Interference and Diffraction

of EM waves

Maxwell Equations in General Form

Differential form Integral Form

Gauss’s Law for E

field.

Gauss’s Law for H

field. Nonexistence

of monopole

Faraday’s Law

Ampere’s Circuit

Law

vD

0 B

t

BE

t

DJH

v

v

s

dvdSD

0s

dSB

sL

dSBt

dlE

sL

dSt

DJdlH

Terms

• E = electric field intensity [V/m]

• D = electric field density

• H = magnetic field intensity, [A/m]

• B = magnetic field density, [Teslas]

Maxwell’s Equations

• Additionally the equation of continuity

• Maxwell added the term to Ampere’s Law so

that it not only works for static conditions but

also for time-varying situations.

• This added term is called the displacement

current density, while J is the conduction current.

tJ v

t

D

Energy (intensity) of electromagnetic waves

• The frequency of light is very high,

• There is no such detector to measure the electric

field changes,

• We are able only to measure the mean value of the square

root of the electric field,

• our eyes can detect only intensity of light, not phase.

Energy (intensity) of electromagnetic

waves

Poynting vector

Intensity of EM wave

• Light intensity I is a mean velue of square root of the

electric field intensity and is defined in W/m²

• Taking into account the spectral characteristic of human eye,

light intensity is defined in candelas, lumens or lux

Two waves interfering with each other

If two monochromatic waves described as:

will overlap in some plane x=const, then:

Responsible for interference

Two waves interfering with each other

Responsible for interference

For : > 0 constructive interference

= 0

destructive interference< 0

The same phases The oposite phases

Constructive interference Destructive interference

The key principle: Huygen’s Principle

Christian Huygens

1629-1695

All points in a wavefront serve as point

sources of spherical secondary waves.

After a time t, the new wavefront will be

the tangent to all the resulting spherical

waves.

Huygen’s Principle

For plane waves entering a single slit, the waves emerging from the slit start spreading out, diffracting

Young’s Double Slit ExperimentFor waves entering two slits, the emerging waves interfere and form an interference (diffraction) pattern

Young experiment in 1801: light is wave phenomenon

First plane wave through a small slit yields coherent spherical wave

Then interposed two slits: interference of two spherical waves on a screen

Interference• Phase difference between two waves can change for paths of different lengths

• Each point on the screen is determined by the path length difference DL of the rays reaching that point

Path Length Difference: sinL d D

If sin integer bright fringeL d D

Maxima-bright fringes:

sin for 0,1,2,d m m

Minima-dark fringes: 12

sin for 0,1,2,d m m

1 1.51 dark fringe at: sinm

d

1 22 bright fringe at: sinm

d

Interference

Two sources can produce an interference that is stable over

time, if their light has a phase relationship that does not

change with time: E(t)=E0cos(wt+f).

Coherent sources: Phase f must be well defined and constant

Sunlight is coherent over a short length and time range

Since laser light is produced by cooperative behavior of atoms,

it is coherent of long length and time ranges

Incoherent sources: f jitters randomly in time, no stable

interference occurs

When the interference is possible

Red laser light (=633nm) goes through two slits d=1cm apart, and produces a

diffraction pattern on a screen L = 55cm away. How far apart are the fringes near

the center?

For the spacing to be 1mm, we need d~ L/1mm=0.35mm

If the fringes are near the center, we can use

sin ~ , and then

m=dsin~d => =m/d is the angle for each

maximum (in radians)

D= /d =is the “angular separation”.

The distance between the fringes is then

Dx=LD=L/d=55cm 633nm/1cm=35 mm

Example of interference

Example 2

In a double slit experiment, we can measure the wavelength of the light if we

know the distances between the slits and the angular separation of the fringes.

If the separation between the slits is 0.5mm and the first order maximum of the

interference pattern is at an angle of 0.059o from the center of the pattern, what

is the wavelength and color of the light used?

d sin=m => =0.5mm sin(0.059o)= 5.15 x 10-7m=515nm ~ green

1 0 2 0sin and sinE E t E E tw w f

E1

E2

2 10 2

4 cosI I f2

sind

f

1 1 12 2 2

Minima when: sin for 0,1,2, (minima)m d m mf

12

2Maxima when: for 0,1,2, 2 sin

sin for 0,1,2, (maxima)

dm m m

d m m

f f

avg 02I I

Intensity in Double-Slit Interference

ExampleA double slit experiment has a screen 120cm away from the slits, which are 0.25cm apart. The slits are illuminated with coherent 600nm light. At what distance above the central maximum is the average intensity on the screen 75% of the maximum?

I/I0=4cos2f/2 ; I/Imax=cos2f/2 =0.75 => f=2cos–1 (0.75)1/2=60o=/3 radf=(2d/)sin => = sin-1(/2d)f0.0022o40 mrad (small!)y=L48mm

Interferometers

Michelson’s

Mach-Zehnder’s

Ring

Temporal coherence

White light

LED

SLED

LD

Gas laser

He-Ne

D

Measuring the distance with Michelson’s interferometer

Optical coherence tomography

Diffraction

Huygen’s Principle

Christian Huygens

1629-1695

All points in a wavefront serve as point

sources of spherical secondary waves.

After a time t, the new wavefront will be

the tangent to all the resulting spherical

waves.

Huygen’s Principle

For plane waves entering a single slit, the waves emerging from the slit start spreading out, diffracting

Young’s Double Slit ExperimentFor waves entering two slits, the emerging waves interfere and form an interference (diffraction) pattern

Young experiment in 1801: light is wave phenomenon

First plane wave through a small slit yields coherent spherical wave

Then interposed two slits: interference of two spherical waves on a screen

• Path length difference between rays

r1 and r2 is /2

• Two rays out of phase at P1 resulting in

destructive interference

• Path length difference is distance from

starting point of r2 at center of the slit to

point b

• For D>>a, the path length difference

between rays r1 and r2 is (a/2) sin

Diffraction by a Single Slit:

Locating the Minima

Repeat previous analysis for pairs of rays, each separated by a

vertical distance of a/2 at the slit.

Setting path length difference to /2 for each pair of rays, we obtain

the first dark fringes at:

(first minimum)sin sin2 2

aa

For second minimum, divide slit into 4 zones of equal widths a/4

(separation between pairs of rays). Destructive interference occurs

when the path length difference for each pair is /2.

(second minimum)sin sin 24 2

aa

Dividing the slit into increasingly larger even numbers of zones, we

can find higher order minima:

(minima-dark fringes)sin , for 1,2,3a m m

Diffraction pattern from a single narrow slit.

Diffraction and the Wave Theory of Light

Central

maximum

Side or secondary

maxima

Light

Fresnel Bright Spot.

Bright

spot

Light

These patterns cannot be explained

using geometrical optics!

Single Slit Diffraction

When light goes through a narrow slit, it spreads out to form a diffraction pattern.

Here we will show that the intensity at the screen due to a single slit

is:

Intensity in Single-Slit Diffraction,

Quantitatively

2

sin (36-5)mI I

1where sin (36-6)

2

a f

, for 1,2,3m m

In Eq. 1 , minima occur when:

sin , for 1,2,3

or sin , for 1,2,3

(minima-dark fringes)

am m

a m m

If we put this into Eq. 2 we find:

(1)

(2)

Diffraction of a laser through a slit

(example)

Light from a helium-neon laser ( = 633 nm) passes through a narrow slit and is seen on a screen 2.0 m behind the slit. The first minimum of the diffraction pattern is observed to be located 1.2 cm from the central maximum.

How wide is the slit?

11

(0.012 m)0.0060 rad

(2.00 m)

y

L

74

3

1 1

(6.33 10 m)1.06 10 m 0.106 mm

sin (6.00 10 rad)a

1.2 cm

Width of a Single-SlitDiffraction Pattern

; 1,2,3, (positions of dark fringes)p

p Ly p

a

2(width of diffraction peak from min to min)

Lw

a

w

-y1 y1 y2 y30

X-band: =10cm

You are doing 137 mph on I-10 and you pass a little old lady doing 55mph when a cop, Located 1km away fires his radar gun, which has a 10 cm opening. Can he tell you from the L.O.L. if the gun Is X-band? What about Laser?

1m

1m

10 m1000m

w 2L

a2 0.1m 1000m

0.1m 2000m w

2L

a2 0.000001m 1000m

0.1m 0.02m

Laser-band: =1mm

Angles of the Secondary Maxima

The diffraction minima

are precisely at the angles

where

sin q = p l/a and a = pp (so

that sin a=0).

However, the

diffraction maxima are not

quite at the angles where

sin q = (p+½) l/a

and a = (p+½)p

(so that |sin a|=1).

p (p+½) /a Max

1 0.00475 0.00453

2 0.00791 0.00778

3 0.01108 0.01099

4 0.01424 0.01417

5 0.01741 0.01735

1

2

34 5

l = 633 nm

a = 0.2 mm

q (radians)

2

max

sinI I

A device with N slits (rulings) can be used to manipulate light, such as separate

different wavelengths of light that are contained in a single beam. How does a

diffraction grating affect monochromatic light?

Diffraction Gratings

Fig. 36-17 Fig. 36-18

sin for 0,1,2 (maxima-lines)d m m

(36-11)

Circular Apertures

When light passes through a circular aperture instead of a vertical slit, the diffraction pattern is modified by the 2D geometry. The minima occur at about 1.22/D instead of /a. Otherwise the behavior is the same, including the spread of the diffraction pattern with decreasing aperture.

Single slit of aperture aHole of diameter D

The Rayleigh Criterion

The Rayleigh Resolution Criterion says that the minimum separation to separate two objects is to have the diffraction peak of one at the diffraction minimum of the other, i.e., D 1.22

/D.

Example: The Hubble Space Telescope has a mirror diameter of 4 m, leading to excellent resolution of close-lying objects. For light with wavelength of 500 nm, the angular resolution of the Hubble is D = 1.53 x 10-7 radians.

Example

A spy satellite in a 200km low-Earth orbit is imaging the Earth in the visible wavelength of 500nm.

How big a diameter telescope does it need to read a newspaper over your shoulder from Outer Space?

D 1.22 /D (The smaller the wavelength or the bigger the telescope opening — the better the angular resolution.)

Letters on a newspaper are about Dx = 10mm apart.Orbit altitude R = 200km & D is telescope diameter.

Formula:Dx = RD = R(1.22/D)

D = R(1.22/Dx)

= (200x103m)(1.22x500x10–9m)/(10X10–3m)

= 12.2m

Example Solution

R

Dx

D

Holography

Brief history of holography

• Invented in 1948 by Dennis Gabor – to improve

the resolution in electron microscopy, before the

invention of the laser (this time light sources

were not coherent)

• Leith and Upatnieks (1962) applied laser light to

holography and introduced an important off-axis

technique (the first holographic picture, laser

was necessary to see the picture)

• The pioneer of holography in Poland –

prof. Mieczysław Wolfke (professor from Faculty

of Physics WUT),

Holos - whole, grapho – drawing

•Holography is a method of producing a three-dimensional (3-D) image of an object. (The three dimensions are height, width, and depth.)

•Later the object can be reconstructed.

•The hologram is actually a recording of the difference between two

beams of coherent light

Can be used as optical store disk, in information processing,

Conventional vs. Holographic picture

• Conventional:

– 2-d version of a 3-d scene

– Photograph lacks depth perception or parallax

– Film sensitive only to radiant energy

– Phase relation (i.e. interference) are lost

Conventional photography

Light

Object

Reflected

wave

Photographic film:

The intensity is

recorded

Conventional photography

Light

• Hologram:– Freezes the intricate wavefront of light that carries all

the visual information of the scene

– To view a hologram, the wavefront is reconstructed

– View what we would have seen if present at the original scene through the window defined by the hologram

– Provides depth perception and parallax

Conventional vs. Holographic picture

Conventional vs. Holographic picture

– Converts phase information into amplitude information (in-phase - maximum amplitude, out-of-phase – minimum amplitude)

– Interfere wavefront of light from a scene with a reference wave

– The hologram is a complex interference pattern of microscopically spaced fringes

How hologram is made?

• Need a laser, lenses, mirror, photographic film,

and an object

• The laser light is separated into two beams,

reference beam and object beam

• Reference beam enlarged and aimed at a piece

of holographic film

How hologram is made?

• Object beam directed at subject to be recorded

and expanded to illuminate subject,

• Object beam reflects off of object and meets

reference beam at film,

• Produces interference pattern which is recorded,

Reference wave

Photographic film.

Interference of reference and

reflected waves is recorded

How hologram is made?

How hologram is made ?

• Film is developed,

• Hologram illuminated at same angle as

reference beam during original exposure

to reveal holographic image,

Types of holograms

Transmission hologram:

reference and object waves

traverse the film from the

same side

Reflection hologram:

reference and object waves

traverse the emulsion from

opposite sidesInvented by Benton,

Can be reconstructed

in normal light

Applications of holography

• Credit cards carry monetary value,

• Supermarket scanners,

• Optical Computers,

• Used in aircraft “heads-up display”,

• Art,

• Archival Recording of fragile museum artifacts,

Holography in the future

• Medical Purposes

• Gaming Systems

• Personal Defense

• Computers

• Artwork

• Amusement Park Rides

• Movie production

– Holodeck from Star Trek Holodeck Clip

– Star Wars Chess Game

Summary

• Interference only for coherent light, i.e., with a phase relationship that is time independent

• Intensity in double-slit interference:

• Use Huygens’ Principle to find positions of

diffraction minima of a single slit by subdividing

the aperture

(minima-dark fringes)sin , for 1,2,3a m m

2 10 2

4 cosI I f2

sind

f

Summary

• Diffraction of light occurs at openings of the order of the wave length of the light

• Double slit experiment:

• Intensity in double-slit interference:

Maxima-bright fringes:

sin for 0,1,2,d m m

Minima-dark fringes: 12

sin for 0,1,2,d m m

2 10 2

4 cosI I f2

sind

f

Summary

• To predict the interference pattern of a multi-slit

system, we must combine interference and

diffraction effects.

• Rayleigh’s Criterion for the separability of two

points

• Intensity in single-slit diffraction:

• Double-slit diffraction:

2

sin (36-5)mI I

1where sin (36-6)

2

a f

2

2 sincos (double slit)mI I