Part 2 Diffraction of light

Diffraction phenomena of light

§17-6 Diffraction of light Huygens-Fresnel’s principle

I.I. Diffraction phenomena of light Diffraction phenomena of light

E

S

E

S

Condition :Condition :

The width of the diffracting obstacle is not very The width of the diffracting obstacle is not very largely compared to the wavelength.largely compared to the wavelength.

a a >> >> ，， Diffraction negligibleDiffraction negligible

a >a >101033 ，， Diffraction is not obviousDiffraction is not obvious

a ~a ~101022 — —1010 ，， Diffraction fringes appearDiffraction fringes appear

a ~a ~ 时， 时， Diffraction is obviousDiffraction is obvious

a a << ，， Scatterance Scatterance (( 散射散射 ))

Fresnel diffractioFresnel diffractionn

E

S

source

screen

obstacle

Fraunhofer diffractionFraunhofer diffraction E

S

----the source or the the source or the screen or both are screen or both are at finite distance at finite distance from the diffracting from the diffracting obstacle.obstacle.

----the source and the the source and the screen are at infinite screen are at infinite distance from the distance from the diffracting obstacle.diffracting obstacle.

II. II. ClassificationClassification

Every elementEvery element dS dS of wave frontof wave front SS is the source ois the source of a secondary spherical wavelet.f a secondary spherical wavelet.

III. Huygens-Fresnel’s principleIII. Huygens-Fresnel’s principle

The amplitude The amplitude dAdA of the secondary spherical wa of the secondary spherical wavelet emitted by velet emitted by dSdS is proportional to the size of is proportional to the size of ddSS,,

dA dS dA r

1

r

dSkdA )(

S

dS n

PrdA

k()—inclination factor

The resultant oscillation of light at The resultant oscillation of light at PP is the is the coherent superpositioncoherent superposition of all spherical of all spherical wavelets emitted by all elements on the wavelets emitted by all elements on the SS

S

rtdAPy )

2cos()(

S

dS n

PrdA

The light oscillation aThe light oscillation at t PP produce by produce by dSdS is is

)2

cos( r

tdAdy

Interference appears.The intensity of light changes in space.

IV. Parallel beams interference—a simplified IV. Parallel beams interference—a simplified discussiondiscussion

1 1 , , 22 – – diffraction anglediffraction angle

Optical axis Optical axis

Wave fron

tW

ave front

Wave Wave rayray

aa

screenscreen

PP11

1111

PP22

22

Coherent superpositionCoherent superposition

E

　I. I. Diffraction deviceDiffraction device

§17-7 Single slit Fraunhofer diffraction

S

II. Distribution law of diffraction fringesII. Distribution law of diffraction fringes　　　　　　　　　　

λ2

λ2λ2

λ2

C

1A

2A

3A

sinaBC

P

E

a

A

Bf

---- Fresnel half wave zone method ---- Fresnel half wave zone method (( 半波带法半波带法))

C

BC=aasinsin=2(/2)

----The wave frontThe wave front ABAB is divided intis divided intoo 2 half wave zones 2 half wave zones(2(2 个半波带个半波带 ))

The optical path difference betThe optical path difference between two corresponding points ween two corresponding points on on A0A0 and and 0B 0B is is /2/2 ，，

--point --point PP is dark. is dark.

a

A

BC

0

--Destructive interference.--Destructive interference.

BC=asin=3(/2)

--The wave frontThe wave front ABAB is divided iis divided intonto 3 half wave zones 3 half wave zones(3(3 个半波带个半波带))The optical path difference betweThe optical path difference between two corresponding points on en two corresponding points on AAAA11 and and AA11AA22 is is /2/2 ，，

--point --point PP is bright.is bright.

a

A

BC

1A

2A

TheyThey produce destructive interference. produce destructive interference.

The light oscillations coming from wave front The light oscillations coming from wave front AA22BB produce constructive interference. produce constructive interference.

　BC=asin=n(/2) 2,1n

IfIf n n is even numberis even number (( 偶数偶数 )) ：：--dark fringes--dark fringes

IfIf n n is odd numberis odd number (( 奇数奇数 )) ：：--bright fringes--bright fringes

2)12(sin ka --bright--bright

2,1k22sin

ka --dark--dark

If asin integral times ofintegral times of /2/2, the intensity o, the intensity of light is between maximum and minimum.f light is between maximum and minimum.

Central diffraction Central diffraction maximum fringemaximum fringe ：： the the region between the first region between the first positive and negative dark positive and negative dark fringesfringes

Half-angle widthHalf-angle width ：：10

a

1sin

DiscussionDiscussion

Half Half widthwidth ：：

tg20 fx f2a

f2

E1

ax

10 --inverse proportion--inverse proportion

The distance between other adjacent fringes (The distance between other adjacent fringes (bright or dark)bright or dark)

kk xxx 1 kk ff tgtg 1

])1(

[a

k

a

kf

a

f 02

1x

The intensity distribution of diffraction The intensity distribution of diffraction fringes:fringes:

a2

3

a2

5a2

3

a2

5

a

a

sin

0I

I

Most of the light intensity is concentrated in Most of the light intensity is concentrated in the broad central diffraction maximum.the broad central diffraction maximum.

[[ExampleExample]In experiment of Fraunhofer diffracti]In experiment of Fraunhofer diffraction from a single slit, on from a single slit, f f = = 0.5m0.5m ，， =5000=5000ÅÅ ，， the the width of the slit width of the slit aa=0.1mm=0.1mm. Find . Find the width of cethe width of central maximum, ntral maximum, the width of the secondary mathe width of the secondary maximum.ximum.

1

1P

0

x

f

Solution Solution the width of the width of central maximumcentral maximum

10 tg2 fx

1sin2 f

The width l of the secondary maximum The width l of the secondary maximum equals to the space of the first minimum and equals to the space of the first minimum and the second minimum.the second minimum.

12 sinsin ffl )2

(aa

f

a

f mm5.2

af

23

10

101.0

1050005.02

mm5

§17-9§17-9 Resolving power of optical instrumentResolving power of optical instrument

I. I. Fraunhofer diffraction by circular apertureFraunhofer diffraction by circular aperture

E

S

Airy diskAiry disk

The diffraction angle of first dark ring,The diffraction angle of first dark ring,

r

61.0sin 1 D

22.1

sin

0II

r

610.0

r

610.0

E

1S

2S1A

2A

distinguishdistinguish

II. Resolution of optical instrumentII. Resolution of optical instrument

E

1S

2S 1A2A

Can’t be distinguished

E

1S

2S 1A2A

Just distinguished

Airy diskMinimum

resolving angle

Rayleigh criterionRayleigh criterion ：： Two images are just resolveTwo images are just resolved when the center of central maximum of one patd when the center of central maximum of one pattern coincides with the first dark ring of another. tern coincides with the first dark ring of another.

爱里斑1 1sinD

22.1

1

R

Resolving power of an optical Resolving power of an optical instrument:instrument:

D

22.1

1

ImproveImprove RR ：： increasing increasing DD—astronomical telescope with l—astronomical telescope with large radiusarge radius

decreasing decreasing --electronic microscope--electronic microscope

Minimum resolving angle

§§17-8 Diffraction grating17-8 Diffraction grating

I. I. GratingGrating An optical device which consists of a large number An optical device which consists of a large number of equally and parallel slits with same distance.of equally and parallel slits with same distance.

classificationclassification ：：Transmitting

gratingReflecting

grating

P

ab

E

0x

sind

d

bad --grating constant

sind ---optical path difference between rays from adjacent slits.

II. II. The formation of grating diffracting fringesThe formation of grating diffracting fringes

diffraction diffraction + + interferenceinterference

1. The interference of multi-slits 1. The interference of multi-slits (( 多缝干涉多缝干涉 ))

A

B

Csin)( ba

The phase difference between rays from adjacent slits is

sin)(

2ba

When =2k ,

The rays coming from all slits are in phase.----constructive interference

kba sin)(

2,1,0k

--grating --grating equationequation

According to According to 2

We get We get

The principle maximum appears at the direction with the diffraction angle .

kd sin

Principle maximum and secondary maximumPrinciple maximum and secondary maximum

I II

N=2 N=4 N=6

Principle maximum

secondary maximum

2.The influence of diffraction by each slit to the 2.The influence of diffraction by each slit to the interference fringesinterference fringes

The diffraction patterns of all slits coincide. The diffraction patterns of all slits coincide.

N=1

N=2

N=3The intensity N2

Interference of multi-slits

Slit diffraction

Missing order

Grating differactin

differactin

+

interference

0 1-1 2 3 4 5 6-6 -5 -4 -3 -2

The missing order phenomenon of grating:The missing order phenomenon of grating:

kba sin)(

2'2sin ka

--constructive interference--constructive interference

--minimum of slit diffraction--minimum of slit diffraction

'ka

bak

,2,1'k

then the then the kk-th principle maximum will disappear.-th principle maximum will disappear.

-- The -- The kk--th fringe isth fringe is missing order.missing order.

On some direction, if diffraction angle On some direction, if diffraction angle satisfies, satisfies,

andand

III. III. The incident ray inclinationThe incident ray inclination

ACB

The optical path difference The optical path difference of two adjacent rays isof two adjacent rays is

ACAB

sinsin dd

The grating function for inclination incidence isThe grating function for inclination incidence is

kd )sin(sin 2,1,0k

IV. Grating spectrumIV. Grating spectrum

When a polychromatic light When a polychromatic light (( 复色光复色光)) is incident a grating, is incident a grating,

First order spectrum

Second order spectrum

Third order spectrum

kd sin

except for the central fringe, all except for the central fringe, all others principle maximum with others principle maximum with different different have different have different for each for each same order. same order.

[[ExampleExample]Two slits with ]Two slits with dd=0.40mm=0.40mm, the width is , the width is aa=0.08mm=0.08mm. A parallel light with . A parallel light with =4800=4800ÅÅ is emitted is emitted on the two slits. A lens with on the two slits. A lens with f f =2.0m=2.0m is put on the slits. is put on the slits. Calculate: Calculate: The distance The distance xx of the interference of the interference fringes on the focal plane of the lens. fringes on the focal plane of the lens. The numbers The numbers of interference fringes located in the width of the of interference fringes located in the width of the central maximum producing by single diffraction.central maximum producing by single diffraction.

f

da

?

SolutionSolution

kd sin

the position of the position of kk-th order bright fringe is -th order bright fringe is

tgfxk sinfd

kf

kk xxx 1d

f m104.2 3

For two beams interference, the bright fFor two beams interference, the bright fringes (principle maxima) satisfies:ringes (principle maxima) satisfies:

the distance between two adjacent bright the distance between two adjacent bright fringes is fringes is

100 x

x

For a single diffraction, the width of For a single diffraction, the width of central maximum iscentral maximum is

4,3,2,1,0

bright fringes appears bright fringes appears in the width of in the width of xx0 0 ..

9N

10 tg2 fx 1sin2 fa

f

2 m104.2 2

012345

-2-1

-3-4-5

xx00

x

[[ExampleExample]A diffraction grating has ]A diffraction grating has 500500 slits per slits per millimeter. It is irradiated by Sodiummillimeter. It is irradiated by Sodium(( 钠钠 )) light light with with =0.59×10=0.59×10-3-3mm. mm. Find Find The maximum ordThe maximum order of spectrum can be observed when the beams er of spectrum can be observed when the beams of Sodium light are incident normally? of Sodium light are incident normally? How How many orders of spectrum can be seen whenmany orders of spectrum can be seen when the ithe incident angle isncident angle is 30300 0 ??

Solution Solution

500

1 ba mm102 3

kba sin)( ,2,1,0k

The grating constant is The grating constant is

i.e., when the beams are incident normally, the i.e., when the beams are incident normally, the maximum order which can be observed is maximum order which can be observed is 33-th.-th.

kba )sin)(sin(

When the incident rays and diffraction rays When the incident rays and diffraction rays are in the same side of the optical axis, are in the same side of the optical axis,

When When 2

， k gets maximum. gets maximum.

ba

k

max 39.3 Take integralTake integral=3=3

)30sin90)(sin( 00

max

bak

When the incident rays and diffraction rays When the incident rays and diffraction rays are in the two side of the optical axis,are in the two side of the optical axis,

)]30sin(90)[sin( 00

max

bak

69.1

08.5

the order numbers which can be observed in tthe order numbers which can be observed in this case is his case is 55-th.-th.

Take integralTake integral=5=5

Take integralTake integral= -1= -1

the order numbers which can be observed in the order numbers which can be observed in this case is this case is 11-th.-th.

The spectrum with order The spectrum with order -1, 0,1,2,3,4,5-1, 0,1,2,3,4,5can can be observedbe observed

ACB

1 order

0-th order

5 orders

[[ExampleExample] A monochromatic light with ] A monochromatic light with ==70007000ÅÅ is incident normally on a grating. The is incident normally on a grating. The grating has grating has dd= = 3×103×10-4 -4 cmcm, , aa==1010-4-4cmcm. Find . Find The The maximum order of the spectrum can be maximum order of the spectrum can be observed ? observed ? Which orders are missing?Which orders are missing?

SolutionSolution kd sin 2,1,0k090For , For ,

ba

k

max 28.4

The maximum order which can be observed is The maximum order which can be observed is 44-th.-th.

Take integralTake integral= 4= 4

for interference bright fringes.for interference bright fringes. kd sin

22sin

ka

for diffraction dark fringes.for diffraction dark fringes.

'ka

dk for same for same , ,

,2,1'kWhen , corresponding to When , corresponding to kk==33,,66… …

As As 4max k

The order The order kk=3=3 is missing. is missing.

i.e., the orders of the spectrum that can be obsi.e., the orders of the spectrum that can be observed are erved are 4-1=34-1=3. They correspond with . They correspond with kk== -4,-2,-1,0,1,2,4 -4,-2,-1,0,1,2,4 ( seven principle maxima)( seven principle maxima)

Missing order

0 1-1 2 3 4-4 -3 -2

I. I. x x -Ray-Ray

§17-10 §17-10 x x –Ray diffraction by Crystal–Ray diffraction by Crystal

It was discovered by W.K It was discovered by W.K Roentgen ( a German physicist) in Roentgen ( a German physicist) in 1895.11.1895.11.

The first The first xx–ray photo: his wife’s –ray photo: his wife’s hand.hand.

He got the first Noble Prize of He got the first Noble Prize of

Physics in 1901 as the discovery Physics in 1901 as the discovery

of of xx-ray. -ray.

x-x-ray: produced by bombarding a target ray: produced by bombarding a target element (Anode) with a high energy beam of element (Anode) with a high energy beam of electrons in a electrons in a xx-ray tube.-ray tube.

It’s a type of electromagnetic waves with It’s a type of electromagnetic waves with wavelength ranges about wavelength ranges about 0.1--1000.1--100ÅÅ, between , between Ultraviolet and Ultraviolet and -ray.-ray.

A AnodeCathode

x-x-ray ray tubetube

K

E

filmfilmcrystalcrystal

Lead plateLead plate

the diffraction pattern of the diffraction pattern of xx-ray was observed by -ray was observed by German physicist M. Von Laue.German physicist M. Von Laue.

Laue Laue (( 劳厄劳厄 )) spots spots

In 1912, a collimated beam of In 1912, a collimated beam of xx-ray which -ray which contain a continuous distribution of wavelengths contain a continuous distribution of wavelengths strikes a single crystal, strikes a single crystal,

xx--ray iray is a wave

xx-ray can be used widely to study the internal -ray can be used widely to study the internal structure of crystals.structure of crystals.

Laue got Noble Prize Laue got Noble Prize of Physics in 1914 becof Physics in 1914 because verified that ause verified that xx--ray is a wave. ray is a wave.

II.II.Bragg equationBragg equation

A BC

O

晶面晶面晶面间距 d

掠射角掠射角W.H.Bragg, and W.L.Bragg, two British physicists ( son and father) took another method to study x-ray diffraction.

BCAC sin2d

The optical path difference of the two xx-ray -ray beams scattering by the atoms that they locate the atoms that they locate in two parallel planes isin two parallel planes is

They found that whenThey found that when

kd sin2

,2,1k

---- ---- Bragg equation

，， the the xx-ray beams-ray beams

produceproduce constructive interference.constructive interference.

W.H.Bragg, W.L.Bragg got Noble Prize of got Noble Prize of Physics in 1915 because they found a new Physics in 1915 because they found a new method to study the properties of method to study the properties of xx-ray.-ray.