Download Chapter 8 Wave Optics

Chapter 8: Wave Optics
Chapter 8
Wave Optics
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We have already seen that ray optics can explain many of the properties of
light, but many other interesting and beautiful effects exist that can only be
explained by wave optics.
Experiments show that light bends around corners. The experiments first
performed in the early years of 19th century. But this seems against our daily
experience.
§8.1 Interference of light
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•
•
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We studied the interference of wave motion in chapter 4. It is the basic
property of wave motion.
Light is actually electromagnetic waves. It is difficult for us to see the light
interference by naked eyes.
From chapter 4, we know that interference can only be produced by waves
come from coherent sources and the two coherent waves should be satisfied
with the three conditions: they should have (1) the same frequency (2) the
same vibrational direction and (3) the same initial phase or constant phase
change.
For interference of light, we still need these conditions.
8.1.1 Young’s double-slit experiments
Generally it is hard to obtain the coherent condition with the common light source.
But we will assume some conditions:
(1) The wave has fixed wavelength. The incident beam should be monochromatic
(单色的).
(2) The secondary wavelets that originate from the two small openings are in
phase at their point of origin in the openings。 Thus，the incident light beam
must reach the barrier as a wave front of constant phase。 Such a beam is
called coherent。
(3) The openings are small compared with the wavelength of the incident light.
(4) The distance between the two openings is not too large compared with the
wavelength of the incident light.
Yong’s apparatus is shown in the figure. Monochromatic light issuing from a
narrow slit S0 is divided into two parts by falling upon a screen in which are cut two
other narrow slits S1 and S2 , very close together. The dimensions in this figure are
distorted for clarity. The distance from the source slit S0 to the screen containing S1
and S2 is 20cm to 100cm. The distance from the double-slit screen to the final screen
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is usually from 1m to 5m. The slits are 0.1mm to 0.2mm wide and the separation d
of the two slits is less than 1mm. In short, all slit widths and slit separations are
fractions of millimeters; all other distances are hundreds or thousands of millimeters.
From the interference of wave motion, we know where the interference occurs should
only depends on the ray path change.
screen
P
S1

S0
O
d
S2

 =d sin = x/L
L
Ray path change is exactly the phase change as their initial phase difference is
zero. It is easy to see that the simplest case of constructive interference is at point O. it
is opposite the midpoint between the two slits. The crests arrive simultaneously at O
to produce a bright fringe because the distance S1O is equal to S2O. The fringe is
called zero-order bright fringe. From the figure we know that the path change at point
P approximately is:
x
  d sin   d
L
Where x is the distance from O to P. It is known that the change of the ray path is
the change of their phase. Whenever the path difference is an integer multiple of the
wavelength,  = m (m = 0, 1, 2, …), the constructive interference or reinforcement
occurs. On the other hand, the opposite phenomenon occurs that the two light
waves cancelled each other. This condition is called destructive interference or
cancellation and in this case, the difference of the light path has to be satisfied with
  (m  12 )
(m = 0, 1, 2, …)
Using the above relations we can calculate the angle  or x for the bright or dark
fringes. So we have
(1) Bright fringes
x
d  m

d sin   m
L
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Chapter 8: Wave Optics
m
m
xm 
L

d
d
Using this formula, the distance between two bright fringes, say, the mth and the
(m+1)th, can be obtained as
(m  1)
m

x  xm 1  xm 
L
L L
d
d
d
It has nothing to do with m. Therefore the bright fringes are equally separated. It is
easy to work out that the spacing between two dark fringes is exactly the same as
above.
The spacing is proportional to wavelength and the distance between the slits and
the screen and has inverse ratio to the interval d of two slits. Therefore if we can
measure the fringe spacing Δx, d and L we can calculate the light wavelength λ.
As the different wavelength corresponds to the different spacing, the fringes will be
not sharp if we use the light which is not monochromatic or even we cannot see the
fringe clearly. Using white light, we should get color fringes on the screen from violet
to red with zeroth-order white.
Example 1: in an interference pattern from two slits, the seventh-order bright fringe is
32.1mm from the zeroth-order bright fringe. The double slit is 5 meters away from
the screen, and the two slits are 0.691mm apart. Calculate the wavelength of the light.
Solution: the data we know are
x7 = 3.21×10-2 m,
d = 6.91×10-4 m, m = 7,
L=5m
So we have:
sin  
xm 
m
L
d


xm d 3.21102  6.91104

 634nm
mL
75
Light interference gives us an important method for measuring the wavelength of
light.
8.1.2 Lloyd’s mirror
Lloyd’s mirror is an optical arrangement for producing interference fringes. A slit
is illuminated by monochromatic light and placed close to a plane mirror. Interference
occurs between direct light from the slit and light reflected from the mirror.
It is found that the reflecting light has a π phase change which is called abrupt
phase change. This phenomenon is called half-wave-lost. This phenomenon occurs
when the light wave initially traveling in a optically thinner medium is reflected by an
interface with a optically denser medium.
Read your books and get more detailed explanation.
8.1.3 Optical distance (also called optical length, optical distance, optical path)
In Chapter 4, we learned that the phase difference of two waves are expressed as
x x
  2  1  2 1 2

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Light is also a part of electromagnetic waves. The light interference follows the rule
of wave interference as we saw in Yong’s double-slit experiment.
It is known that light travels at a different speed in different medium. The speed of
light in a medium depends on the refractive index of the medium. From the definition
of the refractive index, we obtain
c
v
n
From the definition of wavelength, which is the period of the wave multiplying the
speed of the wave, we have
cT 0
  vT 

n
n
Where 0 is wavelength of light in free space. So the wavelength of light becomes
shorter in medium. Optical length is not the geometrical length of the light travel and
it is defined as the product of refractive index and the geometrical distance the light
travels.
Let’s have a look at the Optical length at the same period t when it travels in two
different medium. We choose medium one is free space and medium two with
refractive index n.
In free space: light path length = n0 S = n0 c t = ct
c
In medium: light path length = n  L  n  vt  n   t  ct
n
So it is found that though the lights traveling in different medium have different
geometrical path (S and L) at the same period, they have the same optical length. So
when we calculate the phase in medium n, we can use the formula directly

2
0
nL
Again 0 is wavelength of light in free space and L is the geometrical length in the
medium. Also we can use the similar formula to calculate the phase difference
 
2
0

where  is the optical path difference.
8.1.4 Interference in thin films
It is easy for us to see the colored bands in the reflection of light from a thin film
of oil on water and the colors on the reflection of light from a soap bubble. These
phenomenon share a common feature, the interference of light rays reflected from
opposite surfaces of a thin transparent film.
Ray 1 and ray 2 produce the interference. The light path length difference of 1 and 2
depends on the thickness of the film.
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Chapter 8: Wave Optics
1
a
2
c
n
d
Ray 1 has an abrupt phase change at point a where the light initially travels in a less
optically dense medium and is reflected at the interface with a more optically dense
medium. The phase change of  occurs at the upper surface of the film.
It is supposed that the direction of incident light is more or less perpendicular to the
film surface. So the ray 2 has an extra light path length of approximate n*2*d and
ray 1 lost half-wavelength because of reflection. Therefore, we have
 
2

2nd   
2 
  2

 2nd   
 
2 
The difference of light path length is

  2nd 
2
This explains that why the abrupt phase change has a special relation with the
half-wavelength lost.
The condition of destructive interference is:
  2nd 

1

  m  
2 
2
m
m = 1, 2, …
2n
The condition for constructive interference (bright fringes) is
That is 2nd  m
  2nd 


2
d  (2m  1)
 m
d
m = 1, 2, …

m = 1, 2, …
4n
Example 1: A soap bubble 550nm thick and of refractive index 1.33 is
illustrated at near normal incidence by white light. Calculate the wavelengths of the
light for which destructive interference occurs.
Solution: the condition for destructive interference is given by:
2nd = mλ (m = 1, 2, …)
∴
2nd 2 1.33  550 1.463 103



(nm)
m
m
m
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m  1,
  1463nm
(infrared )
m  2,
  732nm
(red )
m  3,
  488nm
(blue  green )
m4
  366nm
(ultraviolet )
In the visible region, the light from both ends of the spectrum is reflected with
destructive interference. We can not see these wavelengths of visible light. The
wavelengths we can see have to be calculated using the constructive condition of
interference.
2nd  m  12 

2nd
2926

(nm)
1
m  2 2m  1
Please read your English text book and find out what wavelength light you can see.
8.1.5 Air wedge
The abrupt phase change occurs at the lower surface of the air wedge. For the wedge
its refractive index is equal to 1. The condition for destructive interference is simpler
2d  m
m = 0, 1, 2, …
Example 2: two microscope slides each 7.5 long are in contact along one pair of edges
are held apart by a piece of paper 0.012mm thick. Calculate the spacing of
interference fringes under illumination by light of 632nm wavelength at near normal
incidence.
Interfering rays
Incident ray
d
Glass plate
Air wedge
Zero-order dark fringe
Solution: let the air thickness d corresponding the mth-order dark fringe and d1 to the
(m+1)th-order dark fringe. As the refractive index of air is 1, we can write out:
2d = m
2d1 = (m+1) 
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Chapter 8: Wave Optics
x
0.012mm
A
d1
d
B
7.5cm
Zero-order dark fringe
It is easy to find the spacing of two neighbor fringes by deleting m from above
equations. We have

d1  d 
2
From similar triangles, we know
AC
x
AC

 x
(d1  d )  1.97mm
BC d1  d
BC
8.2 Diffraction (衍射) of light
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Diffraction is one of the basic properties of wave motion. Diffraction of light
occurs under some conditions.
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There are two kinds of diffractions;
(1) Light emitted from a small source is diffracted by a barrier and the diffraction
pattern shows on a screen which is not far from the barrier. The diffraction is called
Fresnel diffraction.
(2) The parallel rays pass through a barrier and the diffraction pattern shows on a
screen which is at a great distance. The diffraction is called Fraunhofer diffraction.
We discuss the second one only.
8.2.1 Fraunhofer Single slit diffraction(单缝衍射)
Some aspects of Fraunhofer diffraction from a single slit can be deduced easily.
Parallel mono-chromatic light striking single slit of width a gives an interference
pattern on a distant screen as shown in the figure 8.1.
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Medical Physics
θ
a/2 sin
f2
a/2
a/2
P
Figure 8.1 the wavefront is divided into
a large number of narrow strips
We first consider two narrow strips (带), one just below the top edge of the slip and
one just below its center. The difference in length to point P in Fig. 8.1 is (a/2) sinθ,
where a is the slit width. Suppose this path difference happens to be equal to λ/2:

a
  sin  
2
2
or
sin  

a
Then light from these two strips arrives at point P with half-cycle (π) phase
difference, and cancellation occurs. Similarly, light from two strips just below these
two will also arrive a half-cycle out of phase; and in fact, light from every strip in the
top half cancels out the from the corresponding strip in the bottom half, resulting in
complete cancellation.
It is easy to find that when the light path length difference (a/2)sin is equal to the
integer multiple of λ/2, the dark fringe will occur. Therefore, the dark fringe
positions are determined by the equation.
m
sin  
m = 1, 2, 3, …
a
For example, if the slit width is equal to ten wavelengths, dark fringes occur at
sin   101 , 102 , 103 ,
Midway between the dark fringes are bright fringes. We also note that sinθ = 0 is a
bright band, since then light from the entire slit arrives at P in phase. Thus the central
bright fringe is twice as wide as the others, (drawing a curve line to show its intensity).
Finally we have:
m = 1, 2, 3, …, dark fringes
a sin   m
a sin   (m  12 )
m = 1, 2, 3, …, bright fringes
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Chapter 8: Wave Optics
ym
m
 tan    
f
a
 ym 
m
f
a
ym is the distance from the mth-dark fringe to the center bright fringe.
8.2.2 Diffraction by a circular hole
Instead of a single slit of width a, let us now consider a circular hole of diameter
D. According to Huygen’s principle we must add up the radiation or wavelets
originating from all points inside the open circle in order to get the resultant amplitude.
However, the mathematical treatment about this is very complicated and we just
discuss the result of this diffraction. The picture are given in your Chinese text book
on page 213 and English one on page 199. The diffraction pattern consists of a
number of concentric fringes with a maximum intensity in the center. The condition
for the first-order dark fringe is obtained as:

sin 1  1.22
D
This phenomenon is quite important. The resolution of many instrument like telescope
and microscope depends on this.
8.2.3 The diffraction grating (衍射光栅)
Suppose that, instead of a single slit, or two slits side by side in Yong’s experiments,
we have a very large number of parallel slits, all of the same width and spaced at
regular intervals. Such an arrangement, known as diffraction grating, was first
constructed by Fraunhofer.
d=a+b
d is called grating spacing or grating constant. Let’s assume that the slits are so
narrow that the diffracted beam from each spreads out over a sufficiently wide angle
for it to interfere with all the other diffracted beams. Consider first the light
proceeding from elements of infinitesimal width at the upper edge of each opening,
and traveling in a direction making an angle  with that of the incident beam. A lens at
the right of the grating forms in its focal plane a diffraction pattern similar to that
which would appear on a screen at infinity.
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Medical Physics
a
b
d
d

d
ym
L
A
λ
a
d
A
B
2λ
B
θ
θ
b
1
2
First-order maximum when AB=λ, second order maximum when AB=2λ.
It is easy to find that the diffraction grating equation (bright fringes) is
d sinθ = ±mλ
(m = 0, 1, 2, 3, …)
The results are involved in interferences and diffractions. It is known that for the
interference of the double-slit, the fringes are equally bright. But when we consider
the diffraction by a slit, the final pattern actually observed is a combination of both
effects. The interference pattern locates the position of each bright fringe, and the
diffraction pattern from one slit modifies the intensity of each bright fringe.
Diffraction modifies the interference pattern of a grating in just the same way to the
double-slit. The interference pattern determines the position of each bright fringe,
and the diffraction pattern determines its intensity. On the other hand, a grating
produces much sharper bright fringes than a double slit. If a sharp bright fringe of
interference occurs at the position where the first dark fringe of diffraction happens to
be, the sharp bright fringe will be missing. This phenomenon is called order-missing
phenomenon of a grating.
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Chapter 8: Wave Optics
8.2.4 The resolving power of optical instruments
Every wave undergoes diffraction, which set limit on the ability of an optical
instrument to distinguish objects that are close together.
Consider a situation in which two distant point sources of light, such as two stars,
illuminate a small hole. Each beam of light produces its own diffraction pattern on the
screen.. When the two stars have a relatively large angular separation and the hole is
reasonably wide, the two diffraction patterns are separated. We can easily identify the
two patterns as being distinct.
However, if the angular separation of the two stars is sufficiently small, their
diffraction pattern overlap, and identifying the two patterns as being distinct becomes
a matter of judgment.
Resolving power is the angular separation of the two objects. A condition for
identifying diffraction patterns as being distinct was given by the Rayleigh criterion
(准则).
Rayleigh criterion is that the center of one diffraction pattern must come on closer
than the first dark fringe of the other pattern。 Therefore， Rayleigh’s formula for the
resolving power of a circular `aperture of diameter D is：

sin 1  1.22
D
We know that the aperture often is an aperture of a lens in the optical instrument. For
example, in a microscope the aperture is the objective (物镜). So D is the diameter of
the objective. When is bigger, the theta will be small and we have higher resolving
power! It is the same reason for having a big D of telescope.
Example 1: A coherent beam of light from a hydrogen discharge tube falls normally
on a diffraction grating of 8000 lines per centimeter. Calculate the angular deviation
of each line in the first-order spectrum. Do any lines of the second-order spectrum
overlap the first-order spectrum? ( For the hydrogen discharge tube, λrea = 656.3
nm; λblue ----- 486.1 nm, λviolet 1 = 434.0 nm, λviolet 2 = 410.1 nm.)
SOLUTION: Since the grating has 8000 lines per centimeter, the grating spacing is
given by:
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Medical Physics
1
cm  1.25 10 6 m
8000
We can calculate the deviation of each component in turn by using the diffraction
grating equation. According to the grating equation for bright fringes which is
d
d sin   m
(m  0,1,2,)
It is necessary to calculate the angles of the first order fringes and the second order
fringes for the given lights and to compare them to determine whether they have
overlap. The general idea is to compare the biggest angle θ1max of the first-order
fringes and the smallest angle θ2min of second-order fringes. If 1max > 2min, the
overlap happens. Otherwise, no overlap occurs.
(1). First-order
Violet 2
sin 1 

d

410.0 109
   19.2
1.25 106
screen
Zeroth-order
31.7°
grating
Violet 2
Violet 1
Blue
40.1°
Red
Violet 2
Violet 1
Violet 1
sin 1 
Blue
sin 1 
Red
sin 1 

d

d

d

434.0 10 9
 1  20.3
6
1.25 10

486.1109
 1  19.2
1.25 106

656.3 109
 1  31.7
6
1.25 10
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1th order
Chapter 8: Wave Optics
1max  31.7 
So
To determine whether the second-order spectrum for violet 2 overlaps the first-order,
we need only calculate the position of the least deviated line (the smallest diffracting
angle) in the second-order diffractions.
Violet 2
sin  2 min 
2 2  410.0 109

  2 min  41.0
d
1.25 106
It is obvious to see that the biggest angle for the first-order is less than the smallest
angle in the second-order. Therefore there is no overlap happening in such a case.
From the example we know that diffraction gratings can be used to distinguish very
approximate wavelengths, such as violet 1 and violet 2. The two wavelengths appear
as the same color by human eye.
Problems
1.
2.
3.
4.
5.
In an interference pattern from two slits, the fifth-order bright fringe is 22.1mm from the
zeroth-order bright fringe. The double slit is 6 meters away from the screen, and the two slits
are 0.7 mm apart. Calculate the wavelength of the light and the spacing of the fifth and sixth
bright fringes.
In double slit interference, the two slits spaced 0.2 mm apart, and screen at a distance of 1
meter, the third bright fringe is found to be displaced 7.5mm from the central fringe. Find the
wavelength of the light used.
A soap bubble 560nm thick and of refractive index 1.33 is illustrated at near normal
incidence by white light. Calculate the wavelengths of the light for which destructive and
constructive interferences occur.
The wavelengths of the visible spectrum are approximately 400nm to 700nm. (1) Find the
angular breadth of the first-order visible spectrum produced by a plane grating have 6,000
lines per centimeter, when light is incident normally on the grating. (2) show that the violet
of the third spectrum overlaps the red of the second-order spectrum.
In a single slit diffraction, the spacing of the slit is 0.025mm and the wavelength of light
used in the experiment is 450nm. Find the included angle to the first and fourth order dark
fringe.
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