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Electromagnetic Waves
Lecture 13 Electromagnetic Waves Ch. 33
•
•
•
Cartoon
Opening Demo
Topics
– Electromagnetic waves
– Traveling E/M wave - Induced electric and induced magnetic amplitudes
– Plane waves and spherical waves
– Energy transport Poynting vector
– Pressure produced by E/M wave
– Polarization
– Reflection, refraction,Snell’s Law, Internal reflection
– Prisms and chromatic dispersion
– Polarization by reflection-Brewsters angle
•
•
Elmo
Polling
Electromagnetic Waves
Eye Sensitivity to Color
Production of Electromagnetic waves
Spherical waves
Plane waves
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To investigate further the properties of
electromagnetic waves we consider
the simplest situation of a plane wave.
A single wire with variable current
generates propagating electric and
magnetic fields with cylindrical symmetry
around the wire.
If we now stack several wires parallel
to each other, and make this stack
wide enough (and the wires very close
together), we will have a (plane) wave
propagating in the z direction, with
E-field oriented along x, E = Ex
(the current direction) and B-field
along y B=By (Transverse waves)
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Electromagnetic Wave
How the fields
vary at a Point
P in space as
the wave goes
by
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Electromagnetic Wave
Electromagnetic Wave Self Generation
• Faraday’s Law of
Induction
• Changing electric field
induces a magnetic
field
• Maxwell’s Law of
Induction
• Changing magnetic
field induces a electric
field
d B
r
—
 E.ds   dx
d E
r
—
 B.ds  0 0 dx
c
1
0  0
Summary
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Magnitude of S is like the intensity
dU
S 
Adt
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U is the energy
carried by a wave
dU
S 
Adt
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E0
B0 
c
Relation of intensity and power for a Spherical Wave
A point source of light generates a spherical wave. Light is
emitted isotropically and the intensity of it falls off as 1/r2
Let P be the power of the source
in joules per sec. Then the intensity
of light at a distance r is
I
P
4r 2
Lets look at an example
15 . The maximum electric field at a distance of 10 m from an isotropic
point light source is 2.0 V/m. Calculate
(a) the maximum value of the magnetic field and
(b) the average intensity of the light there?
(c) What is the power of the source?
(a) The magnetic field amplitude of the wave is
2.0 Vm
Em
Bm 

c
2.998  108
 6.7  109T
m
s
(b) The average intensity is
I avg 
2.0 Vm 
2
2
E m

2 0 c 2 4  10 7 T  mA 2.998  10 8


(c) The power of the source is

P  4r 2Iavg  4 10m  5.3  10 3
2
W
m2
  6.7W
m
s

 5.3  10 3 mW2
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Speed of light in Water
c 2.998  10 8
v 
n
1.33
m
s
 2.26  10 8
m
s
Nothing is known to travel faster than light in a vacuum
However, electrons can travel faster than light in water.
And when the do the electrons emit light called
Cerenkov radiation
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Momentum and Radiation Pressure
Momentum = p
What is the change in momentum of the
paper over some time interval?
Light beam
c
p=U/c
paper
p=0
1) Black paper absorbs all the light
p  U / c
2) Suppose light is 100% reflected
p  p  ( p)  2 p
p  2U / c
From this we define the pressure
Radiation Pressure Pr
Want to relate the pressure Pr felt by the paper to the
intensity of light
Power U / t
I

Area
A
F p / t
Pr  
A
A
p  U / c
For 100% absorption of light on the paper
F cU / t
Pr  
 cI
A
A
F 2cU / t
Pr  
 2cI
A
A
For 100% reflection of light
Problem 21
What is the radiation pressure 1.5 m away from a 500
Watt lightbulb?
Another property of light
Radiation pressure: Light carries momentum
I
Pr 
c
2I
Pr 
c
This is the force per unit area felt by an object
that absorbs light. (Black piece of paper))
This is the force per unit area felt by an object
that reflects light backwards. (Aluminum foil)
Polarization of Light
• All we mean by polarization is which
direction is the electric vector vibrating.
• If there is no preferred direction the wave is
unpolarized
• If the preferred direction is vertical, then we
say the wave is vertically polarized
A polarizing sheet or polaroid filter is special
material made up of rows of
molecules that only allow light
to pass when the electric
vector is in one direction.
Pass though a polarizing sheet
aligned to pass only the y-component
Unpolarized light
Resolved into its y and z-components
The sum of the y-components and
z components are equal
Intensity I0
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Pass though a
polarizing sheet
aligned to pass
only the y-component
Malus’s Law
I0
I
2
One Half Rule
Half the intensity out
Ey  E cos
Polarizer P1
I0
Analyzer P2
I  Ey 2
y
Ey  E cos
E
2
 E cos 
2
y
I 2  I1 cos 2 
2
I 2  I1 cos2 
I1 
I0
2
Sunglasses are polarized vertically.
Light reflected from sky is partially polarized
and light reflected from car hoods is polarized
in the plane of the hood
Sunglass 1
Sunglass 2
Rotate sun glass 90 deg and no light gets through
because cos 90 = 0
35. In the figure, initially unpolarized light is sent through three
polarizing sheets whose polarizing directions make angles of 1 = 40o,
2 = 20o, and 3 = 40o with the direction of the y axis. What percentage
of the light’s initial intensity is transmitted by the system? (Hint: Be
careful with the angles.)
Let Io be the intensity of the unpolarized light that
is incident on the first polarizing sheet. The
transmitted intensity of is I1 = (1/2)I0, and the
direction of polarization of the transmitted light is
1 = 40o counterclockwise from the y axis in the
diagram. The polarizing direction of the second
sheet is 2 = 20o clockwise from the y axis, so the
angle between the direction of polarization that is
incident on that sheet and the the polarizing
direction of the sheet is 40o + 20o = 60o. The
transmitted intensity is
I 2  I1 cos 2 60o 
I0
I1
I
2
I3
1
I 0 cos 2 60o ,
2
and the direction of polarization of the transmitted light is 20o clockwise
from the y axis.
35. In the figure, initially unpolarized light is sent through three
polarizing sheets whose polarizing directions make angles of 1 = 40o,
2 = 20o, and 3 = 40o with the direction of the y axis. What percentage
of the light’s initial intensity is transmitted by the system? (Hint: Be
careful with the angles.)
The polarizing direction of the third sheet is 3
= 40o counterclockwise from the y axis.
Consequently, the angle between the direction
of polarization of the light incident on that
sheet and the polarizing direction of the sheet
is 20o + 40o = 60o. The transmitted intensity is
I 3  I 2 cos 2 60 o
I 2  I1 cos2 60 o
1
I1  I 0
2
I3 1
 cos 4 60  3.125  10 2
I0 2
Thus, 3.1% of the light’s
initial intensity is transmitted.
Chapter 33 Problem 34
In Figure 33-42, initially unpolarized light is sent into a system of
three polarizing sheets whose polarizing directions make angles of
θ1 = θ2 = θ3 = 16° with the direction of the y axis. What percentage
of the initial intensity is transmitted by the system?
Fig. 33-42
Chapter 33 Problem 49
In Figure 33-51, a 2.00 m long vertical pole extends from the
bottom of a swimming pool to a point 90.0 cm above the
water. Sunlight is incident at angle θ = 55.0°. What is the
length of the shadow of the pole on the level bottom of the
pool?
Fig. 33-51
Chapter 33 Problem 62
Suppose the prism of Figure 33-55 has apex angle ϕ = 69.0° and index of
refraction n = 1.59.
Fig. 33-55
(a) What is the smallest angle of incidence θ for which a ray can enter the
left face of the prism and exit the right face?
(b) What angle of incidence θ is required for the ray to exit the prism with
an identical angle θ for its refraction, as it does in Figure 33-55?
In Figure 33-51, a 2.00 m long vertical pole extends from the bottom of a
swimming pool to a point 90.0 cm above the water. Sunlight is incident
at angle θ = 55.0°. What is the length of the shadow of the pole on the
level bottom of the pool?