Diffraction

Light bends!

Diffraction assumptions

Solution to Maxwell's Equations

The far-field

Fraunhofer DiffractionSome examples

Diffraction

Light does not always travel in a straight line.

It tends to bend around objects. This tendency is called diffraction.

Any wave will do this, including matter waves and acoustic waves.

Shadow of a hand

illuminated by a

Helium-Neon laser

Shadow of a zinc oxide

crystal illuminated

by aelectrons

Why it’s hard to see diffraction

Diffraction tends to cause ripples at edges. But poor source temporal or spatial coherence masks them.

Example: a large spatially incoherent source (like the sun) casts blurry shadows, masking the diffraction ripples.

Untilted rays yield a perfect shadow of the hole, but off-axis rays blur the shadow.

Screenwith hole

A point source is required.

Diffraction of a wave by a slit

Whether waves in water or electromagnetic radiation in air, passage through a slit yields a diffraction pattern that will appear more dramatic as the size of the slit approaches the wavelength of the wave.

λ ≈ slit size

λ slit size

λ < slit size

Diffraction of ocean water waves

Ocean waves passing through slits in Tel Aviv, Israel

Diffraction occurs for all waves, whatever the phenomenon.

Light passing by edge

Diffraction by an Edge

Electrons passing by

an edge (Mg0 crystal)

xTran

smis

sion

Even without a small slit, diffraction can be strong.

Simple propagation past an edge yields an unintuitive irradiance pattern.

Radio waves diffract around mountains.

When the wavelength is

km long, a mountain peak is

a very sharp edge!

Another effect that occurs is scattering, so

diffraction’s role is not obvious.

Diffraction GeometryWe wish to find the light electric field after a screen with a hole in it.This is a very general problem with far-reaching applications.

What is E(x1,y1) at a distance z from the plane of the aperture?

Incident wave This region is assumed to be

much smaller than this one.

x1x0

P10

A(x0,y0) y1y0

Diffraction SolutionThe field in the observation plane, E(x1,y1), at a distance z from the aperture plane is given by:

( ) ( )

0 0

1 1 1 0 1 0 0 0 0 0

( , )

011 0 1 0

01

2 2201 0 1 0 1

( , , ) ( , , ) ( , )

exp( )1 ( , , )

A x y

E x y z h x x y y z E x y dx dy

ikrh x x y y zi r

r z x x y y

λ

= − −

− − =

= + − + −

∫∫where :

and :

A very complicated result! And we cannot approximate r01 in the exp by z because it gets multiplied by k, which is big, so relatively small changes in r01 can make a big difference!

Spherical wave

Fraunhofer Diffraction: The Far Field

( ) ( ) ( )0 0

2 21 1

1 1 0 1 0 1 0 0 0 0

( , )

exp( ), exp exp ,2

A x y

x yikz ikE x y ik x x y y E x y dx dyi z z zλ

+ = − + ∫∫

We can approximate r01 in the denominator by z, and if D is the size of the aperture, D 2 ≥ x0

2 + y02, so when k D2/ 2z << 1, the quadratic terms <<

1, so we can neglect them:

This condition means going a distance away: z >> kD2/2 = πD2/λIf D = 1 mm and λ = 1 micron, then z >> 3 m.

( ) ( ) ( ) ( )2 2 2 22 2 201 0 1 0 1 0 1 0 11 / 2 / 2r z x x y y z x x z y y z = + − + − ≈ + − + −

( ) ( )2 2 2 201 0 0 1 1 0 0 1 12 / 2 2 / 2kr kz k x x x x z k y y y y z≈ + − + + − +

Independent of x0 and y0, so factor these out.

Small, so neglect these terms.

Fraunhofer DiffractionWe’ll neglect the phase factors, and we’ll explicitly write the aperture function in the integral:

( ) ( )1 1 0 1 0 1 0 0 0 0 0 0, exp ( , ) ( , )ikE x y x x y y A x y E x y dx dyz

∞ ∞

−∞ −∞

∝ − + ∫∫

This is just a Fourier Transform!

Interestingly, it’s a Fourier Transform from position, x0, to another position variable, x1 (in another plane). Usually, the Fourier “conjugate variables” have reciprocal units (e.g., t & ω, or x & k). The conjugate variables here are really x0 and kx = kx1/z, which have reciprocal units.

So the far-field light field is the Fourier Transform of the apertured field!

E(x0,y0) = constant if a plane wave

The Fraunhofer Diffraction formula

where we’ve dropped the subscripts, 0 and 1,

( ) ( ), exp ( , ) ( , )x y x yE k k i k x k y A x y E x y dx dy

∞ ∞

−∞ −∞

∝ − + ∫∫E(x,y) = const if a plane wave

Aperture function

We can write this result in terms of the off-axis k-vector components:

kx = kx1/z and ky = ky1/z

( ) { }, ( , ) ( , )x yE k k A x y E x y∝ F

kz

ky

kx

θx = kx /k = x1/z and θy = ky /k = y1/z

and:

or:

The Uncertainty Principle in Diffraction!

Because the diffraction pattern is the Fourier transform of the slit, there’s an uncertainty principle between the slit width and diffraction pattern width!

If the input field is a plane wave and ∆x = ∆x0 is the slit width,

( ) { }, ( , ) ( , )x yE k k A x y E x y∝ F

1xx k∆ ∆ >

0 1 /x x z k∆ ∆ >

kx = kx1/z

Or:

The smaller the slit, the larger the diffraction angle and the bigger the diffraction pattern!

Fraunhofer Diffraction from a slit

Fraunhofer Diffraction from a slit is simply the Fourier Transform of a rect function, which is a sinc function. The irradiance is then sinc2 .

Fraunhofer Diffraction from a Square Aperture

The diffracted field is a sinc function in both x1 and y1because the Fourier transform of a rect function is sinc.

Diffractedirradiance

Diffractedfield

Diffraction from a Circular Aperture

A circular apertureyields a diffracted"Airy Pattern,"which involves aBessel function.

Diffracted field

Diffracted Irradiance

Diffraction from small and large circular apertures

Recall the Scale Theorem!This is the Uncertainty Principle for diffraction.

Far-field intensity pattern

from a small aperture

Far-field intensity pattern

from a large aperture

Fraunhofer diffraction from two slits

A(x0) = rect[(x0+a)/w] + rect[(x0-a)/w]0 x0a-a

1 0( ) { ( )}E x A x∝F

1 1

1 1

sinc[ ( / ) / 2]exp[ ( / )]sinc[ ( / ) / 2]exp[ ( / )]

w kx z ia kx zw kx z ia kx z

∝ + +−

1 1 1( ) sinc( / 2 ) cos( / )E x wkx z akx z∝ kx1/z

w w

Diffraction from one- and two-slit screens

Fraunhofer diffraction patterns

One slit

Two slits

Diffraction Gratings

•Scattering ideas explain what happens when light impinges on a periodic array of grooves. Constructive interference occurs if the delay between adjacent beamlets is an integral number, m, of wavelengths.

[ ]sin( ) sin( )m ia mθ θ λ− =

where m is any integer.

A grating has solutions of zero, one, or many values of m, or orders.

Remember that m and θm can be negative, too.

Path difference: AB – CD = mλ

Scatterer

Scatterer

a

θi

θm

a

AB = a sin(θm)

CD = a sin(θi)

A

DC

B

Potentialdiffracted wave-front

Incident wave-front

θi

θm

Diffraction orders Because the diffraction angle depends on λ, different wavelengths are separated in the nonzero orders.

No wavelength dependence occurs in zero order.

The longer the wavelength, the larger its deflection in each nonzero order.

Diffraction angle, θm(λ)

Zeroth order

First order

Minus first order

Incidence angle, θi

Real diffraction gratings

Diffracted white light

White light diffracted by a real grating.

m = 0

m=1m =

2

m = -1

The dots on a CD are equally spaced (although some

Diffraction gratings

World’s largest diffraction grating

Lawrence Livermore National Lab