29. Diffraction of waves
Light bends!
Diffraction assumptions
The Kirchhoff diffraction
integral
Fresnel Diffraction
diffraction from a slit
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29. Diffraction of waves, Schemes and Mind Maps of Quantum Physics
Ocean waves passing through, and bending around the edges of, slits (regions between wave ... Diffraction occurs for all waves, whatever the phenomenon.
Typology: Schemes and Mind Maps
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29. Diffraction of waves^ Light bends!Diffraction assumptionsThe Kirchhoff diffraction
integralFresnel Diffractiondiffraction from a slit
Diffraction^ Light does notalways travel ina straight line.It tends to bendaround objects.This tendency iscalled "diffraction.“
Shadow of arazor bladeilluminated bya laser
This is why radio communications doesnot necessarily require a line of sight.
Diffraction of awave by a slit^ Passing light through a smallslit yields a diffraction patternthat depends on the size ofthe slit and the wavelength ofthe wave.This phenomenon is general,and can be observed usingwaves of any kind.
Large slit Smaller slit Very small slit
C. J. Davisson andL. H. Germer, 1927
Diffraction of particles
This experiment was completed only afew years after Louis de Broglie haddescribed the wavelength of a particlein terms of its momentum,
λ^ =^
h/p.
( h^ = Planck’s constant)
Electrons do this too. The observation ofthis fact was one of the first importantconfirmations of quantum physics.
Diffraction Assumptions
This set of assumptions actuallyover-determines the problem.But even so, this is a usefulstarting point. A more accuratetreatment is very complicated! The first thorough treatment of this problem wasdue to Kirchhoff. He made a few assumptions:
Gustav R. Kirchhoff(1824 - 1887)
- Maxwell's equations2) Inside the aperture at
z^ = 0, the field and
its spatial derivative are the same as if thescreen were not present.3) Outside the aperture (in the shadow ofthe screen) at
z^ = 0, the field and its spatial
derivative are zero.
Incidentwave
^ ( , ) E r t
^ ( , ) E r t ∇ ( , )
E r t
^ =^ ( , )^
E r t ∇
E r t
^ =^ ( , )^
E r t ∇
^
z
0
Huygens’ Principle^ Our solution for diffraction uses this idea.
Christiaan Huygens1629 – 1695
Huygens’Principle saysthat everypoint along awave-frontemits aspherical wavethat interfereswith all others.
The Solution: Kirchhoff Diffraction Integral^ The field in the observation plane,
E ( x^0
,y ), at a distance^0
z^ from the
aperture plane is given by a convolution:
1 1
0
0
0
1 0
1
1 1
1 1
Aperture(
,^ )
(^ ,
)
(^
,^
)^ (
,^
)
=^
−^
−
∫∫ x y
E x
y
h x
x^ y
y^ E x
y^
dx dy 01
0
1
0
1
01 exp(
)
1
(^
,^
)^
jkr
h x
x^ y
y^
j^
r λ
−^
−^
=
where : A very complicated result! In order to use this, we must make someapproximations…In the denominator, we can approximate
r by^01
z.
But not in the exponent, because
k^ is large so
kr^01
cannot be neglected.
Paraxial approximation In the spirit of the paraxial approximation, we will assume that the apertureis small compared to the distance
z , so that
z^ >>
x −^0
x and^1
y −^0
y.^1
2
2
0 1
0 1
01
−^
^
^
=^
+^
^
^
^
^
x^ x
y^ y
r^
z^
z^
z
First, we note that we can factor
z^ out of the square root in the
expresson for
r :^01
(^
)^ (
) 2
2
2
2
0 1
0 1
0 1
0 1
01
small corrections
^
^
−^
−^
^
^
^
≈^
+^
+^
=^ +
+^
=^ +
^
^
^
^
^
^
^
^
^
x^ x
y^ y
x^ x
y^ y
r^
z^
z^
z
z^
z^
z^
z
Make use of the Taylor expansion:
+^ ≈
(^
)^
(^
)^
(^
)
2
2
0 1
0 1
0 1
0
1
exp^
exp^
exp 2
(^
,^
x^ x^
y^
y
jkz^
jk^
jk z^
z
h x^
x^ y^
y^
j^
z
λ
^
^
^
−^
^
^
^
^
^
^
^
^
^
−^
−^
Replace
r in the exponent of^01
h(x^0
− x, y^1
− y) 01
The Fresnel Diffraction Integral^ And we’ll usually neglect the factors in front of the integral, to obtain:^ (^
)^
2
2
0 1
0 1
1
1
0 0
1 1
1 1
( 2^
)^ (^
,^
exp^
(x ,^
^
^
−^
−^
^
∝^
^
^
^
^
^
∫∫^
x x^
y y^
x^
y
E^ x^
y^
jk^
Aperture
y
dx dy
z^
z
This is the
Fresnel Diffraction integral.
Even with all the approximations
Usually, instead of writing an integral over an aperture, we willexplicitly write the aperture function in the integral: ( )^ we’ve made, it is usually difficult to evaluate.
(^
2 2
2 2
0 0
0 1
0 1
1 1
0 0
1 1
1 1
( 2^
2
)^ (^
)
,^
exp^
exp^
,
2
2
E^ in
x^ y
x x^
y y^
x^ y
E^ x^
y^
jk^ z^
jk^
Aperture x
y^ dx dy
j^ z^
z^
z^
z
^
^
^
^
^
^
+^
−^
−^
^
=^
+^
^
^
^
^
^
^
^
^
^
^
^
^
∫∫
λ
Consider a uniform plane wave incidenton a metal screen with a slit of width 2bin the x
-direction. A one-dimensional 1 problem, this may be the simplest of allpossible diffraction problems.It’s still not easy. 2b
Before solving it, let’s first try to anticipatewhat we might expect the answer to look like.
observation screen
Destructiveinterference whenthe path lengthdifference is
λ/2,
3 λ/2, 5
λ/2, etc.
Fresnel diffraction example: a slit
Write the Fresnel integral for this one-dimensional problem:
Next step: define new variables
1 and
x = ξ 1^ b
0
x = ξ 0^ b
(^ )^
(^ )^
(^
)
1
2
2
0
0
0 1
1
exp 1
j^ b
E x^
E^
d
π^ z
ξ^ ξ
−
^
→^
∝^
− ^
^
∫
Then:
2 π = k^ λ
since
(^ )^
(^ )
2
2 0
0 1
1
0
1
1
exp^
x^
x x^
x
E^ x^
jk^
Aperture x
dx
z ^
^
−^
^
∝^
^
^
^
^
^
∫
Fresnel integral for a slit^ The aperture function is given by:
(^ )^
1
1
1 0 otherwise
b^ x
b
Aperture x
−^ <
(^ )^
(^
(^2) ) 0 1
0
1
exp^
2
−
^
^
− ^
=^
^
^
^
^
^
^
b ∫ b
x^ x
E x^
jk^
dx z
Thus the integral becomes:
Fresnel diffraction example: a slit
(^ )^
(^
)
1
2
2
0
0 1
1
exp 1
j^ b
E^
d
π^ z
ξ^ ξ
−
^
∝^
− ^
^
∫
As a shorthand, we define a dimensionless quantityknown as the “Fresnel number”:
(^2) b = N^
z λ
(^ )^
(^
) {^
}
1
2
0
0 1
1
exp 1 E^
j^ N^
d
π^
ξ^ ξ
∝^ −
∫
and, of course, we are really interested in the intensity This is not an integral that can be solved in closed form.It must be computed numerically.
(^
)^
(^ )^ 2 0
0 I^
E
∝
Fresnel Diffraction through a slit:numerical results for
I ( ξ
Far from the slit: Closer to the slit:
Fresnel number N
0.5^12481020
(^2) b = N^
z λ
Fresnel number: Example: green light (
λ^ = 0.
μm)
a) slit width b = 1 millimeter = 2000
λ
N = 1 at a distance of 2 meters b) slit width b = 10 microns = 20
λ
N = 1 at a distance of 200
μm
of ripples ~ Fresnel number!
-2b^
2b
Recall: this Fresnel calculation is onlyvalid for z >> b, which is the paraxialapproximation.
Fresnel Diffraction through a slit: far field
(^ )^
(^
) {^
}
1
2
0
0 1
1
exp 1 E^
j^ N^
d
π^
ξ^ ξ
∝^ −
−
∫
In the limit that N << 1 (very far from the aperture), the integral canbe performed analytically. The math is a bit tedious, so we justquote the result here:^ (^
)
0
0
2 0 ^ sin 2
^
^
∝
Nx b
E x^
π Nx π b
Our old friendthe sinc function!
In this regime (the “far field”), thediffraction pattern no longer changesshape as z increases, but merelyexpands in size uniformly.
Fresnel number
0.03 0.02 0.
-100b^
100b