PHY 102: Waves & Quanta

Topic 7

Diffraction

John Cockburn (j.cockburn@... Room E15)

•Interference re-cap

•Phasors

•Single slit diffraction

•Intensity distribution for single slit

Electromagnetic Waves

)sin(0 tkxEEy

)sin(0 tkxBBz

Where E0 and B0 are related by: E0 = cB0

INTENSITY of an EM wave E02

NB. we will see later that EM radiation sometimes behaves like a stream of particles (Photons) rather than a wave………………

Interference

First, consider case for sound waves, emitted by 2 loudspeakers:

Path difference =nλConstructive Interference

Path difference =(n+1/2)λDestructive Interference

(n = any integer, m = odd integer)

Interference

Young’s Double Slit Experiment

•Demonstrates wave nature of light

•Each slit S1 and S2 acts as a separate source of coherent light (like the loudspeakers for sound waves)

Young’s Double Slit Experiment

Constructive interference:

Destructive interference:

nd sin

2

1sin nd

Young’s Double Slit Experiment

Y-position of bright fringe on screen: ym = Rtanm

Small , ie r1, r2 ≈ R, so tan ≈ sin

So, get bright fringe when:

d

nRym

(small only)

Young’s Double Slit Experiment:Intensity Distribution

For some general point P, the 2 arriving waves will have a path difference which is some fraction of a wavelength.

This corresponds to a difference in the phases of the electric field oscillations arriving at P:

tEE sin01

tEE sin02

Young’s Double Slit Experiment:Intensity Distribution

Total Electric field at point P:

tEtEEEETOT sinsin 0021

Trig. Identity:

2

1sin

2

1cos2sinsin

With = (t + ), = t, get:

2sin

2cos2 0

tEETOT

2sin

2cos2 0

tEETOT

So, ETOT has an “oscillating” amplitude:

2cos2 0

E

Since intensity is proportional to amplitude squared:

2cos4 22

0

EITOT

Or, since I0E02, and proportionality constant the same in both cases:

2cos4 2

0

IITOT

differencepath

2

difference phase

sin

2

d

2cos4 2

0

IITOT

sin

cos4 20

dIITOT

For the case where y<<R, sin ≈ y/R:

R

dyIITOT

20 cos4

Young’s Double Slit Experiment:Intensity Distribution

R

dyIITOT

20 cos

2-slit intensity distribution: “phasor” treatment

•Remember from Lecture 1, harmonic oscillation with amplitude A and angular frequency can be represented as projection on x or y axis of a rotating vector (phasor) of magnitude (length) A rotating about origin.

Light•We can use this concept to add oscillations with the same frequency, but different phase constant by “freezing” this rotation in time and treating the 2 oscillations as fixed vectors……

•So called “phasor method”

2-slit intensity distribution: “phasor” treatment

Use phasor diagram to do the addition E1 + E2

tEE cos01

tEE cos02

Using cosine rule:

cos2 20

20

20

2 EEEETOT

cos2 20

20

20 EEE

cos12 20 E

2cos4 22

02

EETOT

Another way: Complex exponentials

)sin(cos iAAei

)sin(cos00 titEeE ti

))sin()(cos(0)(

0 titEeE ti

tieEtE 00 Re)cos(

)(00 Re)cos( tieEtE

Another way: Complex exponentials

Single Slit Diffraction

“geometrical” picture breaks down when slit width becomes comparablewith wavelength

Single Slit Diffraction

observed for all types of wave motion

eg water waves in ripple tank

Single Slit Diffraction

Single Slit Diffraction

•Explain/analyse by treating the single slit as a linear array of coherent point sources that interfere with one another (Huygen’s principle)…………………….

All “straight ahead” wavelets in phase → central bright maximum

Destructive interference of light from sources within slit for certain angles

Fraunhofer (“far-field”)case

Single Slit Diffraction

•From diagram, can see that for slit of width A, we will get destructive interference (dark band on screen) at angles which satisfy…..:

2sin

2

a

2sin

4

a

a

sina

2sin

Choice of a/2 and a/4 in diagram is entirely arbitrary, so in general we have a dark band whenever;

a

m sin (m=±1, ±2, ±3………..)

Position of dark fringes in single-slit diffraction

a

m sin

If, like the 2-slit treatment we assume small angles, sin ≈ tan =ymin/R, then

a

Rmy

min

Positions of intensity MINIMA of diffraction pattern on screen, measured from central position.

Very similar to expression derived for 2-slit experiment:

d

nRym

But remember, in this case ym are positions of MAXIMA

In interference pattern

Width of central maximum

•We can define the width of the central maximum to be the distance between the m = +1 minimum and the m=-1 minimum:

a

R

a

R

a

Ry

2

Ie, the narrower the slit, the more the diffraction pattern “spreads out”

image of diffraction pattern

Intensitydistribution

Single-slit diffraction: intensity distribution

To calculate this, we treat the slit as a continuous array of infinitesimal sources:

Can be done algebraically, but more nicely with phasors………………..

Single-slit diffraction: intensity distribution

E0 is E-field amplitude at central maximum

2

0 2/

)2/sin(

II

= total phase difference for “wavelets” from top and bottom of slit

2/

)2/sin(0

EETOT

Single-slit diffraction: intensity distribution

2

0 2/

)2/sin(

II

How is related to our slit/screen setup?

Path difference between light rays from top and bottom of slit is

sinax

differencepath

2

difference phase

sin

2

d

From earlier (2-slit)

sin2 a

Single-slit diffraction: intensity distribution

2

0 2/

)2/sin(

II

sin2 a

2

0 /)(sin

)/)(sinsin(

a

aII