Tài liệu Chapter XVIII Diffraction of light pdf

Fresnel diffraction :We consider separatly two kinds of diffraction: Near-field diffraction → Fresnel diffraction Far-field diffraction → Fraunhofer diffraction S Screen Screen with a

GENERAL PHYSICS III

Optics

&

Quantum Physics

Chapter XVIII Diffraction of light

§ 1 Fresnel diffraction

§ 2 Fraunhofer diffraction

§ 3 X-ray diffraction from crystals

“Diffraction” of light can be understood as any deviation of light raysfrom their geometrical propagation line (that is straight in a homogeneuosmaterial)

Point source

Area of illuminaton

Geometrical

shadow

Straight edge

An example (shown in the picture):

The edge of shadow is never

perfectly sharp Some light

appears in the geometrical

shadow, and there are dark

and light fringers in the area of

illumination

 Diffraction of light can be considered as an argument for wave

characteristics of light, like other wave processes (sound, etc.)

§1 Fresnel diffraction :

We consider separatly two kinds of diffraction:

Near-field diffraction → Fresnel diffraction

Far-field diffraction → Fraunhofer diffraction

S

Screen

Screen with a circular aperture

1.1 Diffraction through a circular aperture:

According to geometric optics,

the image of a circular aperture

in the screen must be a light circular

dick with a perfectly sharp edge

But it’s not so

 Huygen’s principle (1678):

All points on wavefront are point

sources for spherical secondary

wavelets with speed, frequency

that equal to initial wave The wave front at a

later time is the envelope of these wavelets

Wavefront at

Basing on Huygen’s principle one can interpret diffraction as

interference of light from secondary sources For example, every

point of circular aperture becomes a secondary source, and what wesee in the screen is the interference of secondary sources

But this interpretation is only qualitively, for a quantitive analyze,

we need more It has not been known from Huygen’s principle:

how can determine the amplitude and the phase of secondary waves ?

Fresnel’s complementary statement:

Observation

point Wave

front

 Fresnel states that for the vibration

at P due to waves from the secondary

source dS we have the following formula:

dS

where

a0 & (t + ) → the amplitude &

phase of vibration of secondary sources at dS on the wave front S

K → a coefficient which depends on the angle K decreases

as  increases; K = 0 when  = / 2

The total vibration

at P is

1.3 Analysis of diffraction through circular aperture:

Having Huygen-Fresnel principle we tend to analyze the phenomenon

of diffraction through a circular aperture

1.3.1 The Frsenel method to devide a spherical wave front into adjacent zones (Fresnel zones):

1-th zone 2-nd zone 3-rd zone 4-th zone

Calculate the area of the

m-th zone:

where Sm is the area of

the m-th spherical segment:

(hm – the hight of segment)

hm is defined from the following equation:



Remark: The area of a zone does not depend on m It means that theareas of zones are approximatly the same (for values of m that are notlarge).

We have also the formula for the radius of the m-zone:

rm is proportional to

From the Fresnel formula for dξand all the described properties ofFresnel zones we can lead to the following formula for the amplitudes

of vibrations at P which are sent from Fresnel spherical zones :

from the 1-st zone ,,, from the m-th zone …

Further, the phases of vibrations from two adjacent zones have thephase difference  → we can write for the total amplitude of

vibrations at P:

It equals a half of amplitude due tothe 1-st zone !

1.3.2 Come back to the experiment of diffraction through an aperture:

→ What happens if there is a screen with circular aperture in the

light propagation line ?

Suppose that the part of wave front based on the apertureincorporates m zones:

+ for odd m

- for even m

(odd m)(even m)

Screen with aperture

Conclusions of diffraction picture on the screen:

Depending on the size of aperture, the number of open zones m isodd or even:

• If m is odd → at the center point P there is a light spot

• If m is even → at the center point P there is a dark spot

Besides the center point P, at other points in the screen, the lightintensity has maxima or minima, depending on the distance fromthe center point P

Owing to the symmetry, light & dark fringers on the screen arecircles centered at P

Comparison the Fresnel zone pictures according toa) the axis SP; b) the axis SP’; c) the axis SP”

To understand why the light intensity is different at P, P’ & P”

we describe the Fresnel zones for three cases, and make a

comparison Here we suppose that the aperture exposes 3 zones

Consider a plane light wave that comes to a slit Suppose that

slit is very long (or its length is much larger the width)

2.1.1 Diffraction minima:

First we define the location of minima in the screen

• At this angle the light from the

top and the middle of the slit

destructively interfere Below two

these points we can find

corresponding pairs from which two

waves destructively interfere at P

 The second minimum is at an angle

such that the light from the top and

a point at a/4 destructively interfere:

O

• For small → n-th minimum corresponds to  ,(n 1,2, )

a

n L

fringer with the width 2/a (twice in width than other bright fringers)

2.1.2 Intensity in the single-slit pattern:

By using the phasor diagram method, we can have a detail analysis

 To analyze diffraction, we treat it as

interference of light from strips

 Model the single slit as M strips with

spacing between the strips of a/M

 There are phase differences between

waves coming from adjacent strips

 The phase difference between first and

last source is given by

   a = (asin 

Screen (far away)

a = a sina

 a

L >> a implies rays are

The amplitude at O (in phase)

• Taking M = 10, we draw the phasor diagrams

A0Vibration vectors at O

and at P

 As we let M  , the polygon becomes

the arc of a circle

 The radius of the circle is determined

by the relation between angle and

/ ) sin (

sin 2

/

) 2 /

I I

Intensity is related to amplitude: I = A2 So, here’s the final answer:

Trigonometry: A/2 = R sin(/2)

With R = Ao/, A = (2Ao/sin(/2)

2 /

) 2 /

• The graph of the intensity versus

sinis shown in picture

I 0

I

-2 a – a  a 2 a sin

• Intensity maxima:

Consider the obtained equation

as a function I = I () we can find

maxima by solving dI/d= 0

This is a transcendental equation that has to be solved numerically.Here is the results:

 Example:

Suppose that when we pass red light (= 600 nm) through a slit of

width a, the width of the spot (the distance between the first zeros

on each side of the bright peak) is W = 1 cm on a screen that is L = 2mbehind the slit How wide is the slit?

• multiple slit interference

• single slit diffraction from every slit

2.2.1 Multiple slit interference:

We denote

+ by d the distance between adjacent slits + by a the width of every slit

a

Draw normal lines bisecting the phasors They intersect, defining R asshown:

22

sin R

A 

22



N sin R

A N 

) / sin(

) / N sin(

Result:

2 1

) / N sin(

Combine: Slit Interference + Diffraction

2/

)2/sin(

2 1

) 2 / sin(

) 2 / sin(

I N

2 2

0

) 2 / sin(

) 2 / N sin(

2 /

) 2 / sin(

I I

angle between adjacent phasors

angle between 1st and last phasor

(for plane-wave sources, I1 = constant)

to obtainTotal Interference Pattern,

Single-slit Diffraction,

 Example:

Light of wavelength is incident on an N-slit

system with slit width a and slit spacing d

1 The intensity I as a function of y at a viewing

screen located a distance L from the slits is

shown to the right What is N? (L >> d, y, a)

N is determined from the number of minima between two principal

maxima Here, there are two minima between principal maxima

Therefore, N = 3

2 Now the slit spacing d is halved, but the slit width a is kept constant.Which of the graphs best represents the new intensity distribution?

Decreasing d will increase

spacing between maxima.

The spacing between maxima

is increased , but diffraction profile shouldn’t expand, as seen here.

This one does it all.

Increased spacing between maxima and constant diffraction.

2.3 Diffraction Gratings:

 Diffraction gratings rely on N-slit interference.They simply consist

of a large number of evenly spaced parallel slits

 Recall that the intensity pattern produced by light of wavelength passing through N slits with spacing d is given by:

2 1

) / N

 The position of the first principal maximum is given by sin = /d

(can’t assume small !)  Different colors  different angles.

 Width of the principal maximum varies as 1/N  one can improve

ability to resolve closely spaced spectral lines

By examples we will see how effective are diffraction gratings atresolving light of different wavelengths (i.e separating closely-spaced

min min

(Higher spectral resolution)

 We can squeeze more resolution out of a given grating by working in

“higher order” Remember, the principal maxima occur at sin= m/d,where m = 1,2,3… designates the “order” (min≈/Nd still*) You

can easily show:

* To be precise: min= /(Nd cos), (but  /= 1/Nm is correct.)

Larger Nm Smaller  min

(Higher spectral resolution)

 Numerical example: Angular splitting of the Sodium doublet.

Consider the two closely spaced spectral (yellow) lines of sodium(Na), 1 = 589 nm and 2 = 589.6 nm, mentioned earlier If lightfrom a sodium lamp illuminates a diffraction grating with 4000

slits/cm, what is the angular separation of these two lines in thesecond-order (m=2) spectrum?

4

1 cm

2.5 10 2.54000

 Notes:

Diffraction gratings can be made using special techonology

In applications there are two types of diffraction gratings:

’transmission grating’ and ‘reflection grating’

Reflection grating

One of common example of reflectiongratings is a compact disk (CD)

The rainbow-color reflection are due

to the reflection-grating effect

§ 3.X -Ray Diffraction from Crystals:

 Diffraction gratings are excellent tools for studying visible light

because the slit spacing is on the order of the wavelength of the light(~ few tenths of microns)

 Visible light is a very small part of the spectrum of electromagneticwaves How can we study e-m waves with smaller wavelengths (e.g., X-rays with ~ 10-10 m)?

 We can’t use a standard diffraction grating to do this Why?

Calculate for first order peak for X-rays with = 10-10 m for agrating with d=1000 nm

 = /d = 10-4 = 0.1 mradThe first-order peak is too close to the central maximum to bemeasured!

 Solution? We need a “diffraction” grating with a spacing d that is

on the order of the wavelength of e-m waves we’d like to measure –

i.e., for X-rays: ~10-10m

 Where do we find such a “grating”? Nature helped us !!!

Crystalline solids (having regularly spaced atoms with d ~ 10-10m)

 Illuminate crystal with X-rays

 The X-rays are scattered by the ions

 Enhanced scattering at certain scatteringangles reveal constructive interferencebetween the scattered waves from

different regularly-spaced planes of atoms

 If we know about the grating, we can use the diffraction pattern to

learn about the light source

 If instead we know about the source, we can use the diffraction pattern

to learn about the “grating”

 For this to work, we need to have a source wavelength that is less thanthe grating spacing (otherwise, there are no orders of diffraction)

 Crystals consist of regularly spaced atoms  regular array of scatteringcenters Typical lattice spacing is 5 angstroms = 5 x 10-10 m = 0.5 nm

for constructiveinterference:

2d sin= 

d = lattice spacing

= X-ray angle (withrespect to plane ofcrystal)

= X-ray wavelength

 Applications of X-ray Crystallography:

• could distinguish different cubic lattices

• discovered the crystal structure of diamond

• Lawrence Bragg was the youngest Laureate ever (25)

to receive a Nobel Prize (shared with his father in

1915)

• now standardly used for all kinds of materials

analysis, even biological samples!

• The same multi-layer interference phenomenon is

now used to make highly wavelength-specific mirrors

for lasers

(“distributed Bragg feedback” [DBF])

The Braggs made so many discoveries that Lawrence described thefirst few years as “like looking for gold and finding nuggets lying aroundeverywhere”:

• howed that the sodium and chloride ions were not

bonded into molecules, but arranged in a lattice

Diffraction is any deviation from straight propagation lines

Diffraction can be understood as interference of secondary sources:

• The concept of secondary sources is introduced by Huygen

• Fresnel added in more details the information about theamplitude & the phase of waves from secondary sources

 Near-field diffraction (Fresnel diffraction) concerns spherical waves.For analyzing it we use the method of Fresnel spherical zones:

• By this method we can avoide taking an integral over allsecondary sources Instead of this we have only to take

a sum over Fresnel zones

• We can go further to a simple formula

Far-field diffraction (Fraunhofer diffraction) concerns plane waves.The diffraction pattern is obtained by focusing (with a lens) parallelbeams on a screen.

• By using the method of phasor diagrams we can find formulas

for amplitudes, and then for light intensity as a function of

inclination angles of the beam

Applying multiple-slit diffraction one creates diffraction gratingsthat gives an effective method to obseve and study spectral lineswith high resolution

Some natural crystals can use as diffraction gratings for X-ray It isthe priciple for X-ray crystalography that is a helpfull tool in

the investigation of structure of material