Tutorial 8

Transcription

Tutorial 8

Elec308: Engineering Optics
Tutorial 8
Zeng Yan
[email protected]
5/10/2011
ELEC 308 Engineering Optics
Outline
• Interference of waves
– Two sources interference
– Young’s double slit experiment
• Thin film interference
– Equal inclination
– Equal thickness
– Newton’s rings
• Interferometer
– Michelson, Mach-Zender, Fizeau
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Example
• The figure below shows a laser beam incident on a
wet piece of filter paper atop a sheet of glass whose
index of refraction is to be measured. The photograph
shows the resulting light pattern. Explain what is
happening and derive an expression for ni in terms of
R and d.
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Example
• Wet paper is on optical contact with the glass. The
beam scatters off the wet paper and is mostly
transmitted until the critical angle is attained, at
which point the light is reflected back toward the
source.
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Example
• On the air-glass surface of the second
interface TIR occurs at the critical angle.
TIR condition:
1
ni 
sin  c
ni sin  c  nair  1
tan  c 
• From the geometry of the experiment:
1
1  cot  
sin 2 
2

1
1
 1
sin 
tan 2 
1
1
4d 2
ni 
 1
 1 2
2
sin  c
tan  c
R
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R
2d
Coherent waves
• Waves for which phase difference is
constant in time and space
E1  E 01 sin(t  kx)
E 2  E 02 sin(t  kx   )
  const.
• Conditions for interference
– Identical polarization
– Identical frequency
– Constant phase difference
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Interference of waves



E  E (r ) exp j k r  t 
• Two waves



E  E (r ) exp j k r  t 
1P
01
1 1
2P
02
2 2
• Resultant intensity
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Phase shift
  kasin   sin  i 
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wave summation
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Wave addition (superposition)
• If constructive or
destructive
interference is to
continue occurring
at a point, the
sources of the
waves must be
coherent sources.
combined
waveform
wave 1
wave 2
Two waves in phase
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Two waves 180°out
of phase
• Two sources are
coherent if the
waves they emit
maintain a constant
phase difference.
Young’s double slit experiment
  d sin
Bright fringes
of a double-slit
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Dark fringes
of a double-slit
sin  m
sin   m 

d
1
2

m  0,1,2,3,

d
m  0,1,2,3,
Example 1 Young’s Double-Slit Experiment
Red light (664 nm) is used in Young’s experiment with slits separated
by 0.000120 m. The screen is located a distance 2.75 m from the slits.
Find the distance on the screen between the central bright fringe and
the third-order bright fringe.
Solution:
sin  m

d
m  0,1,2,3,
9
 
1  66410 m 
  0.951
  sin  m   sin  3
4
 d
 1.2010 m 
1


y  L tan   2.75 m  tan 0.951  0.0456 m
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Example 2 White Light and Young’s Experiment
The figure shows a photograph that illustrates the kind of interference
fringes that can result when white light is used in Young’s experiment.
Why does Young’s experiment separate white light into its constituent
colors? In any group of colored fringes, such as the two singled out,
why is red farther out from the central fringe than green is? Why is
the central fringe white?
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New example 1 Young’s Experiment
9.4 Will we get an interference pattern in Young’s Experiment if we replace
the source slit by S by a single long-filament light bulb? What would occur
if we replaced the slit S1 and S2 by these same bulb?
•Bulb-extended source
The source is made up of a
large number of incoherent
point source.
•Pattern overlapping
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9.6 Two 1.0MHz radio antennas emitting in-phase are separated by 600m
along a north-south line. A radio receiver placed 2.0km east is equidistant
from both transmitting antennas and picks up a fairly strong signal. How far
North should that receiver be moved if it is again to detect a signal nearly
as strong?
Like the Young’s Experiment,
Two antennas->two slits
Receiver->screen
Interference for Young' s Experiment : a sin  m  m
Geometric condition : sin   y /(s 2  y 2 )1/ 2
Wavelength :   c / f
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9.5 Figure P.9.5 shows an output pattern that was measured by a tiny
microphone when two small piezo-loudspeakers separated by 15 cm were
pointed toward the microphone at a distance of 1.5 m away. Given that the
speed of sound at 20 degree is 343 m/s, determined the approximate
frequency at which the speakers were driven. Discuss the nature of the
pattern and explain why it has a central minimum.
y
ms

a sin   m   a  m   y 

s
a

f  v/
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9.10 White light falling on two long narrow slits emerges and is observed on
a distant screen. If red light (780nm) in the first order fringe overlaps violet
in the second-order fringe, what is the latter’s wavelength?
Interference for Young' s Experiment : a sin  m  m
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New example 4 Interference of Lloyd’s mirror
9.23 Using Lloyd’s mirror, X-ray fringes were observed, the spacing of
which was found to be 0.0025 cm. The wavelength used was 0.833nm.
If the source-screen distance was 3 m, how high above the mirror plane
was the point source of X-ray placed?
Interference condition : a sin   m
Geometric condition : y1  sin θ1s; y2  sin θ2 s
Spacing : y  y1 - y 2  s
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
a
New example 4 Interference of Lloyd’s mirror
9.24 Imaging that we have an antenna at the edge of a lake picking up a
signal from a distant radio star, which is just coming up above the horizon.
write expression for deta and for the angular position of the star when the
antenna detects its first maximum.
For Lloyd’s mirror:
α
α
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α
  k (r1  r2 )  
    k (a /(2 sin  )  [sin(90  2 )]a /(2 sin  ))  
  ka(1  cos 2 ) /(2 sin  )  
  when  2 ,   sin 1 ( / 2a)
Thin film Interference
• Phenomenon commonly seen as colored
patches on thin layers of leaked oil,
grease and soap bubbles.
• To obtain a nice colored pattern, the
thickness of the film has to be on the
order of the wavelength of light.
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Thin film interference
• In thin film interference the phase
difference is due to reflection at either
side of a thin film of transparent material.
• The phase difference is due to two
factors:
– Path difference through the film (corrected for
the change in speed of light in the material)
– Phase shift at the interface
The wavelength of light in oil though is not the
same as in air:

0 
n
In this case, constructive interference takes place when:
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Example 3: anti-reflection coating
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New example 5 Thin film interference
Q6. The refractive index of a dense flint glass is about 1.75.
Design an antireflection coating to maximally reduce the
reflection from glass surface at 0deg incident angle
(1)The index of the coating material
(2)Phase shift exist?
(3)Maximun interference condition
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9.27 A thin film of ethyl alcohol (n=1.36) spread on a flat glass plate and
illuminated with white light shows a color pattern in reflection. If a region
of the film reflects only green light (500nm) strongly, how thick is it?
(1)Phase shift in the interface
(2) Reflectionenhanced : 2nd  2 /   2m
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Equal inclination interference
The phase difference:
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9.29 Consider the circular pattern of Haidinger’s fringes resulting from a
Film with a thickness if 2mm and an index of refraction of 1.5. For
Monochromatic illumination of λ=600 nm, find the value of m for the central
Fringe (θ=0). Will it be bright or dark?
δ  2nktcosθ  π
 2nktcosθ  2 1.5  2/600 2 1000001  2π 10000
 Order : m  10000,considering the phase shift, minimum
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1)when thin film thickness increases： fringes extend outwardly
i
i
i'
i'
e
λ
δ  2e n  n sin i 
2
2
2
2
1
2
 k
k  1.2.3.
(max)
2)when thin film thickness decreases： fringes contract inwardly
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Equal inclination interference
3) if white light falls on, color
fringes appear from red to violet.
  2e n22  n12 sin 2 i 
Since
e.k

2
 k
fixed,
n22  n12 sin 2 i
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k  1.2.3. (max)

should be larger w. r. t. longer
i

Therefore, it should be closer to
the center.
Thin Air Wedge
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Thin Air Wedge
As the thickness of air d increases, the fringes move down;
While the thickness d decreases, the fringes move up;
When the angle θ increases, the fringes move down;
When the angle θ decreases, the fringes move up;

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
Thin Air Wedge
d

*Measure the
distance change
by the moving
fringes.
e  m

2
d

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Application of equal thickness
interference
Thermal expansion instrument
M：test object
C：made of In, low thermal
expansion coefficient
M
C
We count the number of
passing fringes m when the
temperature Increase t.

l  m
M heighten:
2
Thermal expansion coefficient:
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
l
l t
New example 7 Equal thickness interference
9.34 Suppose a wedge-shaped air film is made between two sheets of
glass, with a piece of paper 7.618*10-5m thick used as the spacer at
their very ends. If light of wavelength 500nm comes down from directly
above, determine the number of bright fringes that will be seen across
the wedge?
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Newton’s rings
---A special case of equal thickness interference
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Newton’s rings
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Newton’s rings
Newton’s rings effect on
scanned picture
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• Newton’s ring is
often used to
measure the
curvature of radius
of a convex lens.
New example 8 Newton ring
9.31 Figure illustrates a setup used for testing lenses. Show that
d  x 2 ( R2  R1 ) / 2 R1 R2
When d1 and d2 are negligible in comparison with 2R1 and 2R2,
Respectively. (Recall the theorem from plane geometry that relates the
Products of the segments of intersecting chords.) Porve that the radius of
The mth dark fringe is then
xm  [ R1 R2 m f /( R2  R1 )]1/ 2
Geometry condition :
R 12  x 2  (R1  d1 ) 2  R 12  x 2  (R 12  2R 1d1  d12 )
 x 2  2R 1d1  d12  d1  x 2 /2R1
The same for R 2
R 22  x 2  (R 2  d 2 ) 2  R 22  x 2  (R 22  2R 2 d 2  d 22 )
 x 2  2R 2 d 2  d 22  d 2  x 2 /2R 2
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New example 8 Newton ring
9.31 Figure illustrates a setup used for testing lenses. Show that
d  x 2 ( R2  R1 ) / 2 R1 R2
When d1 and d2 are negligible in comparison with 2R1 and 2R2,
Respectively. (Recall the theorem from plane geometry that relates the
Products of the segments of intersecting chords.) Porve that the radius of
The mth dark fringe is then
xm  [ R1 R2 m f /( R2  R1 )]1/ 2
Differenceof optical path :
d  d1  d 2  ( x 2 /2)(1 / R 1  1 / R 2 )  x 2 (R 2 - R 1 )/2R1R 2
For the dark fringe :
2d  mλ
 m/2  x 2m (R 2 - R 1 )/2R1R 2  x m  [R 1R 2 mλ f /(R 2  R 1 )]1/ 2
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Problem 4: Newton’s ring
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9.32 Newton rings are observed on a film with quasimonochromatic light
that has a wavelength of 500 nm. If the 20th bright ring has a radius of 1
cm, what is the radius of curvature of the lens forming one part of the
interfering system?
Newton ring approximat ion : r 2  2 yR
Interference condition : y  (m  1/2)0 / 2n
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The Michelson Interferometer
• Precise distance
measurements can
be made with the
Michelson
interferometer by
moving the mirror
and counting the
interference fringes
which move by a
reference point.
• The distance d
associated with m
fringes is
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Haidinger fringes
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Fizeau fringes
White light fringes
• Applications
– The core of
autocorrelator/crosscorrelator,
Fourier transform spectroscopy.
– Precise measurement of the
wavelength.
– Tested the dependence of speed
of light on the motion of the
Earth
An interesting Demo:
http://www.youtube.com/watch?v=V
yePASErr5Q&feature=related
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Mach-Zender Interferometer
• A device used to determine the phase shift caused by a
small sample which is placed in the path of one of two
collimated beams.
• In contrast to the Michelson interferometer, there are two
output ports.
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Fizeau Interferometer
Fizeau interferometers are commonly used for measuring
the shape of an optical surface.
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Fizeau interferometer
Surface test: make use of the principle of thin air wedge.
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New example 9 Michelson Interferometer
9.35 A Michelson Interferometer is illuminated with monochromatic light.
One of its mirrors is then moved 2.53*10-5m, and it is observed that 92
fringe-pairs, bright and dark, pass by in the process. Determine the
wavelength of the incident beam.
9.36 One of the mirrors of a Michelson Interferometer is moved, and 1000
finge-pairs shift past the hairline in a viewing telescope during the process.
If the device is illuminated with 500-nm light, how far was the mirror moved?
For 9.35, a motion of /2 causes a single fringe pair to shift past,
hence 92 / 2  2.53*10  5 and   550nm
For 9.36
d  N(0 /2)  (1000)(5.00 *10 - 7m)/2  2.50 *10 - 4m
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Tutorial 8

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