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Chapter 11
Optics
Shanghai Normal University
Department of Physics
Development of Optics
Geometrical optics
Wave optics
Quantum optics
Wave optics：
Interference?
Diffraction?
Polarization?
本章目录
Catalog of Chapter 11
11-0 Teaching Basic requirements
11-1 Coherent Light
11-2 Young’s Double – slit Interference,
Lloyd Mirror
11-3 Optical Path, Film Interference
11-4 Wedge, Newton’s Ring
11-5 Michelson Interferometer
11-6 The Diffraction of Light
11–7 Single – slit Diffraction
Catalog of Chapter 11
11- 8 Hole Diffraction, the Resolution Capability
of Optical Instruments
11- 9 Diffraction Grating
11-10 The Polarization of Light, Malus Law
11-11 The Polarization of Reflected Light and
Refracted Light
11-12 The Birefringence
*11-13 The Liquid Crystal Display
*11-14 Geometrical Optics
11-0 Teaching Basic requirements
Interference of Light
1. Understand the interference condition and
the method of getting coherent light
2. Master concept of optical path as well as
relationship of the optical path difference and
the phase difference, and understand under what
circumstances the reflected light has phase jump.
11-0 Teaching Basic requirements
3. Analyze position of Young’s double-slit
interference fringes and film thickness
interference fringe .
4. Understand principle of Michelson
interferometer
.
11-0 Teaching Basic requirements
The Diffraction of Light
1. Understand the Huygens-Fresnel
principle and its qualitative interpretation of
diffraction phenomenon of light.
2. Understand using zone method to
analyze distribution of single-slit Fraunhofer
diffraction fringes, and analyze the effect of
the slit width and wavelength on distribution
of diffraction fringes.
11-0 Teaching Basic requirements
3. Understand grating diffraction formula,
determine the position of grating diffraction
line, analyze the effect of grating constant and
the wavelength on the grating diffraction
spectrum.
4. Understand the effect of the diffraction
of optical instrument resolution.
5. Understand the x-ray diffraction
.
phenomenon and the physical meaning of
Prague formula.
11-0 Teaching Basic requirements
The Polarization of Light
1. Understand difference of Nature light
and Polarized light.
2. Understand Brewster law and Marius's
law.
3. Understand birefringence of light.
4. Understand the method of getting and
testing linearly polarized light.
11-1 Coherent Light
I. Light wave is the electromagnetic wave
r
(traverse wave)
E  E cos  (t  )
0
u
Plane Electromagnetic
r
Wave Equation
H  H 0 cos  (t  )
u
Light vector：the component who can
produce the vision of eyes and the sensitization on
the photographic plate is the vector E called light
vector.
E oscillation is called light oscillation.
11-1 Coherent Light
The speed of light
in vacuum
c
1
 00
The speed of light in vacuum c and the
speed of light in transparent medium u are：
1
1
1
c
u

 00

 0  r  0 r
c
n    r  r (Refractive index of the
u
medium)
11-1 Coherent Light
Light source: object of emitting light.
Ordinary light emitting mechanism is the
spontaneous radiation of the atomic.
(Atoms and molecules in the absorption of energy are in
an unstable excited state, even in the absence of any external
forces, they would spontaneously return to low excited state
and ground state, at the same time sending out light.)
11-1 Coherent Light
The light source and the mechanical wave source
detected difference: mechanical wave source is often a
vibrating object, while the light source is thousands on
thousands of atoms which randomly emits light as one
falls.
11-1 Coherent Light
According to the light excited method,
the light source can be divided into：
Heat light source: exciting light utilizing
heat energy (Incandescent lamp)
Cold light source: exciting light utilizing
chemical energy (such as phosphor), power
(such as fluorescent lamps) or light energy (such
as fluorescent, phosphorescent)
Range of visual
light
 : 400 ~ 760 nm
 : 7.5  10 ~ 4.3  10 Hz
14
14
11-1 Coherent Light
The monochromatic light
The monochromatic light： a sine (cosine) light
wave with a constant frequency and an infinite
length
Although yellow light emitting from a Sodium lamp
0
has a wavelength of 5893 A, it is not monochromatic
strictly.
11-1 Coherent Light
Polychromatic light: light composited by various
frequency light ( such as white light ).
Polychromatic light can produce dispersion
phenomenon through the three prism.
11-1 Coherent Light
II. Coherent Light
1. light emitting
Excited
mechanism of the
state
normal light source：
the spontaneous
radiation of atoms
(molecules).
E  h
En
Ground
state
Atomic energy level transition
and luminescence
11-1 Coherent Light
1
2
P
Light-emitting characteristics of normal light
source: is intermittent, each light to form a
short wavAtomic emission e train, the atoms of
each luminous independently of each other,
each wave train irrelevant.
11-1 Coherent Light
Coherent Light: If light vectors of two light beams
satisfy the interference condition, they are the coherent
light.
The relative light sources are called the coherent light
source.
Sodium lamp A
Two beams
of light are
Incoherent !
Sodium
lamp B
11-1 Coherent Light
Two independent light sources may not be a pair
of coherent light source.
Reason: Atomic emission is random as well as
intermittent, so vibration directions of two columns of
light can not be the same, and phase difference may not
be constant.
Sodium lamp A
Two beams
of light are
Incoherent !
Sodium
lamp B
11-1 Coherent Light
The basic idea of realizing optical interferometer:
The light from each light were decomposed into two
sub-light columns, and then let the two sub-light columns
in the same area and produce interference.
。
11-1 Coherent Light
Two kinds of specific practices:
（1） Method of dividing wave front (Fig. a)，
e.g. , Young’s two slit interference.
（2）Amplitude segmentation（Fig. b）
11-1 Coherent Light
(a) Method of dividing wave front (b) method of dividing amplitude
The light through the two slits is
deco posited two sub-light columns.
A wave train is divided into two
coherent wave trains by reflected of
the two interfaces of film.
11-1 Coherent Light
2 Production of coherent light
Oscillation amplitude
segmentation method
Wave front segmentation
method
s1
Source *
s2
11-2 Young’s Double – slit Interference,
Lloyd Mirror
I. Experiment of Young’s Double – slit Interference
experimental
device
d
s1
s
s2
r1

r
x
r2

o
B
o
d'
d '  d
sin   tan   x / d '
x
Wave path
r  r2  r1  d sin   d
difference
p
d'
11-2 Young’s Double – slit Interference,
Lloyd Mirror
d
实
s验
装
置
s1
s2
x
r  d 
d'
r1


r2

o
B
r
 k
p
x
o
d'
Strengthen

 (2k  1) Weaken
2
k  0,1,2,
11-2 Young’s Double – slit Interference,
Lloyd Mirror
d
s1
s
r1

r2

o
s2
d'
k 
d
x
d'

d
r
( 2k  1)
B
p
x
o
d'

2
bright fringes
k  0,1,2,
Dark fringes
11-2 Young’s Double – slit Interference,
Lloyd Mirror
Distribution of the bright and dark fringes
d'
k 
bright fringes
d
k

0
,
1
,
2
,

x
d'

dark fringes
 (2k  1)
d
2
d 
x  ( 2k  1)
k
2d
( k  1,2,3...)
If white light is used, color fringes appear.
Distance between the
centers of adjacent
bright fringes
d'
x 
d
(k  1)
The bright and dark fringes distribute in eaual distance.
11-2 Young’s Double – slit Interference, Lloyd Mirror
d 'are constant, if changes, then how will x
(1) d 、
changes?
Distance
d'
x 
d
Light
intensity of
each fringe
is equal.
11-2 Young’s Double – slit Interference, Lloyd Mirror
(2)、d ' Keep constant, what relationship
between d and x ?
d'
x 
d
11-2
杨氏双缝干涉实验
11-2 Young’s Double – slit
Interference,
Lloyd 劳埃德镜
Mirror
Example 1 In Young’s double – slit interference
experiment, a Sodium light with the wavelength
=589.3 nm is used as the light source. The distance
d’=800 nm between the screen to the double-slit. The
question:
(1)When the distance between two slits is 1 mm, how
much is the distance between the centers of adjacent
bright fringes?
(2)Assuming the distance between two slits is 10 mm,
how much is the distance between the centers of
adjacent bright fringes?
11-2 Young’s Double – slit Interference, Lloyd Mirror
Known =589.3 nm d`=800 nm
Ask
(1) d=1 mm时 x  ?
(2) d=10 mm时 x  ?
Solution (1) As d=1 mm,
d
x    0.47 mm
d
(2) If d=10 mm, then
d
x    0.047 mm
d
11-2 Young’s Double – slit Interference, Lloyd Mirror
Example 2: Two slits with the gap of 0.2 mm are
irradiated by the monochromatic light, and the distance
between double-slit and the screen is 1 m. Question:
(1)The distance is 7.5 mm from the first order bright
fringe to the fourth order bright fringe in the same side,
what the wavelength of the monochromatic light is?
(2)If the wavelength of the incident light is 600 nm, how
much the distance is from the center of the central bright
fringe to the center of the closest adjacent dark fringe?
11-2 Young’s Double – slit Interference, Lloyd Mirror
Known d  0.2 mm
d ' 1 m
Ask(1) x14  7.5 nm λ  ?
(2) λ  600 nm x' ?
d
Solution (1) xk   kλ , k  0 , 1, 2,
d
d
Δx14  x4  x1  k 4  k1 
d
d x14
λ
 500 nm
d ' k 4  k1 
1 d
(2) x' 
λ  1.5 mm
2d
11-2 Young’s Double – slit Interference, Lloyd Mirror
II The effect of the slit width to the fringes,
space coherence
In the experiment, meanwhile the slit width is
increasing gradually, the fringes on the screen is
going to be faintness gradually, and disappear
finally．
space coherence
11-2 Young’s Double – slit Interference, Lloyd Mirror
III Lloyd Mirror
P
P'
s1
d
s2
M
L
d'
Half wave expense: when light eradiate from the media
with higher light velocity into one with lower light
velocity, the phase of reflected light transits comparing
with that of the incidence light.
π
11-2 Young’s Double – slit Interference, Lloyd Mirror
If n 1 < n 2, call medium 1 optically thinner
medium and medium 2 optically denser medium
incidence light
n
n
1
2
The half wave expense will
Reflected light occur when light eradicates
from the optically thinner
medium to optically denser
Refractive light medium and is reflected by
the interface. The refraction
of light is without half wave
expense.
11-2 Young’s Double – slit Interference, Lloyd Mirror
Example 3 as shown in the figure, there is an electromagnetic wave
receiver on the C position where is with the height h=0.5 m above the lake
surface. As a radio star is rising gradually from the ground level on the left
side, the receiver measured brokenly a serious of maximum value. It is
known that the wavelength of the electromagnetic wave emitted by the
radio star is 20.0 cm, what the azimuth angle is between the radio star and
the ground level as the first maximum value is measured?
2
B
1

2

A
C
h
11-2 Young’s Double – slit Interference, Lloyd Mirror
Solution: calculate wave
path-difference
2
B
1

2

A
r  AC  BC 
C
h

2

 AC (1  cos 2 ) 
2
AC  h sin 
h
 The maximum
r 
(1  cos 2 ) 
r  k
sin 
2 value is
11-2 Young’s Double – slit Interference, Lloyd Mirror
2
1

Note
(
2
k

1
)

B
sin  
C
4h
2

h

k  1 1  arcsin
A
4h
-2
20.0 10 m
1  arcsin
 5.74
4  0.5 m
It is necessary here to explain that is all reasonable
to take   / 2 as calculating the additional wave pathdifference. To take k does not affect the essential of the
problem but choosing the value of k.
11-2 Young’s Double – slit Interference, Lloyd Mirror
IV Optical path, optical path-difference
Light speed in vacuum c  1
Light speed in the medium u  1
u   '
Wavelength in the  '  
n
medium
 0 0 u

1

c n
c  
Wavelength in vacuum
Refractive index of
medium
11-2 Young’s Double – slit Interference, Lloyd Mirror
s1*
s*
2
r1
r2
P
n
t r1
E1  E10 cos 2π (  )
T 
t r2
E2  E20 cos 2π(  )
T '
11-2 Young’s Double – slit Interference, Lloyd Mirror
Wavelength in
the medium
'

n
s1*
s*
2
 wave path- r  r  r
2
1
r1
r2
P
n
difference
 phase
difference
t r2
t r1
  2π (  )  2π (  )
T '
T 
11-2 Young’s Double – slit Interference, Lloyd Mirror
  2π(
r
2
r
 )
1
' 
nr2  r1
 2π (
)

(1) optical path
s1 *
s*
2
r1
r2
P
n
The product of the refractive index of the
medium and the geometry of the light path = nr
The physical meaning: The optical path is the r nr

distance in vacuum which is equally calculated
' 
from the geometric distance of light in the medium
according to the phase difference .
11-2 Young’s Double – slit Interference, Lloyd Mirror
(2) optical path-difference
optical pathΔ  nr2  r1
difference
s1 *
s*
2
r1
r2
Δ
Phase
Δ  2π
difference
λ
P
n
11-2 Young’s Double – slit Interference, Lloyd Mirror
 constructive Δ   k , k  0,1,2,
interference
  2kπ ,k  0,1,2, 

 destructive
interference
Δ  (2k  1) , k  0,1,2,
2
  (2k  1)π , k  0,1,2,
11-2 Young’s Double – slit Interference, Lloyd Mirror
r1
S1
d


P
r2
φ
φ
S2
x
O
D
Young’s Double – slit experiment：when light obliquely
incidents to double slits, optical path is:


Bright

2
k

2
fringes
  d sin   d sin   
（k  0,1, 2...)

x

(2k  1) Dark fringes
d sin ~ d D

2
11-2 Young’s Double – slit Interference, Lloyd Mirror
n1 t1
S1
d O`
S2
φ
φ
r1
P
r2
n2 t2
d`
x
O
When the optical paths 1,2 are respectively inserted
into two transparent medium sheets with the refractive
index of n1 and n2, thickness of t1 and t2, the optical
path-difference will be:
  [( r2  t 2 )  n 2 t 2 ]  [( r1  t 1 )  n1 t 1 ]
 [r2  (n 2  1)t 2 ]  [r1  (n1  1)t 1 ]
11-2 Young’s Double – slit Interference, Lloyd Mirror
(3) The lens does not introduce the additional
optical path difference
A
o
F
B
Focal plane
A
F
B
'
11-3 Optical path thin film interference
Optical path =Refractive index Geometrical distance
= nr
Optical path-difference
=
= n2r
2
n1r
1


Bright fringes

2
k

= 2
（k  0,1, 2...)

(2k  1)  Dark fringes


2
11-3 Optical path thin film interference
In our daily life，the soap bubbles under the Sun and
the oil films on the water show colorful pattern. The
phenomena can be attributed to the interference by the
light beam reflected on upper and bottom surfaces of the
film. Because of the energies of the reflected and
transmitted are from the incidence，so it is amplitude
segmentation interference.
S
i
n1
A
n2
1
i
2
D
r r
C
B
e
n1
n2  n1
11-3 Optical Oath, Film Interference
screen
Monochromatic
light sources
Equal inclination
interference
lense
S1 *
S2 *
S3 *
e
n1
Film
n2 n2 > n1
n1
11-3 Optical Oath, Film Interference
I Film Interference
L
n2  n1
CDAD
sin i n2

sin  n1
1
M1
n1
n2
M2
n1
P
2
i
D
A

3
C
d

B
4
E
5
11-3 Optical Oath, Film Interference
Δ32  n2 ( AB  BC )  n1 AD 
AB  BC  d cos γ
AD  AC sin i
n2  n1
1
 2d  tan   sin i
M1
n1
n2
M2
n1

2
L
P
2
3
D
i
C
A 

E
B
4
5
i
d
11-3 Optical Oath, Film Interference
2d


2
Δ32 
n2 1  sin    2n2 d cos  
cos 
2
2
Optical path-difference

2
2
2
Δr  2d n2  n1 sin i 
of reflected light
2

Δr 

k constructive
(k  1,2,)
(2k  1)

destructive
2
(k  0,1,2,)
n2  n1
2
1
M1
n1
n2
M2
n1
i
C
d

B
4
P
3
D
A 
L
E
5
11-3 Optical Oath, Film Interference
Δr  2d n  n sin i   / 2
2
2
2
1
2
Optical path-difference
of the transmission light
n2  n1
2
1
M1
n1
n2
M2
n1
i
C
d

B
4
E
5
Δt  2d n  n sin i
2
2
2
1
2
P Note: the interference of the
3
D
A 
L
According to
specific situation
transmission light and of the
reflected light is complement
with each other. This satisfies
the law of conservation energy.
11-3 Optical Oath, Film Interference
When light vertically incidents, i  0

If
n2  n1
Δr  2dn2 

2
n1
n2
n1
If n3  n2  n1
Δr  2dn2
n1
n2
n3
11-3 Optical Oath, Film Interference
  =  ( i ) Equal inclination interference fringe
The optical path difference is just related
with the incident angle, which means that the
same level fringe of the same level has the same
incident angle. Such interference is equal
inclination interference.
 Additional items /2 ：
Half wavelength loss exists only on the upper or bottom surface
of thin films, and the optical path difference should include the
additional item /2；If there is no half wavelength loss on both the
surfaces or both have half wavelength loss, then no additional item
/2 should be calculated.
11-3 Optical Oath, Film Interference
Example The oil (refractive index n1=1.20) leaked
out by an oil tanker pollutes an ocean area, and forms a
thin dirty oil film on the surface of ocean (2=1.30).
(1) If the sun is just in the sky above this ocean area, a
pilot observes the just down direction from the helicopter,
and the thickness of the oil layer faced to him is 460 nm,
what is the color of the oil layer he watched?
(2) If a diver dives under the water in this area, what is
the color of the oil layer he observed when he looks a his
up direction?
11-3 Optical Oath, Film Interference
Known n1=1.20
Solution (1)
n2=1.30
d=460 nm
Δ  2dn  k
r
1
2n1d

, k  1,2,
k
k  1,
  2n1d  1104 nm
k  2,
  n1d  552 nm
k  3,
2
  n1d  368 nm
3
Green
11-3 Optical Oath, Film Interference
amaranthine
(2)the optical path difference of the transmission
light is Δt  2dn1   / 2
2n1d
k  1,

 2208 nm
1 1/ 2
2n1d
k  2,

 736nm Red
2 1/ 2
2n1d
k  3,  
 441.6nm Purple
3 1/ 2
2n1d
k  4,

 315.4 nm
4 1/ 2
11-3 Optical Oath, Film Interference
The reflection and transmission increasing film
Thin film interference can improve
the optical transmittance.
11-4 Wedge, Newton’s Ring
I Production of a wedge
As shown in the Figure，a wedge can be formed by two
transparent glass. If the two glass are in air ambient，an air
wedge is formed; If it is placed in translucent liquid, then it is
a liquid wedge. The interferences by wedges are called wedge
interference which is a kind of equal thickness interference.
A wedge formed by two inclined glass
11-4 Wedge, Newton’s Ring
II Optical path of
n
T
n1
n1
L
S
M
Angle of
wedge 
D
b
d
Optical path difference of the
two reflected light by the up
and down surfaces of the
wedge with thickness d is：
Δ  2nd 

2
11-4 Wedge, Newton’s Ring
The interference condition of wedge
Δ  2nd 
Δ

2
k, k  1,2,
n
n1
n1
Bright fringe

(2k  1) , k  0,1,Dark fringe
2
θ
A
d
 ： Angle of wedge
b
dk
θ
dk+1
The dark (light) Kth fringe
11-4 Wedge, Newton’s Ring
Disscussion
n1  n
b
(1) The edge d  0
n
D
n / 2

L
n1
b
Wedge interference
 dark fringe.
Δ
2 1 
（Bright fringe
d
(k  )
2 2n
k 2n
（dark fringe）
11-4 Wedge, Newton’s Ring
(2)Thickness difference of
the adjacent bright
(dark)fringes
n1  n
b
kλ
dk =
2n
n
D
n / 2

L
n1
dk+1 =
Wedge interference
2n
d d 
k 1
b
( k +1 )λ
k
 D L

2n


n
2
n 2

b
11-4 Wedge, Newton’s Ring
(3)Fringe distances
n1  n
b
The distance of two adjacent
dark (bright) fringes
n
λ
b
 2n sin θ
L
D
n / 2
n1
b
Wedge Interference

b
2 n
n

D L
L
2b
2nb
11-4 Wedge, Newton’s Ring
(4 )The shift of the fringes
λ
b
2n sin θ
11-4 Wedge, Newton’s Ring
Discussion：
λ
b
2n sin θ
1. With wedge interference, small angle θ、small
thickness and the wavelength of incidence can be
meausured.
2.
 = ( e )
Optical difference is a function
of film thickness
Equal thickness interference：To the same order fringe，
the film thickness should be the same.
3. Question：dark fringe or bright fringe？
11-4 Wedge, Newton’s Ring
The applications of wedge interference
l  N
(1)Interference of dilatometer
l
l0

2
(2)Measure film thickness
n1
n2
SiO2
Si
eN

2n1
e
11-4 Wedge, Newton’s Ring
(3) The surface inspection of the optical component
e
b
b'
b 
e 
b 2
'
1  
  
3 2 6
11-4 Wedge, Newton’s Ring
(4) Diameter of thin wires
Air n  1
n1
n1
n
L
b
d
 L
d 
2n b
11-4 Wedge, Newton’s Ring
三
Newton’s Ring
A convex A with one flat surface, which
curvature R is very big, touches a flat glass B，as
shown in the figure，so that forms an air wedge
which is called Newton’s ring film.
Optical path
difference
Δ  2d 
A
d
B

2
11-4 Wedge, Newton’s Ring
The experimental setuup of Newton’s Ring
Microscope T
L
S
R
r
M beam
splitting
lens
d
Interference
pattern
11-4 Wedge, Newton’s Ring
With monochromic parallel light incidence, concentric circles with
their center O can be observed on the convex surface，which are
called Newton’s ring.
A small wedge can be regarded in every part of the wedge,
however, their inclinations are not the same at different places,
leading to unequal distance of the fringes which are denser in
inner circles and sparser in outer part.
λ
b
2n sin θ
2
The Newton’s ring is still equal thickness
interference, which means the film thicknesses of
bright and dark fringes obey the equal thickness
interference rules.
11-4 Wedge, Newton’s Ring
Optical
path
difference
Δ
Δ  2d 
k (k  1,2,)

2
Bright
fringe
1
(k  ) (k  0,1,) Dark
2
fringe
R
r
d
11-4 Wedge
r  R  ( R  d )  2dR  d
2
2
2
Newton’s Ring
2
 R  d  d  0
2
R
r

r  2dR  ( Δ  ) R
2
1 Radius of bright
r  (k  ) R
fringe
2
r  kR Radius of dark
fringe
d
11-4 Wedge, Newton’s Ring
Diss
cuss
ion
Radius of
bright ring
Radius of
1
r  (k  ) R (k  1,2,3,)
2
r  kR (k  0,1,2,)
Dark ring
(1) On the reflection side，It is dark or
bright of the center point？On the transmission
side，how about the center point?
(2)It belongs to equal thickness，but the
fringe distances are not uniform，why？
11-4 Wedge, Newton’s Ring
(3) Put the setup in liquids with
n  1,
how about the fringes?
(4) Application example: measuring the
optical wavelength，convex quality inspection，
and curvature radius etc..
Workpiece
Standard unit
11-4 Wedge, Newton’s Ring
If the Newton’s Ring setup is put into other ambient (e.g. water), then
k 
  2nd   
2 (2k  1)

Bright
Dar k
r
d= R
2
2
r
r
k
k
R
 (2k  1)
2n

kR
n
(k  1, 2, 3...) Bright
(k  0,1, 2,...) Dark
11-4 Wedge, Newton’s Ring
Measuring the curvature radius
R
r  kR
2
k
r
rk2m  (k  m) R
R
r
m
2
k m
r
2
k
2r
11-4 Wedge, Newton’s Ring
Summary
(1) The interference pattern are tracing point
aggregation with the same optical path difference,
i.e., those with the same film thickness.
k  1
d
d 

2n
11-4 Wedge, Newton’s Ring
(2)The fringe distance increases
linearly with the film thickness.
(3)Dynamic analysis of the fringers
（ n,  , variation）
11-4 Wedge, Newton’s Ring
（4） The additional optical path needs to be
analysis specifically.
n
n
n1
n3
n2
n1  n2  n3
11-5 Michelson Interferometer
一
The structure drawing of Michelson interferometer
Reflect mirror
M1  M 2
M1
M1movable rails
Monochromic
light source
Reflec
tion
Mirror
M2
Beam spliting G1
G1//G 2
Compensation plate G 2
M1 , M 2
45
11-5 Michelson Interferometer
M 2 image M'2
d
Reflection M 1
mirror
M1  M 2
Monochromic
light source
Reflec
tion
Mirror
G1
G2
Optical path difference
M2
Δ  2d
11-5 Michelson Interferometer
M'2
Reflection M 1
mirror
when M 1 is not
perpendicular to M 2 ,
The setup generates
wedge patterns.
Monochromic
light source
Reflec
tion
Mirror
G1
G2
M2
11-5 Michelson Interferometer
二 The main features of Michelson
Interferometer
(1)seperation of two light beam；
(2)tunable of the optical path differentce.
M'2
M1
Moving M1
d
d
d  k
M1
Moving
Distance
G1
G2
M2

2
The
number of
moving
fringes
11-5 Michelson Interferometer
 Moving of the
interference fringes
when the distance
between M1 and M2 ，the
circular interference fringes
grows from the center of the
circle and the patterns
become denser; when the
distance decreases，the
pattern shrinks and becomes
sparser.
11-5 Michelson Interferometer
M'2
M1
d
n
The optical
difference
Δ  2d
The optical path after the
M 2 dielectric slice insertion
Δ  2d  2(n  1)t
'
G1
G2
t
The change of the optical
difference
Δ  Δ  2(n  1)t
'
11-5 Michelson Interferometer
M'2
M1
n
G1
2(n  1)t  k
d
M2
The moving number of the
interference fringe
The thickness of the
inserted slice
G2
t
k 
t

n 1 2
11-6 The Diffraction of Light
一、The diffraction phenomena of light
Screen
Screen E
Diffracted
light
Shadowed
K
S
*
Large slit (much larger
than wavelength)，light
is propagating
straightly
a
b
Small slit
(comparable with
light wavelength)，
Diffraction
11-6 The Diffraction of Light
二、
Huygens-Fresnel principle
We introduced Huygens principle which can explain refraction,
reflection and scattering of mechanical waves. Huygens principle
can explain diffraction of light qualitatively, however, it can not
quantitatively demonstrate the light intensity distribution.
After studying light interference, Fresnel, according to the
principle of the wave superposition and interference, raised the
concept of ‘wavelet coherent addition’. He thought, wavelets
emitted by each point on a same wave front are coherent, and as
propagating on the certain point in the space, the result of which
each wavelet makes the coherent addition determine the wave
amplitude on the place. This developed Huygens principle is named
as Huygens-Fresnel principle.
11-6 The Diffraction of Light
e
S

S
r
P
*
St wave front
S ：unit part on
wave front
(wavelet)
s
Wavelet dS induce the amplitude 
at P point,
r
which is also related with 。
Huygens-Fresnel principle：the wavelet from
the same wave front can interfere with each
other when they meet at some point in space.
11-6 The Diffraction of Light
Light osillation at P point（the mathematic form）is：
(
)
K
θ
C
E (P ) = 
(ω t
cos
S
r
2π r ) dS
λ
The formula is Fresnel diffraction theory
integration equation. Generally speaking,
the integration is very complicated. In
college physics, we only require to master
the Fresnel half wave zone approach.
Key of Huygens-Fresnel: interference of
wavelets. Diffraction originates from
interferences of infinite wavelets.
11-6 The Diffraction of Light
三
Fresnel diffraction and Fraunhofer diffraction
The setup to observe light diffraction includes three parts:
source、diffraction objects（slit or hole etc.）、screen。The
diffraction can be classified into two types according to
position of the three parts.
Fresnel diffraction
S
Slit
P
Source、screen and slit: finite
distance
（At least one non-parallel light
beam of incidence and emergence）
Fraunhofer diffraction
Slit
P
Source、screen and slit: infinite
distance
（Parallel light beam of incidence
and emergence）
11-6 The Diffraction of Light
Setup of single-slit diffraction
Screen
屏幕
L1
K
L2
S
*
Single-slit Fraunhofer diffraction experimental setup.
The light from source S is changed to parallel light
beam via convex L1 and perpendicular incident on the
single-slit. The diffracted light is then focused on the
screen, where diffraction fringes can be observed, via
convex L2.
11-7 Single slit interference
Fraunhofer single-slit
diffraction
b
L
R

A
f
P
Q
o
B
C
b sin 
The wavefronts of incidence reach the slit，then the points in the slit forms
new wavelet sources which emit wavelets with the same initial phase. The
wavelets propagate in different directions and then focuse on the screen. For
example, the wavelet in θ direction focuse on screen Q point.
Θ is diffraction angle。The optical paths from different wavelet
propagating in θ direction is different. There exists optical path
differences which determine the intensity at Q point.
11-7 Single slit interference
Fraunhofer single-slit
diffraction
b
R
L

Diffraction
angle
f
P
A
Q
o
B C
（Diffraction
b sin 
angle
 ：above the
horizontal position，otherwise negative）
Fresnel wave
zone approach
BC  b sin    k

2
(k  1,2,3,)
11-7 Single slit interference
Half wave zone approach
R
A
A
b

L
P
Q
A1
Slit length
b sin   2k 
A
o
C
B
2
B
/2
R
 L
A
P
Q
A1
b
B
b sin   (2k  1) 
2
k  1,2,3,
A2
B
o
C
/2
11-7 Single slit interference
R
A

L
A1
A2
B
C
/2
P
Q BC  b sin 

 k
o
2
（ k number half wave
zones）
Center of central bright fringe
b sin   0

b sin   2k   k (dark fringe) 2k half zones
2

b sin   (2k  1) (bright fringe) 2k  1
wave zones
2

b sin   k （between dark and bright）(k  1,2,3,)
2
11-7 Single slit interference
二
Light intensity distribution

b sin   2k   k （dark fringe）
2 
b sin   (2k  1)
（bright fringe）
2
I


3 2
b
b


b
o

b
2

b
3

b
sin 
11-7 Single slit interference
二
Light intensity distribution
b
S
L1
 L2
R

I
x
P
f
O
x
When is  small，sin   
x  f


2
3
b
b


3 f 2 f
b
b


b

 f
b
o

b

f
b
2

b

2 f
b
 sin 
3
b

3 f
b
x
11-7 Single slit interference
Discussion
b sin   2k

  k （dark fringe）
2 
b sin   (2k  1)
（bright fringe）
2
11-7 Single slit interference
(1)The distance between
1st dark fringe and the
central bright fringe
x1  f 

b
f
R  L
The diffraction
angle of 1st dark
fringe.
1  arcsin

b
P

b
f
o
x
11-7 Single slit interference
1  arcsin
The diffraction angle of
1st dark fringe
 一定
b increase， 1
decrease

b

 0, 1  0
b
Straight propagation
b
1
decrease， increase
π
b  ,  
2
1
Maximum diffraction
11-7 Single slit interference
st
(2)Central bright fringe （between the 1 dark fringes）


  sin  
b
b


Distance range 
f x f
b
b
Angle range
The width of central fringe
l0  2 x1  2

b
f
11-7 Single slit interference
How about the central bright fringe if the slit
width changes?
11-7 Single slit interference
How about the diffraction if the incidence
wavelength changes?
With larger  and bigger  1 ，it is
obvious to observe diffraction.
11-7 Single slit interference
(3)the width of fringe（the distance of adjacent orders）
b sin   2k

  k （dark fringe）
2 
b sin   (2k  1)
（bright fringe）
2
l   k 1 f   k f 
 f
b
The bright fringe
width except the
central part
11-7 Single slit interference
(4) Dynamic changes of single-slit diffraction
Up and down movements of single-slit，the diffraction
pattern does not change according to the convex imaging
principle.
R
f
o
With up movement of the
slit，the zero order
bright fringe is still
on axis of the convex.
11-7 Single slit interference
(5)the optical path calculation non-normal incidence
Δ  DB  BC
 b(sin   sin  )


A
b
D
（The central fringe
move downwards）
B
C
11-7 Single slit interference
Δ  BC  DA
 b(sin   sin  )
D
b

（ The central fringe
move upwards ）

A
B
C
11-8 Diffraction grating
Double-slit interference and single-slit diffraction can
not be used in high-accuracy spectra measurement due
to the small distance between fringes and low intensity.
If we have a fence-tape optical element which includes
many slits with equal distances and is identified as
diffraction gratings. We can achieve larger distance, very
tiny and higher intensity fringes which is advantageous
for measurement with high-accuracy.
d
b`
b
11-8 Diffraction grating
一
Gratings
Slits with equal width、equal distance.
The width of transparent parts（slits ）in gratings are
represent by b, and that of opaque parts is b`. And b+b`，
i.e. the central distance of the slit is grating constant
which is d（as shown in the Figure）
In practical grating，several dozens and
even thousands of grooves are etched
every millimeter (N/cm)，which leads d
the order of microns.
d=1/N
d=( b + b` ) ~
~ 10-5 ~ 10-6 m
。
b
b`
d=b+b`
11-8 Diffraction grating
二、Grating equation (Diffraction pattern is mixed effect of
interference of
diffraction)
(b +b` ) sinθ
Screen
θ
b`
b
0
θ
x
p
f
(b+b`)sin
The optical path difference of
adjacent slits.
11-8 Diffraction grating
Optical path difference of
adjacent slits
Δ  (b  b' ) sin 
Locations of bright fringes
(b  b' ) sin    k
(k  0,1,2, )
Grating equation
The optical path difference of two adjacent slits
equals to multiple number of wavelength, leading
to interference enhancement and bright fringes.
11-8 Diffraction grating
Light intensity distribution
Slit number N = 4 the diffraction intensity
distribution of grating
The envelope line is light
intensity of single-slit
Central
diffraction
bright
fringe
k=-6
Bright fringe
Dark fringe
k=-4
k=-2
k=0
k=2
k=4
k=6
k=-1
k=3
k=-5
k=1
k=5
k=-3
11-8 Diffraction grating
More slits in gratings，much thinner of the fringes.
(a)1条缝
(d)5条缝
(b)2条缝
(e)6条缝
(c)3条缝
(f)20条缝
11-8 Diffraction grating
When N is large, i.e. many slits
The light intensity distribution:
I
 3  2  
0

2
3
(b  b' ) sin 
11-8 Diffraction grating
(b  b' ) sin   k
(k  0,1,2,)
the maximum order of the fringes
k
sin  k  
b  b'
π
  ,
2
k  kmax
b  b'


11-8 Diffraction grating
(b  b' ) sin    k
(k  0,1,2,)

k  1, sin  k 1  sin  k 
b  b'
Thinner the grating is, narrower width is, farer the
bright fringes is .
Fixed  ，b  b' decreases， k 1   k
increases.
Larger wavelength is ，farer seaperation of the
bright fringes.
'
 increases，
Fixed b  b，
increases．
k 1
 k
11-8 Diffraction grating
With oblique incidence，the grating equation changes to：
  (b  b)(sin   sin  )  k
When incidence and diffracted light are at the different side
of the grating normal，add；otherwise minus.
P
A
φ φ
d=b+b`
θ
B
θ
O
三、Grating spectra
Monochromic light is diffracted by
gratings, forming a series of linear main
bright fringes called linear spectrum. If
the incident light are polychromatic, the
positions of main bright fringes of the
same order at different wavelength are
different，the diffraction intensity which
is called grating spectrum expands
according to wavelength.
Third spectrum
Second spectrum
First spectrum
(b  b' ) sin    k
Polychromatic
( k  0,1,2,)
Screen
0
φ
x
f
11-8 Diffraction grating
Classification of diffraction spectra
Continuous spectrum：Broiling solid
Linear spectrum：the gas in a
discharge tube
Zonary spectrum：molecule spectrum
11-8 Diffraction grating
四 、Order shortage
Due to single-slit diffraction，there is no bright fringes at
the positions where should have interference enhancement. The
phenomena is called order shortage.
The two requirements should be satisfied for order shortage：
( b +b` ) sin = kλ
b sin = k´λ
Interference enhancement by slits
Diffraction minimum by single-slit
The condition of order shortage：
( b + b` ) k
=n
=
b
k´
11-8 Diffraction grating
if (b+b`) equal integer
，the position of kth bright
b
fringe coincide with that of k’th dark fringe，then
kth main bright fringes disappear，leading to order
shortage.。
(b+b`)
，2、4、6、8…… order shortage
b =2
(b+b`)
b =3
，3、6、9、12…… order shortage
Order shortage
1st dark
fringe
k=-4
k=-6
k=-5
If：
3rd main bright
fringe of
grating
diffraction
(b
k=-2
k=-3
+b ) =
b
k=0
k=-1
3
1
Order shortage
k=2
k=1
k=4
k=3
k=6
k=5
k Order
= k´ ，shortage k = ± 3,6,9,...
：
The end