Chapter 37 - Interference
and Diffraction
A PowerPoint Presentation by
Paul E. Tippens, Professor of Physics
Southern Polytechnic State University
©
2007
Objectives: After completing this
module, you should be able to:
• Define and apply concepts of constructive
interference, destructive interference,
diffraction, and resolving power.
• Describe Young’s experiment and be able to
predict the location of dark and bright fringes
formed from the interference of light waves.
• Discuss the use of a diffraction grating, derive
the grating equation, and apply it to the
solution of optical problems.
Diffraction of Light
Diffraction is the ability of light waves to bend
around obstacles placed in their path.
Ocean
Beach
Light rays
Fuzzy Shadow
Water waves easily bend around obstacles, but
light waves also bend, as evidenced by the lack
of a sharp shadow on the wall.
Water Waves
A wave generator sends periodic water waves
into a barrier with a small gap, as shown below.
A new set of waves is observed
emerging from the gap to the wall.
Interference of Water Waves
An interference pattern is set up by water
waves leaving two slits at the same instant.
Young’s Experiment
In Young’s experiment, light from a monochromatic
source falls on two slits, setting up an interference
pattern analogous to that with water waves.
Light
source
S1
S2
The Superposition Principle
• The resultant displacement of two simultaneous waves (blue and green) is the
algebraic sum of the two displacements.
• The composite wave is shown in yellow.
Constructive Interference
Destructive Interference
The superposition of two coherent light waves
results in light and dark fringes on a screen.
Young’s Interference Pattern
s1
Constructive
Bright fringe
s2
s1
s2
s1
Destructive
Dark fringe
s2
Constructive
Bright fringe
Conditions for Bright Fringes
Bright fringes occur when the difference in path Dp
is an integral multiple of one wave length l.
p1
p2
l l l
p3
p4
Path difference
Dp = 0, l , 2l, 3l, …
Bright fringes:
Dp = nl, n = 0, 1, 2, . . .
Conditions for Dark Fringes
Dark fringes occur when the difference in path Dp
is an odd multiple of one-half of a wave length l/2.
l
2
p1
p2
p3
l
l
2
n = odd
n=
1,3,5 …
p3
Dark fringes:
Dp  n
l
Dp  n
l
2
n  1, 3, 5, 7, . . .
Analytical Methods for Fringes
x
s1
d q
s2
Path difference
determines light
and dark pattern.
d sin q
p1
p2
y
Dp = p1 – p2
Dp = d sin q
Bright fringes: d sin q = nl, n = 0, 1, 2, 3, . . .
Dark fringes:
d sin q = nl/2 , n = 1, 3, 5, . . .
Analytical Methods (Cont.)
s1
d q
s2
From geometry,
we recall that:
x
d sin q
p1
p2
Bright fringes:
dy
 nl , n  0, 1, 2, ...
x
y
y
sin q  tan q 
x
So that . . .
dy
d sin q 
x
Dark fringes:
dy
l
 n , n  1, 3, 5...
x
2
Example 1: Two slits are 0.08 mm apart, and
the screen is 2 m away. How far is the third
dark fringe located from the central maximum if
light of wavelength 600 nm is used?
x = 2 m; d = 0.08 mm
l = 600 nm; y = ?
d sin q = 5(l/2)
The third dark fringe
occurs when n = 5
Dark fringes:
dy
l
 n , n  1, 3, 5...
x
2
x
s1
s2
q
d sin q
y
n = 1, 3, 5
dy 5l

x
2
Example 1 (Cont.): Two slits are 0.08 mm apart,
and the screen is 2 m away. How far is the third
dark fringe located from the central maximum if
l = 600 nm?
x = 2 m; d = 0.08 mm
l = 600 nm; y = ?
dy 5l

x
2
x
s1
s2
5l x 5(600 x 10-9 m)(2 m)
y

2d
2(0.08 x 10-3m)
q
d sin q
y
n = 1, 3, 5
y = 3.75 cm
The Diffraction Grating
A diffraction grating consists of thousands of
parallel slits etched on glass so that brighter and
sharper patterns can be observed than with
Young’s experiment. Equation is similar.
d sin q
d q
d sin q  nl
n = 1, 2, 3, …
The Grating Equation
The grating equation:
d sin q  nl n  1, 2, 3, ...
d = slit width (spacing)
l = wavelength of light
q = angular deviation
n = order of fringe
3l
2l
l
1st
order
6l
4l
2l
2nd
order
Example 2: Light (600 nm) strikes a grating ruled
with 300 lines/mm. What is the angular deviation
of the 2nd order bright fringe?
To find slit separation,
we take reciprocal of
300 lines/mm:
Lines/mm  mm/line
n=2
300 lines/mm
1
d
 0.00333 mm/line
300 lines/mm
mm  103 m 
d  0.00333


line  1 mm 
d  3 x 10 m
-6
Example (Cont.) 2: A grating is ruled with 300
lines/mm. What is the angular deviation of the
2nd order bright fringe?
l = 600 nm
d  3 x 10-6 m
d sinq  nl
n2
2l 2(600 x 10-9 m)
sin q 

;
-6
d
3.33 x 10
Angular deviation of
second order fringe is:
n=2
300 lines/mm
sinq  0.360
q2 = 21.10
A compact disk acts as a diffraction grating. The colors
and intensity of the reflected light depend on the
orientation of the disc relative to the eye.
Interference From Single Slit
When monochromatic light strikes a single slit,
diffraction from the edges produces an interference
pattern as illustrated.
Relative intensity
Pattern Exaggerated
The interference results from the fact that not all
paths of light travel the same distance some
arrive out of phase.
Single Slit Interference Pattern
a
sin q
2
For rays 1 and 3
and for 2 and 4:
a/2
a
a/2
Each point inside slit
acts as a source.
1
2
3
4
5
a
Dp  sin q
2
First dark fringe:
a
l
sin q 
2
2
For every ray there is another ray that differs by
this path and therefore interferes destructively.
Single Slit Interference Pattern
a
sin q
2
First dark fringe:
a/2
a
a/2
a
l
sin q 
2
2
1
2
3
4
5
sin q 
l
a
Other dark fringes occur
for integral multiples of
this fraction l/a.
Example 3: Monochromatic light shines on a
single slit of width 0.45 mm. On a screen 1.5 m
away, the first dark fringe is displaced 2 mm
from the central maximum. What is the
wavelength of the light?
l=?
sin q 
l
x = 1.5 m
q
a
y
sin q  tan q  ;
x
a = 0.35 mm
y l
 ;
x a
(0.002 m)(0.00045 m)
l
1.50 m
ya
l
x
l = 600 nm
y
Diffraction for a Circular Opening
D
Circular diffraction
The diffraction of light passing through a circular
opening produces circular interference fringes
that often blur images. For optical instruments,
the problem increases with larger diameters D.
Resolution of Images
Consider light through a pinhole. As two objects
get closer the interference fringes overlap,
making it difficult to distinguish separate images.
Clear image of
each object
d1
Separate images
barely seen
d2
Resolution Limit
Images are just resolved
when central maximum
of one pattern coincides
with first dark fringe of
the other pattern.
Separate images
Resolution
limit
d2
Resolution Limit
Resolving Power of Instruments
The resolving power of
an instrument is a
measure of its ability to
produce well-defined
separate images.
D
q
Limiting angle
For small angles, sin q  q, and the limiting
angle of resolution for a circular opening is:
Limiting angle of
resolution:
q 0  1.22
l
D
Resolution and Distance
p
so
q
D
q
Limiting angle qo
l s0
Limiting Angle
q 0  1.22 
of Resolution:
D p
Example 4: The tail lights (l = 632 nm) of an
auto are 1.2 m apart and the pupil of the eye
is around 2 mm in diameter. How far away can
the tail lights be resolved as separate images?
p
so
q
D
q
Eye
Tail lights
l
s0
q 0  1.22 
D p
(1.2 m)(0.002 m)
p
1.22(632 x 10-9 m)
s0 D
p
1.22l
p = 3.11 km
Summary
Young’s
Experiment:
Monochromatic
light falls on two
slits, producing
interference fringes
on a screen.
s1
d q
s2
Bright fringes:
dy
 nl , n  0, 1, 2, ...
x
x
d sin q
p1
p2
y
dy
d sin q 
x
Dark fringes:
dy
l
 n , n  1, 3, 5...
x
2
Summary (Cont.)
The grating equation:
d sin q  nl n  1, 2, 3, ...
d = slit width (spacing)
q = angular deviation
l = wavelength of light
n = order of fringe
Summary (Cont.)
Interference from a single slit of width a:
Relative Intensity
Pattern Exaggerated
Dark Fringes: sin q  n
l
a
n  1, 2, 3, . . .
Summary (cont.)
The resolving power of instruments.
p
so
q
D
q
Limiting angle qo
l s0
Limiting Angle
q 0  1.22 
of Resolution:
D p
CONCLUSION: Chapter 37
Interference and Diffraction