Transcript Document

Lecture 13 Electromagnetic Waves Chp. 34
Thursday Morning
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Cartoon -. Opening Demo - Warm-up problem
Physlet
Topics
– Light is a Electromagnetic wave
– eye sensitivity
– Traveling E/M wave - Induced electric and induced magnetic amplitudes
– 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
Electromagnetic spectrum
Eye Sensitivity to Color
Production of Electromagnetic waves
How the fields
vary at a Point
P in space as
the wave goes
by
Electromagnetic Wave
Properties of a plane wave of light
E  E m sin(kx  t)
B  Bm sin(kx  t)
c

k

2

k
1
00
Em
c
Bm
 speed of light

E and B are always perpendicular to the direction of travel
E is perpendicular to B
E X B = the direction of travel
E and B both vary with the same frequency and in phase.
Speed of wave is independent of speed of observer
Wave doesn’t need a medium to travel in.
A point source of light is one that emits 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r2
The Poynting vector S
S
1
0
E B
S has units of energy/time per unit area or Watt/m2
S
1
0
EB
E is perp. to
B and in the energy flows in the
direction of the wave. Since B=E/c, we get for
the instantaneous power of the wave
S
1
0
EB 
1
c 0
E2
Intensity I of the wave is defined as Savg
1  2
1
2
I  Savg 
E rms 
Em
c0
2c0
E rms
1

Em
2
17. 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
Bm 
Em
2.0V m
9


6.7
10
T
8
c
2.99810 m s
(b) The average
intensity is 2

Iavg
2.0V m
E m

3
2



5.3
10
W
m
20c 24  10 7 T  m A2.998 10 8 m s
2
(c) The power of the source is



P  4r 2 I avg  4 10m 5.3 103 W m2  6.7W
2
Radiation pressure
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
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 cos2 60o 
1
I 0 cos2 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 
1
I 0 cos 4 60 o  3.110  2.
2
Thus, 3.1% of the light’s initial intensity
is transmitted.
Reflection and refraction
1   '1

n1
n2

Snells Law
n1 sin 1  n 2 sin  2
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?
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
x = l1tan1 = (0.50m)tan35o = 0.35 m.
According to the law of refraction, n2sin2 =
n1sin1. We take n1 = 1 and n2 = 1.33 (from
Table 34-1). Then,
o




sin

sin
35
1
1
1
  sin 
  25.55o
 2  sin 
 1.33 
 n2 
L is given by
1
l1
air
water
2
L  l2 tan 2  (1.50m) tan25.55  0.72m.
o
The length of the shadow is 0.35m + 0.72 m
= 1.07 m.
l2
shadow
L
x
Dispersion: Different wavelengths
have different indices of refraction
Total Internal reflection
n1 sin 1  n 2 sin  2
(1.33)sin1  (1.00)sin90  1.00