Lecture Notes
A. La Rosa
APPLIED OPTICS
_________________________________________________________________________________
Effects of polarization on the reflected wave
Ref: The Feynman Lectures on Physics, Vol-I, Section 33-6
Z
Plane of
incidence
Plane of
interface
Fig. 1
Y
Y
-1
b
-1
B
a
A
i
i
X
i
Air
1
Glass
X
i
Air
1
Glass
Fig. 2 Reflection and refraction with
s-polarized radiation
Fig. 3 Reflection and refraction with
p-polarized radiation

E perpendicular to the plane of incidence

E parallel to the plane of incidence
Transverse Electric (TE) radiation
also called s- polarized radiation
Transverse Magnetic ™ radiation,
also called p- polarized radiation
a: Amplitude of the refracted wave
b: Amplitude of the reflected wave
A: Amplitude of the refracted wave
B: Amplitude of the r reflected wave
2
2
Objective: To figure out the reflection coefficients b and B as a function of the angles of
incidence and refraction
Principle: The currents that are generated in the glass produce two waves
a) They produce the reflected wave
b) If there were no currents generated in the glass, the incident wave would
continue straight into the glass; i.e. producing a field of unit amplitude, which
would go along the dotted line as shown in the figure. This field is not observed.
Therefore the currents generated in the glass must produce a field of amplitude
-1 that moves along the dotted line.
Using these principles we will find the reflection coefficients
In figure 2:
A field of amplitude b is radiated by the motion of charges inside the glass which
are responding to a field a inside the glass. Therefore it is plausible to assume that,
b is proportional to a.
(1)
In figure 3:
Except for the polarization of the fields, this figure is identical to Fig. 2.
Can we say then that the ratio B/A is the same as b/a?
The answer is no.
The reason is because in Figure 3 the direction of the polarizations (of the reflected
and transmitted waves) are not parallel to each other.
How to find B/A?
To find a proper relationship between B/A and b/a let’s take a look at Fig 4.
Notice only the component of A along the direction SN is effective in generating the reflected
wave. Such a component is A cos(). The figure also reveals that,
= ( i + r).
Accordingly,
B
b

A cos (i  r ) a
(attained from Fig. 2 and Fig. 4)
(2)
Y
-1
B
N
90o
S

SN = MN cos 
N’

i
R
M
r
A
M’
i
X
Only its component along
SN is effective in
generating the reflected
wave.
1
Fig. 4 Reflection and refraction from p-polarized radiation.
On the other hand, in both cases it happens that the oscillations of charges in the glass
generate a field of amplitude -1 (so that when added to the incident field of amplitude 1 both
will cancel out).
In Figure-5 we notice that the component of A along the PR direction is the effective
component in generating a transmitted wave along the dotted line. That component is
A cos(). The figure also reveals that,
=( i - r).
Accordingly
-1
-1

A cos (i  r ) a
(from Fig. 2 and Fig. 5)
(3)
From (2) and (3) we eliminate the ratio a/A and obtain,
cos (i  r ) B

cos (i  r ) b
Verification of the formula: Notice when i  r  90 o expression (4) gives B=0, as expected
(Brewster condition). No reflected p-polarized reflected wave.
(4)
Y
-1
Q
PR = PQ cos 
B
R


i
A
Q’
r
P’
P
i
X
Only its component along
PR is effective in
generating a transmitted
wave of amplitude -1
Fig. 5
2
2
Finding the reflection coefficients b and B .
Expression (4) so far can give us only the ratio of the reflection coefficients. To find each
coefficient individually we resort to the conservation of energy.
The energy in each wave (incident, reflected, or transmitted) is proportional to the
corresponding amplitude.
In figure 2:
Conservation of energy implies that,
the energy of the transmitted wave
(5)
2
must be proportional to 1 - b .
[Whatever coefficient that must have been included in front of the (amplitude)2 it
will be the same for the incident and reflected waves, since both are in the same
medium. For that reason the energy of the transmitted wave is simply proportional
2
to 1 - b ]
In figure 3:
Conservation of energy implies that,
the energy of the transmitted wave
(6)
2
must be proportional to 1 - B .
Notice that the constant of proportionality in (5) and (6) must be the same.
2
2
[Basically, we are saying that the same positive constant h plays in 1 = b + h a and 1 =
2
2
B + hA ]
On the other hand,
the ratio of the energies of the transmitted waves in
2
Fig. 2 and Fig. 3 must be equal to a / A
(7)
2
From (5), (6), and (7),
1 b
2
1 B
2

a
A
2
(8)
2
Using (3), A cos (i  r )  a , and (4),
1 b
1
cos (i  r ) B
 , in expression (8) one obtains,
cos (i  r ) b
2
cos (i  r )
b
cos (i  r )
2
 cos 2 (i  r )
1  b  cos 2 (i  r )  cos 2 (i  r ) b
2
1  cos 2 (i  r )  b  cos 2 (i  r ) b
2
2
2
sin 2 (i  r )  sin 2 (i  r ) b
sin 2 (i  r )
b 
sin 2 (i  r )
2
Using (4),
cos (i  r ) B
 , one obtains,
cos (i  r ) b
2
(9)
B 
cos 2 (i  r ) sin 2 (i  r )
cos 2 (i  r ) sin 2 (i  r )
B 
tan 2 (i  r )
tan 2 (i  r )
2
2
(10)
(4) in (2) gives,
A
B
1


a bcos (i  r ) cos (i  r )
(5)