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.
r
P’
Q
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’ can be calculated (after finding a proper Green function
G) by the fields existent at an arbitrarily given boundary enclosing that point P’.
P’
S
( P )   ( x' )
S
Surface
Field
at P’
G
dA
x'
Field at the
surface
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 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.
nr
T
E
Q
P’
D
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 CT line (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). The next section shows the exact calculation.
2. Calculation of the phase-difference between the sources
As a reference point, let’s take source D located oppsite to P’ (fig.4).
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 different in phase and magnitude, compared to
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 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 EP’-DP’ in terms of h (Fig.5).
E

(P’E)2
h
D
= (P’D)2 + h2
(P’E)2 - (P’D)2 = h2
P’
(P’E+P’D) (P’E-P’D) = h2
(P’E-P’D) = h2 / (P’E+P’D)
For a point E close to D:
(P’E-P’D) ~ h2 / (2 P’D)
For a point E far away from D:
(P’E-P’D)  h
Fig. 5 Figure shows that the path length EP’ varies quadratically with the
distance DE.
Hence, starting with a phasor representing the field at P due to the source located at D,
the additional phasors representing the contribution from the other sources start
changing its phase proportional to h2 first; for point far away the phase changes linearly.
The previous analysis suggests the following way to evaluate the total field phasor E at
different points on the screen:

Total field at point C’ on the screen:
This would be the case if
the phase were to change
linearly with the position
distance of the sources
E ( at C’ )
C
C’
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
B
Opaque
obstacle
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
D
R
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.
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, peculiar accumulation of charges establishes around the edges of the
aperture. Hence, the exact field near those borders may be a bit complicated.
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
For analysis purposes, 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.
Still, charges on the metal are being driving by the incident light from the source. It
might be 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
(2)
Wall
Plug
P
E =E Source
E =E Source + E ‘wall + E ‘plug = 0
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 may occur). It is for this
reason that we write E ‘wall instead E wall . (For points P located away from those
borders these distributions of charges may have no net significant effect.)
Subtracting (2) from (1),
E (at P) = - E ‘wall - E ‘plug + E wall
= (E wall - E ‘wall ) - E ‘plug
(3)
As pointed out above, in general E wall ≠ E ‘wall . But for point away from the borders
one obtains,
E (at P) = - E ‘plug
(4)
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 (in Fig 11) or away from the borders of the plug (in Fig.13).
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”.