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Chapter 17 Wave Optics. Part 1 Interference of light. Colorful interfering phenomena. Soap film under sunlight. Oil film under sunlight. Interference fringes produced by wedge-shaped air film between two flat glass plates. Newton’s rings (Equal thickness fringes). - PowerPoint PPT Presentation
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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.