-x) +(y’-y) +z ) =z ((x’-x) /z +(y’-y) /z +1) ) 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 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 = 2 2 ∞ ∞ Exp [-i k ½ (x’ +y’ )/f)] -∞ ∫ -∞ ∫ g(x,y) Exp[i k r1] dx dy =Exp [-i k ½ (x’2+y’2)/f) Exp[i k z] ∞ ∞ 2 2 -∞ ∫ -∞ ∫ g(x,y) Exp[i ½(x’-x) /z+ ½ (y’-y) /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

2 2 E (x’,y’) after lens = Exp [-i k ½ (x’ +y’ )/f) ∞ ∞ 2 2 -∞ ∫ -∞ ∫ g(x,y) Exp[i ½(x’-x) /z+ ½ (y’-y) /z] dx dy Lec 2 :29 Huygen’s Wavelets and Fourier Optics 2 ∞ ∞ G(kx,ky)= 1/(2) -∞ ∫ -∞ ∫ g(x,y) Exp[i(kx x +ky y) dx dy 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.

Example r1

K is in the z f direction f only Diffracted light propagates freely for a Example r distance f= focal length 2

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

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)

∞ ∞ 2 2 2 2 2 2 (-∞ ∫ -∞ ∫ Exp[i(k/2f) (- (x’ +y’ )+((x’-x) +(y’-y) +(x’’-x’) +(y’’-y’) ]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 :30 Huygen’s Wavelets and Fourier Optics 2 ∞ ∞ G(kx,ky)= 1/(2) -∞ ∫ -∞ ∫ g(x,y) Exp[i(kx x +ky y) dx dy

Example r1

K is in the z f direction f only Diffracted light propagates freely for a Example r distance f= focal length 2

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

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

∞ ∞ 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 :31 The End

Lec 2 :32