Topic 7 wave interference

Topic 4.1 Waves, Interference and Optics

1

UEEP1033 Oscillations and Waves

Topic 7:

Interference and Diffraction


2


• Interference e.g. rainbow colours produced by a thin film of oil on a wet road, where the light reflected off the surface of the oil interferes with the light reflected off the water surface underneath

• Diffractione.g. the waves spread out in a semicircular fashion after passing through the narrow mouth of a harbour

• Both result from the overlap and superposition of waves

Interference and Diffraction of Waves


3


Interference

two waves are out of phase

destructive interference

two waves are in phase

constructive interference

amplitude of their superposition is zero

amplitude of the superposition (ψ1 + ψ2) = 2A

A is the amplitude of the individual waves


4


Figure (a)• Two monochromatic waves ψ1 and ψ2 at a

particular point in space where the path difference from their common source is equal to an integral number of wavelengths

• There is constructive interference and their superposition (ψ1 + ψ2) has an amplitude that is equal to 2A where A is the amplitude of the individual waves.

Figure (b)• The two waves ψ1 and ψ2 where the path

difference is equal to an odd number of half wavelengths

• There is destructive interference and the amplitude of their superposition is zero

Interference


5


Point source of light is illuminating an opaque object, casting a shadow where the edge of the shadow fades gradually over a short distance and made up of bright and dark bands, the diffraction fringes. Shadow fades gradually

>> Bright and Dark Bands

= Diffraction Fringes

Diffraction


6


Francesco Grimaldi in 1665 first accurate report

description of deviation of light from rectilinear propagation (diffraction)

The effect is a general characteristics of wave phenomena occurring whenever a portion of a wavefront is

obstructed in some way

Diffraction


7


Plane wavefronts approach a barrier with an opening or an obstruction, which both the opening and the obstruction are large compared to the wavelength

Opening(size = d)

Obstruction (size = d)

wavelength, d >>


8


• If the size of the opening or obstruction becomes comparable to the wavelength

• The waves is not allowed to propagate freely through the opening or past the obstruction

• But experiences some retardation of some parts of the wavefront

• The wave proceed to "bend through" or around the opening or obstruction

• The wave experiences significant curvature upon emerging from the opening or the obstruction

curvatured


9


As the barrier or opening size gets smaller, the wavefront experiences more and more

curvature

More curvature

Diffraction

d


10



11


Historical Background


12


Light source

ApertureObservatio

n plane

Screen

Arrangement used for observing

diffraction of light

Corpuscular Theoryshadow behind the

screen should be well defined, with sharp

borders

Observations• The transition from light to shadow was gradual rather than

abrupt

• Presence of bright and dark fringes

extending far into the geometrical shadow of the

screen


13


Christian Huygens

Huygens’s Principle

Each point on the wavefront of a disturbance were considered to be a new source of a “secondary” spherical disturbance, then the wavefront at a later

instant could be found by constructing the “envelope” of the secondary wavelets”


14


Huygens’s PrincipleEvery point on a propagation wavefront serves as the source of spherical secondary wavelets, such

that the wavefront at some later time is the envelope of these wavelets

Plane wave Spherical wave


15



16



Plane waveSpherical wave

Every point on a propagation wavefront serves as the source of spherical

secondary wavelets

the wavefront at some later time is the envelope of these wavelets


17


• When a wavefront encounters an aperture in an opaque barrier, the barrier suppresses all propagation of the wave except through the aperture

• Following Huygen’s principle, the points on the wavefront across the aperture act as sources of secondary wavelets

• When the width of the aperture is comparable with the wavelength, the aperture acts like a point source and the outgoing wavefronts are semicircular

Huygen’s Principle


18


18

• Ignores most of each secondary wavelet and only retaining the portions common to the envelope

• As a result, Huygens’s principle by itself is unable to account for the details of the diffraction process

• The difficulty was resolved by Fresnel with his addition of the concept of interference



19


Augustin Jean Fresnel

• 1818, Fresnel brought together the ideas of Huygens and Young and by making some arbitrary assumptions about the amplitude and phases of Huygens’ secondary sources

• Fresnel able to calculate the distribution of light in diffraction patterns with excellent accuracy by allowing the various wavelet to mutually interfere

Huygens-Fresnel Principle


20


Huygens-Fresnel Principle

Every unobstructed point of a wavefront, at given instant, serves as a source of spherical secondary

wavelets (with the same frequency as that of the primary

wave)The amplitude of the optical field at any point

beyond is the superposition of all these wavelets (considering their amplitudes and relative phases)


21


James Clerk Maxwell

Electromagnetic Wave

t

HE

t

EH

0 E

0 H


22


Gustav Kirchhoff

• Kirchhoff's diffraction formula can be used to model the propagation of light in a wide range of configurations, either analytically or using numerical modelling

• It gives an expression for the wave disturbance when a monochromatic spherical wave passes through an opening in an opaque screen

• The equation is derived by making several approximations to the Kirchhoff integral theorem which uses Green’s theorem to derive the solution to the homogeneous wave equation

• In 1882, Kirchhoff developed a more rigorous theory based directly on the solution of differential wave equation

Kirchhoff's diffraction formula


23


Although the problem was physically somewhat unrealistic , i.e. it involved an infinitely thin yet opaque, perfectly conducting plane screen, the result was nonetheless extremely valuable, providing a good deal of insight into the fundamental processes involved

Arnold Johannes Wilhelm Sommerfeld

In 1896, Sommerfeld published the first exact solution for a particular diffracting configuration , utilizing the electromagnetic theory of light

Rigorous solutions of this sort do no exist even today for many of the configurations of practical interest

The approximation treatments of Huygens-Fresnel and Kirchhoff are adequate for many purposes


24


InterferenceYoung’s Double-Slit Experiment


25


Young’s Double-Slit Experiment

L >> a

a = slits separation


26


• A monochromatic plane wave of wavelength λ is incident upon an opaque barrier containing two slits S1 and S2

• Each of these slits acts as a source of secondary wavelets according to Huygen’s Principle and the disturbance beyond the barrier is the superposition of all the wavelets spreading out from the two slits

• These slits are very narrow but have a long length in the direction normal to the page, making this a two-dimensional problem

• The resultant amplitude at point P is due to the superposition of secondary wavelets from the two slits



27


• Since these secondary wavelets are driven by the same incident wave there is a well defined phase relationship between them

• This condition is called coherence and implies a systematic phase relationship between the secondary wavelets when they are superposed at some distant point P

• It is this phase relationship that gives rise to the interference pattern, which is observed on a screen a distance L beyond the barrier



28


The secondary wavelets from S1 and S2 arriving at an arbitrary point P on the screen, at a distance x from the point O that coincides with the mid-point of the two slits

Distances: S1P = l1 S2P = l2 Since L >> a it can be assumed that the secondary wavelets arriving at P have the same amplitude A

The superposition of the wavelets at P gives the resultant amplitude:


)cos()cos( 21 kltkltAR

ω = angular frequencyk = wave number

(5)



29


This result can be rewritten as:

Since L >> a, the lines from S1 and S2 to P can be assumed to be parallel and also to make the same angle θ with respect to the horizontal axis


2/)(cos[]2/)(cos2 1212 llkllktAR

The line joining P to the mid-point of the slits makes an angle θ with respect to the horizontal axis

21 cos/ lLl

cos/212 Lll

(6)



30


When the two slits are separated by many wavelengths, θ is very small and cos θ 1. Hence, we can write the resultant amplitude as:


)2/cos()cos(2 lkkLtAR

= path difference of the secondary wavelets

The intensity I at point P = R2

12 lll

)2/(cos)(cos4 222 lkkLtAI

This equation describes the instantaneous intensity at PThe variation of the intensity with time is described by the cos2(ωt − kL) term

(7)

(8)


31


• The frequency of oscillation of visible light is of the order of 1015 Hz, which is far too high for the human eye and any laboratory apparatus to follow.

• What we observe is a time average of the intensity• Since the time average of cos2(ωt − kL) over many

cycles = 1/2

the time average of the intensity is given by:


)2/(cos20 lkII

20 2AI = intensity observed at a maximum of the interference pattern

described how the intensity varies with l)2/(cos2 lk

(9)


32


I = maximum whenever l = n (n = 0,±1, ±2, …)I = 0 whenever l = (n + ½)


From figure on slide-25: l a sin θSubstituting for l in Equation (9), we obtain:

(10) )2/sin(cos)( 20 kaII

When θ is small so that sinθ θ, we can write:

)/(cos)(

)2/(cos)(2

0

20

aII

kaII(11)

/2where ka = slits separation


33


separation of the bright fringes

If there were no interference, the intensity would be uniform and equal to Io/2 as indicated by the horizontal dashed line


Light intensity I (θ) vs angle θ



34



Intensity maxima: .....,2,1,0,

na

n

.....,2,1,0,

na

LnLx

(12)

(13)

(14)

(15)

The bright fringes occur at distances from the point O given by:

Minimum intensity occur when:

The distance between adjacent bright fringes is:

.....,2,1,0,2

1

n

a

Lnx

a

Lxx nn

1



35


Fraunhofer and Fresnel Diffraction


36


Observationscreen


S

Lens

Plane waves

Opaque shield , with a singlesmall aperture of width a is

being illuminated by plane wave of wavelength from a distant

point source S

Case-1observation screen is very

close to

Image of aperture is projected onto the screen


37


Observationscreen


S

Lens

Plane waves Case-2

observation screen is moved farther away from

Image of aperture become increasingly more structured as the

fringes become prominent

Fresnel or Near-Field Diffraction


38



S

Lens

Plane waves Case-3

observation screen is at very great distance away from

Projected pattern will have spread out considerably, bearing a little or

no resemblance to the actual aperture

Observationscreen

Thereafter moving the screen away from the aperture change

only the size of the pattern and not its shape

Fraunhofer or Far-Field Diffraction


39



S

Lens

Plane waves Case-4

If at that point, the wavelength of the incoming radiation is reduce

Observationscreen

the pattern would revert back to the Fresnel case

If were decreased even more, so that → 0The fringes would disappear, and the image

would take on the limiting shape of the aperture


40



If a point source S and the observation screen are very far

from

S

Lens

Plane waves

Observationscreen

Fraunhofer Diffraction

If a point source S and the observation screen are

too near Fresnel Diffraction


41



S

Lens

Plane waves

Observationscreen

Fraunhofer Diffractiond

R R

R is the smaller of the two distances from S to and to

2dR

d = slit width


42


Practical realization of the Fraunhofer condition

F1 F2


43


Diffraction

• Any obstacle in the path of the wave affects the way it spreads out; the wave appears to ‘bend’ around the obstacle

• Similarly, the wave spreads out beyond any aperture that it meets. such bending or spreading of the wave is called diffraction

• The effects of diffraction are evident in the shadow of an object that is illuminated by a point source. The edges of the shadow are not sharp but are blurred due to the bending of the light at the edges of the object

• The degree of spreading of a wave after passing through an aperture depends on the ratio of the wavelength λ of the wave to the size d of the aperture

• The angular width of the spreading is approximately equal to λ/d; the bigger this ratio, the greater is the spreading


44


The Mechanism of Diffraction• Diffraction arises because of the way in which waves propagate as

described by the Huygens-Fresnel Principle

• The propagation of a wave can be visualized by considering every point on a wavefront as a point source for a secondary radial wave

• The subsequent propagation and addition of all these radial waves form the new wavefront

• When waves are added together, their sum is determined by the relative phases as well as the amplitudes of the individual waves, an effect which is often known as wave interference

• The summed amplitude of the waves can have any value between zero and the sum of the individual amplitudes

• Hence, diffraction patterns usually have a series of maxima and minima


45


• A monochromatic plane wave is incident upon an opaque barrier containing a single slit

• Replace the relatively wide slit by an increasing number of narrow subslits

• Each point in the subslits acts as a point source for a secondary radial wave

• When waves are added together, their sum is determined by the relative phases and the amplitudes of the individual waves, an effect which is often known as wave interference

• The summed amplitude of the waves can have any value between zero and the sum of the individual amplitudes

• Hence, diffraction patterns usually have a series of maxima and minima

Single Slit Diffraction


46


Diffraction at a Single Slit

The resultant amplitude at point P is due to the

superposition of secondary wavelets from the slit

monochromatic

x = 0

x

each of these strips acts as a source of secondary wavelets

d = slit width


47



Figure in slide-46:

• A monochromatic plane wave of wavelength λ is incident upon an opaque barrier containing a single slit

• The slit has a width d and a long length (>> d) in the direction normal to the page, reducing this to a two-dimensional problem

• The resultant amplitude at point P is due to the superposition of secondary wavelets from the slit


48



• The centre of the slit is at x = 0

• We divide the slit into infinitely narrow strips of width dx

• Following Huygen’s principle, each of these strips acts as a source of secondary wavelets and the superposition of these wavelets gives the resultant amplitude at point P


49


• We consider the case in which P is very distant from the slit

• Consequently, all the wavelets arriving at P can be assumed to be plane waves and to have the same amplitude

• In addition, we can assume that the lines joining P to all points on the slit make the same angle θ to the horizontal axis

• The amplitude dR of the wavelet arriving at P from the strip dx at x is proportional to the width dx of the strip

• its phase depends on the distance of P from the stripi.e. on (l − x sin θ), where l is the distance of P from the midpoint of the slit



50


Hence dR is given by:


ω = angular frequency k = wave number α = constant

)]sin(cos[ xlktdxdR

The resultant amplitude at P due to the contributions of the secondary wavelets from all the strips is

2/

2/)]sin(cos[

d

dxlktdxR

)cos(]sin)2/sin[(sin)2/(

kltkdkd

dR

d = slit width


51


Instantaneous intensity I at P:

2

22222

]sin)2/[(

]sin)2/[(sin)(cos

kd

kdkltdRI


Since the time average over many cycles of cos2(ωt − kl) = 1/2

the time average of the intensity is given by:

2

2

02

2

0

sin

]sin)2/[(

]sin)2/[(sin)(

Ikd

kdII

2/220 dI = maximum intensity of the diffraction pattern

This equation describes how an incident plane wave of wavelength λ spreads out from a single slit of width d in terms of the angle θ

2/sin kd


52


The diffraction pattern of a single slit

The zeros of intensity in the diffraction pattern occur at θ = ±nλ/d, where n = ±1,±2, . . . , under the small angle approximation sin θ θ

This figure is a plot of I(θ) against θ for a value of kd/2 = 10π

2

2

0sin

)(

II

sin)2/(kd

θ

d = slit width


53


The first zeros in the intensity occur when sin)2/(kd


dk /sin/2but

the degree of spreading depends upon the ratio λ/d

When λ << d, as in the case of light, sin θ θ, giving the first zeros in the diffraction pattern at:

...),2,1(

nd

n

d

In general, zeros in intensity occur when:

d = slit width


54


Double slits of finite width

• Consider each of the two slits to be composed of infinitely narrow strips that act as sources of secondary wavelets

• Then the resultant amplitude R at a point P is the superposition of the secondary wavelets from both slits

2/2/

2/2/

2/2/

2/2/

)]sin(cos[

)]sin(cos[

da

da

da

da

xlktdx

xlktdxR

d = the width of each slit a = separation of the slits


55


]sin)2/cos[(sin)2/(

]sin)2/sin[()cos(2

kakd

kdkltdR

Double slits of finite width

]sin)2/[(cos]sin)2/[(

]sin)2/[(sin)( 2

2

2

0

kakd

kdII

Resultant Intensity:

• This result is the product of two functions. • The first is the square of a sinc function corresponding to

diffraction at a single slit• The second is the cosine-squared term of the double-slit

interference pattern


56


Diffraction at a Double Slit

)2/sin(cos)( 20 kaII

2

2

0 )2/sin(

)2/sin(sin)(

kd

kdII

)2/sin(cos)2/sin(

)2/sin(sin)( 2

2

2

0

kakd

kdII



57


Diffraction at a Double Slit

double-slit interference maxima occur at angles:

...),2,1(

na

n

zeros in the diffraction pattern occur at angles:

...),2,1(

nd

n



58




59


Observationscreen


S

Lens

Plane waves

Opaque shield , with a single

small aperture of width a is being illuminated by

plane wave of wavelength from a

distant point source S

Case-1observation screen is very close

to

Image of aperture is projected onto the screen


60


Observationscreen


S

Lens

Plane waves

Case-2observation screen is moved

farther away from

Image of aperture become increasingly more structured as the fringes become

prominent

Fresnel or Near-Field

Diffraction


61



S

Lens

Plane waves

Case-3observation screen is at very great distance away from

Projected pattern will have spread out considerably, bearing a little or

no resemblance to the actual aperture

Observationscreen

Thereafter moving the screen away from the aperture change

only the size of the pattern and not its shape

Fraunhofer or Far-Field

Diffraction


62



S

Lens

Plane waves

Case-4If at that point, the wavelength of the

incoming radiation is reduce

Observationscreen

the pattern would revert back to the Fresnel case

If were decreased even more, so that → 0The fringes would disappear, and the image

would take on the limiting shape of the aperture


63



S

Lens

Plane waves If a point source S and the

observation screen are very far from

Observationscreen


If a point source S and the observation screen are

near to Fresnel Diffraction


64



S

Lens

Plane waves

Observationscreen


a

R R

R is the smaller of the two distances from S to and to

2aR


65


Practical realization of the Fraunhofer condition

F1 F2


66


coherent line source

Introduction

yiA linear array of N in-phase coherent

point sources

e.g. secondary sources of the Huygens-

Fresnel Principle

a long slit whose width D is much

less than

Each point emits a spherical wavelet equal to:

A0 = source strength

)sin(0 krtE rA


67


Introduction

The sources are very weak and their number N is tremendously large and the separation between is vanishing small

Finite segment of array yi contain yi (N/D) sources

Assume that the array is divided into M such segments (i.e. i goes from 1 to M)

The contribution of electric field intensity at P from the i-th segment is:

DyN

irA

ii

ikrtE )sin(0


68


Introduction

The contribution of electric field intensity at P from the i-th segment is:


Finite segment of array yi

contain yi (N/D) sources

)sin(0 krtE rA

DyN

irA

ii

ikrtE )sin(0

Each point emits a spherical wavelets:


69


Introduction

Net field at P from all M segments is:

AL = source strength per unit length

For continuous line source (yi 0,i.e. M )

M

iiir

A ykrtEi

L

1

)sin(

NAD

AN

L 0lim1

dyr

krtAE

D

DL

2

2

)sin(


70


Introduction

3rd term

Can be ignored as long as its contribution to the phase is

insignificant

22 cos)2/(sin RyyRr

For continuous line source (yi 0,i.e. M )

dyr

krtAE

D

DL

2

2

)sin(


71


Single-Slit Fraunhofer Diffraction


Point of observation is very distant from the coherent line source and R >> D

r(y) never deviates appreciably from its midpoint value R

Quantity (AL/R) at P is essentially constant for all elements dy


72


Point of observation is very distant from the coherent line source and R >> D

PD

r1

rM

R

P

r1 rM R



73



(AL/R)dy = amplitude of the wave

The field at P due to the differential segment of the source dy is:

The phase is more sensitive to variation of r(y) than is the

amplitude

dyr

krtAE

D

DL

2

2

)sin( dykrtR

AdE L )sin(


74



Even when y = ± D/2

i.e. (D2/4R)cos2 = negligible

This is true for all values of when R is adequately large


22 cos)2/(sin RyyRr


75


Fraunhofer conditions: The distance r is linear in y


Therefore the phase can be written as a function of the aperture variable

sinyRr

dyyRktR

AE

D

D

L 2

2

)]sin(sin[

)sin( yRkkr


76



Simplify:

where:

dyyRktR

AE

D

D

L 2

2

)]sin(sin[

)sin(sin)2/(

]sin)2/sin[(kRt

kD

kD

R

DAE L

)sin(sin

kRtR

DAE L

sin)2/(kD


77


22sin

2

1)(

R

DAI L


Irradiance:

where:

When = 0, sin/ = 1

Principle maximum

TEI 2)(

21)sin(

TkRt

2

2

1)0(

R

DAI L


78



The Irradiance resulting from an idealized coherent line source in the Fraunhofer approximation is:

When D >> the irradiance drop extremely rapidly as deviates from

zero

Or:

Where:

2sin

)0()(

II

2

2

1)0(

R

DAI L sin)2/(kD

sin)/( D


79



An aperture of this sort might typically have a width of several hundred and a length of few centimeter


80



l

Each strip is a long coherent line source and can be replaced by a point emitter on the z-axis, which radiates a circular wave in the y = 0 (xz-plane)

Usual Analysis ProcedureDivide the slit into a series of long differential strips (dz by l) parallel to the y-axis


81



Our Problem is:To find the field in the xz-plane due to the infinite

number of point sources extending across the width of the slit along the z-axis

Only evaluate the integral of the contribution of dE from each element dz in the Fraunhofer

approximation


82



The complete solution for the slit is:

Here the line source is short, D = b, thus is not large

Or:Where:

Although the irradiance fall off rapidly, higher-order subsidiary maxima will be observable

2sin

)0()(

II

sin)2/(kD sin)/( D


83



The extreme of I() occur at values of that cause dI/d = 0

The irradiance has minima (equal to zero) when sin = 0

Also when:

0)sincos(sin2

)0(3

I

d

dI

......,3,2,

tani.e.

0sincos


84



The extreme of I() occur at values of that cause dI/d = 0

The irradiance also has minima when:

0)sincos(sin2

)0(3

I

d

dI

tani.e.

0sincos


85



The irradiance also has minima when:

Only one such extremum exist between adjacent minima

I() must have subsidiary maxima at

......,4707.3

,4590.2,4303.1

tani.e.

0sincos


86



Easy Way of Analysis

Every point in the aperture emitting rays in all direction in the xz-plane

Undiffracted beam arrive on the viewing screen in phase and a central bright spot will be formed by them


87



Rays coming off at an angle 1

Path length difference between the ray from the

very top and bottom bsin1 =

Easy Way of Analysis

bsin1 =


88



A ray from the middle of the slit will then lag ½ behind a ray from the top and exactly cancel it

A ray just below center will cancel a ray from just below the top, and so on …

All across the aperture ray-pairs will cancel, yielding a minimum

The irradiance has dropped from its high central maximum to the first zero on either size at bsin1 = ±


89



As the angle increases further, some small fraction of the rays will again interfere constructively, and the irradiance will rise to form a subsidiary peak.

A further increase in the angle produces another minimum

bsin2 = 2


90



In general,

the zero irradiance will occur when

bsinm = m

where, m = ±1, ±2, ±3, …

Equivalent to: = m = (kb/2)sinm

wave number, k = 2/


91



The Fraunhofer diffraction pattern of a single slit

Irradiance drop: 1.0 0.047 0.017 0.008

2.463.47

1.43subsidiary maxima

2sin

)0()(

II


92


Double-Slit Fraunhofer Diffraction

Two long slits of width b and center-to-center

separation is a

Each aperture by itself would generate the same

single-slit diffraction pattern on the viewing

screen

At any point on the viewing screen, the

contributions from the slits overlap


93



If the primary plane wave is incident on the double-slit aperture

at normal incidence, the wavelets are all emitted in-phase

The interference fringe at a particular point is determined by

the differences in the optical path lengths traversed by the

overlapping wavelets from the two slits

The flux-density distribution =rapidly varying double-slit

interference system modulated by a single-slit diffraction

pattern


94



Double-slit pattern for

a = 3b

a = slits separationb = individual slit width


95



(a)

(b) (c)

Intensity plot for a double-slit

interference

Intensity plot for diffraction by a typical by a single-slit of width

a

Intensity plot for diffraction by double-slit

of width a

The curve of (b) acts as an envelope, limiting the intensity of the double-slit fringes in (a)


96



The total contribution to the electric fieldin the Fraunhofer approximation

where:

constant-amplitude factor C =

secondary source strength per unit

length along the z-axis divided by R

2/

2/

2/

2/)()(

ba

ba

b

bdzzFCdzzFCE

)sin(sin)( zRktzF


97


)2sin()sin(sin

kRtkRtbCE


Integration of the equation yield:

where:

Just the sum of the two fields at P, one from each slit

Distance from the first slit to P is R

phase contribution = -kR

Distance from the second slit to P

is (R - asin ) or (R - 2/k)

phase contribution = (-kR + 2)

sin)2/(;sin)2/( kakb


98



2 = phase difference between two nearly parallel rays arriving P from the edges of one of the slit

2 = phase difference between two ways arriving at P, one having originated at any point in the first slit, the other coming from the corresponding point in the second slit

Simplifying theprevious equation

Irradiance

)sin(cossin

2

kRtbCE

22

2

0 cossin

4)( II


99



In the = 0 (i.e. when = = 0)

I0 = the flux-density contributing from either slit

I(0) = 4I0 = total flux density

In the case b becomes vanishing small (i.e. kb << 1)

sin/ 1 I( ) = 4I0cos2 = flux-density for a pair of long line sources

(Young’s double-slit interference

experiment)

22

2

0 cossin

4)( II


100



In the case a = 0

The two slits coalesce into one, = 0

I() = 4I0(sin2 )/ 2 = flux-density for single-slit diffraction with

the source strength doubled

The total expression as being generated by a

cos2 interference term

modulated by a

(sin2)/2 diffraction term

22

2

0 cossin

4)( II


101



At angular positions where

= ±, ±2, ±3, …Diffraction effects are such that no light reaches the viewing

screen, and none is available for interference

22

2

0 cossin

4)( II


102



At points on the viewing screen where

= ±/2, ±3 /2, ±5 /2, …The various contributions to the electric field will

be completely out-of-phase and will cancel

22

2

0 cossin

4)( II


103



Double-slit pattern for a = 3b (i.e. = 3)

If a = mb, there will be 2m bright fringes within the central diffraction peak

2 3 = 6 bright fringes within the central diffraction peak

1

3

5

4

2

# 6: ½ + ½


104



No light is available at that precise position to partake in the interference process and the suppressed peak is said to be a missing-order

An interference maximum and a diffraction minimum may correspond to the same -value


105


Diffraction by Many Slits

Procedure for obtaining the irradiance function for diffraction by many slits is

the same as that used when considering two slits

N long, parallel, narrow slits, each of width b and center-to-center

separation a

The total optical disturbance at a point on the viewing screen is given by:

2

2

2

2)()(

ba

ba

b

bdzzFCdzzFCE


106



The total optical disturbance at a point on the viewing screen is given by:

Where:

2)1(

2)1(

22

22

2

2

2

2

)(

.....)(

)()(

baN

baN

ba

ba

ba

ba

b

b

dzzFC

dzzFC

dzzFCdzzFCE

)sin(sin)( zRktzF


107



The contribution from the j-th slit: (by evaluating only that one integral in the previous equation)

After some manipulation:

where:

2)1(

2)1(

22

22

2

2

2

2

)(

.....)(

)()(

baN

baN

ba

ba

ba

ba

b

b

dzzFC

dzzFC

dzzFCdzzFCE

)2sin(sin

jj kRtbCE

sin)2/(;sin)2/( kakb


108



Rj = R - ja sin

-kR + 2j = -kRj

Total optical disturbance:

)2sin(sin1

0

1

0

j

N

j

N

jj

kRtbCE

EE


109



Total optical disturbance written as the imaginary part of a complex exponential:

Geometric series

1

0

2)(sinIm

N

j

jikRti eebCE

][

][

1

12

21

0

2

iii

iNiNiN

i

NiN

j

ji

eee

eee

e

ee

sin

sin)1(1

0

2 Nee Ni

N

j

ji


110



Total optical disturbance written as the imaginary part of a complex exponential:

1

0

2)(sinIm

N

j

jikRti eebCE

])1(sin[sin

sinsin

NkRtN

bCE


111



Flux-density distribution function:

I0 = flux density in the = 0 direction emitted by any one of the slits

I(0) = N2I0

The waves arriving at P in the forward direction are all in-

phase, and their fields add constructively yield a multiple-

wave interference system modulated by the single-slit

diffraction envelope

22

0 sin

sinsin)(

N

II


112




If the width of each aperture were shrunk to zero

i.e. 0 (or N)

Principal maxima occur when (sinN / sin) = N

i.e. when: = 0, ± , ± 2, …

Or since = kasin/2 a sinm= m

m = 0, ± 1, ± 2, …

22

0 sin

sinsin)(

N

II

2sin

)0()(

II


113




Minima or zero flux density exist whenever (sinN / sin)2 = 0

i.e. when:

Between consecutive principal maxima, there will be (N-1) minima

Between each pair of minima there will have to be a subsidiary maximum

22

0 sin

sinsin)(

N

II

.....,)1(

,)1(

.....,,3

,2

,N

N

N

N

NNN


114



Subsidiary Maximum are located

approximately at point where

(sin N) has its greatest value, i.e.

The dark regions become wider than the bright bands as N increase and the secondary peaks fade out

As N increases, the Principal Maxima maintain their relating spacing (/a) while becoming increasingly narrow

.....,2

5,

2

3

NN


115



The dark regions become wider than the bright bands

as N increase and the secondary peaks fade out

As N increases, the Principal Maxima maintain their relating spacing (/a)

while becoming increasingly narrow


116


Diffraction GratingDefinition

A repetitive array of diffracting elements that has the effect of producing periodic alterations in the phase, amplitude, or

both of an emergent wave

An idealized grating consisting of only

five slits

Opaque surface with narrow parallel grooves

e.g. made by ruling or scratching parallel notches into the surface of a flat, clean glass plate

Each of the scratches serves as a source of scattered light, and together they form a regular array of parallel line sources


117


Diffraction Grating

Grating Equation: a sinm = m

m = specify the order of the various principal maxima

The intensity plot produced by a diffraction grating consists of narrow peaks, here label with their order number m

The corresponding bright fringes seen on the screen are called lines

The maxima are very narrow and they separated by relatively wide dark region

a = grating spacing (spacing between rulings or slits)

N rulings occupy a total width w, then a = w/N


118


Diffraction Grating

Grating’s ability to revolve or separate lines of different wavelengths depends on the width of the lines

Half-width hw of the central line is measured from the center of that line to the adjacent minimum on a plot of intensity

The path length difference between the top and bottom rays is Na sin hw

Na sin hw = Na

The first minimum occurs where

Since hw is small, then sin hw = hw

hw = / Na (Half-width of central line)

Half-width of line at : hw = / Na cos


119


Diffraction Grating

Application: Grating Spectroscope

collimator

Plane waveDiffraction grating

telescope

Visible emission lines of cadmium

Visible emission lines from hydrogen

The lines are farther apart at greater angles


120


Reflection and Refraction


121


ri Law of Reflection

Law of Refraction (Snell’s law)

ttii nn sinsin

Interface

Incident medium ni

Refractingmedium ni

Surface normal


122


Law of ReflectionWhen a ray of light is reflected at an interface dividing two uniform media, the reflected ray remains within the plane of incidence, and the angle of reflection equals the angle of incidence. The plane of incidence includes the incident ray and the normal to the point of incidence

Law of Refraction (Snell’s law)When a ray of light is refracted at an interface dividing two uniform media, the transmitted ray remains within the plane of incidence and the sine of the angle of refraction is directly proportional to the sine of the angle of incidence


123


Huygens’ construction to prove the law of reflection

Narrow, parallel ray of light

Plane of interface XY

Angle of incidence

Angle of reflection


124



• Since points along the plane wavefront do not arrive at the interface simultaneously, allowance is made for these differences in constructing the wavelets that determine the reflected wavefront

• If the interface XY were not present, the Huygens construction would produce the wavefront GI at the instant ray CF reached the interface at I

• The intrusion of the reflecting surface, means that during the same time interval required for ray CF to progress from F to I, ray BE has progressed from E to J and then a distance equivalent to JH after reflection


125



• Wavelet of radius JN = JH centered at J is drawn above the reflecting surface

• Wavelet of radius DG is drawn centered at D to represent the propagation after reflection of the lower part of the light

• The new wavefront, which must now be tangent to these wavelets at points M and N, and include the point I, is shown as KI in the figure

• A representative reflected ray is DL, shown perpendicular to the reflected wavefront

• The normal PD drawn for this ray is used to define angles of incidence and reflection for the light


126


The Law of Refraction

Use Huygen’s principle to derive the law of refraction

The refraction of a plane wave at an air-glass interface

Figures show three successive stages of the refraction of several wavefronts at a plane interface between air (medium 1) and glass (medium 2)

1 = wavelength in medium 1v1 = speed of light in medium 1v2 = speed of light in medium 2 < v1 1 = angle of incidence


127


As the wave moves into the glass, a Huygens wavelet at point e will expand to pass through point c, at a distance of 1 from point e.

The time interval required for this expansion is that distance divided by the speed of the wavelet = 1/v1

In the same time interval, a Huygens wavelet at point h will expand to pass through point g, at the reduced speed v2 and with wavelength 2, i.e. the time interval = 2/v2

2

2

1

1

vv

2

1

2

1

v

v


128


According to Huygens’ principle, the refracted wavefront must be tangent to an arc of radius 2 centered on h, say at point g

the refracted wavefront must also be tangent to an arc of radius 1 centered on e, say at point c

2 = angle of refraction

h c

e

h c

g

hc1

1sin

hc2

2sin

2

1

2

1

2

1

sin

sin

v

v


129


Define: refraction index for a medium

c = speed of lightv = speed of light in the medium

Speed of light in any medium depends on the index of refraction of the medium

11 v

cn e.g.

22 v

cn

v

cn

1

2

2

1

2

1

2

1

/

/

sin

sin

n

n

nc

nc

v

v

2211 sinsin nn


130


The wavelength of light in any medium depends on the index of refraction of the medium

Let a certain monochromatic light:Medium refraction index wavelength

speed vacuum 1 c medium n

n v2

1

2

1

v

v

From slide-8:c

vn

The greater the index of refraction of a medium, the smaller the wavelength of light in that medium

nn


131



132


Frequency Between Media

• As light travels from one medium to another, its frequency does not change.

– Both the wave speed and the wavelength do change.

– The wavefronts do not pile up, nor are they created or destroyed at the boundary, so ƒ must stay the same.


133


nn

vf

Frequency of the light in a medium with index of refraction n

fv

fc

n

ncfn

/

/

f = frequency of the light in vacuum

The frequency of the light in the medium is the same as it is in vacuum


134


The fact that the wavelength of light depends on the index of refraction is important in situations involving the interference of light waves

Example: Two light rays travel through two media having different indexes of refraction

• Two light rays have identical wavelength and are initially in phase in air (n 1)

• One of the waves travels through medium 1 of index of refraction n1 and length L

• The other travels through medium 2 of index of refraction n2 and the same length L


135


• When the waves leave the two media, they will have the same wavelength – their wavelength in air

• However, because their wavelengths differed in the two media, the two waves may no longer be in phase

The phase difference between two light waves can change if the waves travel through different materials having different indexes of refraction

How the light waves will interfere if they reach some common point?


136


Number N1 of wavelengths in the length L of medium 1

11 / nn wavelength in medium 1:

1

11

LnLN

n

wavelength in medium 2: 22 / nn

2

22

LnLN

n

)( 1212 nnL

NN

Phase difference between the waves

21 nn


137


Example: phase difference = 45.6 wavelengths

• i.e. taking the initially in-phase waves and shifting one of them by 45.6 wavelengths

• A shift of an integers number of wavelengths (such as 45) would put the waves back in phase

• Only the decimal fraction (such as 0.6) that is important• i.e. phase difference of 45.6 wavelengths 0.6 wavelengths

• Phase difference = 0.5 wavelength puts two waves exactly out of phase

• If the two waves had equal amplitudes and were to reach some common point, they would then undergo fully destructive interference, producing darkness at that point


138


• With the phase difference = 0 or 1wavelengths, they would undergo fully constructive interference, resulting brightness at that common point

• In this example, the phase difference = 0.6 wavelengths is an intermediate situation, but closer to destructive interference, and the wave would produces a dimly illuminated common point


139


Example:

= 550 nm

Two light waves have equal amplitudes and re in phase before entering media 1 and 2

Medium 1 = air (n1 1)

Medium 2 = transparent plastic (n2 1.60, L = 2.60 m)

Phase difference of the emerging waves:

o

9

6

1212

1020 rad17.8

swavelength84.2

)00.160.1(10550

1060.2

)(

nnL

NN


140


Effective phase difference = 0.84 wavelengths = 5.3 rad 300o

• 0.84 wavelengths is between 0.5 wavelength and 1.0 wavelength, but closer to 1.0 wavelength.

• Thus, the waves would produce intermediate interference that is closer to fully constructive interference,

• i.e. they would produce a relatively bright spot at some common point.


141


Fermat’s Principle

• The ray of light traveled the path of least time from A to B

• If light travels more slowly in the second medium, light bends at the interface so as to take a path that favors a shorter time in the second medium, thereby minimizing the overall transit time from A to B

Construction to prove the law of refraction from Fermat’s principle


142



• Mathematically, we are required to minimize the total time:

ti v

OB

v

AOt

22 xaAO 22 )( xcbOB

ti v

xcb

v

xat

2222 )(


143



0)( 2222

xcbv

xc

xav

x

dx

dt

ti

• minimize the total time by setting dt / dx = 0

22sin

xa

xi

• From diagram:

22 )(sin

xcb

xct

0sinsin

t

t

i

i

vvdx

dt

0/

sin

/

sin

t

t

i

i

ncnc ttii nn sinsin

Science

Topic 7 wave interference