Transcript polarized

Chapter 24
Wave Nature of Light:
© 2006, B.J. Lieb
Some figures electronically reproduced by permission of Pearson
Education, Inc., Upper Saddle River, New Jersey Giancoli,
PHYSICS,6/E © 2004.
Ch 24
1
Huygens’ Principle
•Wave motion is described by some rather complex equations but Huygens’
principle provides a simple way to predict wave propagation.
•Every point on a wave is a source of tiny wavelets that spread out in the
forward direction at the speed of the wave itself.
•The new wave is the envelope of all the wavelets.
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2
Diffraction
•Diffraction occurs when a wave encounters an obstacle. Applying
Huygens principle results in a bending of the wave behind the obstacle.
•Note that if light were a stream of particles, there would be no light in
the shadow of the obstacle.
•Note that a small opening (usually it’s a slit) results in a nearly spherical
wave.
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3
Interference
•We will study several situations where a wave is split in half, travels over
different paths and then is reunited.
Constructive Interference
results when the amplitudes of
the two waves add to give a
larger amplitude.
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Destructive Interference results
when the amplitudes of the two
waves add to give a smaller
amplitude.
4
Young’s Double-Slit Interference
•Waves passing through two slits will diffract and spread out. The two
waves will then interfere with each other when they reach the screen.
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5
Young’s Double-Slit Interference
•In the figure below, the path taken by the lower wave is greater by
“d sin ”
•The waves interfere constructively when the path difference is “m ”
where m is an integer.
Condition for Constructive interference
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d sin   m
m  0, 1, 2.....
6
Double Slit Pattern
The double slit interference pattern is a series of bright
lines with the m = 0 line the brightest. The value of m is
often called the order of the interference fringe.
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7
Small Angle Approximation
In working these problems it is often convenient to
use the small angle approximation which is true when
   7o
sin   tan  
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8
Example 24-1: If 520-nm and 640-nm light passes through two slits 0.50 nm apart,
calculate the angle of the second order fringes for these two wavelengths.
m2
d sin   m 
m  
  sin 1 
9
 d 

(
2
)(
640

10
m) 
1
 640  sin 

3
9
0
.
50

10
m


(
2
)(
520

10
m
)


520  sin 1 

3
 0.5010 m 
 0.150
 0.120
How far apart are the second-order order fringes for these two wavelengths on a
screen 1.5m away?
For small angles
d sin   m 
m
sin  
d
y  L tan
tan   sin  
m  
y  L

 d 
y520
m
d
(1.5m) (2) (520 109 m)

0.50103 m
y

L
 3.12 m m
y640  3.84 mm
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y  3.84 mm 3.12 mm  0.72 mm
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Diffraction Grating
A diffraction grating has many slits. As additional slits are
added, the interference maxima become sharper and
narrower as shown below (a) for two slits and (b) for six
slits.
If “d” is the distance
between adjacent slits, the
equation for the grating is
the same as for double slits.
d sin   m 
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Using a Diffraction Grating to Produce an
Atomic Spectrum
If the light striking a diffraction grating is not monochromatic
(single color) then the light is spread out into its component
wavelengths. The resulting pattern is called a spectrum. We can
the determine the wavelength from
m   d sin  m  1, 2, 3....
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Types of Spectrum
•Continuous Spectrum: includes all wavelengths
•Line Spectrum: contains only certain discrete wavelengths
characteristic of the atom. Only emitted by gases.
•Absorption Spectrum: continuous spectrum with dark lines
characteristic of the atoms absorbing the light. Example is the
solar absorption spectrum shown below. Note that double dark
lines correspond to the sodium spectrum.
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Example 24-2: What is the highest spectral order that can be seen if a
grating with 6000 lines per cm is illuminated with white light?
White light spans the visible range
red  750 nm
violet  400 nm

 1 m 
1
 

d  
6000
lines
/
cm
100
cm



 1.667  10 m
6
d sin   m 
m
d sin 

The limit of the spectra is θ = 90º and in order to see the full spectra, you can figure
m for each λ
mviolet 
mred
d sin 90
violet
1.667106 m

400109 m
1.667106 m

 2.2
750109 m
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 4.2
So only the second order of red would be
seen and so m = 2
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Dispersion of a Prism
The index of refraction n depends slightly on wavelength with n being
highest for short wavelengths. This allows a prism to produce a
spectrum with red being deviated the least.
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14
Rainbows
Rainbows are produced by dispersion in spherical water
droplets. Red light is bent the least, so it is seen from
droplets higher in the sky than violet light. This
separation produces a spectrum of the sunlight.
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15
Single Slit Diffraction
This picture shows the diffraction pattern
caused by monochromatic light on a
narrow slit
•The single-slit pattern is produced by interference of light
from different parts of the slit as shown below.
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Single Slit Diffraction II
D sin   m
m 1, 2, 3, .. .
•This equation gives the angular position  of the
minima of the intensity curve.
•D is the width of the slit.
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•Note that if the width of the slit is made narrower, the
above pattern gets wider.
17
Thin Film Interference
•Light reflected off of the air/oil interface interferes with
light reflected off of the oil/water interface.
•The situation can be complicated by a phase shift that can
occur during reflection.
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Phase Shift of Reflected Ray
n2 > n1
180o phase shift
n2 < n1
no phase shift
•There is no single equation to solve these problems.
•For constructive interference, the path difference
( usually 2 t where t is the thickness) must equal m  if
there is no phase shift
•If there is a phase shift then for constructive interference
2 t = (m+1/2)
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Example 24-4: Solar cells are often coated with a transparent thin film
such as silicon monoxide SiO, n = 1.45) to minimize losses due to
reflection. A silicon solar cell (n = 3.50) is coated with silicon monoxide
for this purpose. Determine the minimum thickness of film that will
produce the least reflection for light of wavelength 500nm.
Note that there is a phase change at each reflection.
In order to cancel. The path difference (2 t ) must
equal half of the wavelength in SiO.
t
1
n
2
1
t  n
4
( 2t ) 
n 
air
1.45
 1  500nm
 86 nm
t  
 4  1.45
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Polarization
•We pictured light as a wave with transverse electric
fields (E) perpendicular to transverse magnetic fields
(B).
•Most light is a jumble of photons with different
orientations of E fields (always with B field
perpendicular to E). This is unpolarized light.
•With polarized light, all of the E fields have the same
orientation.
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Polaroid Materials
•Plane-polarized light can be made using special materials with
oriented long molecules. These molecules act like slits that pass
light with E fields in the slit direction but eliminate light with
perpendicular E fields.
•The action of polaroid materials is illustrated below
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Intensity
I  I0 cos2 
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First Polarizer
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Intensity
I  I0 cos 
2
Note: The first polarizer reduces the intensity by 1/2 , so the
intensity after a pair of polarizers is
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I0
I  cos 2 
2
25
Example 24-5 Unpolarized light passes through two polaroids. The axis of the
first polaroid is vertical and the second is at 450 to the vertical. What percent of
the light intensity is transmitted through the two polaroids.
The first polaroid eliminates half of the light
1
I1  I 0
2
I 2  I1 (cos45)2
I2 
1
I0
4
Answer = 25%
A third polaroid is added at 900 to the vertical. What percent is transmitted? (Note
that none would be transmitted without the polaroid at 45º.)
Light emerging from the second polaroid is polarized at 45º to the vertical
and this is 450 to the third polaroid.
I3  I 2 (cos45)2
1
1
I 0 (cos 45) 2  I 0
8
4
Answer = 12.5%
I3 
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Polarization by Reflection
Reflected light is partially polarized. It is 100 % plane polarized
when the angle between the reflect and refracted ray is 90o.
For light from air on a medium of index of refraction n, reflected light
is 100 % plane polarized when:
tan  P  n
Polarized with E 
page
Ch 24
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