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Chapter 28 Physical Optics: Interference and Diffraction
Chapter Outline
28-1 Superposition and Interference
28-2 Young’s Two-Slit Experience
28-4 Diffraction
28-5 Resolution
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28-4 Diffraction
Figure 28-17Diffraction of Water Waves
The water wave demonstrates that the wave can spread out in all direction after it goes through a small gap/slit: Huygens’s Principle for light diffraction.
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Figure 28-18Single-Slit Diffraction
Similar, when the light goes through narrow slit, the light is spread out in all directions, and a diffraction pattern of bright - dark fringes is formed.
Diffraction Pattern on screen
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Figure 28-19Locating the First Dark Fringe in Single-Slit Diffraction
To find the first dark fringe, we divide the slit into two regions.First dark fringe happens when:
sin..,2
sin2
weiw
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To find the second dark fringe, we divide the slit into four regions.The second dark fringe happens, when:
Figure 28-20Locating the Second
Dark Fringe in Single-Slit Diffraction
2sin..,2
sin4
weiw
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To find the third dark fringe, we divide the slit into six regions. The third dark fringe happens, when:
3sin..,2
sin6
weiw
And so on: divide the slit as even number regions….
Condition for Dark Fringe in Single-Slit Interference
)1228(,3,2,1,sin mmW
M=1, first dark fringe;
M=2, second dark fringe;
……
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The central bright fringe of diffraction slit has a angular width:
Unit: radian
)1328(2 WfringeCentral
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Exercise 28-2
Monochromatic light passes through a slit of width 1.2 x 10-5 mm. If the first dark fringe of the resulting diffraction pattern is at angle =3.25º, what is the wavelength of the light?
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Solution:
5
7
since, sin ,(1.2 10 )sin 3.25 (1)
6.80 10 680
W mmm
mm nm
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Example 28-5
Light with a wavelength of 511 nm forms a diffraction pattern after passing through a single slit of width 2.20x10-6m. Find the angle associated with (a) the first and (b) the second dark fringe above the central bright fringe.
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Solution:
(a) First Dark Fringe, m=1
(b) Second Dark Fringe, m=2
6 9
since sin ,(2.2 10 )sin (1)(511 10 )
13.4
W mm m
6 9
since sin ,(2.2 10 )sin (2)(511 10 )
27.7
W mm m
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28-5 Resolution
Diffraction pattern of a circular aperture
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The angle of First Dark Fringe for the Diffraction Pattern of a Circular Aperture is (from the bright point center to the first dark fringe)
)1428(22.1sin D
It can be shown:
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Rayleigh’s Criterion for two-point sources:
If the first dark fringe of one circular diffraction pattern passes through the center of a second diffraction pattern, the two sources will appear to be a single source, and therefore the two objects can not be seen as separate source and can not be resolved.
Two objects can be seen as separate source only if their angular separation is great than the following minimum:
Rayleight’s Criterion (minimum angular resolution)
Unit: radian
)1528(22.1min D
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Active Example 28-3Resolving Ida and Dactyl
Active Example 28-3
The asteroid Ida is orbited by its own small “moon” called Dactyl. If the separation between these two asteroids is 2.5 km, what is the maximum distance at which the Hubble Space Telescope (aperture diameter = 2.4m) can still resolve them at 550-nm wavelength?
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Solution
1) The minimum angular resolution is
radianm
mD
79
min 108.2)4.210550(22.122.1
kmrad
kmyLn
67 109.8
)108.2tan(5.2
tan
2) The maximum distance is
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)1228(,3,2,1,sin mmW
Summary
Condition for Dark Fringe in Single-Slit Interference
Two object can be seen as separate source only if their angular separation is great than the following minimum:
Rayleight’s Criterion
Unit Radian
)1528(22.1min D