Transcript Lecture

Lecture 13 Electromagnetic Waves Ch. 33
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Cartoon
Opening Demo
Topics
– Electromagnetic waves
– Traveling E/M wave - Induced electric and induced magnetic amplitudes
– Plane waves and spherical waves
– Energy transport snd Intensity of a wave Poynting vector
– Radiation Pressure produced by E/M wave
– Polarization
– Reflection, refraction,Snell’s Law, Internal reflection
– Prisms and chromatic dispersion
– Polarization by reflection-Brewsters angle
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Elmo
Polling
Electromagnetic Waves
Electromagnetic Waves
Production of Electromagnetic waves
Spherical waves
Plane waves
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)
Traveling Electromagnetic Wave
How the fields
vary at a Point
P in space as
the wave goes
by
Electromagnetic Wave Self Generation
• Faraday’s Law of
Induction


r v
d
E .ds   B
dt
• Maxwell’s Law of
Induction

r v
d
B.ds  00 E
dt
c
1
0  0
• Changing electric field
induces a magnetic
field
d B
r
—
 E.ds   dx
• Changing magnetic
field induces a electric
field
Summary
Magnitude of S is like the intensity
dU
S 
Adt
U is the energy
carried by a wave
dU
S 
Adt
E0
B0 
c
Relation of intensity and power for a
Spherical Wave - Variation of Intensity
with distance
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
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
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
F p
U
I
Pr  


A At cAt c
F 2p 2U 2I
Pr  


A At cAt c
For 100% absorption of light on the paper
For 100% reflection of light
Problem 21
What is the radiation pressure 1.5 m away from a 500
Watt lightbulb?
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
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
Malus’s Law
I 2  I1 cos2 
Ey  E cos
Light comes
from here
Intensity is proportional to
square of amplitude
I0

I  Ey 2
Ey  E cos
y

E 2 y  E 2 cos 2 
I 2  I1 cos 2 
I 2  I1 cos2 
I0
I1 
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 2 90 deg and no light gets through
because cos 90 = 0
Polaroid Material
Transmits
Light
Yes
Light
N
o
Chapter 33 Problem 33
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 cos 2 60 o
1
I1  I 0
2
I3 1
 cos 4 60  3.125  10 2
I0 2
Chapter 33 Properties of Light Continued
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Law of Reflection
Law of Refraction or Snell’s Law
Chromatic Dispersion
Brewsters Angle
Lab06
Lab07
Lab06
Lab08
33
Reflection and Refraction of Light
Dispersion: Different wavelengths have different velocities and
therefore different indices of refraction. This leads to different
refractive angles for different wavelengths. Thus the light is dispersed.
The frequency does not change when n changes.
v  f
 changes when medium changes
f does not change when medium changes
Snells Law
n1 sin 1  n2 sin 2
Red
1.00sin 90
 0.75
1.33
 2  48.75deg
sin  2 
n1=1.00

n2=1.33
θ1
θ2

v1 sin 1  v2 sin 2
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?
Chapter 33
47. In the figure, a 2.00-m-long vertical pole extends from
the bottom of a swimming pool to a point 50.0 cm above
the water. What is the length of the shadow of the pole on
the level bottom of the pool?
1
l1
air
water
2
l2
shadow
L
x
Consider a ray that grazes the top of the
pole, as shown in the diagram below. Here
1 = 35o, l1 = 0.50 m, and l2 = 1.50 m.
The length of the shadow is x + L.
x is given by
1
x = l1tan1 = (0.50m)tan35o = 0.35 m.
l1
air
L is given by
water
2
l2
L=l2tan 
Use Snells Law to find 
shadow
L
x
Calculation of L
According to the law of refraction, n2sin2 =
n1sin1. We take n1 = 1 and n2 = 1.33
o




sin

sin
35
1
1
1
  sin 
  25.55o
 2  sin 
 1.33 
 n2 
L is given by
L  l2 tan  2  (1.50m) tan 25.55o  0.72m.
1
l1
air
water
2
The length of the shadow is L+x.
l2
L+x = 0.35m + 0.72 m = 1.07 m.
shadow
L
x
Fiber Cable
Why is light totally reflected inside a fiber
optics cable? Internal reflection
n1 sin 1  n2 sin 2
(1.509)sin 1  (1.00)sin 90  1.00

1  sin
1
1
1.509
 41.505 deg
Total Internal Reflection
Chromatic Dispersion of light with a prism
Equilateral prism
dispersing sunlight
late afternoon.
Sin θrefr= 0.7nλ
n increases with frequency
Blue light bends more than red
45
How much light is polarized when reflected from a surface?
Polarization by Reflection: Brewsters Law
What causes a Mirage
eye
sk
y
1.09
1.09
1.08
Index of refraction
1.08
1.07
1.07
1.06
Hot road causes gradient in the index of refraction that increa
as you increase the distance from the road
Inverse Mirage Bend