sin 1 . 22

d
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22
These objects are just resolvable.
Chapter 42
Diffraction
Rayleigh’s criterion Two objects are just resolved when the maximum of one is at the minimum of the other.*
2

2
)0
1
secondary maxima
(m )
2 a sin ( 2m 1)
2012-10-26

2
, m 1,2,3
15
Chapter 42
sin
2
Diffraction
The intensity of secondary maxima
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Chapter 42
Diffraction
Example: A slit of width a = 0.5 mm is illuminated by a monochromic light. Behind the slit there placed a lens (f = 100 cm), and the first maximum fringe is observed at a distance of 1.5 mm from the central bright fringe on the focal plane (screen). Find the wavelength of the light, and the width of the central bright fringe.
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Chapter 42
Intensity Distribution
Diffraction
2012-10-26
21
Chapter 42
1st diffraction minimum
Diffraction
Central maximum

Diameter d
light
First diffraction minimum is at
Chapter 42
Diffraction
Content of this Chapter
Diffraction and the wave theory of light Single-slit diffraction
Chapter 42 Diffraction Intensity in single-slit diffraction
Evidence for the wave nature of light*
2012-10-26 2
Chapter 42
Diffraction
Fresnel’s theory of light (1788-1827)
2
dE C
P
dS
K ( )cos( cos( 2
E C
P S
Eθ
E 2 R sin

2
2
Em

sin

2
E1 E1
Em
Em
let 2 E Em
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sin

,
I I m
sin
2

2
where

a sin
14
Chapter 42
I I m sin
2
Diffraction

a sin
a N sin

2
, ,
for adjacent rays
a sin N

2
if N is even, a sin m , min ima
if N is odd, a sin (2m 1) max ima
2012-10-26 8

2
,
Chapter 42
Diffraction pattern
I Im

2

1 (2m 1) 2
0.045; 0.0083
2
,
where m 1,2,3
for m 1, for m 3,
I Im I Im
for m 2,
I Im
0.016
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16
Chapter 42
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5
Chapter 42
Diffraction
Fraunhofer’s Single-slit Diffraction
P0
All rays arriving at P0 are in phase, they interfere constructively.
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r K ( ) r

r t )
0

r t )dS
0
Makes it feasible for calculation of intensity at any point on screen
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3
Chapter 42
Two cases of diffraction
Solution:
maxima : asin (2m 1) 2ax 3f 1500nm 3

2
, m 1 and sin tan
x f


500nm
min ima : x0 2 f
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a sin m , m 1

a
2mm
xm f a f a ( m 1) center of minima m center of minima
xm 1
x f
2012-10-26

a
,
width of other fringes
Sample problem 42-1,211
Chapter 42
19
Chapter 42
Diffraction
42-4 Diffraction at a Circular Aperture
The lens behaves like a circular aperture in an opaque screen.
Airy spot
Diffraction effects often limit the ability of optical instruments to form precise images.
resolved
just resolved
1
not resolved
R min sin (1.22

d
)
R 1 / R resolution ability
Diffraction
Quick quiz-1 If a classroom door is open slightly, you can hear sounds coming from the hallway. Yet you cannot see what is happening in the hallway. Why is there this difference? (1) light waves do not diffract through the single slit of the open doorway. (2) Sound waves can pass through the walls, but light waves cannot. (3) The open door is a small slit for sound waves, but a large slit for light waves. (4) The open door is a large slit for sound waves, but a small slit for light waves.
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Optical path difference of ajacent rays is x sin
13
Chapter 42
for adjacent rays
2
Diffraction

x sin
2
2 a
N
sin
φ
N

a sin

2
2
,
where
0,
sin

2
1, I I m
central maxima mixima
m 1,2,3
when sin 0, 0, I 0
sin 0, m , a sin m ,
d d ( sin
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12
Chapter 42
Diffraction
42-3 Intensity in Single-Slit Diffraction
The whole slit is divided into N strips with x = a/N, which can be regarded as Huygen’s wavelet, therefore,
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Chapter 42
Diffraction
Width of the bright fringe at the center: