Lecture Notes
A. La Rosa
APPLIED OPTICS
________________________________________________________________________
I. Diffraction by an edge obstacle
II. Diffraction of light by a screen
Reference: Feynman, "Lectures on Physics," Vol I, Addison Wesley.
Section 30-6 and Section 31-6.
I. Diffraction by an edge obstacle
Figure 1 shows light travelling from left to right, obstructed by an opaque obstacle.
We want to evaluate the light intensity profile on a screen located at a finite distance
from the obstacle, particularly around the geometrical projection of the edge.
Incident
field
P’
Opaque
obstacle
Screen
Fig. 1 Monochromatic light obstructed by an opaque object. The edge will
produce diffraction of the light, to be observed on the screen.
1. Help from math theorems
To evaluate the fields at the screen, we will use a theorem in electromagnetism theory
that the field at a given point P’ , let’s call it (P' ) , can be calculated (after finding a
proper Green function G) in terms of the fields existent at an arbitrarily given boundary
S enclosing the point P’.
P’
S
( P' )   ( x' )
S
Surface
Field
at P’
G
dA
x'
Field at the
surface S
Fig. 2 The field at a given point is calculated based on the values of the field at an
arbitrarily chosen boundary surface S enclosing the point P’.
Choosing a proper surface S (one that fits to your problem)
For the particular case illustrated in Fig.1, we can choose an infinitely extended syrface
(as shown in Fig.3). For any practical situation the fields would decay at infinity.
Accordingly, when applying the surface integral (the one indicated in Fig.2 ), only the
contributions from the side CT will effectively contribute.
T
C
To infinity
P’
Surface
S
To infinity
Opaque
obstacle
To infinity
Screen
T
S
Q
P’
C
Opaque
obstacle
Screen
Fig. 3 The contributions to the field at P’ from from the surface sections
located at infinity would be zero. Hence, the field at P’ will depend only on
the values of the field on the surface that passes through C and T.
We will solve the problem illustrated in Fig.1 by assuming that we are dealing with a
two-dimensional case (to simplify the math). The field at an arbitrary point P’ will be
calculated by assuming that we have a set of effective sources uniformly distributed
over the infinite line that passes through CT, as shown in Fig.4.
Fields
produced
by the
Incident
field
T
E
D
P’
C
Opaque
obstacle
Fig. 4 Experimental setup equivalent to the one presented in Fig. 1, based on the
theoretical analysis presented in figures 2 and 3. The field at P’ will have
contributions from the multiple synchronous sources located along the vertical
line along the CT direction.
To find the field at P’ (located at a finite distance from the obstacle), we need to
calculate the contribution from all the individual sources located along the line CT (Fig.
4).
Since the point P’ is at a finite distance, it turns out (as we will see below) that the phase
difference between the fields arriving to P’ from two arbitrary sources (D and E for
example) is not proportional to their separation distance (DE). (Such as proportionality
between phase difference and the separation-distance between the sources would
occur if the screen were at infinity).
2. Calculation of the phase-difference between the sources
As a reference point, let’s take source D located oppsite to P’ (see Fig. 4 and Fig. 5).
Let’s take another point E located a litle bit up (at a height h from D):
Source E will produce a field at P’ that has different phase and magnitude than the
field produced by the source D.
But, for sources close to D, the magnitude will not be affected as drastically as the
phase. Hence, let’s concentrate our effort for calculating only the phase difference
between fields arriving from D and E.
Such a phase difference is due to the difference in path length: =EP’-DP’
We want to calculate  in terms of h (Fig.5).
E

h
Triangle EXD similar to
X
D
P’
triangle EDP’

h

h EP'
h2

EP'
h2
h2

EP'
h 2  DP' 2
h2
h << DP’ then  
DP'

For a point E close to D:
For a point E far away from D:
h >> DP’ then
h
(3)
Fig. 5 Figure shows that the path length  varies quadratically with h for points
close to E; but linearly with h for point far away from D.
Electric field at C’
(1)
(2)
To find the total field at the point C’ on the screen (due to the incident fields coming
from C, D, E, T, … etc.),
i) we start with a phasor representing the field at C’ due to the source located at C;
subsequently,
ii) we add phasors representing the contribution to the field at C’ from the other sources
(D, E, T, etc). Notice, according to expression (2) above, that the phase of the new
phasors increases proportional to h2 for points closer to C. But for point far away
from C the phase changes linearly with h, according to expression (3).
This analysis suggests the following graphical way to evaluate the total field phasor E at
C:

Total field at point C’ on the screen:
This would be the path if
the phase were to change
linearly with the position
distance of the sources
E ( at C’ )
T
D
C
C
This path correspond to the
fact that the phase
changes quadratically with
h (i.e. for small h the phase
practically does not
change)
Opaque
obstacle
C’
screen
Fig. 6 Total field phasor at point C’ on the screen is indicated by the thick oriented segment
in the figure. [Notice the figure is missing the small phasors that circle around in smaller and
smaller circumferences. Please complete them on your own.]

Total field at point B’:
C
E ( at B’ )
C
B
C
Opaque
obstacle
B
B’
screen
Fig. 7 Total field phasor at point B’ on the screen is indicated by the thick oriented segment
in the figure. [Notice the figure is missing the small phasors that circle around in smaller and
smaller circumferences. Please complete them on your own.]

Total field at point A’:
C
E ( at A’ )
C
B
C
A
A
A’
Opaque obstacle
Fig. 8 Total field phasor at point A’ on the screen is indicated by the thick oriented segment
in the figure. [Notice the figure is missing the small phasors that circle around in smaller and
smaller circumferences. Please complete them on your own.]
This analysis leads to the following intensity profile. Here Io is the intensity of the light at
one point on the screen when no obstacle is present. The intensity is proportional to
2
E .
D
A
C
R
D
Fig. 9 Profile of the light intensity of the screen (see also Fig. 4 above).
II. Diffraction of light by a screen
Field at the other side of an opaque screen that has an aperture
Metallic
wall
P
E =E Source
E =E Source + E wall
Fig. 10 Light incident on a metallic wall that has an aperture at the
center.
Objective: In this section, we illustrate that the settings displayed in Fig. 10
and Fig. 13 are equivalent when the observation point P is away
from the rims of the aperture
Light incident on the metal place the charges in oscillation, which re-emit light also
reaching the point P.
The total field at P has contributions from the source and from the radiators on the wall.
E (at P) =E source + E wall
(1)
Hence, the setting in Fig. 10 could be depicted a bit more accurate by the one in Fig. 11.
I general, a peculiar accumulation of charges occur around the edges of the aperture.
Hence, the exact field at points very near those borders may be “a bit” complicated
(actually it is a lot complicated).
E wall is a bit complicated to calculate
So let’s consider a point P away from the aperture.
For points P located away from those borders these
peculiar charge distributions around the rim of the
aperture may have no net significant effect.
Metallic
wall
P
E =E Source
E (at P) =E Source + E wall
Fig. 11 Light incident on a metallic wall that has an aperture at
the center. The incident like drives fee electrons on the metal,
which produce an additional field at point P.
An opaque screen
Now it comes a trick to evaluate E wall is in an approximate way.
Let’s block the aperture with a metal plug (similar material to that of the wall). The total
field at P will then be equal to zero. But, still charges on the metal are being driving by
the incident light from the source. It must occur, then, that the field from the wall E’ wall
plus the field from the plug E’ plug cancel out;
E Source + E ‘wall + E ‘plug = 0
Wall
Plug
P
E =E Source
E =E Source + E ‘wall + E ‘plug = 0
(2)
Fig. 12 Light incident on a metallic wall whose aperture has been
covered with a metallic plug.
Notice, by placing a perfect plug, the peculiar charge distribution around
the wall aperture may have disappeared (a new charge distribution, a
more uniform one, will occur). It is for this reason that we write E ‘wall
instead E wall .
For points P located away from those borders these peculiar distributions of charges
around the rim of the aperture may have no net significant effect. We will make such an
approximation at the end.
Subtracting (2) from (1),
E (at P) = - E ‘wall - E ‘plug + E wall
= (E wall - E ‘wall ) - E ‘plug
(3)
where E wall correspond to the original field that takes into account the
E ‘wall
E ‘wall
complicated charge distribution around the aperture rim
correspond to a uniform charge distribution on the rim (because of
the presence of the metallic plug)
correspond to a uniform charge distribution on the plug.
As pointed out above, in general E wall ≠ E ‘wall . But for a point P located away from
those borders these peculiar distributions of charges around the rim of the aperture
may have no net significant effect.) For such a point, E wall ~ E ‘wall. Expression (3) then
becomes,
E (at P) = - E ‘plug
for points P away from the borders (4)
of the aperture
Plug
P
E (at P) =- E ‘plug
Fig. 13 Experimental setting equivalent to the one depicted in Fig. 11
when evaluating the field at points P away from the borders of the
aperture.
I short, the setting in Fig. 11 shows that the field at P is the superposition of fields
produced by the light source plus the field from the charges in the metallic
screen.
We have shown that, an equivalent way to calculate the filed at P is to consider fictitious
sources located inside the aperture (except for a minus sign). Curiously, we
place sources right where Fig 11 indicate that in fact there is none.
References:
Feynman, "Lectures on Physics," Vol I, Addison Wesley.
- Section 30-6 “Diffraction by opaque screens”.
- Section 31-6 “Diffraction of light by a screen”.