17. Electromagnetic waves
From the Maxwell’s work it follows that optics is a branch of
electromagnetism. A beam of light is a traveling wave of electric
and magnetic fields – EM wave.
The spectrum of light as a part
of electromagnetic spectrum
~
From HRW 4
1
17.1. Generation of EM waves
Emission of EM waves is effectively realised when we use an electric dipole
(antenna). Starting from the simple resonance LC circuit and „opening” the circuit
with increasing frequency, one can schematically show how the radiating dipole is
formed.
Realistic shape of the
EM wave at a distant point
Generation of EM wave in the shortwave radio region
2
17.2. Equation of EM wave
From the Maxwell’s equations it is easy to obtain for both electric and magnetic
fields the expressions having the shape of wave equations (11.5) (Chapter 11)


2E
2E
 00 2
x 2
t
(17.1)


 2B
 2B
  0 0 2
x 2
t
(17.2)
Comparing (17.1) and (17.2) with the wave equation (11.5)
 2
1  2

x 2 v 2  t 2
it can be concluded that in the case of an EM wave
v2 
or
v
1
 0 0
1
 0 0
c 
1
8,910
12
4 10
7
 3  108 m / s
(17.3)
where c is the speed of EM wave in vacuum (all EM waves, including light, have
the same speed in vacuum).
3
Equation of EM wave, cont.
Similarly to (17.3) the speed of light in a medium (with neglected absorption) is
given by
1
(17.4)
v

where ε, μ are electric permittivity and magnetic permeability of a given material,
respectively.
The solutions of equations (17.1) and (17.2) are simple for the case of plane waves.
Assuming for electric and magnetic vectors the following components:
E x  Ez  0
E y x   0
Bx  By  0
Bz x   0
equations (17.1) and (17.2) can be written as
 2Ey
 2Ey
  0 0
x 2
t2
(17.5)
 2 Bz
 2 Bz



0 0
x 2
t2
(17.6)
It is easy to prove that solutions of above equations are
B z  B0z sin kx  t 
E y  E0 y sin kx t 
where
k

c
(17.7)
4
Equation of EM wave, cont.
The sinusoidal electric and magnetic fields are perpendicular to each other and
perpendicular to the direction of propagation indicated by the wave vector k (EM
waves are transverse).
Configuration of fields E and B for
a plane wave propagating in x direction
at speed c.
The instantaneous values of fields E and
B as functions of x. The wave components
E and B are in phase.
The two fields continuously create each
other via induction.
5
17.3. Polarization
The wave for which the E vector oscillates only in one plane is
polarized (plane-polarized). Polarized waves are emitted by
e.g. radio and TV transmitters, microwave antennas.
The light waves emitted by natural light sources (the Sun)
or common sources as a bulb are unpolarized.
This is connected with the mechanism of radiation.
Vector E of EM wave is parallel
to the dipols axes oriented
vertically. Turning the receving
dipole by 90o makes the received
signal disappear.
Natural unpolarized light is polarized by the
polarization sheet (a Polaroid filter) and passes
through another filter – an analyser.
6
Polarization, cont.
The intensity of polarized light transmitted by the filter can
be found from the analysis of the transmitted component
of electric field Ey
Ey  E cos 
θ – angle between E and Ey (17.8)
As the intensity of light is proportional to the square of
electric field amplitude one obtains from (17.8)
I  Im cos 
2
The vector of electric field can be
(17.9)
resolved into two components.
Component Ey parallel to the
From (17.9) it follows that I = Im for θ = 0 or
polarizing direction is transmitted
by the sheet, component Ez is
θ = π and I = 0 when polarizer and analyzer
attenuated.
are crossed.
When the light incident on the filter is unpolarized, the angle θ varies randomly and
in this case the intensity of transmitted light is
(17.10)
1
2
2
I   I 0 cos  
avg
 I 0  cos  
avg
 I0
2
which follows from the averaging: cos  
This is called the one-half rule.
2
avg



1
1

1
  cos  d 
d 
cos 2 d 
0


0
2 0
2 0
2
2
1
2
7
17.4. Reflection and refraction of light
In geometrical optics we consider the light waves as
straight lines (rays).
Law of reflection:
The reflected ray, incident ray and normal to the surface
lie in one plane. The angle of reflection is equal to the
angle of incidence
1 '  1
The incident ray of light is reflected
Law of refraction (Snell’s law):
from the interface separating two
An angle of refraction θ2 is related to the angle of
media and refracted. All angles are
measured relative to the normal.
incidence θ1 by
n2,1 – relative index of refraction
sin 1 n2
sin  2

n1
 n2 ,1
The index of refraction for a given medium is defined as
n1 
c
v1
n2 
c
v2
v – speed of light in a medium
c – speed of light in vacuum
medium
water
fused quartz
crown glass
flint glass
diamond
n
1.33
1.46
1.52
1.65
2.42
8
Chromatic dispersion
The index of refraction depends on the wavelength. For the light consisting of
different wavelengths one observes different angles of refraction for these
wavelengths what is called chromatic dispersion.
Figures
from HRW 4
prism
A prism separates white light
into component colors
9
Total internal reflection
When light travels from a medium of larger index of refraction to a medium with
a smaller index of refraction, then the total internal reflection may occur (when the
angles of incidence are greater than the critical angle).
n1 sin c  n2 sin 90  n2
sin  c 
From HRW 4
n2
n1
θc – critical angle
The refraction angle for ray e is 900.
Rays f and g only reflect.
Total internal reflection is used
to guide light in optical fibres
10
Polarization by reflection
The reflected light in general is partially polarized because the electric fields along
one direction have greater amplitudes than those oscillating along other directions.
However at particular angle of incidence,
called Brewster angle θB , the reflected light
has only perpendicular components.
The reflected light is then fully polarized.
Experimentally it was found that in this case
the reflected and refracted rays are
perpendicular to each other.
In this case θB + θr = 90o .
From HRW 4
From the law of refraction
n1 sin θB = n2 sin θr and then
n1 sin θB = n2 sin (90o- θB )
tan θB = n2/n1.
Finally one obtains:
Brewster angle B  tan 1
n2
n1
17.5. Interference from thin films
Rays 1 and 2 are the result of reflections by the front and back sides of the film,
respectively. These waves interfere and the result of interference depends on their
phase shift.
The phase shift depends not only on the
thickness d and refractive index n2 of the
thin film but also it has to be taken into
account that during reflection at the
interface the change in phase depends
on the refractive index of the medium
from which the ray reflects..
1.
If the ray reflects from the medium of higher refractive index, it undergoes a
phase shift of  rad ( half of a wavelength).
2. If the ray reflects from the medium of lower refractive index it does not
undergoe a change in phase.
For nearly perpendicular incidence of the ray shown in the figure above and for the
condition n3 > n2 > n1 in order to observe fully constructive interference of rays 1
and 2, the following condition must be fulfilled
2d= m ln2 , where m – integer number, ln2 - wavelength in medium n2.
Interference from thin films, cont.
Antireflective coatings
The structure shown in the figure is
an example of suppression of
unwanted reflections from glass by
deposition of thin film of magnesium
fluoride with adequate thickness.
In this case the waves of selected
wavelengths reflected from the two
film interfaces should be exactly
out of phase
2d  (2m  1)
l
2n2
for m  0,1, 2, ...
l - wavelength in air.
If we want the least thickness of the coating, one selects m = 0. In this case one
obtains for l = 550 nm (the middle of the visible spectrum)
d
l
4n2

550
 99.6 nm
4  1.38
Professional antireflective coatings consist of many layers with proper thicknesses
and indices of refraction to reduce the reflection in a desired region on wavelength.