Chapter XVIII Diffraction of light
“Diffraction” of light can be understood as any deviation of light rays
from their geometrical propagation line (that is straight in a homogeneuos material)
An example (shown in the picture):The edge of shadow is never
perfectly sharp. Some light
appears in the geometrical
shadow, and there are dark
and light fringers in the area of
illumination.
GENERAL PHYSICS III
Optics
&
Quantum Physics
Chapter XVIII
Diffraction of light
§1. Fresnel diffraction
§2. Fraunhofer diffraction
§3. X-ray diffraction from crystals
“Diffraction” of light can be understood as any deviation of light rays
from their geometrical propagation line (that is straight in a homogeneuos
material) Area of
illuminaton
An example (shown in the picture):
The edge of shadow is never
Point source
perfectly sharp. Some light
appears in the geometrical
shadow, and there are dark
and light fringers in the area of Straight
Geometrical
edge
illumination. shadow
Diffraction of light can be considered as an argument for wave
characteristics of light, like other wave processes (sound, etc.)
We consider separatly two kinds of diffraction:
Near-field diffraction → Fresnel diffraction
Far-field diffraction → Fraunhofer diffraction
§1. Fresnel diffraction:
1.1 Diffraction through a circular aperture:
According to geometric optics,
S
the image of a circular aperture
in the screen must be a light circular
dick with a perfectly sharp edge.
But it’s not so. Screen with a
Screen
circular aperture
a b
For analyzing this phenomenon, we use the Huyghen-Fresnel principle.
1.2 Huygen-Fresnel principle:
For analyzing properties of wave processes, Huygen introduced the
concept of wave front, and the rule how to draw a wave front from
known sharp of at some former time
Huygen’s principle (1678):
Wavefront at Wavefront at
All points on wavefront are point t=0 time t
sources for spherical secondary
wavelets with speed, frequency
that equal to initial wave. The wave front at a
later time is the envelope of these wavelets.
Basing on Huygen’s principle one can interpret diffraction as
interference of light from secondary sources. For example, every
point of circular aperture becomes a secondary source, and what we
see in the screen is the interference of secondary sources.
But this interpretation is only qualitively, for a quantitive analyze,
we need more. It has not been known from Huygen’s principle:
how can determine the amplitude and the phase of secondary waves ?
Fresnel’s complementary statement:
Fresnel states that for the vibration
at P due to waves from the secondary dS
source dS we have the following formula:
Observation
point
where Wave
front
0 & ( +
a t ) → the amplitude &
phase of vibration of secondary sources at dS on the wave front S
→ a coefficient which depends on the angle K decreases
K
as increases; K = 0 when = 2
/
The total vibration
at P is
1.3 Analysis of diffraction through circular aperture:
Having Huygen-Fresnel principle we tend to analyze the phenomenon
of diffraction through a circular aperture.
1.3.1. The Frsenel method to devide a spherical wave front into
adjacent zones (Fresnel zones):
1-th zone
2-nd zone
3-rd zone
4-th zone
Calculate the area of the
m-th zone:
where Sm is the area of
the m-th spherical segment:
(hm – the hight of segment)
hm is defined from the following equation:
Remark: The area of a zone does not depend on m. It means that the
areas of zones are approximatly the same (for values of m that are not
large).
We have also the formula for the radius of the m-zone:
rm is proportional to
From the Fresnel formula for dξand all the described properties of
Fresnel zones we can lead to the following formula for the amplitudes
of vibrations at P which are sent from Fresnel spherical zones :
from the 1-st zone ,,, …
from the m-th zone
Further, the phases of vibrations from two adjacent zones have the
phase difference we can write for the total amplitude of
→
vibrations at P:
It equals a half of amplitude due to
the 1-st zone !
1.3.2 Come back to the experiment of diffraction through an aperture:
→ What happens if there is a screen with circular aperture in the
light propagation line ?
Screen with aperture y y
Screen
Suppose that the part of wave front based on the aperture
incorporates m zones:
+ for odd m
- for even m
(odd m)
(even m)
Conclusions of diffraction picture on the screen:
Depending on the size of aperture, the number of open zones m is
odd or even:
• If m is odd → at the center point P there is a light spot
• If m is even → at the center point P there is a dark spot
Besides the center point P, at other points in the screen, the light
intensity has maxima or minima, depending on the distance from
the center point P.
Owing to the symmetry, light & dark fringers on the screen are
circles centered at P.
a) b) c)
Comparison the Fresnel zone pictures according to
a) the axis SP; b) the axis SP’; c) the axis SP”
To understand why the light intensity is different at P, P’ & P”
we describe the Fresnel zones for three cases, and make a
comparison. Here we suppose that the aperture exposes 3 zones.
§2. Fraunhofer diffraction:
We have studied an example of Fresnel diffraction. In Fresnel (or
near-field) diffraction, we deal with spherical wave (wave front is
spherical).
Now we consider Fraunhofer (or far-field) diffraction that concerns
plane waves. An example is diffraction of a beam of prallel rays on
a single slit, or multiple slits.
2.1 Single-slit Diffraction:
Consider a plane light wave that comes to a slit. Suppose that
slit is very long (or its length is much larger the width)
2.1.1 Diffraction minima:
First we define the location of minima in the screen
a
sin P
2
min
2
a/2
Incident Wave
(wavelength ) y
sin
min
a
• At this angle the light from the
top and the middle of the slit
a
destructively interfere. Below two
O
these points we can find
corresponding pairs from which two
L
waves destructively interfere at P.
The second minimum is at an angle
such that the light from the top and
a point at a/4 destructively interfere: Location of nth-minimum:
a n
sin sin n
a/4 4
min,2
2 min, (n =±1, ±2,…)
a
2
sin
min,2
a
n
• For small → n-th minimum corresponds to , ( n 1, ,...)
n 2
a
and the corresponding y- coordinates in the screen are
n
yn L , ( n 1, ,...)
2
a
a
a
L
• Note that for n = 0, that is = 0, it is the point O in the screen,
y
Light from entire slit arrives at O in phase. Here we have a bright
fringer with the width 2 (twice in width than other bright fringers).
/a
2.1.2 Intensity in the single-slit pattern:
By using the phasor diagram method, we can have a detail analysis
To analyze diffraction, we treat it as = a sina
a Screen
interference of light from strips (far away)
Model the single slit as M strips with
spacing between the strips of a/M.
a
There are phase differences between
waves coming from adjacent strips
P
The phase difference between first and
last source is given by L
L >> a implies rays are
(asin
a =
parallel.
• Taking M = 10, we draw the phasor diagrams
A0
Vibration vectors at O AP
A0
The amplitude at O (in phase)
and at P
As we let M the polygon becomes
,
the arc of a circle.
The radius of the circle is determined A
by the relation between angle and R Ao
arc length: Ao/R.
=
2
a
Trigonometry: A/2 = R sin(/2)
With R = Ao/, A = (2Ao/ /2)
sin(
sin( / 2)
A A0
/ 2
Intensity is related to amplitude: I = A2. So, here’s the final answer:
a sin /
)
2 2
/ 2)
sin( sin (
I I 0
/ 2 I 0 ( sin /
a
)
Remember: ( sin
a
2 a 2
I
• The graph of the intensity versus
I0
sinis shown in picture
• Intensity maxima:
Consider the obtained equation
as a function I = I ( we can find
)
-2a –a
a 2a
sin
maxima by solving dI/d 0.
=
This is a transcendental equation that has to be solved numerically.
Here is the results:
