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Part 2 Diffraction of light
Diffraction phenomena of light
§17-6 Diffraction of light
Huygens-Fresnel’s principle
I. Diffraction phenomena of light
E
E
S
S
Condition :
The width of the diffracting obstacle is not very
largely compared to the wavelength.
a >>  ，Diffraction negligible
a >103，Diffraction is not obvious
a ~102 —10，Diffraction fringes appear
a ~时， Diffraction is obvious
a < ， Scatterance (散射)
II. Classification
obstacle
Fresnel diffraction
S
--the source or the
screen or both are
at finite distance
source
from the diffracting
obstacle.
Fraunhofer diffraction
--the source and the
screen are at infinite
distance from the
diffracting obstacle.
S
E
screen
E
III. Huygens-Fresnel’s principle
Every element dS of wave front S is the source
of a secondary spherical wavelet.
The amplitude dA of the secondary spherical
wavelet emitted by dS is proportional to the size
of dS,
1
dA  dS
dS
S

r
n
dA

P
dA 
r
dS
dA  k ( )
r
k()—inclination
factor
dS
S
The light oscillation
at P produce by dS is
n
dA
2r

dy  dA cos( t 
)

P

r
The resultant oscillation of light at P is the
coherent superposition of all spherical
wavelets emitted by all elements on the S
y( P )   dA cos( t 
S
2r

)
Interference appears.
The intensity of light
changes in space.
IV. Parallel beams interference—a simplified
discussion
screen
Coherent superposition
P1
Wave front
Wave
ray
1
a
2
1
Optical axis
1 , 2 – diffraction angle
P2
§17-7 Single slit Fraunhofer diffraction
I. Diffraction device
E
S
II. Distribution law of diffraction fringes
---- Fresnel half wave zone method (半波带法)
BC  a sin 
a
A
A1
A2
A3
B
P

C

f
E
 BC=asin =2( /2)
--The wave front AB is divided
into 2 half wave zones(2个半波带)
A
a 0
The optical path difference
between two corresponding
points on A0 and 0B is /2，
--Destructive interference.
--point P is dark.

B
C
 BC=asin=3( /2)
--The wave front AB is divided
into 3 half wave zones(3个半波带)
The optical path difference
between two corresponding points
on AA1 and A1A2 is /2，
A
A1
A2
a
B

C
They produce destructive interference.
The light oscillations coming from wave front
A2B produce constructive interference.
--point P is bright.
n  1,2
BC=asin =n( /2)

If n is even number (偶数)： --dark fringes
If
n is odd number (奇数)： --bright fringes

--dark
a
sin



2
k


2
k  1,2

 a sin   ( 2k  1)   --bright
2
 If asin integral times of /2, the intensity of
light is between maximum and minimum.
Discussion
Central diffraction
maximum fringe：the
region between the first
positive and negative dark
fringes
Half-angle width：
1
E

 0  1  sin1 
a
2 f
Half width：x  2 f tg   2 f 
0
a
1
--inverse proportion
 x0 
a
 The distance between other adjacent fringes
(bright or dark)
x  xk 1  xk  f tg  k 1  f tg  k
( k  1) k
 f[
 ]
a
a
f 1
 x0

2
a
The intensity distribution of diffraction
fringes:
Most of the light intensity is concentrated in
the broad central diffraction maximum.
I
I0
5

2a
3
2a
3

2a



a
a
5
2a
sin 
[Example]In experiment of Fraunhofer
diffraction from a single slit, f = 0.5m，=5000Å，
the width of the slit a=0.1mm. Find the width
of central maximum, the width of the
secondary maximum.
Solution  the width of
central maximum
x0  2 f tg 1
 2 f sin1
P1
x
0
1
f
2f

a
2  0.5  5000  10

0.1  103
10
 5 mm
The width l of the secondary maximum
equals to the space of the first minimum and
the second minimum.
2 
l  f sin 2  f sin1  f (  )
a a
f

 2.5 mm
a
§17-9 Resolving power of optical instrument
I. Fraunhofer diffraction by circular aperture
Airy disk
E
S
The diffraction angle of first dark ring,
sin 1  0.61

r
 1.22

D
I
 0.610

r
I0
0.610

sin 
r
II. Resolution of optical instrument
E
S1
S2
A2

A1
distinguish
E
S1
S2
S1
A2
A1
Can’t be
distinguished
Minimum
resolving angle

S2
Just
distinguished
E
Airy disk
A2
A1
Rayleigh criterion：Two images are just resolved
when the center of central maximum of one
pattern coincides with the first dark ring of
another.      sin  1.22 
1
1
爱里斑
D
Minimum resolving angle
Resolving power of an optical
1 D
instrument: R  1


1.22 
Improve R：
 increasing D—astronomical telescope with
large radius
 decreasing --electronic microscope
§17-8
Diffraction grating
I. Grating
An optical device which consists of a large number
of equally and parallel slits with same distance.
classification：
Reflecting
Transmitting
grating
grating
d sin 
b
da

E
0
x
P
d  a  b --grating constant
d sin  ---optical path difference between
rays from adjacent slits.
II. The formation of grating diffracting fringes
diffraction + interference
1. The interference of multi-slits (多缝干涉)
The phase difference
between rays from
adjacent slits is
2
  (a  b) sin 

When  =2k ,

A
B
(a  b) sin 
C
The rays coming from all slits are in phase.
----constructive interference
According to  
We get
2


  (a  b) sin    k
d sin    k
k  0,1,2
--grating equation
The principle maximum appears at the direction
with the diffraction angle .
Principle maximum and secondary maximum
N=2
N=4
N=6
Principle maximum
I
I
secondary maximum
I
2.The influence of diffraction by each slit to the
interference fringes
The diffraction patterns of all slits coincide.
The intensity N2
N=3
N=2
N=1
Slit diffraction
Interference of multi-slits
differactin
+
interference
Missing
order

Grating
differactin
-6 -5 -4 -3
-2 -1 0 1 2
3
4
5 6
The missing order phenomenon of grating:
On some direction, if diffraction angle  satisfies,
(a  b) sin    k
--constructive interference
and
a sin   2k '

ab
k
k'
a
2 --minimum of slit diffraction
k' 1,2, 
then the k-th principle maximum will disappear.
-- The k-th fringe is missing order.
III. The incident ray inclination
The optical path difference
of two adjacent rays is

A
B
C

  AB  AC
 d sin  d sin 
The grating function for inclination incidence is
d (sin  sin  )   k
k  0,1,2
IV. Grating spectrum
Third order spectrum
Second order spectrum
First order spectrum
When a polychromatic light (复色光)
is incident a grating, d sin    k
except for the central fringe, all
others principle maximum with
different  have different  for each
same order.
[Example]Two slits with d=0.40mm, the width is
a=0.08mm. A parallel light with =4800Å is emitted
on the two slits. A lens with f =2.0m is put on the slits.
Calculate: The distance x of the interference
fringes on the focal plane of the lens.  The numbers
of interference fringes located in the width of the
central maximum producing by single diffraction.
?
d
a
f
Solution
 For two beams interference, the bright
fringes (principle maxima) satisfies:
d sin    k
the position of k-th order bright fringe is
k
xk  f tg   f sin   f
d
the distance between two adjacent bright
fringes is
f
3
 x  xk 1  xk 
 2.4  10 m
d
 For a single diffraction, the width of
central maximum is
x0  2 f tg 1  2 f sin1 2 f
x0
 10
x
0,1,2,3,4
bright fringes appears
in the width of x0 .
N 9
x0
5
4
3
2
1
0
-1
-2
-3
-4
-5

a
2
 2.4  10 m
x
[Example]A diffraction grating has 500 slits per
millimeter. It is irradiated by Sodium(钠) light
with =0.59×10-3mm. Find  The maximum
order of spectrum can be observed when the
beams of Sodium light are incident normally? 
How many orders of spectrum can be seen when
the incident angle is 300 ?
Solution  The grating constant is
1
3
ab
 2  10 mm
500
(a  b) sin    k
k  0,1,2, 
When  

2
，k gets maximum.
ab
 kmax 
 3.39
Take integral=3

i.e., when the beams are incident normally, the
maximum order which can be observed is 3-th.

 (a  b)(sin   sin )   k
When the incident rays and diffraction rays
are in the same side of the optical axis,
( a  b)(sin 90  sin 30 )
kmax 

0
0
 5.08
Take integral=5
the order numbers which can be observed in
this case is 5-th.
When the incident rays and diffraction rays are
in the two side of the optical axis,
( a  b)[sin 90  sin( 30 )]
kmax 

0
 1.69
0
Take integral= -1
the order numbers which can be observed in
this case is 1-th.
The
spectrum with order -1, 0,1,2,3,4,5can
be observed
1 order

0-th order
A
B
C

5 orders
[Example] A monochromatic light with
=7000Å is incident normally on a grating. The
grating has d= 3×10-4 cm, a=10-4cm. Find
The maximum order of the spectrum can be
observed ?  Which orders are missing?
d sin    k
Solution 
k  0,1,2
For   90 ,
0
kmax 
ab

 4.28
Take integral= 4
The maximum order which can be observed is
4-th.
d sin    k

for interference bright fringes.

a sin   2k 
2
for diffraction dark fringes.
d
for same  , k  k '
a
When k' 1, 2,  , corresponding to k=3,6…
As
kmax  4
The order k=3 is missing.
i.e., the orders of the spectrum that can be
observed are 4-1=3. They correspond with
k= -4,-2,-1,0,1,2,4 ( seven principle maxima)
Missing
order
-4 -3
-2 -1 0 1 2
3
4
§17-10 x –Ray diffraction by Crystal
I. x -Ray
It was discovered by W.K
Roentgen ( a German physicist) in
1895.11.
first x–ray photo: his wife’s
hand.
The
He
got the first Noble Prize of
Physics in 1901 as the discovery
of x-ray.
x-ray:
produced by bombarding a target
element (Anode) with a high energy beam of
electrons in a x-ray tube.
Anode
Cathode
K
A

x-ray tube
It’s
a type of electromagnetic waves with
wavelength ranges about 0.1--100Å, between
Ultraviolet and -ray.
In 1912, a collimated beam of x-ray which
contain a continuous distribution of wavelengths
strikes a single crystal,
x-ray is a wave
the diffraction pattern of x-ray was observed by
German physicist M. Von Laue.
Lead plate
E
crystal
film
Laue (劳厄) spots
x-ray
can be used widely to study the internal
structure of crystals.
Laue
got Noble Prize
of Physics in 1914
because verified that
x-ray is a wave.
II.Bragg equation
and
W.L.Bragg, two
British physicists
( son and father)
took another
method to study
x-ray diffraction.
掠射角
W.H.Bragg,

O 
A
晶面间距d
B
C
晶面
The optical path difference of the two x-ray
beams scattering by the atoms that they locate
in two parallel planes is
  AC  BC  2d sin
They found that when
2d sin  k ，the x-ray beams
produce constructive interference.
k  1,2, 
---- Bragg equation
W.H.Bragg,
W.L.Bragg got Noble Prize of
Physics in 1915 because they found a new
method to study the properties of x-ray.