Physics 2102
Jonathan Dowling
Lecture 40: MON 27 APR
Ch. 36: Diffraction
Things You Should Learn from
This Lecture
1. When light passes through a small slit, is spreads out and produces a
diffraction pattern, showing a principal peak with subsidiary maxima
and minima of decreasing intensity. The primary diffraction maximum
is twice as wide as the secondary maxima.
2. We can use Huygens’ Principle to find the positions of the diffraction
minima by subdividing the aperture, giving θmin = ±p λ/a, p = 1, 2, 3, ... .
3. Calculating the complete diffraction pattern takes more algebra, and
gives Iθ=I0[sin(α)/α]2, where α = π a sin(θ)/λ.
4. To predict the interference pattern of a multi-slit system, we must
combine interference and diffraction effects.
Single Slit Diffraction
When light goes through a
narrow slit, it spreads out
to form a diffraction
pattern.
Analyzing Single Slit Diffraction
For an open slit of width a, subdivide the opening into segments and
imagine a Hyugen wavelet originating from the center of each segment. The
wavelets going forward (θ=0) all travel the same distance to the screen and
interfere constructively to produce the central maximum.
Now consider the wavelets going at an angle such that λ = a sin θ ≅ a θ.
The wavelet pair (1, 2) has a path length difference Δr12 = λ/2, and
therefore will cancel. The same is true of wavelet pairs (3,4), (5,6), etc.
Moreover, if the aperture is divided into p sub-parts, this procedure can be
applied to each sub-part. This procedure locates all of the dark fringes.
p
!
= sin " p # " p ;
a
p = 1, 2, 3, ! (angle of the p th dark fringe)
Conditions for Diffraction
Minima
p
!
= sin " p # " p ; p = 1, 2, 3, !
a
(angle of the p th dark fringe)
Pairing and Interference
Can the same technique be used to find the maxima, by choosing pairs
of wavelets with path lengths that differ by λ?
No. Pair-wise destructive interference works, but pair-wise
constructive interference does not necessarily lead to maximum
constructive interference. Below is an example demonstrating this.
Calculating the
Diffraction Pattern
We can represent the light through the
aperture as a chain of phasors that “bends” and
“curls” as the phase Δβ between adjacent
phasors increases. β is the angle between the
first and the last phasor.
Calculating the
Diffraction Pattern (2)
E! = 2r sin (" / 2 )
! = Emax / r ; r = Emax / !
E! =
Emax
sin "
sin (# / 2 ) = Emax
# /2
"
#&
! "a
=
sin $
2
%
I = CE 2
# sin " $
I! = I max %
&
' " (
2
Minima : ! = ± m" or a sin # = ± m$
Diffraction Patterns
1
λ = 633 nm
0.8
a = 0.25 mm
0.6
0.5 mm
0.4
1 mm
0.2
- 0.03
!a
"=
sin #
$
2 mm
- 0.02
- 0.01
# sin " $
I! = I max %
&
' " (
0
θ (radians)
2
Blowup
0.01
0.02
0.03
The wider the slit opening a, or the
smaller the wavelength λ, the narrower
the diffraction pattern.
Radar: The Smaller The Wavelength the Better The Targeting Resolution
X-band:
λ=10cm
Ka-band:
λ=1cm
K-band:
λ=2cm
Laser:
λ=1 µm Angles of the Secondary Maxima
The diffraction
minima are precisely at
the angles where
sin θ = p λ/a and α = pπ
(so that sin α=0).
However, the
diffraction maxima are
not quite at the angles
where sin θ = (p+½) λ/a
and α = (p+½)π
(so that |sin α|=1).
0.05
0.04
0.03
1
λ = 633 nm
a = 0.2 mm
# sin " $
I! = I max %
&
' " (
2
0.02
0.01
2
0.005
3
0.01
p
(p+½) λ/a
θMax
1
0.00475
0.00453
2
0.00791
0.00778
3
0.01108
0.01099
4
0.01424
0.01417
5
0.01741
0.01735
4
0.015
5
0.02
θ (radians)
0.025
0.03
To find the maxima, one must look near sin θ = (p+½) λ/a, for places
where the slope of the diffraction pattern goes to zero, i.e., where
d[(sin α/α)2]/dθ = 0. This is a transcendental equation that must be solved
numerically. The table gives the θMax solutions. Note that θMax < (p+½) λ/a.