Diffraction

Diffraction
Deviation of light from rectilinear propagation caused by the
obstruction of light waves, by a physical obstacle.
There is no significant physical distinction between interference and
diffraction. However, interference generally involves a superposition
of only a few waves whereas diffraction involves a large number of
waves.
Huygens-Fresnel Principle: Every unobstructed point of a wavefront
serves as a source of spherical secondary wavelets. The amplitude of
the optical field at any point is the superposition of these wavelets.
Suppose light strikes a screen containing an aperture. The effects of
diffraction can be understood by considering the phasor addition of
the electric field for an array of point sources emitting E-M waves
that are unobstructed by the aperture.
The principles of diffraction can be understood by considering the
interference of water waves whose wavelength is made to vary in
comparison to the size of the aperture.
To illustrate diffraction consider water waves in
a ripple tank: (OPD )max = Λ max = AP − BP < AB
Destructive
interference of
most phasors
for large
angles; N large
shadow region
Effective number
of point sources
N for phasor
construction
(a ) λ < Λ max
Intermediate
region; N
medium, mixed
interference
N=5
(b) λ ≅ Λ max
Constructive
interference of
most phasors for
all angles; N
small Approx.
spherical wave
(c) λ > Λ max
Light waves striking an aperture
a
Fresnel Diffraction (near-field
diffraction) when incoming and
outgoing waves are both nonplanar.
R
Source S and Screen C are moved
to a large distance R, resulting in
Fraunhofer diffraction (Far-field
diffraction). As a rule-of-thumb:
R > a2/λ for Fraunhofer diffraction.
Both incoming and outgoing
wave contributions are
considered as plane waves,
and θ = const.
Fraunhofer diffraction conditions
produced by lenses, leaving source
S and screen C in their original
position. Note that the lenses act to
filter non-plane wave components
from striking point P on the
observation screen C.
R6 > R5 > R4 > R3 ….
R6
Far field
a
Fraunhofer
R5
The diffraction pattern in the near-field
(bottom) is sensitive to variations in R whereas
the shape is independent of distance for R
large (top).
The sharp structure shown in the near-field
pattern is a result of rapidly changing phasor
(E-field) orientations whose summations at
each point on the screen are sensitive to
distance, slit geometry, and angular spread of
the waves. As shown, small changes in R in
the near-field can cause large changes in the
resulting phasor addition and irradiance
distribution on the screen. If a >> λ, the
R4
R3
R2
R1
Consider a series of diffraction measurements
on a screen whose distance (R) from the slit
varies from close (bottom) to far (top).
Near field
Fresnel
pattern will resemble a sharp geometric
shadow for near-field distances.
For large distances (e.g., R6 > a2/λ), the parallel
nature of the plane waves will result in phasor
additions yielding smooth distribution in
irradiance, whose shape is independent of R.
As a bridge between interference and diffraction, consider a linear array of
equally spaced N coherent point oscillators. We examine the superposition of
the fields from each source at a point P sufficiently far such that rays are nearly
parallel.
The field of each wave is equal
so that E0(r1) = E0(r2) = … =
E0(rN) = E0(r).
In order to evaluate the field, we
employ a phasor sum, as follows:
~
E = E0 (r )ei ( kr1 −ωt ) + E0 (r )e i ( kr2 −ωt )
+ ...E0 (r )ei ( krN −ωt )
= E0 (r )ei ( kr1 −ωt ) [1 + eik ( r2 − r1 ) + eik ( r3 − r1 )
+ ...eik ( rN − r1 ) ]
Due to a difference in OPL, there is a phase difference
between adjacent sources of δ = k0Λ, where Λ = ndsinθ,
and δ = kdsinθ. From the figure, δ = k(r2-r1), 2δ = k(r3-r1),
3δ = k(r4-r1), …
Therefore, at point P, the phasor sum is obtained using a geometric series
(similar to what was done for multiple beam interference) and yields:
[
~
E = E0 (r )e −iωt e ikr1 1 + e iδ + e 2iδ + e 3iδ + ... + e ( N −1)iδ
= E0 (r )e −iωt e ikr1
(e δ − 1) = E (r )e
(e δ − 1)
iN
i
0
−iωt ikr1
e
]
[
e iNδ / 2 e iNδ / 2 − e −iNδ / 2
e iδ / 2
[e
iδ / 2
− e − iδ / 2
]
]
~ ~*
2
(
)
( Nδ / 2 )
sin
/
2
δ
sin
N
E
E
= E0 (r )e −iωt e ikr1 e i ( N −1)δ / 2
, I=
= I0
sin (δ / 2 )
2
sin 2 (δ / 2 )
Note that as δ → 0, I → N 2I0,as expected for N coherent sources in-phase.
Also note that for N = 2, I = 4I0cos2(δ/2) (interference of two slits)
sin 2 Nδ / 2
2
lim
=
N
This is true for δ → 0 and for δ = 2mπ, m = 0, ±1, ±2, …
δ →0 sin 2 δ / 2
2
constructive interaction
kd / 2 ⋅ sin θ m = mπ
when
d sin θ m = mλ
Im = N I 0
If d< as expected from a line source of electron oscillators, there is only one
maximum at m=0
Fraunhofer diffraction from a single slit
Consider now a line source of oscillators, where each point in a slit of width
D emits a spherical wavelet as shown in the figure. The electric field for a
spherical wavelet emitted from a point is E = ε 0 sin (ωt − kr )
r
where ε0 is the source strength for a point emitter.
(90° - θ )
y
Y
Y
sin θ =
R
Let εL = source strength per unit length. Then for
each differential segment of source length dy, the
contribution of the spherical wavelet at point P is
ε dy
dE = L sin (ωt − kr ) From geometry,
r
r 2 = y 2 + R 2 − 2 yR cos(90° − θ )
= y 2 + R 2 − 2 yR sin θ ,
y sin θ
y2
2 y sin θ
y2
+1−
≅ R 1−
+ 2 + ...
r=R
2
R
2R
R
R
≅ R − y sin θ + ... where R >>D and D > y.
Note that in order to fulfill Fraunhofer (far-field) diffraction conditions, we also
require R > D 2/λ. Thus r ≈ R - ysinθ is the resulting approximation that we will
use. We can calculate the total field appearing on the screen at angle θ, by
summing or integrating all contributions of dE on the linear source:
E=
+D/2
dE =
−D / 2
+D/2
dy
−D / 2
εL
r
sin (ωt − kr ) ≅
+D/2
dy
−D /2
εL
R
sin[ωt − k ( R − y sin θ )]
Note that we are also ignoring the presence of ysinθ in the denominator since
its presence in the phase of the sine in the numerator has a much greater
contribution to the integral.
We can further separate terms in the sine by using the following identity:
sin (ωt − kR + ky sin θ ) = sin (ωt − kR ) cos(ky sin θ ) + cos(ωt − kR )sin (ky sin θ )
Since sin(kysinθ) is an odd function in y, the last term cancels in the integral.
Therefore, the integral is as follows:
εL
+D/2
2ε
E=
dy sin (ωt − kR ) cos(ky sin θ ) = L
R −D / 2
R
=
+D/2
dy cos(ky sin θ )sin (ωt − kR )
0
2ε L sin ((kD / 2) sin θ )
ε D sin ((kD / 2) sin θ )
sin (ωt − kR )
sin (ωt − kR ) = L
(kD / 2) sin θ
R
R
k sin θ
So let β ≡ (kD / 2) sin θ
E=
ε L D sin β
sin (ωt − kR ),
R
β
and the time − averaged irradiance is
I (θ ) =
1 εLD
2 R
2
sin β
β
2
= I (0)
sin β
2
β
Consider the image of a point source S on a screen passing through a slit of
width b and length l. Lenses L1 and L2 produce condition of Fraunhofer
diffraction through the slit as seen in the next slide.
Clearly, I(θ) have minima when sinβ = 0 and β = mπ, m = ±1, ±2, ±3,…
Note β = (kb / 2) sin θ = mπ
2π b
sin θ = mπ
λ 2
b sin θ = mλ
(minima)
Note that the condition for diffraction minima, bsinθ = mλ, should
not be confused with a similar expression for two-slit interference
maxima, asinθ = mλ.
Fig. 10.6 (a) Single-slit Fraunhofer diffraction.
(b) Diffraction pattern of a single vertical slit
under point-source illumination.
Fraunhofer diffraction using lenses so that the
source and fringe pattern can both be at
convenient distances from the aperture.
We can determine the conditions for maxima in the diffraction pattern by
setting the derivative of the Irradiance to zero:
I = I (0 )
sin β
2
β
dI
sin β
= I (0 )2
dβ
β
= 2 I (0 )sin β
sin β −
1
β
2
+
1
β
cos β
β cos β − sin β
=0
3
β
(i ) sin β = 0 or (ii ) β cos β = sin β
tan β = β
The first condition (i) is just the condition
for minima that we have already seen (i.e.
β = mπ, m = ±1, ±2, ±3, …)
The second condition (ii) results in a
transcendental equation whose graphical
solutions can be observed on the left. The
solutions can easily be found numerically:
β ≈ ±1.43π, ±2.46π, ±3.47π…which
approaches (m+1/2)π for large β.
Principal or Central
Maximum
Subsidiary Maximum (SM)
Minima (I = 0)
1st SM
2nd SM 3rd SM
For the figure on the far
left, the conditions for
minima of mλ = bsinθ are
understood by dividing the
phasors of rays in two and
four equal segments in (b)
and (c), respectively. In
(b), for each ray in the top
half, a phasor 180° out-ofphase can be found in the
bottom half, resulting in
zero E-field. Likewise, in
(c), each successive quarter
contains phasors that are
180° out-of-phase,
resulting in zero E-field
and zero irradiance.
Phasor constructions of the E-field for
(a) the central maximum, (b) a
direction slightly displaced from the
central maximum, (c) the first
minimum, and (d) the first maximum
beyond the central maximum for
single-slit diffraction. The example
corresponds to N = 18 phasors.
The Double Slit
Consider now the set-up for a double-slit diffraction measurement. The
rectangular geometry of the slit is shown such that l >> b, where l and b are the
height and width of each slit. The screen on which the diffraction pattern is
observed is located a distance R from the slits, where R >> a and R >> b2/λ.
The point P is located on the screen. We will need to calculate the resulting Efield contribution at point P from all points (sources) located in both slits, using
essentially the same procedure performed for the single-slit case.
Figure 10.13 (a) Double-slit
geometry. Point P on the screen
σ is essentially infinitely far
away. (b) A double-slit
diffraction pattern for a =3b.
The E-field is obtained, as before, by integration over all sources in each slit:
E =C
b/2
F ( z )dz + C
−b / 2
a +b / 2
F ( z )dz where C =
a −b / 2
εL
R
and
F ( z ) = sin[ωt − k (R − z sin θ )]
Integration procedure is similar to that of the single slit and results in:
E = bC
sin β
β
[sin (ωt − kR ) + sin (ωt − kR + 2α )] where α ≡ (ka / 2)sin θ
and β ≡ (kb / 2 )sin θ
Note that the definition for β is precisely as before for single-slit diffraction.
Using the identity: sin x + sin (x + 2y) = 2cos y sin(x + y), we can express E as
E = 2bC
sin β
β
cos α sin (ωt − kR + α )
After squaring and time-averaging, we obtain the irradiance as:
I (θ ) = 4 I 0
sin 2 β
β2
in which β = α = 0 represents θ = 0 and I(0) = 4I0
cos α
since E0 = 2bC and twice that relative to a single slit.
Note that I0= (bC)2/2 = (E0/2)2 is the contribution from
one slit.
2
Examine some limits:
If kb << 1, (sinβ /β ) ≈ 1 and we get I(θ) = 4I0 cos2α = 4I0 cos2(δ/2), which
yields the expected result for Young’s two slit interference when the slit
width b is very small.
If a = 0, α = 0, then I(θ) =4I0 (sinβ /β )2 = I0’ (sinβ /β )2 , which is just the
situation of two slits coalescing into a single slit.
Note that the expression I(θ) =4I0 (sinβ /β )2 cos2α is that of an interference
term modulated by a diffraction effect.
Note that in general for two slits, minima occur when β = mπ, m = ±1, ±2,
±3, … and α = ±(2m+1)π/2, m = 0, 1, 2, 3, …
As a general rule-of-thumb, for a = mb, we observe 2m bright fringes within
the central diffraction peak, including fractional fringes. So, in Fig. 10.13,
for a =3b, we have 5 + 2(1/2) = 6 bright fringes within the central
maximum.
When an interference maximum overlaps with a diffraction minimum (as
shown in Fig. 10.13c) it is often referred to as a missing order (when
m=multiple of b/a)
(a)
(b)
Figure 10.14
Single- and double-slit Fraunhofer
patterns. (a) Photographs taken with
monochromatic light. The faint
cross-hatching arises entirely in the
printing process. (b) When the slit
spacing equals b, the two slit
coalesce into one (of width 2b) and
the single-slit pattern appears (i.e.,
that’s the first curve closest to you).
The farthest curve corresponds to
the two slits separated by a =10b.
Notice that the two-slit patterns all
have their first diffraction minimum
at a distance from the central
maximum of Zo. Note how the
curves gradually match Fig.
10.13(b) as the slit width b gets
smaller in comparison to the
separation a.
Diffraction by many Slits
For diffraction from many slits (i.e., the N-slit problem), we generalize the
( N −1) a + b / 2
2 a +b / 2
a +b / 2
b/2
2-slit problem:
E = C F ( z )dz + C
F ( z )dz + C
F ( z )dz + .... + C
F ( z )dz
−b / 2
a −b / 2
where C =
2 a −b / 2
εL
R
( N −1) a −b / 2
and F ( z ) = sin[ωt − k (R − z sin θ )]
The integrals can be evaluated as
before for the single- and double-slit
cases using the same identities. The
result is
E=
N −1
j =0
bC
sin β
β
sin[ωt − kR + 2 jα ]
where R j = R − ja sin θ , α =
ka
sin θ
2
− kR + 2 jα = −kR j (see figure for Rj)
The summation can be evaluated most easily using phasors, as before:
E = bC
sin β
β
N −1
j =0
sin[ωt − kR + 2 jα ] = bC
sin β
β
Im e
i (ωt − kR )
N −1
j =0
(e )
i 2α j
Again, the phasor sum is evaluated as a geometric series, as was done
for the case of multiple beam interference. Therefore, we can write
E = bC
sin β
β
sin Nα
ka
sin[ωt − kR + ( N − 1)α ], α = sin θ and
sin α
2
kb
β = sin θ
2
Therefore, the time-averaged irradiance can be expressed as
I (θ ) = I 0
sin β
β
2
sin Nα
sin α
2
Note that I0 is the irradiance emitted by one slit and I(0) = N2I0, for θ = 0 which
is seen by taking the limit for θ = 0 of the above expression. Also note that if kb
<< 1 (i.e. very narrow slits), we get the same expression for a linear coherent
array of oscillators.
Diffraction term
Therefore, I (θ ) = I 0
multiple-slit interference term
sin β
β
2
sin Nα
sin α
2
I (0) sin β
= 2
β
N
2
sin Nα
sin α
2
The principal maxima are given by the following positions: α = 0, ±π,
±2π, ±3π, … = mπ (m = 0, ±1, ±2, ±3 ..) and α = (ka/2)sinθ a sinθm = mλ
and (sinNα/sinα)2 = N2 at these positions.
Minima are determined by (sinNα/sinα)2 = 0 α = ±π/N, ±2π/N, ±3π/N, …,
±(N-1)π/N, ±(N+1)π/N,… Therefore, between consecutive principle maxima
we have (N – 1) minima (see figures on next slide).
Subsidiary maxima: Between consecutive principle maxima we have (N-2)
subsidiary maxima when |sinNα| ≈1. This occurs for α ≈ ±3π/2N, ±5π/2N,
±7π/2N, …For large N, sin2α ≈ α2 and the intensity of the first subsidiary
peak is approximately
2
2
2
sin β
2
I (0) sin β
I (θ ) ≈ I (0)
≈
3π
22
β
β
Figure 10.17 multiple-slit pattern with a = 4b, N = 6
Finally, note that for N = 2 slits, we observe that
I (0) sin β
I (θ ) = 2
N
β
= I (0)
sin β
β
2
sin Nα
sin α
2
I (0) sin β
=
4
β
2
2 sin α cos α
sin α
2
cos 2 α
where α = (ka / 2) sin θ = δ / 2
because δ = ka sin θ = kΛ = k (OPD).
2
The Rectangular Aperture
Consider Fraunhofer (far-field) Diffraction from an arbitrary aperture
whose width and height are about the same.
Let εA = the source strength per unit area. Then each infinitesimal area
element dS emits a spherical wave that will contribute an amount dE to the
field at P (X, Y, Z) on the screen
ε
dE =
The distance from dS to P is
A
r
e i (ωt − kr ) dS
r = X 2 + (Y − y ) 2 + ( Z − z ) 2
which must be very large compared
to the size (a) of the aperture and
greater than a2/λ in order to satisfy
conditions for Fraunhofer
diffraction. Therefore, as before, for
OP → ∞, we can expect εA/r ≈ εA/R
as before (i.e., the behavior is
approximated as that of a plane
wave far from the source).
R2 = X 2 + Y 2 + Z 2
At point P (X, Y), the complex field is
calculated as follows:
~ ε
E ≅ A ei (ωt − kR ) e ikYy / R dy eikZz / R dz
R
−b / 2
−a / 2
b/2
β′ ≡
~ Aε Aei (ωt − kR ) sin α ′
E=
α′
R
kbY
2R
a/2
and α ′ ≡
sin β ′
β′
and the time − averaged irradiance is
(
~
I (Y , Z ) = Re E
)
2
T
sin α ′
= I ( 0)
α′
2
sin β ′
β′
2
kaZ
2R
For x, y << R, we can approximate r as follows:
(
r = X 2 + (Y − y ) 2 + ( Z − z ) 2
)
1/ 2
(
= X 2 + Y 2 − 2 yY + y 2 + Z 2 − 2 zZ + z 2
)
1/ 2
= [ R 2 + y 2 + z 2 − 2(Yy + Zz )]1/ 2 = R [1 + ( y 2 + z 2 ) / R 2 − 2(Yy + Zz ) / R 2 ]1/ 2
[
]
≅ R 1 − (Yy + Zz ) / R 2 + ...
Therefore,
~
E=
εA
dE =
Aperture
Aperture
r
ei (ωt − kr ) ds ≅
εA
R
ei (ωt − kR )
eik (Yy + Zz ) / R ds
Aperture
Thus, for the specific geometry of the rectangular aperture:
~ ε
E ≅ A ei (ωt − kR ) eikYy / R dy eikZz / R dz
R
−b / 2
−a / 2
b/2
Let
β′ ≡
kbY
2R
a/2
and α ′ ≡
kaZ
2R
iβ ′
− iβ ′
e
e
sin β ′
−
Then
eikYy / R dy = b
=b
β′
2iβ ′
−b / 2
b/2
iα ′
− iα ′
e
e
sin α ′
−
eikZz / R dz = a
=a
α′
2iα ′
−a / 2
a/2
and
Therefore, the resulting complex field at point P on the screen for a rectangular
aperture having area A = ab is given by
~ Aε Ae i (ωt − kR ) sin α ′
E=
R
α′
(
~
I (Y , Z ) = Re E
)
2
T
sin β ′
β′
sin α ′
= I (0)
α′
and the time − averaged irradiance is
2
sin β ′
β′
2
where I(0) is the intensity at the center of the screen at point P0 (Y = 0, Z = 0).
A typical far-field diffraction pattern is shown in Fig. 10.20. Note that when α′
= 0 or β′ = 0, we get the familiar single slit pattern.
The approximate locations of the secondary maxima along the β′-axis (which
is the Y- axis when α′ = 0 or Z = 0) is given by β′m ≈ ±3π/2, ±5π/2, ±7π/2...
Since sin β′ = 1 at these maxima, the relative irradiances along the β′-axis are
approximated as
I (Y ,0) ≅
I (0)
(β m′ )2
I (0)
(α m′ )2
1
1
1
I →1→ ~
→ ~
→ ~
122
62
22
and similarly for the α ′ or Z − axis, I (0, Z ) ≅
where α′m ≈ ±3π/2, ±5π/2, ±7π/2...
Square aperture in which a = b.
kaZ
kbY
β′ ≡
and α ′ ≡
2R
2R
where α′m= β′m ≈ ±3π/2, ±5π/2, ±7π/2... are the
positions of the secondary maxima and
1
1
1
I →1→ ~
→ ~
→ ~
122
62
22
Distribution of
irradiance, I(Y,Z)
α′ (Z)
β′ (Y)
Distribution of
electric field, E(Y,Z)
The Circular Aperture
A very important aperture shape is the circular hole, as this involves the natural
symmetry for lenses. Such a symmetry suggests the need for cylindrical
coordinates:
z = ρ cos φ ,
y = ρ sin φ , Z = q cos Φ, Y = q sin Φ
ds = dydz = ρdφdρ
~ ε Ae i (ωt − kR )
E=
R
a
2π
and the complex field is
ρdρdφei ( kρq / R ) cos(φ −Φ )
ρ =0 φ =0
ik(Yy+Zz)/ R
e
We are
calculating the
field E on the
screen σ as a
function of the
screen’s radial
coordinate q.
Note that the term eik (Yy + Zz ) / R involves :
yY + zZ = ρq (sin φ sin Φ + cos φ cos Φ )
= ρq cos(φ − Φ )
Fig. 10.21 Circular
Aperture Geometry
The circular shape of the aperture results in complete axial symmetry.
Therefore, the solution must be independent of Φ. Therefore, we are permitted
to set Φ = 0. We are therefore led to evaluate the integral:
2π
ei ( kρq / R ) cosφ dφ
0
1
J 0 (u ) =
2π
2π
eiu cos v dv,
with v = φ
and u = kρq / R
0
where J0(u) is the Bessel function (BF) of order zero. More generally the BF
of order m, Jm(u), is represented by the following integral:
J m (u ) =
Fig. 10.22
Therefore the field is expressed as
i
( − m ) 2π
2π
0
ei ( mv +u cos v ) dv,
Bessel functions are slowly
decreasing oscillatory
functions very common in
mathematical physics.
~ ε Aei (ωt − kR )
2π J 0 (kρq / R) ρdρ
E=
R
ρ =0
a
Another important property of the Bessel
Function is the recurrence relation that
connects BFs of consecutive orders:
[
]
d m
u J m (u ) = u m J m −1 (u )
du
d
when m = 1,
u J1 (u ) = u J 0 (u )
du
u
u
d
θ J 0 (θ )dθ =
θ J1 (θ ) dθ
dθ
0
0
[
]
[
[
= θ J1 (θ )
Let
a
]
u
0
θ − dummy var .
= u J1 (u )
w = kρq / R, then dρ =
J 0 (kρq / R )ρdρ = ( R / kq) 2
ρ =0
]
R
dw and
kq
kaq / R
J 0 ( w) wdw
w=0
We can express the field, using the recurrence relation, as
~ ε Ae
E=
=
i (ωt − kR )
R
a
2π
ρdρdφei ( kρq / R ) cos(φ −Φ ) =
ε Aei (ωt −kR )
ρ =0 φ =0
ε Aei (ωt −kR )
R
2πa
R
2π
a
J 0 (kρq / R) ρdρ
ρ =0
R
J1 (kaq / R )
kaq
2
Therefore, the irradiance at point P on the screen is
I (q) =
1 ~ ~*
ε A
EE = 2 A
R
2
2
J1 (kaq / R )
(kaq / R )
2
; A = πa 2
It is useful to examine the series representation of the Bessel Functions:
n+2k
(−1) ( x / 2)
J n ( x) =
k = 0 k! Γ ( n + k + 1)
∞
k
where Γ(n) = (n − 1)!
x 1 x
J1 ( x) = −
2 2 2
3
+ ...
Consider the limit near x = 0 : lim
x →0
J 1( x) 1
=
x
2
Therefore, the irradiance at P0 when q = 0 is
I (0) = 2
εAA
2
R
I (q) = I (0)
1
2
2
1 εAA
=
2 R
2 J1 (kaq / R )
(kaq / R )
I (θ ) = I (0)
2
and the Irradiance becomes
2
; and with sin θ =
2 J1 (ka sin θ )
ka sin θ
q
R
(Fig. 10.21)
2
We usually express the irradiance as a function of the angular deviation θ
from the central maximum at point P0. The large central maximum is called
the Airy Disk, which is surrounded by the first dark ring corresponding to
the first zero of J1(u). J1(u) = 0 when u = 3.83 or kaq1/R = kasinθ1= 3.83.
Rλ
2π aq1
3.83 Rλ
q1 =
= 3.83
≅ 1.22
λ R
π 2a
2a
sin θ1 = q1 / R ≅ 1.22λ / D
kaq1 / R =
D = 2a is the diameter
of the circular hole
Note that ~84% of the light is contained in the Airy Disk (i.e. 0 ≤ kasinθ ≤ 3.83)
Airy Disk
Airy Disk
Airy Rings with different
hole diameters
D = 0.5 mm
Suppose that the aperture is a lens which
focuses light on a screen:
Entrance Pupil
(Aperture)
screen
Converging
lens
f
Airy
Disk
D = 1.0 mm
Since f ≅ R
D
q1 ≈ 1.22 fλ / D
which gives the radius of the Airy disk on
the screen.
High f/# of a lens means high speed lens
and poor resolution and vice versa.
Analysis of overlapping images
using Airy rings
Rays arriving from two
stars and striking a lens
Suppose that we image two equal irradiance point sources
(e.g., stars) through the objective lens of a telescope. The
angular half-width of each image point is q1/f = sin ∆θ ≈ ∆θ .
If the angular separation of the stars is ∆ϕ and ∆ϕ >> ∆θ the
images of the stars will be distinct and well resolved.
Fig. 10.24 Overlapping Images
Rayleigh’s criterion for the minimum resolvable
angular separation or angular limit of resolution
Half-angle of an
airy disk:
1.22λ
∆θ1 ≅
D
f
∆l
min
Fig. 10.25
Rays from two stars
If the stars are sufficiently close in angle so that the center
of the Airy disk of star 1 falls on the first minimum (dark
ring) of the Airy pattern of star 2, we can say that the two
stars are just barely resolved. In this case, we have ∆θ1 =
(∆ϕ )min = q1/f ≈ 1.22 λ / D
(∆l)min ≈1.22 fλ/D . This is
Rayleigh’s criterion for angular or spatial resolution.
Another criterion for resolving two objects has
been proposed by C. Sparrow.
At the Rayleigh limit there is a central
minimum between adjacent peaks.
Further decrease in the distance between two
point sources will cause this minimum (dip) to
disappear such that
Sp
d 2 I (r )
dr 2
=0
r <Sp
The resultant maximum will therefore have
a broad flat top when the distance between
the peaks is r = Sp, and serves as the
Sparrow criterion for resolving two point
objects.
The Diffraction Grating
Transmission Grating is made by
scratching rulings or notches onto a clear
flat plate of glass. Each notch serves as a
source of scattering that affects radiating
secondary sources, in much the same way
as for a multiple-slit diffraction array.
When the phase conditions are met
through OPD = Λ = mλ, constructive
interference is observed.
Oblique incidence (θi > 0)
a sin θ i = CD
a sin θ m = AB
from the geometry.
Therefore AB − CD = a (sin θ m − sin θ i ) = mλ
m = 0 (zeroth order), m = ±1 (first order),
m = ±2 (second order), m = ±3 (third order)
The diffraction grating can also be constructed as a
reflection grating. The principals and conditions for
constructive interference are the same as that for a
transmission grating.
AB − CD = a (sin θ m − sin θ i ) = mλ
Most commercial gratings for spectroscopy are constructed
with a Blaze angle γ to control the efficiency of diffraction
for a particular λ and order m.
Controlling the irradiance distribution of diffracted orders using a Blazed grating.
Most of the incident light undergoes specular reflection, similar to a plane mirror, and
this occurs when θi = θm and thus for the zeroth order beam, which is independent on .
The problem is that most of the irradiance is wasted for the purpose of spectroscopy. It
is possible to shift the reflected energy distribution into a higher order (m = 1) in which
θm depends on λ. It is possible to change the distribution of the specular reflection by
changing the blaze angle γ so that the first order diffraction is optimized for a particular
range of wavelengths.
m-th order
Consider the situation such that the light
enters at θi = 0 with respect to the grating
plan, and an angle with respect to the
mirrors. For specular reflection θi - θr = 2γ
and so most of the diffracted irradiance is
concentrated near θr = -2γ. This will
correspond to a particular non-zero order
in which θm = -2γ and asin(-2γ) = mλ.
can be constructed for particular
wavelength and m.
Features of the grating spectrometer:
The angular width of the grating is defined by the angular distance between zeros
on either sides of the principal maximum, thus the angular width of a line is:
α = ka / 2(sin θ m − sin θ i )
∆α = (ka / 2) cos θ (∆θ ) = 2π / N
2λ
∆θ =
Na cos θ
The angular dispersion is defined as:
D ≡ dθ / dλ ;
a (sin θ m − sin θ i ) = mλ
D = m / (a cos θ m )
The chromatic resolving power is defined as:
R ≡ λ /(∆λ ) min
(∆θ )min = λ / (Na cosθ m ); D = (∆θ )min / (∆λ )min = m / (a cosθ m )
λ / (∆λ )min = mN
For the condition for autocollimation = i=- r ,
D=2tan i/ and R=2Nasin i/
Schematic from the Fluorescence Group, University
of California, Santa Barbara, USA
Fresnel (Near-Field) Diffraction
The basic idea is to start again with the Huygen’s-Fresnel principle for
secondary spherical wave propagation. At any instant, every point on the
primary wavefront is envisioned as a continuous emitter of spherical secondary
wavelets. However, no reverse wave traveling back toward the source is
detected experimentally.
1
K (θ ) = (1 + cos θ )
2
Therefore, in order to introduce a
realistic radiation pattern of
secondary emitters, we introduce
the inclination factor, K(θ) =
(1+cosθ)/2 which describes the
directionality of secondary
emissions. K(0) = 1 and K(π) = 0.
A rectangular aperture in the near-field (Fresnel Diffraction)
The monochromatic point source S and the point P on a screen are placed
sufficiently close to the aperture where far-field conditions are no longer
applicable. Consider a point A in the aperture whose coordinates are (y,z).
The location of the origin O is determined by a perpendicular line from
the source S to the aperture Σ. The field
contributions at P from the secondary
sources on dS (area element at point A)
is given by
K (θ )ε A
cos[k ( ρ + r ) − ωt ]dS
r
K (θ )ε 0
=
cos[k ( ρ + r ) − ωt ]dS ,
ρrλ
dEP =
where ε0 is the source strength at
S, εA is the secondary wavelet
source strength per area, and
εAρλ = ε0 is obtained from the
Huygen’s-Fresnel formalism.
In the case where the dimensions of the aperture are small compared to ρ and r,
we can assume primarily forward propagation in the secondary spherical waves
so that K(θ) ≈ 1 and 1/ρr ≈ 1/ρ0r0. Also, from the figure the geometry yields:
ρ = (ρ + y + z
2
0
2
)
2 1/ 2
and
(
r= r +y +z
2
0
2
)
2 1/ 2
Expand both terms in a binomial series for small y and z:
ρ 0 + r0
ρ + r ≅ ρ 0 + r0 + ( y + z )
2 ρ 0 r0
2
2
Note that this approximation contains quadratic terms that appear in the phase
whereas the Fraunhofer approximation contains only linear terms. Thus, we
can expect a greater sensitivity in the phase of the cosine for this near-field
treatment. The complex field at point P on the screen is therefore:
− i ωt y 2 z 2
εe
~
EP = 0
ρ 0 r0 λ
y1 z1
e ik ( ρ + r ) dydz
Let u ≡ y
2(ρ 0 + r0 )
λρ 0 r0
1/ 2
1/ 2
2(ρ 0 + r0 )
λρ 0 r0
; v≡z
Change of Variables Gives
~
EP =
~
Eu =
ε0
e i [k ( ρ
2(ρ 0 + r0 )
ε0
(ρ 0 + r0 )
0 + r0 ) −ωt
e iπ
2
u /2
(
v2
du e iπ
u1
e i [k ( ρ0 + r0 ) −ωt ]
w
]
u2
v1
)
w
and
0
Since e
iπ w′ 2 / 2
~ u2
E
/2
dv = u eiπ
2 u1
2
u /2
v2
du eiπ
v2 / 2
dv
v1
This is the unobstructed disturbance at P.
Let A( w) ≡ cos πw′2 / 2 dw′
w
v
2
(
)
C ( w) ≡ sin πw′2 / 2 dw′
0
dw′ = A( w) + iC ( w)
0
~
Eu
~
[A(u ) + iC (u )] uu12 [A(v) + iC (v)] vv12 ,
Therefore at P, EP =
2
where A(w) and C(w) are called Fresnel Integrals; note that both are odd
functions of w.
Very often, we work in the limit of incoming plane-waves striking the aperture.
For example, a laser beam could strike the aperture. In this limit we let the radius
from the source to the aperture ρ0 → ∞. This results in an immediate
simplification for the change of variables:
1/ 2
2(ρ 0 + r0 )
2(ρ 0 + r0 )
2
; v=z
Then u = y
=y
λρ 0 r0
λr0
λρ 0 r0
~ ~
2
2
E P* EP I 0
[A(u2 ) − A(u1 )] + [C (u2 ) − C (u1 )]
=
IP =
2
4
{
×
1/ 2
{ [A(v ) − A(v )]
2 ~
2
I ~
= 0 B12 (u ) B12 (v)
4
2
1
}
2
+ [C (v2 ) − C (v1 )]
2
}
u
1/ 2
[
2
=z
λr0
2
~
~
where B12 (u ) = [A(u ) + iC (u )] = B (u )
u1
]
1/ 2
u2
u1
Cornu Spiral
Elegant geometrical description of the Fresnel Integrals (Fig. 10.50).
Fig. 10.50 The Cornu Spiral
for a graphical representation
of the Fresnel integrals.
C(w)
A(w)
w
A(w) C(w)
A(w) C(w)
A(w) C(w)
w
A(w) C(w)
Let
~
B ( w) = A( w) + iC ( w) − ∞ < w < +∞
Arc length along the curve : dl 2 = dA2 + dC 2
From the Integrals :
dl 2 = cos
2
π
w
2
2
+ sin
2
π
w
2
2
dl = dw
dw2
Therefore, values of w in B(w) correspond to arc length on the Cornu spiral.
(-1 mm, 1 mm)
Consider a 2-mm square aperture hole:
(y1, z2)
We are given that λ = 500 nm, r0 = 4 m,
plane wave approx. is valid. Find the
O
irradiance at a point P on the screen along
the axis x, directly behind the center of
(y1, z1)
the aperture.
(-1 mm, -1 mm)
2
Then u1 = y1
λr0
1/ 2
=
(− 1×10 m) 5 ×10 2m ⋅ 4m
−3
u2 = 1.0, v1 = −1.0, v2 = 1.0
−7
1/ 2
z
(1 mm, 1 mm)
(y2, z2)
y
(y2, z1)
r0
(1 mm, -1 mm)
= −1.0
P
For u1 = −1, u 2 = 1
~
~
B (u 2 ) = 0.7799 + i 0.4383 and B (u1 ) = −0.7799 − i 0.4383
~
~
B 12 (u ) = B 12 (v) = 2(0.7799 + i 0.4383)
Therefore
[
2 ~
2
I0 ~
I
2
2
IP =
B 12 (u ) B 12 (v) = 0 (4 2 ) (0.7799 ) + (0.4383)
4
4
]
2
= 4(0.64) I 0 = 2.56 I 0 > I 0
Notice that there is an increase of the irradiance at the center point P on the
screen by 256% relative to the unobstructed intensity due to a redistribution of
the energy.
z
(-0.9
mm,
1
mm)
(1.1 mm, 1 mm)
In order to find the irradiance 0.1
(y1, z2)
(y2, z2)
mm to the left of center, move
the aperture to the right relative
y
to the OP line. While y1 and y2
O′
are shifted, z1 and z2 remain
unchanged. Then we have u2 =
(y2, z1)
(y1, z1)
1.1, u1 = -0.9, v2 = 1.0, v1 = -1.0.
(1.1 mm, -1 mm)
(-0.9 mm, -1 mm)
r0
u1 = y1
2
λr0
1/ 2
=
(− 0.9 ×10 m) 5 ×10 2m ⋅ 4m
−3
u2 = 1.1, v1 = −1.0, v2 = 1.0
−7
1/ 2
= −0.9
P′
~
~
Then B (u1 ) = B (−0.9) = −0.7648 − i 0.3398
~
~
B (u2 ) = B (1.1) = 0.7638 + i 0.5365
~
B12 (u ) = 1.5286 + i 0.8763
~
~
Also B12 (v) remains unchanged : B12 (v) = 2(0.7799 + i 0.4383)
Thus
I P′ =
[
2 ~
2
I0 ~
I
2
2
B 12 (u ) B 12 (v) = 0 (4) (0.7799 ) + (0.4383)
4
4
] [(1.5286) + (0.8763) ] = 2.485I
2
2
The decrease in the irradiance (2.485I0 < 2.56I0) for a small 0.1 mm shift to
the left (or right) of center on the screen shows that the center position is a
relative maximum (see Cornu spiral on the next slide).
Note that if the aperture is completely opened:
u1 = v1 = −∞, u2 = v2 = +∞
I0
2
IP =
2
2
4
2
2
2
2
2
= I0
which must equal to the
unobstructed intensity as a check.
0
1.1
C(w)
~
B12 (u ′) = 1.762
~
B12 (u ′)
~
B12 (u )
A(w)
-0.9
~
B12 (u ) = 1.789
The decrease in the complex vector
length from the position of the central
peak shows that the central position is a
maximum.
We can apply this formalism for Fresnel diffraction by a long narrow slit in which
y1 , u1 → −∞, y2 , u 2 → +∞
Therefore
I P′ =
~
~
B12 (u ) = 2eiπ / 4 , B12 (u ) = 2
2
2 ~
2
I0 ~
I ~
B 12 (u ) B 12 (v) = 0 B 12 (v)
2
4
b = z1 – z2 = slit width and let ∆v = v2 – v1 which is a string of length ∆v
lying along the Cornu spiral (next slide).
Suppose that ∆v = 2. At point P, opposite point O in the aperture, the
aperture and the screen are centered symmetrically and the string is centered
at point Os. If the aperture is moved up or down, the arc length of the string
remains constant, but the length of the vector B12(v) changes, as before.
~
B12 (v′)
~
B12 (v)
It should be apparent that the length of
B12 (and the intensity at point P on the
screen) will oscillate as the string slides
around one of the spirals, which is
equivalent to the slit moving up or down
with respect to a reference point on the
screen, as shown in the previous slide.
~
B12 (3.5)
It is also possible to visualize a clear
minimum at the center of the near field
diffraction pattern on the screen by
considering the an arc-length of ∆w = 3.5.
Any change in the slit position will give
and increase in B12 and therefore an
~
B12 (2.5) increase in irradiance.
It is apparent that the slit width has a
marked effect on whether the central
position is a maximum or local
minimum. Also note the oscillation
in irradiance for positions beyond
the width of the slit in both cases.