# wug.physics.uiuc.educcIAStatePhys222falllectu

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```					  Lecture 41
Single and double slit
diffraction.
No sharp edges

should have a sharp boundary.

This is instead what is seen
(with monochromatic light)
Single-slit diffraction

Slit of width a
• consider it made of a large number of point sources
• interference by N slits

N slits   N phasors
P
θ
a                                                EP

R >> a                                       β
E0

2
Angle between first and last phasors:         ka sin          a sin 

Length of the arc is NE0 = Emax
r
(amplitude when β = 0)
β/2
β/2                                         E max      (β in radians)
 
r               EP                               r
       EP
From the triangle:      sin         
2       2r
β
    E max     
EP  2r sin     2       sin
E0                                             2             2
Emax
r 


2
                             
sin                          sin     
E P  E max         2                           2
IP  I max          
                                  
2                                   
 2       
Minima:
  2m m  1, 2...
a sin  m m  1, 2...                                 2
        
 sin     
I            2
         
Imax            
Maxima:                                                
 2       
(
b » 2m + 1 p )
(and b = 0)

θ

This is the             This is light
DEMO:
Single-slit                “normal” part          “bending around
diffraction                                          the corner”!
In-class example: Single slit

A single slit is cut in a dark film and placed R = 4.00 m away from a screen. A
laser with λ = 700 nm shines on the slit. The first dark fringe is y = 1.00 cm above
the center of the most intense (central) bright fringe. What is a, the width of
the slit?
Destructive interference: a sin   m m  1, 2...
A. 350 nm
y
B. 700 nm            From the triangle: tan  
R
C. 28 μm
For small angles,   sin  tan
D. 70 μm
E. 0.28 mm                                             y
        a 
R
y
R  700  10 m  4.00 m 
9
θ
a                                       a                             2.8  10   4
m
y      1.00  10 m 2

R

Visible light shines on a slit 1 cm by 5 cm. Qualitatively, which is the
approximate picture of light on the screen?

A                                           B

a ~ 10-2 m        First maximum after center:   2
λ ~ 10-9 m              2
        a sin   2  107 sin   2         sin   107

This is geometrical optics (light traveling only in a straight line)… because a >>λ
Diffraction and wavelength

Diffraction effects (= light bending around the corner) are
most important when a ~λ

Diffraction pattern
of a square slit (with
both sides a ~λ)
Light bends around the corner

Huygens’ model: each point in the wavefront emits a wavelet.

obstacle
• Block the lower sources of induced fields with an obstacle
• Induced fields from upper sources/wavelets are still spherical
– reach beyond blockage
– waves effectively “bend” around corners                DEMO:
Arago (or
Poisson’s) spot

Examples:

AM signal with f = 1000 kHz        λ = 300 m
FM signal with f = 100 MHz         λ=3m

In mountainous areas, AM radio station reception is much
better because AM waves are diffracted “around” the
mountain tops and into the valleys

FM waves propagate “in a straight line” and cannot reach
the valleys (unless an antenna on top of the mountain is
used)

Like interference, diffraction applies to ALL waves (not just
light)

Human ear perceives f ~ 20 Hz – 20 kHz         λ ~ 2 cm – 2 m

A door left ajar (opening ~ 1-5 cm) protects you from most of
your roomie’s awful music, but not from the bass track.

Also: That’s why all home theater speakers must be “in line of
sight”, but subwoofers can be behind the sofa.
Realistic double-slit

We need to take diffraction into account!
Superposition of interference and diffraction effects:

                  2
Interference: IP  Imax cos2                          d sin 
2                 
2
        
 sin     
2                2
Diffraction:    IP  I max                            a sin 
 
         


 2       

2
        
 sin     
2         2
The whole nine yards:     IP  Imax cos              
 2  
         

 2       
 d

Two slit interference
m
MAXIMUM when sin 
d

Each slit diffraction                  a
m
MINIMUM when sin 
a

Interference pattern modulated
by diffraction pattern
(with d >>a )

DEMO:
Double-slit
Example: Two-slit interference-diffraction

Light of λ = 550 nm illuminates two slits of width 0.03 mm and
separation 0.15 mm. How many interference maxima fall within
the full width of the central diffraction maximum?

First diffraction minimum: a sin 1  
 550  10 9 m
sin 1                    1  1.05
a 0.03  10 3 m
d sin 1 d
Interference maxima: d sin  m           m          5
     a
Central maxima + 5 on each side = 11

(But those at   1.05 are not visible
because of the diffraction minimum)
0

–1       1

–2            2

–3                     3

–4                          4
(–5)                                   (5)

```
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