Download Chapter 8a Wave Optics

Chapter 8
Wave Optics (1)
(May 11, 2005)
A brief summary to the last lecture
1. The structure of the eye
The eye is nearly spherical in shape & 1inch in diameter.
Cornea (角膜)
(1.376)
Crystalline lens
Aqueous humor
Retina
Yellow spot
(水状液，1.336)
Pupil
Iris (虹膜)
Vitreous humor
Ciliary muscles (睫状肌)
The cross-section of human eye
Optic nerve O
2. Vision of the human eye
vision of the human eye 
1

from 0.1 ~ 1.5
 is the viewing angle and in unit of minute.
3. Corrective eyeglasses for visual defects
(1) Myopia (nearsighted);
(2) Hyperopic eye (farsighted);
(3) Astigmatic eye (散光眼)
4. Optical instruments used in medicine
(1) Magnifying glass
(2) Compound microscope
(3) Fibrescope
25

f
s 25
M 
f1 f 2
Wave optics (part 1)
1. The corpuscular (微粒) theory of light (Until
the middle of 17th century , Newton (1642-1727)
supported).
2. Ray optics can explain many of the properties
of light, but there exist many other interesting
and beautiful effects that cannot be explained
by the geometric optics. For example,
Experiments show that light bends around
corners.
3. The wave theory of light (Huygens (1629-95))
4. Interference effects of light were first observed
by Thomas Yong in 1800.
8.1 Interference (干涉) of light
• Interference
of wave motion, What is the
phenomenon?
• Two coherent waves should be satisfied
with the three conditions: what are they?
(1) the same frequency
(2) the same vibrational direction
(3) the same initial phase or constant
phase change.
A chart of electromagnetic spectrum
Visible light (very
approximately):
400~450 nm Violet
450~500 nm Blue
500~550 nm Green
550~600 nm Yellow
600~650 nm Orange
650~700 nm Red
8.1.1 Optical length (optical length, distance, path)
In Chapter 4, we learned that the phase difference
of two waves are expressed as
   2  1  2
x1  x2

Light is also a part of electromagnetic waves. The
light interference follows the rule of wave
interference as you know 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. Light path
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 light path 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:
opitacl length = n0 S = n0 c t = ct
In medium:
c
optical 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
light path 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, L is
the geometrical length in the medium. Also we can
use the similar formula to calculate the phase
difference
2
 
0

8.1.2 Young’s double-slit experiments
Slits are 0.1~0.2 mm wide, separation of two slits < 1mm
20~100cm
1~5m
1. Coherent conditions of light:
(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.
(3) The openings are small in comparison with the
wavelength of the incident light.
(4) The distance between the two openings is not too
big compared with the wavelength of the incident
light.
2. Yong’s formulas for bright and dark
fringes
P
S1
S0

d

x
O
S2 B
L
In the above figure, S1P = BP, the light path
difference is S2B = . Therefore,
OP
x
  d sin   d tan   d
d
L
L
• what is the constructive conditions for
two waves?
• by phase difference
• by path difference
• what is the initial phase change in doubleslit interference experiment ?
•how about total phase change? What does it
depend on?
Total phase change is therefore caused by
the light path changes.
• Constructive interference
According to the interference theory of wave
motion, whenever the path difference is an
integer multiple of the wavelength,  = m, the
constructive interference or reinforcement
interference should occurs as long as light is
wave. Therefore,
For bright fringes:
d sin   m
x
d  m
L
∴
m
sin  
d
m
xm 
L
d
• Destructive interference
On the other hand, the opposite phenomenon occurs
that the two light waves are cancelled each other.
This condition is called destructive interference or
cancellation and in this case, what is the light path
difference should be equal to? 陈文灯
  d sin   (m  )
1
2
(m = 0, 1, 2, …)
So for dark fringes:
1
sin    m  12 
d

x
1
d   ( m  2 )
L
1
xm    m   L
d
2
• The spacing of two bright or dark fringes:
(m  1)
m

x  xm 1  xm 
L
L L
d
d
d
Yong’s experiments show that all the above
formulas can describe the phenomena
observed in his experiments very well, so the
wave property of light is proved.
• The analysis of the results
x  xm 1  xm 

d
L
1. The spacing between two dark or
bright fringes is independent from m,
so they are equally spaced.
2. As  is small, so d cannot be too big,
otherwise, they cannot be distinguished.
3. What will you get if you use the sun
(white) light as a light source? 陈善源
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,
5 m, So we have:
m
xm 
L
d

d = 6.91×10-4 m, m = 7, L =
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.
8.1.3 Lloyd’s mirror
L
Lloyd’s mirror is an optical instrument 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-wavelength
lost.
• It occurs when the light wave initially
traveling in an optically thinner medium (光
疏介质) is reflected by an interface with an
optically denser medium (光密介质).
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
shares a common feature, the interference of light rays
reflected from opposite surfaces of a thin transparent film.
Observer
Ray 1 and ray 2 produce the
1
interference. The light path length
2
difference of 1 and 2 depends on
a
c
the thickness of the film.
n
d
b
Ray 1 has an abrupt phase change at point a where the light
initially travels in an optically thinner medium and is reflected
at the interface with an optically denser 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
2 
  2
 
2nd   

 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:

That is
1

  2nd    m  
2 
2
2nd  m

m
d
2n
m = 1, 2, …
The condition for constructive interference is
  2nd 

2
d  (2m  1)
 m

4n
m = 1, 2, …
m = 1, 2, …
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: what is the condition for
destructive interference in such a case? *
2nd  m
m = 1, 2, …
2nd 2 1.33  550 1.463 103



(nm)
m
m
m
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
)
m  12 2m  1
m  1,   2926nm; m  2,   975nm(orange)
m  3,   585nm; m  4,   418nm(violet )
m  5,   325nm;
8.1.5 Equal thickness interference,
Generally speaking, the abrupt phase change occurs at
one of the surface of the wedge. So it is easy to get the
difference of light path length.
  2ne 

Interfering rays
Incident ray
2
Glass plate
The condition for destructive
interference is simpler
2ne  m
e
Zero-order dark fringe
m = 0, 1, 2, …
Air wedge
e
L
,
sin 
e  e k 1  ek 


L

2n sin  2n

2n
Example 2: two microscope slides each 7.5cm
long are in contact along one pair of edges while
the other 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.
Solution: let the air thickness e corresponding
the mth-order dark fringe and e1 to the (m+1)thorder dark fringe. As the refractive index of air
is 1, we can write out:
2e = m,
2e1 = (m+1) 
It is easy to find the
spacing
of
two
neighbor fringes by
deleting m from
above equations. We
have
e1  e 

2
C
x
A
e1
e
0.012mm
B
7.5cm
Zero-order dark fringe
From similar triangles, we
know
AC
x
AC

 x
(e1  e)  1.97mm
BC e1  e
BC
Newton’s rings
If the convex surface of a lens with large
radius is placed in contact with a plane glass
plate, a thin film of air is formed between the
two surfaces. The thickness of this film is
very small at the point of contact, gradually
increasing as one proceeds outward. The loci
(locus, 轨迹) of points of equal thickness are
circles concentric with point of contact.
At such a case, the difference of lightpath-length is
  2ne 

2
Where e is the thickness of air film, /2 is
from the half-wavelength lost for the two
rays considered. The condition for bright
fringes is
The condition for bright fringes is

1
2ne   k  ek  2k  1
2
2
The condition for dark fringes is


k
2ne   ( 2k  1)  ek 
2
2
2n
On the other hand, we could also
calculate the radii of bright and dark
rings.
C
r  R  ( R  e)
2
2
 2 Re  e
2
2
As R>>e, e2 can be
dropped. So we
have:
R
R-e
r
e
O
The radius for kth bright ring is
(2k  1) R
rk 
2
k  1, 2, ...
The radius for kth dark ring is
rk  kR
k  1, 2, ...