Diffraction

See Chapter 10 of Hecht

Diffraction

• Send light through apertures, slits or gratings

• Predict intensity distribution of light

• Further examples of interference.

Huygens – Fresnel Principle

Every unobstructed point of a wavefront serves as a source of spherical wavelets. Resulting amplitude is the superposition of all of these waves.

Fraunhofer Diffraction – Far Field

Fresnel – Near Field

Kirchhoff – Derivation of diffraction from wave equation

In the limit -> 0 recover geometric optics

Line of point sources (pinholes), all in phase with same amplitude

d=distance between sources

r2

r1

r10

Note that:

sin1sin3

sin2sin

114

1312

dnrrdrr

drrdrr

n

In[2]:= Intensity@Ns_, d_, l _,q_D:=ikSinANs2 pdSin@qD2 lE

SinA2 pdSin@qD2 lEy{^2

In[9]:= Plot@Intensity@10, 10^- 6, 0.632810^- 6,qD,8q, - 0.2, .2<, PlotRange®88- .2, .2<,80, 10^2<<D

-0.2 -0.15 -0.1 -0.05 0.05 0.1 0.15 0.2

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Out[9]= … Graphics …In[16]:= Plot@Intensity@20, 10^- 6, 0.632810^- 6,qD,8q, - 0.2, .2<, PlotRange®88- .2, .2<,80, 20^2<<D

-0.2 -0.15 -0.1 -0.05 0.05 0.1 0.15 0.2

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Out[16]= … Graphics …

In[13]:= Plot@Intensity@10, 410^- 6, 0.632810^- 6,qD,8q, - 1, 1<, PlotRange®88- 1, 1<,80, 10^2<<D

-1 -0.75 -0.5 -0.25 0.25 0.5 0.75 1

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Out[13]= … Graphics …

Single Slit & Fraunhofer

x

y = D/2

y = -D/2

rR

P

Using Huygens, treat slit (length D) as a line of point radiators.Point radiator at y is a distance r from observation point P; R is distance from slit center to P.

y

...cos2

sin 22

R

yyRr

In[21]:= IntensityS1@d_, l _, q_D:=ikSinA2 pdSin@qD2 lEI2 pdSin@qD

2 lMy{^2

In[31]:= Plot@IntensityS1@10^- 2, 0.632810^- 6, qD,8q, - 0.0002, .0002<,PlotRange®88-0.0002, 0.0002<,80, 1<<D

-0.0002 -0.00015 -0.0001 -0.00005 0.00005 0.0001 0.00015 0.0002

0.2

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Out[31]= … Graphics …

x

P

z

Single Slit, other dimension. Slit width of b

Same derivation as before:

2

2

0

2/

2/

sin2

sin2

sin

sinexp

kb

kb

II

ikzdzeR

AE

b

b

tkRi

Double Slit

x

P

z

y

z

y = D/2

y = -D/2

Slits of widthb, separated by distance a.

In[32]:= IntensityS2@a_, b_, l _, q_D:=ikSinA2 pbSin@qD2 lEI2 pbSin@qD

2 lMy{^2CosA2 paSin@qD2 l

E2In[42]:= Plot@IntensityS2@310^- 2, 10^- 2, 0.632810^-6, qD,8q, -0.0002, .0002<,

PlotRange®88-0.0002, 0.0002<,80, 1<<D

-0.0002 -0.00015 -0.0001 -0.00005 0.00005 0.0001 0.00015 0.0002

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1

Out[42]= … Graphics …

Line of point sources (pinholes), all in phase with same amplitude

d=distance between sources

r2

r1

r10

Note that:

sin1sin3

sin2sin

114

1312

dnrrdrr

drrdrr

n

Circular Aperture (of radius a)

z

y

x

P

R

Source is at an angle and at radius .

In[7]:= IntensityCircle@k_, a_, q_D:=ik2 BesselJ@1, akSin@qDDkaSin@qDy{^2In[24]:= Plot@IntensityCircle@2 pH632.810^-9L, .001, qD,8q, - .001, .001<, PlotRange®88- .001, .001<,80, 1<<D

-0.001 -0.00075 -0.0005 -0.00025 0.00025 0.0005 0.00075 0.001

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Out[24]= … Graphics …

Diffraction Gratings

Used to separate light of different wavelengths with high resolution.

Diffraction grating applications: measuring atomic spectra in laboratory instruments and telescopes.

A large number of parallel, closely spaced slits constitutes a diffraction grating.

The condition for maximum intensity is the same as that for the double slit or multiple slits, but with a large number of slits the intensity maximum is very sharp and narrow, providing the high resolution for spectroscopic applications.

Transmission Grating

Example of a transmission grating.

Use Fraunhofer to model a transmission grating of N-slits

x

P

a

b

N-slits, b – wide, separated by distance a.

22

0

sin2

sin

sin2

sin

sin2

sin2

sin

ka

Nka

kb

kb

II

Reflection Grating

differencelength path sinsin ni dd

Angular Dispersion

How the diffraction angle changes with wavelength.

The effective width of a spectral line?

Phase difference between minima (zeros) is:

DispersionAngular

lkN 2

Resolving Power

minPower Resolving R

changeh wavelengtresolvableLeast min

im

NdR

NdR

sinsin

sin

Transmission

Reflection

A hole in a opaque screen. Model hole as a distribution of point-like spherical radiators; integrate over hole.

If that were true, and how nature works, then not only would light propagate forward through the hole, but light would also propagate back towards source of light.

Light is not observed propagating back to source!?

Solution – Kirchhoff & Fresnel

Kirchhoff’s Diffraction Theory

0 If

equation waveA 1

22

2

2

22

keE

t

E

cE

ikct

Fraunhofer diffraction: The source wavefront is assumed to be planar, the different elements of the wavefront have a constant phase difference.

In the Fresnel diffraction the curvature of the wavefront is included, the relative phase is not constant..

r0

r

SourceP

2

cos1exp0

rikr

dE

FactorObliquity 2

cos1 K

Semi-Infinite Opaque Screen

x

y

z

Screen from

0 y

No screen for

0y

x

y

z

r0

r

In[13]:= PlotAAbsAIntegrateAãä pu2‘2,8u, y, Infinity<EE2,8y, - 8, 8<, PlotRange®88- 8, 8<,80, 3<<E

-8 -6 -4 -2 2 4 6 8

0.5

1

1.5

2

2.5

3

Out[13]= … Graphics …

Babinet’s Principle

versus

A hole in a screen

A small opaque screen, which would fill hole above

fieldERadiation 1 E

fieldERadiation 2 E

If hole filled in, no radiation out

21

21 0

EE

EE

Two cases produce the same field, but 180o out of phase.

=> Same intensity pattern