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CHAPTER 26 INTERFERENCE AND DIFFRACTION
INTERFERENCE
CONSTRUCTIVE
DESTRUCTIVE
YOUNG’S EXPERIMENT
THIN FILMS
NEWTON’S RINGS
DIFFRACTION
SINGLE SLIT
MULTIPLE SLITS
RESOLVING POWER
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IN PHASE 1800 OUT OF PHASE
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OR THE BOOKS DISCRIPTION
CONSTRUCTIVE INTERFERENCE
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DESTRUCTIVE INTERFERENCE
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When two waves arrive at a point where path
differences from the sources differ by one
wavelength, two wavelengths, three wavelengths,
etc. there will be constructive interference.
Or mathematically:
�� −�� = ��ℎ���� = 0,±1,±2,±3,….
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When two waves arrive at a point where path
differences from the sources differ by ����������ℎ, ����������ℎ, ����������ℎ, ���. (Or half integer number of wavelengths) there will
be destructive interference.
Or mathematically:
�� −�� = �� +��! �
Where ℎ���� = 0,±1,±2,±3,….
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Many ways to produce two rays of coherent light.
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Consider path differences
"��ℎ#$%%������ = #&$�'
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Constructive Interference
#&$�' = �� ℎ���� = 0,±1,±2,±3,….
Destructive Interference
#&$�' = �� + ��! �
ℎ���� = 0,±1,±2,±3,….
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We can obtain the distance ( a bright line is below
(or above) the center of the screen.
Use
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If ) ≫ #
tan ' ≈ ()
Therefore
( = ) tan '
And for the �/0 bright line
(1 ≈ ) tan '1
For small angles
tan ' ≈ sin '
Therefore
(1 ≈ ) sin '1
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Use this with
#&$�' = ��
Or
&$�'1 = ��#
To get
(1 ≈ )��#
ℎ���� = 0,±1,±2,±3,…
This is the equation that Young used to measure
the wave length of light.
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As was said many ways of producing two rays of
coherent light. We have been discussing
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Can also get two coherent rays from thin films.
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We will discuss the interference between rays
coming from the two surfaces on opposite sides of
a space filled with liquid (thin film) or air.
First: we need to know that when light reflects
from a surface sometimes it undergoes a phase
shift of 1800.
For example if a ray is in air 4� = 1.005and reflects
at the surface with water 4� = 1.335back into the
air it does not undergo a phase shift.
But if a ray is in glass 4� = 1.335and reflects at the
surface with air 4� = 1.005 back into the glass it
undergoes a 1800 phase shift.
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Rule
If ray goes from low n material and reflects at
surface of high n material no phase shift.
If ray goes from high n material and reflects at
surface with low n material 180 degree phase
shift.
6%�7 89:;9</9=:8>1?@@@@@@@@@@@A�B��# �B > �7
No phase shift.
But
6%�B 89:;9</9=:8>1?@@@@@@@@@@@A�7��# �B > �7
1800
(half wavelength) phase shift
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Now consider we have a film as shown
Assume perpendicular.
Film thickness where shown �.
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No phase shift at either surface or phase shift at
both surfaces
Constructive interference if
2� = ��� = 0, 1, 2, 3, …
But if one surface has phase shift and other does
not
Destructive interference.
Or
No phase shift at either surface or phase shift at
both surfaces
Destructive interference if
2� = �� +��! �� = 0, 1, 2, 3, …
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Example 26.3
Find separation of interference fringes.
One reflection (top) – no phase shift other
(bottom) – phase shift.
Glass
Air
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Equation for destructive interference
2� = ��� = 0, 1, 2, 3, …
Where glass touches – destructive interference.
From figure
�D =
ℎ�
So
� = Dℎ�
Therefore
2 EDℎ� F = ��
D = � E �2ℎF � = ���2ℎ
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D = � G40.1�54500D10IJ�5240.02D10IK�5 L
D = �41.25��5
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EXAMPLES OF INTERFERENCE PHENOMONO
Newton’s Rings
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Nonreflective Coatings
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DIFFRACTION
Light goes in straight line.
Therefore would expect sharp shadow of object on
a screen.
But
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Consider light going through single slit.
This shows a ray from the top of the slit and one
from half way down the slit (middle).
Or looking only at the slit
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The difference in path distance to the screen is
"��ℎM$%%. = �2 sin '
If the path difference is one half wave length –
destructive interference.
Therefore the requirement for a dark fringe is
�2 sin ' = ±
�2
sin ' = ± ��
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If you divide the slit into 4 regions
If the path difference is one half wave length –
destructive interference.
Therefore the requirement for a dark fringe is
�4 sin ' = ±
�2
sin ' = ±2 ��
Becomes a/4
Becomes O/Q sinθ
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Divide the slit into more sections and can show
The requirements for dark fringes are
sin ' = � ��4� = ±1,±2,±3,… 5
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If the wavelength of light is much smaller than the
slit width (and normally it is) then
��� ≪ 14���(&����5
Therefore
' ≪ 14���(&����5
And
sin ' ≈ '
Therefore for dark fringes
' ≈ ��� 4� = ±1,±2,±3,… 5
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Also for
Use for distance to screen ) = D
Then tan ' = ST
And for the ��ℎ dark band
tan '1 =(1)
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If the screen is a distance from the slit such that
) ≫ (1
And it normally is
Then ' ≪ 14���(&����5
And tan '1 ≈'1
Therefore
'1 ≈(1)
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Then using
' ≈ ��� 4� = ±1,±2,±3,… 5
'1 ≈(1) ≈ ��� 4� = ±1,±2,±3,… 5
Solve for (1
(1 ≈ )��� 4� = ±1, ± 3, ± 5,… 5
This is equation 26.10 in the book.
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Multiple Slits
Path difference will be # sin '
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Constructive interference occurs for
# sin ' = ��4� = 0, ±1,±2,±3,… . 5
More complicated than two slit diffraction
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Eight Slit
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More – 16 slits
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DIFFRACTION GRATINGS
Prisms refract light different amounts depending
on the wavelength of the light.
Note short wavelengths (blue) are refracted more
than he long wavelengths (red).
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Diffraction Gratings will do much the same.
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The equation for the maxima is the same as for
multiple slits.
# sin ' = ��4� = 0, ±1,±2,±3,… . 5
Note
sin ' = ��#
Thus long wavelengths diffracted more than short
wavelengths. (Opposite to prisim.)
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X-RAY DIFFRACTION
First experiments in 1912.
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Crystal is uniform arrangement of atoms forming
planes for diffraction.
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Gives pattern
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Consider the atoms as all lined up on the sites and
represented by the dots shown.
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Look at top row for example
Path lengths are equal when 'B ='8
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Therefore when the two angles are equal get
constructive interference from diffraction from
rows.
Then considering two rows one beneath the other.
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Additional constructive interference when path
lengths are equal to integer times wavelength of
x-rays.
2#&$�' = ��4� = 1, 2, 3, … . . 5
When both conditions are satisfied
1. 'B ='8
And
2. 2#&$�' = ��
When these conditions are not satisfied radiation
interferes destructively.
Therefore obtain diffraction pattern characteristic
of crystal.
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Scattering and thus diffraction can occur from
various planes within the crystal.
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CIRCULAR APERTURES
Light travelling through a small circular aperture is
diffracted also.
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The angular locations of the dark rings are given
by:
sin '� = 1.22 �M
sin '� = 2.23 �M
sin 'K = 3.24 �M
OBSERVING OBJECTS THROUGH VARIOUS SIZE
APERTURES.
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If the aperture in (a) were smaller 3 and 4 would
appear as one object.
The larger the aperture the better the resolving
power.
Lord Rayleigh proposed a criterion for telescopes
(and other optical instruments) to give a measure
for their capability.
Two objects can only be resolved if the center of
the second just falls at a distance of the center of
the first dark ring of the first.
The limit of resolution, '89U for a telescope (or
other optical instrument):
'89U = 1.22 �M