Download Unit 18 - HKU Physics

Unit 18
Electromagnetic Waves
18.1
Electromagnetic waves
18.2
The electromagnetic spectrum
18.3
Polarization
18.4
Polarization by reflection
18.5
Polarization by scattering
18.1 Electromagnetic waves
Electromagnetic waves are traveling waves of oscillating electric and magnetic fields.
The electric and magnetic fields in an electromagnetic wave are perpendicular to each
other, and to the direction of propagation. They are also in phase. To find the direction
of propagation of an electromagnetic wave, point the fingers of your right hand in the
direction of E, then curl them toward B. Your thumb will be pointing in the direction
of propagation. Electromagnetic waves can be generated by an antenna, which is
connected to an AC generator.
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Direction of propagation for electromagnetic waves can be shown by using the righthand rule: Point the fingers of your right hand in the direction of E, curl your fingers
toward B, and your thumb will point in the direction of propagation.
In fact, another way to generate electromagnetic waves is to accelerate electric charge.
As accelerated charges radiate electromagnetic waves, the intensity of radiated
electromagnetic waves depends on the orientation of the acceleration relative to the
viewer. For example, viewing the antenna perpendicular to its length, so that the
charges accelerate at right angles to the line of sight, results in maximum intensity.
Conversely, viewing the antenna straight down from above, in the same direction as
the acceleration, results in zero intensity.
According to Maxwell’s work in electric and magnetic fields, he proposed the light is
an electromagnetic wave. Visible light is of course an electromagnetic wave. Unlike
sound or a wave on a string requires a medium through which it can propagate, light
can propagate through a vacuum. In fact, electromagnetic waves travel through a
vacuum with the maximum speed, c, that any form of energy can have. It is also the
speed of light.
c = 3.00 × 10 8 m / s
A beam of light can travel around the world about seven times in a single second.
Light travels slower in a denser medium, e.g. glass and water.
Example
An electromagnetic wave propagates in the positive y direction, as
shown in the figure. If the electric field at the origin is in the positive z
direction, is the magnetic field at the origin in (a) the positive x
direction, (b) the negative x direction, or (c) the negative y direction?
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Answer
According to the right hand, which has your thumb pointing the propagation direction,
the direction of magnetic field should be along positive x.
18.2 The electromagnetic spectrum
18.2.1 Radio waves
The lowest-frequency electromagnetic waves of practical importance are the radio and
television waves. The frequency ranges from 106 Hz to 109 Hz.
18.2.2 Microwaves
Electromagnetic radiation with frequencies from 109 Hz to about 1012 Hz is referred
to as microwaves. Waves in this frequency range are used to carry long distance
telephone conversations, as well as to cook our food. Microwaves, with wavelengths
of about 1 mm to 30 mm, are the highest-frequency electromagnetic waves that can be
produced by electronic circuitry.
18.2.3 Infrared waves
Electromagnetic waves with frequencies just below that of light –
roughly 1012 Hz to 4.3×1014 Hz – are known as infrared rays. Most
remote controls operate on a beam of infrared light with a
wavelength of about 1000 nm. This infrared light is so close to the visible spectrum
and so low in intensity that it cannot be felt as heat.
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18.2.4 Visible light
The full range of colors in a rainbow as seen by human eyes is visible light. The
frequency range is 4.3×1014 Hz to 7.5×1014 Hz.
18.2.5 Ultraviolet light
When electromagnetic waves have frequencies just above that of violet light – from
about 7.5×1014 Hz to 1017 Hz – they are called ultraviolet or UV rays. More prolonged
or intense exposure to UV rays can have harmful consequences, including an
increased probability of developing a skin cancer.
18.2.6 X-ray
The frequency range of X-ray is about 1017 Hz to 1020 Hz. Typically, the X-rays used
in medicine are generated by the rapid deceleration of high-speed electrons projected
against a metal target. These energetic rays, which are only weakly absorbed by the
skin and soft tissues, pass through our bodies rather freely, except when they
encounter bones, teeth, to other relatively dense material. This property makes X-rays
most valuable for medical diagnosis, research, and treatment.
18.2.7 Gamma rays
Electromagnetic waves with frequencies above about 1020 Hz are generally referred to
as gamma (γ) rays. These rays, which are even more energetic than X-rays, are often
produced as neutrons and protons rearrange themselves
within a nucleus, or when a particle collides with its
antiparticle, and the two annihilate each other. Gamma
rays are highly penetrating and destructive to living
cells. It is for this reason that they are used to kill
cancer cells and, more recently, microorganisms in food.
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18.3 Polarization
To understand what is meant by the polarization of light, or any other electromagnetic
wave, consider the electromagnetic waves pictured in the figure. Each of these waves
has an electric field that points along a single line. For example, the electric field that
points in either the positive or negative z direction. We say, then, that this wave is
linearly polarized in the z direction.
A wave of this sort might be produced by a straight-wire antenna oriented along the z
axis. In general, the polarization of an electromagnetic wave refers to the direction of
its electric field. Polarization is an evidence for electromagnetic waves being a
transverse wave.
A beam of unpolarized light can be polarized in a number of ways,
e.g. by passing it through a polarizer. To be specific, a polarizer is
a material this is composed of long, thin, electrically conductive
molecules oriented in a specific direction. When a beam of light
strikes a polarizer, it is readily absorbed if its electric field is parallel to the molecules;
light whose electric field is perpendicular to the molecules passes through the material
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with little absorption, As a result, the light that passes through a polarizer is
preferentially polarized along a specific direction.
A simple mechanical analog of a polarizer is shown in the following figure.
Remarks:
Polaroid is made from tiny crystals of quinine iodosulphate all lined up in the same
direction in a sheet of nitrocellulose. Crystals which transmit light vibrations, i.e.
electric field variations, in one particular plane and absorb those in a mutually
perpendicular plane are said to be dichroic.
We now consider what happens when a beam of light polarized in one direction
encounters a polarizer oriented in a different direction. In the left figure in next page,
we see light with a vertical polarization and intensity I 0 passes through a polarizer
with its preferred direction – its transmission axis – at θ an angle to the vertical. The
component of E along the transmission axis is E cosθ . Recalling that the intensity of
light is proportional to the electric field squared, we see that the intensity, I, of the
transmitted beam is reduced to
I = I 0 cos 2 θ .
When an unpolarized beam of intensity I 0 passes through a polarizer (the right figure
in next page), the transmitted beam has an intensity of
1
I 0 and is polarized in the
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direction of the polarizer.
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An unpolarized beam of intensity I 0 is polarized in the vertical direction by a polarizer
with a vertical transmission axis. Next, it passes through another polarizer, the
analyzer, whose transmission axis is at an angle θ relative to the transmission axis of
the polarizer. The final intensity of the beam is I =
1
I 0 cos 2 θ .
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Example
Vertically polarized light with an intensity of 515 W/m2 passes through a polarizer
oriented at an angle θ to the vertical. Find the transmitted intensity of the light for (a)
θ = 10.0o , (b) θ = 45.0o , and (c) θ = 90.0o .
Answer
We apply the formula I = I 0 cos 2 θ .
(a)
=
I (515
=
W / m 2 ) cos 2 10.0o 499 W / m 2
(b)
=
I (515
=
W / m 2 ) cos 2 45.0o 258W / m 2
(c)
=
I (515
=
W / m 2 ) cos 2 90.0o 0 W / m 2
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Example
In a polarization experiment, the final intensity
of the beam is 0.200 I 0 . What is the angle θ
between the transmission axes of the analyzer
and polarizer?
Answer
The intensity of the light is first reduced to
I
=
1
I 0 and then to 0.200 I 0 . That is,
2
1
I 0 cos 2 θ 0.200 I 0 . We therefore have
=
2
1
cos 2 θ = 0.200 .
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Solving for θ, we obtain θ = 50.8o .
Since the analyzer absorbs part of the light as the beam passes through, it also absorbs
energy. Thus, in principle, the analyzer would experience a slight heating.
The applications of crossed polarizers include photoelastic stress analysis and LCD.
The operation of a liquid-crystal display (LCD) is simply “ON” and “OFF”. The
structure of LCD is shown in the figure.
(a) Off
(b) On
There are basically three essential elements to an LCD – two crossed polarizers and a
thin cell that holds a fluid formed of long, thin molecules known as a liquid crystal.
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The liquid crystal is selected for its ability to rotate the direction of polarization, and
the thickness of the liquid-crystal cell is adjusted to give a rotation of 90o. Thus, in its
“OFF“ state, the liquid crystal rotates the direction of polarization, and light passes
through the crossed polarizers. In this state, the LCD is transparent. When a voltage is
applied to the liquid crystal it no longer rotates the direction of polarization, and light
is no longer transmitted. Thus, it is in the “ON” state. The advantage of LCD is its
energy efficient. A very little energy is required to give the voltage necessary to turn a
liquid crystal cell “ON”. The light that the LCD uses is already present in the
environment.
18.4 Polarization by reflection
Unpolarized light can be partially polarized by reflection. When unpolarized light
strikes a reflecting surface between two optical materials, preferential reflection
occurs for those waves in which the electric-field vector is parallel to the reflecting
surface. At one particular angle of incidence, called the polarization angle θ p , only the
light for which the E vector is perpendicular to the plane of incidence is reflected. The
reflected light is therefore linearly polarized perpendicular to the plane of incidence.
θp θp
Sir David Brewster noticed that when the angle of incidence is equal to the polarizing
angle θ p , the reflected ray and the refracted ray are perpendicular to each other, so
r = 90 − θ p or sin r = cosθ p . From the law of refraction,
n1 sin θ p = n2 sin r ,
which gives
n1 sin θ p = n2 cosθ p ,
and thus the Brewster’s law: tan θ p =
n2
.
n1
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Remark:
When light is incident at the polarizing angle, none of the E field component parallel
to the plane of incidence is reflected; this component is transmitted 100% in the
refracted beam. So the reflected light is completely polarized. The refracted light is a
mixture of the component parallel to the plane of incidence, all of which is refracted,
and the remainder of the perpendicular component; it is therefore partially polarized.
Example
Sunlight reflects off the smooth surface of an unoccupied swimming pool. (a) At what
angle of reflection is the light completely polarized? (b) What is the corresponding
angle of refraction for the light that is transmitted (refracted) into the water? (c) At
night an underwater floodlight is turned on in the pool. Repeat parts (a) and (b) for
rays from the floodlight that strike the smooth surface from below.
Answer
(a),(b) The light is first in the air, then in the water. Take n a = 1.0 (air) and n b = 1.33
(water). The Brewster’s law tan θ p =
nb
gives
na
−1  nb 
−1  1.33 
o
=
=
θ p tan
  tan=

 53.1 .
 1.0 
 na 
The reflected and refracted rays are perpendicular, so θ p + r =
90o and thus
r = 90o − θ p = 90o − 53.1o = 36.9o .
(c)
Now the light is first in the water, then in the air. Take n a = 1.33 (water) and
n b = 1.00 (air). The Brewster’s law tan θ p =
nb
gives
na
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−1  nb 
−1  1.0 
o
=
θ p tan
=
  tan=

 36.9 .
n
1.33


 a
The reflected and refracted rays are perpendicular, so θ p + r =
90o and thus
r = 90o − θ p = 90o − 36.9o = 53.1o .
18.5 Polarization by scattering
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