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PES 2130 Fall 2014, Spendier
Lecture today: Chapter 33 Electromagnetic Waves
1) Total Internal Reflection
2) Polarization
Announcements:
- Exam 2 Wednesday October 29
- Poynting Vector
S
1
0
E B
I  Sav 
I
Emax 2
2 c 0
(Poynting vector in vacuum)
(intensity of sinusoidal wave in vacuum)
P
power
 s2
area
4 r
Ps = wave's power
4πr2 = surface area of sphere
- Reflection and Refraction
θ1 = θ1 '
n2 sin(θ2) = n1 sin(θ1)
- Chromatic dispersion (Rainbows)
Lecture 17/Page 1
PES 2130 Fall 2014, Spendier
Lecture 17/Page 2
1) Total Internal Reflection
We have described how light is partially reflected and partially transmitted at an interface
between two materials with different indexes of refraction. Under certain circumstances
all the light can be reflected back from the interface, with none of it being transmitted.
(Even though the second material is transparent.)
DEMO
Light introduced at the outer end bend piece of glass
undergoes repeated total internal reflection within
the glass so that, even though the glass provides a
curved path, most of the light ends up exiting the
other of the bend glass pipe (this is how optical
fibers work)
The angle of incidence giving this situation is called the critical angle θc. For angles of
incidence larger than θc there is no refracted ray and all the light is reflected; this effect is
called total internal reflection.
To find θc, we use Snell’s law and set θ2 = 90o
n1 sin(θc) = n2 sin(θ2) = n2 sin(90) = n2
θc = sin-1(n2/n1)
…… critical angle
NOTE:
Because the sine of an angle cannot exceed unity, n2 cannot exceed n1 in this equation.
This restriction tells us that total internal reflection cannot occur when the incident light
is in the medium of lower index of refraction. If source S were in the air in Fig. 33-23a,
all its rays that are incident on the air – glass interface (including f and g) would be both
reflected and refracted at the interface.
PES 2130 Fall 2014, Spendier
Lecture 17/Page 3
Example 1:
Consider a fish (or a diver) swimming in a clear pond. If the fish looks upwards it sees
the sky, but if it looks at too large an angle to the vertical it sees the bottom of the pond
reflected on the surface of the water. What is the angle to the vertical at which the fish
first sees the reflection of the bottom of the pond?
This must be the critical angle:
nwater = n1 = 1.33
nair = n2 = 1.00
θc = sin-1(n2/n1) = sin-1(1.00/1.33) = 48.8o
2) Polarization of light
Only transverse waves can be "polarized". (The word comes from "poles" - meaning, can
have an orientation.)
We define the direction of polarization of an EM wave to be the direction of the electricfield vector, not the magnetic field vector, because many common EM-wave detectors
respond to the electric force on electrons in materials, not the magnetic forces.
Electric field oscillates vertically  waves is polarized vertically
Figure shows an electromagnetic wave with its electric field oscillating parallel to the
vertical y axis. The plane containing the vectors is called the plane of oscillation of the
wave (hence, the wave is said to be plane-polarized in the y direction).We can represent
the wave’s polarization (state of being polarized) by showing the directions of the
electric field oscillations in a head-on view of the plane of oscillation.
PES 2130 Fall 2014, Spendier
Lecture 17/Page 4
Unpolarized (randomly polarized) light:
such as light from the Sun or a light bulb
The electric field at any given point is always perpendicular to the direction of travel of
the waves but changes directions randomly. There is equal probability to measure E in
any direction in the y-z plane.
This is a lot of arrows to draw so we (your book) will us the following to symbolize
unpolarized light.
Polarizing Sheets:
Waves emitted by a radio transmitter are usually linearly polarized. We can transform
unpolarized visible light into polarized light by sending it through a polarizing sheet.
Electric field components along one direction pass through the sheet, while components
perpendicular to that direction are absorbed by the molecules and disappear. Thus, the
electric field of the light emerging from the sheet consists of only the components that are
parallel to the polarizing direction of the sheet; hence the light is polarized in that
direction
An electric field component parallel to the polarizing direction is passed (transmitted) by
a polarizing sheet; a component perpendicular to it is absorbed.
PES 2130 Fall 2014, Spendier
Lecture 17/Page 5
Intensity of Transmitted Polarized Light
We now consider the intensity of light transmitted by a polarizing sheet
one-half rule; we can use it only when the light reaching a polarizing sheet is unpolarized.
I
1
I0
2
intensity I of the emerging polarized light
intensity I0 of the original light
1) DEMO: What happens when unpolarized light basses through 2 polarizes at 90o to one
another?
no light will pass through
2) What happens when linearly polarized light passes through a polarizer at some angle
θ?
We can resolve E into two components relative to the polarizing direction of the sheet:
parallel component Ey is transmitted by the sheet, and perpendicular component Ez is
absorbed. Since θ is the angle between and the polarizing direction of the sheet, the
transmitted parallel component is
E y  E cos
Recall that the intensity of an electromagnetic wave (such as our light wave) is
proportional to the square of the electric field’s magnitude
I  E2
PES 2130 Fall 2014, Spendier
Lecture 17/Page 6
and one can show that
I  I 0 cos2 
θ = 0 ==> all light gets through I = I0
θ = 90o ==> no light gets through I = 0
Example: A beam of unpolarized light of intensity I0 passes through a series of ideal
polarizing filters with their polarizing directions turned to various angles as shown in the
figure.
(a) What is the intensity at each of the 3 points?
(b) If the middle filter was removed, what will be the light intensity at point C?
a)
b)
PES 2130 Fall 2014, Spendier
Lecture 17/Page 7
Polarization by Reflection
Polarization by reflection is the reason polarizing filters are widely used in sunglasses or
put in front of a camera lens.
This is just one example of polarization by reflection. Although the light from the sun is
not polarized, it can be separated into two polarized components that are reflected and
transmitted in different amounts by the surface of the water for example (Fresnel laws).
Figure below shows a ray of unpolarized light incident on a glass surface. We can resolve
the electric field vectors of the light into two components. The perpendicular components
are perpendicular to the plane of incidence and thus also to the page. The parallel
components are parallel to the plane of incidence and the page. Because the light is
unpolarized, these two components are of equal magnitude.
In general, the reflected light also has both components but with unequal magnitudes.
This means that the reflected light is partially polarized—the electric fields oscillating
along one direction have greater amplitudes than those oscillating along other directions.
However, when the light is incident at a particular incident angle, called the Brewster
angle θB, the reflected light has only perpendicular components. The reflected light is
then fully polarized perpendicular to the plane of incidence. The parallel components of
the incident light do not disappear but (along with perpendicular components) refract into
the glass.
PES 2130 Fall 2014, Spendier
Lecture 17/Page 8
For light incident at the Brewster angle θB, we find experimentally that the reflected and
refracted rays are perpendicular to each other. Because the reflected ray is reflected at the
angle θB in Fig. 33-25 and the refracted ray is at an angle θR, we have
θB + θR = 90o
using Snell's law:
n1 sin(θB) = n2 sin(θR)
n1 sin(θB) = n2 sin(90o - θB) = n2 cos(θB)
n2 / n1 = sin(θB) /cos(θB) = tan(θB)
θB = tan-1(n2 / n1)
(Brewster angle)
Application: Brewster's window
Gas lasers typically use a window tilted at Brewster's angle to allow the beam to leave the
laser tube. Since the window reflects some perpendicular-polarized light but no parallelpolarized light, the round trip loss for the perpendicular-polarization is higher than that of
the parallel-polarization. This causes the laser's output to be parallel-polarized due to
competition between the two modes
lasers:
perpendicular-polarization = s-polarization
parallel- polarization = p-polarization