Wave Optics 1

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Chapter 8
Wave Optics (1) (May 11, 2005)
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A brief summary to the last lecture

• The structure of the eye

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2. Vision of the human eye
vision of the human eye
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 (???)
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4. Optical instruments used in medicine (1)
Magnifying glass (2) Compound microscope (3)
Fibrescope
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• Wave optics (part 1)
• The corpuscular (??) theory of light (Until the
  middle of 17th century , Newton (1642-1727)
  supported).
• 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.
• The wave theory of light (Huygens (1629-95))
• Interference effects of light were first observed
  by Thomas Yong in 1800.

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8.1 Interference (??) 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.
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A chart of electromagnetic spectrum
Visible light (very approximately) 400450 nm
Violet 450500 nm Blue 500550 nm
Green 550600 nm Yellow 600650 nm
Orange 650700 nm Red
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8.1.1 Optical length (optical length, distance,
path)
In Chapter 4, we learned that the phase
difference of two waves are expressed as
Light is also a part of electromagnetic waves.
The light interference follows the rule of wave
interference as you know in Yongs double-slit
experiment.
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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
From the definition of wavelength, which is the
period of the wave multiplying the speed of the
wave, we have
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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.
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Lets 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 optical length
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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
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
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8.1.2 Youngs double-slit experiments
Slits are 0.10.2 mm wide, separation of two
slits lt 1mm
20100cm
15m
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• 1. Coherent conditions of light
• The wave has fixed wavelength. The incident beam
  should be monochromatic (???).
• The secondary wavelets that originate from the
  two small openings are in phase at their point of
  origin in the openings.
• The openings are small in comparison with the
  wavelength of the incident light.
• The distance between the two openings is not too
  big compared with the wavelength of the incident
  light.

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2. Yongs formulas for bright and dark fringes
In the above figure, S1P BP, the light path
difference is S2B ?. Therefore,
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• what is the constructive conditions for two
  waves?
• by phase difference
• by path difference

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• what is the initial phase change in double-slit
  interference experiment ?

• how about total phase change? What does it depend
  on?
• Total phase change is therefore caused by the
  light path changes.

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• 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
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?

• 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? ???

(m 0, 1, 2, )
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So for dark fringes
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• The spacing of two bright or dark fringes

Yongs 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.
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• The analysis of the results

• The spacing between two dark or bright fringes is
  independent from m, so they are equally spaced.
• As ? is small, so d cannot be too big, otherwise,
  they cannot be distinguished.
• What will you get if you use the sun (white)
  light as a light source? ???

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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 3.2110-2 m, d 6.9110-4
m, m 7, L 5 m, So we have
Light interference gives us an important method
for measuring the wavelength of light.
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8.1.3 Lloyds mirror
L
Lloyds 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.
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• It is found that the reflecting light has a p
  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 (????)
  is reflected by an interface with an optically
  denser medium (????).

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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.
Ray 1 and ray 2 produce the interference. The
light path length difference of 1 and 2 depends
on the thickness of the film.
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.
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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 n2d and ray 1 lost
half-wavelength because of reflection. Therefore,
we have
The difference of light path length is
This explains that why the abrupt phase change
has a special relation with the half-wavelength
lost.
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The condition of destructive interference is
That is
The condition for constructive interference is
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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?
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m 1, 2,
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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.
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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.
Interfering rays
Incident ray
Glass plate
e
The condition for destructive interference is
simpler
Zero-order dark fringe
m 0, 1, 2,
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(No Transcript)
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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 (m1)th-order dark fringe. As the refractive
index of air is 1, we can write out 2e
m?, 2e1 (m1) ?
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It is easy to find the spacing of two neighbor
fringes by deleting m from above equations. We
have
e
B
From similar triangles, we know
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Newtons 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, ??) of points of equal thickness are
circles concentric with point of contact.
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At such a case, the difference of
light-path-length is
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
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The condition for bright fringes is
The condition for dark fringes is
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On the other hand, we could also calculate the
radii of bright and dark rings.
As Rgtgte, e2 can be dropped. So we have
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The radius for kth bright ring is
The radius for kth dark ring is