Part 1 Interference of light

Chapter 17 Wave Optics

Colorful interfering phenomena

Soap film under sunlight

Oil film under sunlight

Interference fringes produced by wedge-shaped air film between two flat glass plates

Equal inclination fringes

Newton’s rings(Equal thickness fringes)

§17-1 Monochromaticity and Coherence of Light

Stimulation Radiation

The persistent of light isThe persistent of light is

The life time of electroThe life time of electrons on the excited state ns on the excited state (( 激发态激发态 ) is) is

s811 1010~

s810~

1.The emitting process of light1.The emitting process of light

StimulatingStimulatingpumpingpumping

Lighting is intermittentLighting is intermittent(( 间歇）间歇）

Optical wave trainOptical wave train

independent(from different atoms)·

·independent(from different time of same atom)

Different wave trains Different wave trains are not coherent light.are not coherent light.

f= (E2-E1)/h

E1

E2

ff

fIdentical wave train (frequency, vibrating direction, phase)

2. Laser (from stimulated radiation 受激辐射 )

Coherent light

•Monochromatic light ： has single frequency

•Spectrum curve ( 光谱曲线 ) ： The intensity distribution of light with wavelength ( or frequency )

•Frequency width( 频宽 ) f•wavelength width( 线宽 )

3. 3. MonochromaticityMonochromaticity(( 单色性单色性 ))

Good monochromaticity: f, are smaller.

I

I

Bad monochromaticityGood monochromaticity

4. 4. CoherencyCoherency

Coherent lights ： Identical frequency, vibrating direction, phase

Can produce coherent superposition

The intensity distribution of resultant light in crossing space ：

cos2 2121 IIIII

If I1=I2, 2

cos4 21

II

The intensity of light varies between bright and dark alternately.

21 III

If I1 and I2 are not coherent light, then

No varying

5. 5. The methods to obtaining coherent light:

•principle ： By using some arrangements, divide the light wave emitted by an identical point from monochromatic source into two beams. And superposing these two beams in the space.

•methods ：

The way of division of wavefront ：

The way of division of amplitude. ：

S1S

2S

d

The way of division of wavefront ：

The way of division of amplitude. ：

§§17-2 Two beams interference17-2 Two beams interferenceI. Young’s double-silt experimentI. Young’s double-silt experiment

1. The experiment arrangement and principle.1. The experiment arrangement and principle.

2.The positions of bright fringes and dark fringes.2.The positions of bright fringes and dark fringes.The path differenceThe path difference ：： ? rThe distribution characteristics of the fringes The distribution characteristics of the fringes and the distance between them.and the distance between them.

3. Analyze the distribution distribution characteristics of the frincharacteristics of the fringes of the white light.ges of the white light.

Study by yourself

II. Fresnel’s double-mirror experimentII. Fresnel’s double-mirror experiment

III. Lloyd’s mirror experimentIII. Lloyd’s mirror experiment

S

S’

M M’d

EE’

A

B

C

S

S

Half-wave lossHalf-wave loss

I. Optical path nrS

=Distance of light traveling through vacuum at the same time

§17-3 Optical path and optical path differe§17-3 Optical path and optical path difference, Property of thin lensnce, Property of thin lens

nvt

tn

cn ct

v--speed of light in medium

For a monochromatic light, f is same in different medium, but and v are different.

Let in vacuum ： , c

In the medium with refractive index n ： ' ,v

n

cv

n ,'

Then

II. Optical path difference 1122 rnrn

The phase The phase differencedifference ：：

2

：： the wavelength in vacuumthe wavelength in vacuum

III. The half-wave loss of reflecting light 　　

The transmitting ( refracting ) light never have half-wave loss.

n1<n2

Has half-wave loss

n1>n2

No half-wave loss

Phase shift

IV. IV. A lens does not A lens does not cause any additional cause any additional optical path difference optical path difference or phase shift.or phase shift. F

A

B

CAFAF, , CFCF travel a larger travel a larger

distance in the air and distance in the air and a shorter distance in a shorter distance in the lens.the lens.

BF BF is inverse. is inverse.

AFAF= = CFCF= = BFBF ，， they are converging on they are converging on

point point FF and forming a bright image point. and forming a bright image point.

1n

12 nn

1ne

§17-4 Interference by division of amplitude§17-4 Interference by division of amplitude

I. Equal-inclination interferenceI. Equal-inclination interference

iA

B

C

D

The optical path The optical path difference is difference is

22 12

ADnABn

1. The interference of reflecting light1. The interference of reflecting light

cos

eAB

iACAD sin

itge sin2

And And

sinsin 21 nin

2sin2

cos2 12

ietgne

n

2)sin1(

cos

2 22

en

1n

12 nn

1ne

iA

B

C

D

2sin2 22

122

inne

The conditions that occur bright and dark The conditions that occur bright and dark fringes arefringes are

k

2

12

k --Dark fringes--Dark fringes

--Bright fringes--Bright fringes,2,1k

,2,1,0k

222 sin1cos nn inn 2

12

2 sin

2cos2 2

en

S

ii

i i

DiscussionsDiscussions ：：

--Equal inclination --Equal inclination interferenceinterference

，， As long As long as the lights coming fras the lights coming from different area of thom different area of the source have same ince source have same incident angle ident angle ii,, they havthey have same e same , and belong t, and belong to same interference leo same interference levelvel kk

)(i

The interference fringes are series of conThe interference fringes are series of con

centric circles. They are alienation(centric circles. They are alienation( 稀疏稀疏)near the center, and dense near the edge. )near the center, and dense near the edge.

R larger, i larger, k smallerR smaller,

i smaller, k larger

S

filmfilm

Reflecting plateReflecting plate

lenslens

screenscreen

ii

S

film

Reflecting plate

lens

screen

ii

When When e e ，， we have we have kk , the circle fringes , the circle fringes are produced from the center. are produced from the center.

At the center,At the center,

22 2

en----Bright----Bright

----Dark----Dark

2

)12(

k

k

When When e e ，， we have we have kk ，， the circle fringes the circle fringes are swallowed at the center.are swallowed at the center.

2. 2. The interference of The interference of transmitting lighttransmitting light

1n

12 nn 1n

e

B

C

inne 221

22 sin2'

The optical path The optical path difference is difference is

2,1,0k

----bright----bright

----dark----dark

2)12(

k

k

The reflecting light and transmitting light are The reflecting light and transmitting light are compensative each other.compensative each other.

[[ExampleExample] A thin oil film (] A thin oil film (n=n= 1.301.30) is illuminated ) is illuminated by the white light. Someone observes the by the white light. Someone observes the reflected light by the film.reflected light by the film. When the observing When the observing direction has the angle direction has the angle 30300 0 with respect to the with respect to the normal direction of the film, the film appears normal direction of the film, the film appears blue ( blue ( 48004800ÅÅ). Find the minimum thickness of ). Find the minimum thickness of the oil film. If the film is observed at the the oil film. If the film is observed at the normal direction of the film, what color does normal direction of the film, what color does the film appear?the film appear?

Solution Solution : According to : According to

2sin2 22

122

inne k

ine

22minsin4

)12(

22

7

5.03.14

108.4)12(

k=1 e=emin

m100.1 7

The film is observed at the normal direction of The film is observed at the normal direction of the film, the film, ii=0:=0: kne 22

12

4

k

ne12

10205 7

k

.

For For kk =1=1,, m1020.5 7

For For kk =2=2,, m10733.1 7

--green--green

--Ultraviolet

II. ApplicationII. Application1.Transmission 1.Transmission

enhanced filmenhanced film

10 n

5.1'n glass

e 38.1n MgF2

The two beams reflected by the upper and The two beams reflected by the upper and bottom interface of the bottom interface of the MgFMgF2 2 film all have half-film all have half-

wave loss.wave loss.

(anti-reflecting film)(anti-reflecting film)

(( 增透膜增透膜 ))

n0 < n < n

ne2The optical path difference between The optical path difference between andand is is

2)12( k

--reflecting beams are destructive.

The minimum thickness of The minimum thickness of MgFMgF22 ：：n

e4min

oror4min

ne ne --optical thickness--optical thickness

--transmitting --transmitting beams are cobeams are constructive.nstructive.

10 n

5.1'n glass

e 38.1n MgF2

2. Reflection enhanced film(2. Reflection enhanced film( 增反膜增反膜 ))40.2Hn38.1Ln

Considering the reflected Considering the reflected beams by each interface.beams by each interface.

For the first film,For the first film, kenH 22 1

5.1'n

HnLn

Ln

Ln

Hn

Hn

ZnS 2MgF

,2,1k

kk=1 =1 41 enH

For the second film,For the second film, kenL 22 2

,2,1k

kk=1 =1 42 enL

medium wedge

n

1. Wedged interference1. Wedged interferenceAssume the incident beams are Assume the incident beams are

perpendicular perpendicular to the interfaces of the film.to the interfaces of the film.

22 ne

--bright--brightk

--dark--dark2

)12(

k

At the contact At the contact edge(edge(ee=0=0) , ) , kk =0, =0, appearsappears dark fringe. dark fringe.

III. Equal-thickness interference

22 e

For air wedge,For air wedge,

air wedgeair wedge

The points with identical thickness The points with identical thickness e e have the have the same interference level same interference level kk..

DiscussionDiscussion

--Equal thickness interference--Equal thickness interference

The fringes are the straight lines parallel to thThe fringes are the straight lines parallel to the edge. They have same distance. And there is e edge. They have same distance. And there is a dark fringe at the contact edge(a dark fringe at the contact edge(e e =0=0) because ) because of half-wave loss.of half-wave loss.

The thickness difference between two adjaThe thickness difference between two adjacent bright(or dark) fringes:cent bright(or dark) fringes:

ke 1ke el

kk eee 1

knek 22 )1(22 1 knek

n2

For air wedge,For air wedge,2

e

The distance between two bright (or darkThe distance between two bright (or dark) fringes:) fringes:

ke 1ke el

sin

el

e

n2

--identical distance--identical distance

When the upper interface of the wedge film iWhen the upper interface of the wedge film is moved upwards, the distance of the fringes s moved upwards, the distance of the fringes is constant, but all the fringes move toward tis constant, but all the fringes move toward the contact edge.he contact edge.

eA A

eB

[[ExampleExample]To determine the thickness ]To determine the thickness dd of the of the SiSiOO22 over the over the Si Si precisely, it is usually corroded to precisely, it is usually corroded to a wedge shape. The light with a wedge shape. The light with =5893=5893ÅÅ is incide is incident normally from above. There are nt normally from above. There are 77 bright fring bright fringes over the length of the film. Calculate es over the length of the film. Calculate dd=? =? ( (SiSi: : nn11=3.42=3.42 ，， SiOSiO22: : nn22=1.50 =1.50 are known.)are known.)

d1n2n

SiO2Si

At the contact edge, At the contact edge, kk=0=0 ，， the bright fringe the bright fringe at at dd should correspond to should correspond to kk=6=6

26 2

6

ned

m6101786.1

SolutionSolution ：：

en22 ,1,0kk22n

kek

d1n2n

SiO2Si

The optical difference The optical difference between two rays between two rays reflected by the upper reflected by the upper and lower surface of and lower surface of SiOSiO2 2 isis

Plane glass

RPlane-convex

0n

2. Newton’s ring2. Newton’s ring

r e

At points with At points with thickness thickness e e ,,

22 ne

--brightk

--dark2

)12(

k

At the center(At the center(e e =0=0)): : corresponding to a dark corresponding to a dark point with point with k k =0=0..

The fringes are series of concentric circles. The fringes are series of concentric circles.

-- -- Newton’s ringNewton’s ring the radius the radius rr of the circle: of the circle:

222 eRRr 22 eeR R

0rn

Re asas

R

re

2

2

Discussion:Discussion:

is determined by is determined by ee--equal-thickness interference--equal-thickness interference

Bright circle,Bright circle, n

Rkrk 2

12

Dark circle, Dark circle,

n

kRrk

2,1k

2,1,0k

the distance between two the distance between two adjacent circles :adjacent circles :

n

kR

n

Rkr

)1(

n

R

kk

1

1 -- -- alienation(alienation( 稀疏稀疏 )near the )near the center, dense near the edge. center, dense near the edge.

When the When the Plane-convex lens lens moves upwards, moves upwards, the circle fringes are swallowed at the center.the circle fringes are swallowed at the center.

A B

2n3nFind Find Is the bright or dark Is the bright or dark

fringe at the edge of the oil?fringe at the edge of the oil? The maximum thickness The maximum thickness of the oil film =? of the oil film =? If the oil If the oil spreads gradually, How do spreads gradually, How do the fringes change? (oil: the fringes change? (oil: nn22=1.60=1.60, glass: , glass: nn33=1.50=1.50 ））

[[ExampleExample] A drop of oil is on a plane glass. When ] A drop of oil is on a plane glass. When a monochromatic light with a monochromatic light with =5760=5760ÅÅ is incident is incident on it normally, the interference fringes produced on it normally, the interference fringes produced by the reflected lights are shown in figure. The by the reflected lights are shown in figure. The center point of the oil is dark.center point of the oil is dark.

SolutionSolution

i.e.,i.e.,2

2 2

en

So So There is a There is a dark circle fringedark circle fringe at the oil edge, at the oil edge, corresponding to corresponding to kk=0=0..

Because Because nn11<n<n22 ，， nn22>n>n33 ，， there is half-wave loss inthere is half-wave loss in ..

At the edge of oil, At the edge of oil, e=0e=0, we have, we have

2)12( k

i.e., i.e., satisfies dark fringes condition, satisfies dark fringes condition, 2

When the oil spreads gradually, the dark cthe oil spreads gradually, the dark circle fringe located in the edge spreads graircle fringe located in the edge spreads gradually, the center point changes from dark dually, the center point changes from dark to bright, and to dark alternately, the level to bright, and to dark alternately, the level kk of the fringe becomes less and less. of the fringe becomes less and less.

The center dark point cThe center dark point corresponds toorresponds to kk=4=4..

2max 2n

ke

6.12

1057004 10

m102.7 7

1G 2G

§17-5 Michelson’s interferometer§17-5 Michelson’s interferometer

1M

2M

Compensating plate

Beam spliter

I. InstrumentI. Instrument

MM11 is fixed, is fixed, MM22

can be movedcan be moved

'2Md

Assume the thickness of the air film between Assume the thickness of the air film between MM11and and MM22 is is dd

d2

DiscussionDiscussionMM11and and MM22are perpendicular to each other--are perpendicular to each other--

equal inclination interference produced by air equal inclination interference produced by air film.film. -- concentric circles-- concentric circles

kMM22 moves moves ：d )(2' dd 'k

2)'(

kkd 2

n

--The number of the fringes --The number of the fringes out of out of or inor in the center. the center.

n

Then, at the center of the fringes Then, at the center of the fringes

II. ApplicationII. Application

Measure length Measure length dd—known —known , , read outread out nn ，，Measure Measure ----read out ----read out n n and and dd ，，

2

nd

Measure refractive index Measure refractive index nn—known —known ，， read out read out dd end 1

MM11and and MM2 2 are not perpendicular to each are not perpendicular to each

other -- equal thickness interference produced other -- equal thickness interference produced by air wedge.by air wedge.

Look for “etherLook for “ether （以太）（以太）”” -- “ zero”result-- “ zero”result

Michelson’s interferometer Michelson’s interferometer be used to measure the refractive index of gas.

Michelson’s interferometer Michelson’s interferometer be used to measure flow field.