Chapter 11 Fraunhofer Diffraction

Chapter 11
Fraunhofer Diffraction
Lecture Notes for Modern Optics based on Pedrotti & Pedrotti & Pedrotti
Instructor: Nayer Eradat
Instructor: Nayer Eradat
Spring 2009
4/30/2009
Fraunhofer Diffraction
1
Diffraction
•
•
•
•
•
Diffraction is any deviation from geometrical optics that results from obstruction of the wavefront of light. Obstruction causes local variations in the amplitude or phase of the wave
Diffraction can cause image blurriness. So the aberrations. An optical component that is free of aberrations is called diffraction‐
p
p
limited optics and is still subject to blurriness due to diffraction.
Huygens‐Fresnel principle: every point on a wavefront can be considered as a source of secondary spherical wavelets(Huygens). The field beyond a y p
( yg )
y
wavefront is result of the superposition of these wavelets taking into account their amplitudes and phases (Fresnel).
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Fraunhofer Diffraction
2
Diffraction vs. interference:
Diffraction phenomena is calculated from interference of the waves
from interference of the waves originating from different points of a continuous source. Interference phenomena is calculates interference of the beams origination
interference of the beams origination from discrete number of sources. 4/30/2009
Fraunhofer Diffraction
3
Mathematical treatment of Diffraction
Mathematical treatment of Diffraction
•
•
•
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•
Far‐field or Fraunhofer Diffraction or : when both the source and observation screen are far from the diffraction causing aperture, so that the waves arriving at the aperture and screen can be approximated by plane waves. Near‐field or Fresnel diffraction: when the curvature of the wavefronts an not be ignored
We will only investigate the far‐field diffraction in this course using the Huygens‐
Fresnel principle with one approximation. When EM waves hit the edges of the aperture there is oscillations of the electrons
When EM waves hit the edges of the aperture there is oscillations of the electrons in the matte that cause a secondary field (edge effect). In HF approximation we ignore edge effect. So beyond the aperture there is no field. This holds only if the observation point is far from the aperture.
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Fraunhofer Diffraction
4
Diffraction from a single slit
We simulate the ggeometrical arrangement
g
for the Frounhofer diffraction by
y placing
p
g a point
p
source at the
focal point of a lens and a screen at the focal point of a lens after the amerture. The light reaching point
P on the screen is from interference of the parallel rays from different points on the aperture.
We consider each interval of ds as a source and calculate contribution of the all of these sources.
r : optical path length from the ds to the P
EL strength of the electric field contribution from each point.
r = r0 for the field from the center of the slit.
C
Contribution
ib i off eachh interval
i
l ds
d to the
h wavefront
f
at point
i P is
i a spherical
h i l wavelet
l off dEP :
⎛ E ds ⎞
dEP = ⎜ L ⎟ ei( kr -ωt )
r ⎠
⎝
Spherical
p
wave
amplitude
⎛ E ds ⎞ i k ( r +Δ )-ωt )
dEP = ⎜ L ⎟ e ( 0
⎝ r0 + Δ ⎠
⎛ E ds
d ⎞ i( kr0 −ωt ) ik Δ
dEP = ⎜ L ⎟ e
e
r
+
Δ
0
Path difference
⎝
⎠ affets
the phase
Path difference
affets the amplitude
We ignore Δ in the denuminator
since Δ << r0
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Fraunhofer Diffraction
5
Diffraction from a single slit
⎛ E ds ⎞ i kr −ωt
dEP = ⎜ L ⎟ e ( 0 ) eik Δ
⎝ r0 ⎠
The total electric field at the point P
EP =
∫
dEP =
slit
EL i( kr0 −ωt ) b/2 iks sinθ
e
∫−b / 2 e ds
r0
EL i( kr0 −ωt ) ⎛ eiks sin θ ⎞
EL i( kr0 −ωt ) e( ikb sinθ ) / 2 − e −( ikb sinθ ) / 2
=
EP =
e
e
⎜
⎟
sin
θ
r0
ik
r
ik sin θ
⎝
⎠−b / 2
0
1
With β = kb sin θ and using Euler's formula
2
E b sin β i( kr0 −ωt ) EL b sin cβ
=
EP = L
e
r0 β
r0
b/2
β varies
i with
i h θ andd that
h varies
i with
i h the
h distance
di
from
f
the
h screen.
b
2
1
2
β can be interpreted as phase difference kΔ =k sin θ → β = k Δ = kb sin θ
β shows the magnitude of the phase difference at point P between the points from the center and
either endppoint of the slit.
The irradiance I at P is proportional to the E02 :
2
2
ε 0 c ⎛ EL b ⎞
i 2β
sin
i 2β
⎛ ε 0 c ⎞ 2 ε 0 c ⎛ EL b ⎞ sin
with I0 =
I =⎜
E0 =
= I 0 sin c 2 β
⎜
⎟
⎜
⎟ we have I = I 0
⎟
2
2
2 ⎝ r0 ⎠ β
2 ⎝ r0 ⎠
β
⎝ 2 ⎠
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Fraunhofer Diffraction
6
Diffraction from a single slit
The diffraction pattern of the light from a silt by width of b at a distanc screen I = I 0
sin 2 β
β
2
= I 0 sin c 2 β where
β = k Δ = ( kb sin θ ) /2 is the phase difference betwwen the light from the center of the slit the endpoints.
⎛ sin β
Let's plot the I. For the central maximum: lim sin c ( β ) = lim ⎜
β →0
β →0
⎝ β
For zeros of the sinc function:
⎞
⎟ =1
⎠
1
( kb sin θ ) = mπ → b sin θ = mλ where m = ±1, ±2,...
2
m = 0 does
d
nott lead
l d to
t a zero be
b cause off th
the special
i l property
t off the
th sinc
i function.
f ti
if f is the distance of the slit from the screen, then the location of the minima on the screen can be found using
sin β = 0 → β =
small-angle approximation:
sin θ ≅ tan θ = y / f →
mλ ym
mλ f
=
→ ym =
b
f
b
d ⎛ sin β ⎞ β cos β − sin β
=0
⎜
⎟=
β2
dβ ⎝ β ⎠
Cental lobe: image of the slit with roundes edges.
For the other maxima:
Side lobes: what causes blurriness in the image.
Angular width of the central lobe: b sin θ = λ →
θ1 ≅
λ
b
, θ −1 ≅
4/30/2009
λ
b
, → Δθ =
2λ
The central lobe will sprad as the slit-size gets smaller.
b
Fraunhofer Diffraction
7
Maxima of the sinc function
The maxima of the sinc function are solutions of this equation:
β cos β − sin
i β
= 0 → tan β = β
2
β
We can solve this equation by parametrically.
A angle
An
l equall to its
i tangant is
i intersection
i
i
⎧y = β
of the ⎨
⎩ y = tan β
We see that the maxima are not exactly half
way between the minima. They occur slightly
erlier and as β increases the maxima shift
towards the center.
Ratio of the irradiances at the central peak
maximum to
to the first of the secondary maxima ?
I β =0
I β =1.43π
( sin β / β )β =0
=
2
( sin β / β )β =1.43π
2
2
=
1
( sin β / β )β =1.43π
2
⎛ 1.43π ⎞
20.18
=⎜
=
⎜ sin (1.43π ) ⎟⎟ 0.952
⎝
⎠
I β =1.43π = 0.047 I β =0 or only 4.7% of the I0
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Fraunhofer Diffraction
8
Beam spreading
The angular spread of the central maximm Δθ =2λ /b is independent of the distance between the slit
aandd screen.
sc ee . So as screen
sc ee moves
oves away from
o tthee slit
s t the
t e nature
atu e of
o the
t e diffraction
d act o patte
pattern does not
ot change
c a ge..
W is the width of the central maximum,
W=LΔθ =
2 Lλ
b
All of the beams spread according to diffraction as they propagate due to the finite size of the source .
Even if we make them parallel with a lens still they will spread because of the diffraction.
Example:
What is the width of a parallel beam of λ = 546nm and width of b=0.5mm after propagation of
10 meter?
2 Lλ 2 ×10 × 546 ×10−9
W=
=
b
0.5 ×10-3
W = 21.8mm
This treatment of beam spreading is
correct for far-field where L >>
or more generally
generall L >>
4/30/2009
b2
λ
p
area of aperture
λ
Fraunhofer Diffraction
9
Rectangular aperture
For a slit of length a and width b we calculate the diffraction pattern. We assumed a>>b in previous section.
When a and b are comparable and both small we have large contriibutions to the diffractio
diffraction
n pattern from
2
⎛ sin β ⎞
⎛ sin α ⎞
⎛k⎞
⎛k⎞
both dimensions: I = I 0 ⎜
⎟ where α = ⎜ ⎟ a sin θ and I = I 0 ⎜
⎟ where β = ⎜ ⎟ b sin θ
⎝ α ⎠
⎝2⎠
⎝2⎠
⎝ β ⎠
2
mλ f
nλ f
andd x n =
m,n= ± 1,
1 ± 2,...
2
b
a
b/2
E i kr −ωt a / 2
Using a bit more analysis we can write EP = ∫ dEP = L e ( 0 ) ∫ eikXx / r0 dx ∫ eikYy / r0 dy
−a / 2
−b / 2
r0
slit
Z
Zeros
off the
th pattern
tt
due
d occur at:
t ym =
X , Y are the
h coordinates
di
off the
h observation
b
i point
i on the
h screen andd x,y are the
h coordinates
di
off the
h surface
f
element on the aperture. The total irradiance turns out to be the product of the irradiance functions on each
(
)(
dimension: I = I 0 sin c 2 β sin c 2α
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)
Fraunhofer Diffraction
10
Circular aperture
For finding the diffraction pattern caused by a circular aperture we assume the incremental electric
field amplitude at point P due to the surface element dA = dxdy (on the aperture) is E A dA / r0 .
Then we integrate the incremental field over the entire aperture area. The resulting E at point P is
⎛ E dA ⎞ i k ( r +Δ )-ωt )
E A i( kr0 −ωt )
Δ<< r0
dEP = ⎜ A ⎟ e ( 0
E
e
eisk sinθ dA
⎯⎯⎯⎯
→
=
P
Δ= s sin θ
∫∫
r0
⎝ r0 + Δ ⎠
A
Area
By choosing the elemental area the rectangular area of
dA = xds we can reduce the double integral to a single one.
2
⎛x⎞
2
2
2
2
⎜ ⎟ +s = R → x =2 R −s
⎝2⎠
2 E A i( kr0 −ωt ) + R isk sinθ R 2 − s 2
EP =
e
ds
∫− R e
r0
Substituting v = s / R and γ = k sin θ
2 E A R i( kr0 −ωt ) +1 iγ v 1−v2
EP =
e
dv
∫−1 e
r0
From the intergral table:
∫
+1
−1
eiγ v
1− v 2
dv =
π J1 ( γ )
γ
where J1 ( γ ) is the first order Bessel function of the first kind.
J1 ( γ ) =
γ
2
4/30/2009
(γ / 2 )
−
12.2
3
(γ / 2 )
+
5
12.22.3
J1 ( γ )
1
γ →0
γ
2
Fraunhofer Diffraction
− " and lim
=
11
Field from an arbitrary shape aperture on a distant screen For finding the diffraction pattern caused by an aperture of arbitrary shape we assume the incremental
electric field amplitude at point P due to the surface element dA = dxdy (on the aperture) is E A dA / r0 .
Then we integrate the incremental field over the entire aperture area. The resulting E at point P is
Ep =
EA
r0
∫∫
ei(ωt − kr ) dA
where dA = dxdy
Aperture
1/ 2
r = ⎡( X − x ) + ( Y − y ) + Z ⎤
⎣
⎦
2
2
(
2
)
2
⎡ 2 ( Xx + Yy ) ⎤
r0 2 >> x 2 + y 2 → r r0 ⎢1 −
⎥
2
r
0
⎣
⎦
1/ 2
E
EP = A
r0
∫∫
Aperture
e
( Xx + Yy )
r0
⎛
⎛
( Xx +Yy ) ⎞ ⎞
i ⎜ ωt − k ⎜⎜ r0 −
⎟⎟ ⎟⎟
⎜
r0
⎝
⎠⎠
⎝
Y
x
P
1/ 2
⎡
x2 + y 2
2 ( Xx + Yy ) ⎤
⎥
we have: r = r0 ⎢1 +
−
2
2
r0
r0
⎢⎣
⎥⎦
r0 >> x + y → r r0 −
y
2 1/ 2
with r0 = ⎡⎣ X + Y + Z ⎤⎦
2
only first two terms.
dA → EP =
Δ
dy
r
θ
dx
Aperture
r0
P0
Screen
EL i(ωt − kr0 )
ik ( Xx +Yy ) / r0
dxdy
e
e
∫∫
r0
Aperture any shape
For specific shape of aperture the integral limits and relationship between x and y
will change. We will look at two examples of rectangular and circular apertures.
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Fraunhofer Diffraction
12
X
Rectangular aperture
EP =
E A i(ωt − kr0 )
ik ( Xx +Yy ) / r0
e
e
dxdy
∫∫
r0
Aperture
any shape
a / 2 ik Xx / r
E A i(ωt − kr0 ) b / 2 ik (Yy ) / r0
( ) 0
EP =
e
e
dy
e
dx
∫
∫
b
a
−
−
/
2
/
2
r0
⎞ ⎛ sin α ⎞
⎟a⎜
⎟
α
⎝
⎠
⎠
AE A i(ωt − kr0 ) ⎛ sin β ⎞ ⎛ sin α ⎞
Ep =
e
⎜
⎟⎜
⎟
r0
β
α
⎠
⎝
⎠⎝
Ep =
E A i(ωt − kr0 ) ⎛ sin β
e
b⎜
r0
⎝ β
⎛ AE A i(ωt − kr0 ) ⎞
I0 = ⎜
e
⎟
r
⎝ 0
⎠
⎛ sin β ⎞ ⎛ sin α ⎞
I ( X ,Y ) = I ( 0) ⎜
⎟ ⎜
⎟
β
α
⎠
⎝
⎠ ⎝
4/30/2009
y
x
r
Δ
θ
dy
dx
x
P
r0
P0
2
2
Where α =
y
2
Aperture
k X k
k Y k
a = asinθ x and β = b = bsinθ y
2 r0 2
2 r 2
Fraunhofer Diffraction0
Screen
13
Circular aperture I
For circular aperture we introduce the spherical coordinates because of the symmetry
of the problem. On plane of the aperture: x = ρ cos φ , y = ρ sin φ
On plane of the screen: X = q cos Φ, Y = q sin Φ
Differential surface element: dA = ρ d ρ dφ
R
2π
E i ωt − kr
E i ωt − kr
i k ρ q / r0 ) cos(φ −Φ
Φ)
EP = A e ( 0 ) ∫∫ eik ( Xx +Yy ) / r0 dA = L e ( 0 ) ∫ ∫ e (
ρ d ρ dφ
r0
r0
Aperture
ρ =0 φ =0
any shape
Wher we used: Yy + Xx = ρ q ( sin φ sin Φ + cos φ cos Φ ) = ρ q cos (φ − Φ )
Because of the axial symmetry of the problem, the solution must be independent of Φ and
since point P is an arbitrary point on the screen we choose to have Φ = 0 to simplify things.
a
E i ωt − kr
EP = A e ( 0 ) ∫ ρ d ρ
r0
ρ =0
2π
(
) ( )
∫φ =0 e 0 dφ
i k ρ q / r cos φ
y
A tabulated integral. Solution is a
Bessel function of the first kind
i − m 2π i( mv +u cos v )
dx Δ
Bessell function
f
i 1st kind:
ki d J m ( u ) =
e
d
dv
∫
2π 0
dy
ρ
1 2π iu cos v
For m=0: J 0 ( u ) =
e
dv and with u = k ρ q
∫
0
2π
R
E A i(ωt − kr0 )
EP =
e
J 0 ( k ρ q / r0 ) ρ d ρ
∫
r0
ρ =0
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Fraunhofer Diffraction
y
x
r
θ
P
φ
r0
P0
14
x
Bessel functions of the first kind (MATLAB)
Bessel functions of the first kind (MATLAB)
u = (0:0.1:15)
BJ0=besselj(0,u);
BJ1=besselj(1,u);
BJ2=besselj(2 u);
BJ2=besselj(2,u);
BJ3=besselj(3,u)
plot(u,BJ0,u,BJ1,u,BJ2,u,BJ3);
legend('J0','J1','J2','J3')
title('Bessel
title(
Bessel functions of first kind');
kind );
xlabel('u'),ylabel('J')
grid on
Bessel functions of first kind
1
J0
J1
J2
J3
J
0.5
0
-0.5
05
0
5
10
15
u
15
Circular aperture II
R
E i ωt − kr
E P = A e ( 0 ) 2π ∫ J 0 ( k ρ q / r0 ) ρ d ρ
r0
ρ =0
Using the one of many properties of the
bessel functions:
R
∫ρ
∫
u
0
u ' J 0 ( u ' ) du ' = uJ1u we get
J 0 ( k ρ q / r0 ) ρ d ρ = ( r0 / kq )
2
=0
EP =
∫
kRq/r0
u '= 0
J 0 ( u ') u ' du ' = ( kRq / r0 ) J1 ( kRq / r0 )
⎛ kRq ⎞
⎛ kRq ⎞
r
2 AE A i(ωt − kr0 ) r0
E A i(ωt − kr0 )
E
2π R 2 0 J1 ⎜
→
=
e
e
J
⎟
⎟
P
1⎜
r0
kRq ⎝ r0 ⎠
r0
kRq ⎝ r0 ⎠
we use γ = kRq / r0 = kR sin θ and calculate the irradiance at:
I=
( Re( EP )
2
=
2 E A ⎡ J1 ( γ ) ⎤
1
EP EP* → I =
⎢
⎥
2
r
γ
⎣
⎦
p
of the J1 ( γ ) =
Series expansion
2 2
A
2
0
γ
2
(γ / 2 )
−
12.2
2
3
2
(γ / 2 )
+
5
12.2
22.3
3
− " → lim
γ →0
J1 ( γ )
γ
=
1
2
J1 behaves like a damping sin function and J1 ( γ ) / γ behaves like the sinc function.
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Fraunhofer Diffraction
16
Diffraction pattern of the circular aperture with Bessel functions in MATLAB
with Bessel functions in MATLAB
x=(-15.1:0.5:14.9);
y=(-15.1:0.5:14.9);
A=y.'*x;
i_index=0;
for i=-15.1:0.5:14.9
j_index=0;
i_index=i_index+1;
for j=-15.1:0.5:14.9
j_index=j_index+1;
(
j )
r=sqrt(i^2+j^2);
if r <=5
A(i_index,j_index)=1;
else A(i_index,j_index)=0;
end
end
e d
end
subplot(2,1,1);
mesh(x,y,A);
title('Circular Aperture')
axis([-15.1 14.9 -15.1 14.9 0 1]);
a=1;
kx=(-15
kx
( 15.1:0.5:14.9);
1:0 5:14 9);
ky=(-15.1:0.5:14.9);
[kax,kay]=meshgrid(kx,ky);
ka=sqrt(kax.^2+kay.^2);
Gka=2*pi*a^2.*besselj(1,ka)./(ka*a);
subplot(2 1 2);
subplot(2,1,2);
mesh(kx,ky,Gka);
xlabel('kx'); ylabel('ky');
axis([-15.1 14.9 -15.1 14.9 -1 4]);
title('Fourier Bessel of Circular Aperture')
Fraunhofer diffraction pattern is the Fourier transform of the aperture function.
We can show this by the following plots using MATLAB. This topic will be
covered in PHYS 258.
⎧⎪1
g ( x, y ) = ⎨
⎪⎩0
x2 + y 2 ≤ a
⎧1 r ≤ a
→ g R (r ) = ⎨
⎩0 r > a
x2 + y 2 > a
⎡ J (k a) ⎤
Diffraction pattern: G0 (kα ) = F ( kα ) = 2π a 2 ⎢ 1 α ⎥
⎣ kα a ⎦
17
Diffraction pattern of the circular aperture with B
Bessel functions in MATLAB
lf
i
i MATLAB
18
Diffraction pattern of the circular aperture with ( )
Fast Fourier Transformation(FFT) in MATLAB
%PHYS 258 spring 07, Nayer Eradat
%A program to plot a circular aperture function
%and its Fourier transform using fft and shift fft function
x=(-2:0.05:2);
y=(-2:0 05:2);
y=(-2:0.05:2);
A=y.'*x;
i_index=0;
for i=-2:0.05:2
j_index=0;
i_index=i_index+1;
_
_
for j=-2:0.05:2
j_index=j_index+1;
r=sqrt(i^2+j^2);
if r <=0.2
A(i_index,j_index)=1;
else
l A(i
A(i_index,j_index)=0;
i d j i d ) 0
end
end
end
subplot(2,1,1);
mesh(x y A); %3D plot
mesh(x,y,A);
xlabel('x'); ylabel('y'); zlabel('E');
title('Circular aperture');
fft_v=abs(fft2(A));
fft_val=fftshift(fft_v);
%shift zero-frequency component to center of spectrum
subplot(2,1,2);
mesh(x,y,fft_val);
xlabel('fx'); ylabel('fy'); zlabel('E');
title('fft of Circular aperture');
19
Circular aperture diffraction pattern
Circular aperture better than slit or rectangular
aperture for imaging.First
imaging First zero of the circular aperture:
γ =3.832
Thus for an aperture of diameter D we have
1
γ = kD sin θ = 3.832 → D sin θ = 1/ 22λ
2
Airy Disc: the diffracted image of a circular aperture
or the central lobe of the diffraction pattern.
For far-field sinθ θ and the angular half-width of
the Airy disc is: Δθ1/2 =
1.22λ
D
Example:
The beam spread for a beam (λ =564nm) from an
amperure D=0.5
D
0 5 mm at L
L=10m.
10
JJJJJJJJJJJM
The diameter of the Airy Disc is:
Δθ1/ 2 = 1.22λ / D = 1.33 ×10-3 rad and then
rd = LΔθ1/ 2 = 13mm
That is why we pay high $$ for the large lenses. They ca provide better image resolution.
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Fraunhofer Diffraction
20
Resolution and Rayleigh criterion
Rayleigh'ss criterion for just-resolvable
Rayleigh
just resolvable
images requires that the angular
separation of the centers of the image
pattern
tt
be
b no less
l than
th the
th angular
l
radius of the airy disc.
1.22λ
D
Maximum of one pattern falls directly
Δθ min =
under minimum of the next pattern.
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Fraunhofer Diffraction
21
Diffraction by the eye Diffraction
by the eye
Pupil size limits the resolution of image of the objects subtended by Δθmin
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Fraunhofer Diffraction
22
Double‐slit diffraction I
Diffraction pattern of a plane wavefront that is obstructed everywhere but at the two slits shown in fig.
We follow our analysis for the single slit. The diffraction pattern for two slits is going to be superposition
of the patterns by each slit. Therefore we write:
EP =
∫ dE
P1
+
slit1
EP =
∫
− (1/ 2)( a −b )
dEP 2
slit 2
(1/ 2)( a + b )
E i kr −ωt
E i kr −ωt
= L e ( 0 ) ∫ eiks sin θ ds + L e ( 0 ) ∫ eiks sinθ ds
r0
r0
(1/ 2)( a −b )
− (1/ 2)( a + b )
EL i( kr0 −ωt ) 1 ⎡ (1/ 2)ik ( − a +b ) sin θ
(1/ 2 )ik ( − a −b ) sin θ
(1/ 2 )ik ( a + b ) sin θ
(1/ 2 )ik ( a −b ) sin θ ⎤
−
+
+
e
e
e
e
e
⎦
r0
ik sin θ ⎣
With β = (1/ 2 ) kb sin θ and α = (1/ 2 ) ka sin θ
(
)
(
)
EP =
EL i( kr0 −ωt ) b ⎡ iα iβ
iβ
− iα
iβ
iβ
⎤
−
+
−
e
e
e
e
e
e
e
⎣
⎦
r0
2i β
EP =
EL i( kr0 −ωt ) b ⎡ iα
− iα
+
e
e
e
2i β ⎣
r0
(
)( e
iβ
)
− eiβ ⎤⎦
Using Euler's equation:
E i kr −ωt b
E
2b sin β
⎡⎣( 2 cos α )( 2i sin β ) ⎤⎦ = L ei( kr0 −ωt )
EP = L e ( 0 )
cos α
2i β
β
r0
r0
EP = E0 e (
i kr0 −ωt )
where E0 =
EL 2b sin β
cos α
β
r0
The irradiance at point P is:
2
2
2
⎛ sin β ⎞
⎛ε c⎞
⎛ ε c ⎞ ⎛ 2bEL ⎞ ⎛ sin β ⎞
⎛ ε 0 c ⎞ ⎛ 2bEL ⎞
2
2
→
=
=
I = ⎜ 0 ⎟ E02 = ⎜ 0 ⎟ ⎜
α
I
I
α
I
cos
4
cos
where
⎟ ⎜
0⎜
0
⎟
⎟
⎜ 2 ⎟⎜ r ⎟
⎝ 2 ⎠
⎝ 2 ⎠ ⎝ r0 ⎠ ⎝ β ⎠
⎝
⎠⎝ 0 ⎠
⎝ β ⎠
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23
Double‐slit diffraction II
With the modified interference pattern for double-slit due to diffraction of each slit we have:
2
⎛ sin
i β⎞
bEL ⎞
⎛ ε 0 c ⎞ ⎛ 22bE
where
I
=
I = 4 I 0 cos 2 (α ) ⎜
0
⎜ 2 ⎟⎜ r ⎟
⎝ β ⎟⎠
⎝
⎠⎝ 0 ⎠
Interference pattern of the double-slit
2
Diffraction pattern
of a single slit
ka sin θ π a sin θ
=
and β = kb sin θ
2
λ
= 4 I 0 which is expected for the coherent sources.
where α =
Now I max
I fi
In
figure below
b l a=6b
6b andd α =66β thus
th the
th cosα varies
i muchh more rapidly
idl than
h sinc
i 2β
We say the interference pattern of the doublle slit is modulated by the single slit diffraction pattern.
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Fraunhofer Diffraction
24
Diffraction from many slits I
Diffraction pattern of a plane wavefront that is obstructed everywhere but att the two slits
shown in fig. We follow our analysis for the single slit. The diffraction pattern for two slits
is going to be superposition of the patterns by each slit. Therefore we write:
⎡−( 2 j −1) a + b ⎦⎤ / 2
⎣⎡( 2 j −1) a + b ⎦⎤ / 2
⎫⎪
EL i( kr0 −ωt ) N/2 ⎧⎪ ⎣
E
ω
i
kr
−
t
( 0 )
iks sin θ
iks sin θ
L
EP = ∑ ∫ dEPi =
e
e
ds +
e
e
ds ⎬
⎨
∑
∫
∫
r0
r
j =1 slit ( j )
jj=
=1 ⎪ ⎡−
0
⎡⎣ ( 2 j −1) a −b ⎤⎦ / 2
⎡⎣( 2 j −1) a −b ⎤⎦ / 2
⎪
⎩
⎭
N /2
K
Here we are considering the pairs of slits that are symmetrical with respect to the center of the grating.
J = 1, double slit, j = 2, 4 slits, etc. With β = (1/ 2 ) kb sin θ and α = (1 / 2 ) ka sin θ
(
)
(
)
K=
EL i( kr0 −ωt ) b ⎡ iα iβ
iβ
− iα
iβ
iβ
⎤
e
e
e
e
e
e
e
−
+
−
⎣
⎦
2i β
r0
K=
sin β
b
Re ⎡⎣ei( 2 j −1)α ⎤⎦
( 2i sin β ) ( 2 cos ( 2 j − 1) α ) = 2b
2i β
β
N/2
S = ∑ 2b
j =1
sin β
β
Re ⎡⎣ei( 2 j −1)α ⎤⎦ = 2b
sin β
β
Re ⎡⎣eiα + ei 3α + ei 5α + ... + ei( N −1)α ⎤⎦
Geometric series
⎛ r n −1 ⎞
For a geometric series a + ar + ar + ..... + ar = a ⎜
⎟
⎝ r −1 ⎠
⎡
⎛ e 2iα N / 2 − 1 ⎞ ⎤
sin β
⎟⎥
first tem a = eiα and ratio r = e 2iα then S = 2b
Re ⎢e 2iα ⎜
2
i
α
⎢
⎜ e − 1 ⎟⎥
β
⎝
⎠ ⎦⎥
⎣⎢
2
n
( )
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25
Diffraction from many slits II
Continued fromlast page:
( )
⎛ e 2iα N / 2 − 1 ⎞
⎛ eiNα − 1 ⎞
2 iα ⎜
⎟
e
=
=
⎜ e 2iα − 1 ⎟ ⎜⎝ eiα − e − iα ⎟⎠
⎝
⎠
cos Nα + i sin Nα − 1 − sin Να + i ( cos Nα − 1)
=
2i sin
i α
−2sin
2 i α
⎡ − sin Να + i ( cos Nα − 1) ⎤
sin β
sin β sin Να
Re ⎢
S = 2b
⎥→S =b
−2sin α
β
β sin α
⎣
⎦
2
Secondaryy
maxima
⎛ sin β ⎞ ⎛ sin Να ⎞
E i kr −ωt ⎧ sin β sin Να ⎫
EP = L e ( 0 ) b ⎨
⎬ and I = I 0 ⎜
⎟
sin
r0
β ⎠⎟ ⎜⎝
α ⎠
⎩ β sin α ⎭
⎝
Diffraction by a
single slit
2
Interference
between N slits
Principal maxima
With β = (1/ 2 ) kb sin θ and α = (1/ 2 ) ka sin θ
sin Να
N cos Να
lim
= lim
= ± N this resembles a series of
a → mπ sin α
a → mπ
cos
α
Limiting diffraction envelope
i d
indeterminate
i
sharp maxima that we call principal maxima.
I principal maxima ∝ N 2 cantered at values α = 0, ±π , ±2π , ±3π
B t
Between
successive
i peaks
k there
th are N − 2 secondary
d peaks.
k
The full irradiance is product of the diffraction pattern and interference pattern.
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26
Formation of secondary maxima
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