Transcript Lecture 27

Physics 1402: Lecture 27
Today’s Agenda
• Announcements:
– Midterm 2: Monday Nov. 16 …
– Homework 07: due Friday this week
• Electromagnetic Waves
– Energy and Momentum in Waves
• Optics
– Waves, Wavefronts, and Rays
– Reflection
– Index of Refraction
f( x
f( x )
x
x
z
y
E & B in Electromagnetic Wave
• Plane Harmonic Wave:
where:
y
x
z
Note: the direction of propagation
where
is given by the cross product
are the unit vectors in the (E,B) directions.
Nothing special about (Ey,Bz); eg could have (Ey,-Bx)
Note cyclical relation:
Velocity of Electromagnetic Waves
• The wave equation for Ex:
(derived from Maxwell’s Eqn)
• Therefore, we now know the velocity of
electromagnetic waves in free space:
• Putting in the measured values for m0 & e0, we get:
• This value is identical to the measured speed of light!
– We identify light as an electromagnetic wave.
The EM Spectrum
• These EM waves can take on any wavelength from
angstroms to miles (and beyond).
• We give these waves different names depending on the
wavelength.
10-14
10-10
10-6
10-2 1 102
Wavelength [m]
106
1010
Energy in EM Waves / review
• Electromagnetic waves contain energy which is stored in E
and B fields:
=
• Therefore, the total energy density in an e-m wave = u, where
• The Intensity of a wave is defined as the average power
transmitted per unit area = average energy density times wave
velocity:
Momentum in EM Waves
• Electromagnetic waves contain momentum.
• The momentum transferred to a surface depends on the area of the
surface. Thus Pressure is a more useful quantity.
• If a surface completely absorbs the incident light, the momentum
gained by the surface is,
• We use the above expression plus Newton’s Second Law in the
form F=dp/dt to derive the following expression for the Pressure,
• If the surface completely reflects the light, conservation of
momentum indicates the light pressure will be double that for the
surface that absorbs.
The Poynting Vector
• The direction of the propagation of the electromagnetic wave is
given by:
• This wave carries energy. This energy transport is defined
by the Poynting vector S as:
– The direction of S is the direction of propagation of the wave
– The magnitude of S is directly related to the energy being
transported by the wave:
• The intensity for harmonic waves is then given by:
The Poynting Vector
•
Thus we get some useful relations for the Poynting
vector.
1. The direction of propagation of an EM wave is along
the Poynting vector.
2. The Intensity of light at any position is given by the
magnitude of the Poynting vector at that position,
averaged over a cycle.
I = Savg
3. The light pressure is also given by the average value
of the Poynting vector as,
P = S/c
Absorbing surface
P = 2S/c
Reflecting surface
Generating E-M Waves
• Static charges produce a constant Electric
Field but no Magnetic Field.
• Moving charges (currents) produce both a
possibly changing electric field and a static
magnetic field.
• Accelerated charges produce EM radiation
(oscillating electric and magnetic fields).
• Antennas are often used to produce EM
waves in a controlled manner.
•
A Dipole Antenna
V(t)=Vocos(t)
+
+
-
E
E
+
+
-
• time t=0
x
z
y
• time t=p/2
• time t=p/
one half cycle
later
dipole radiation pattern
proportional to sin(t)
• oscillating electric dipole generates e-m radiation that is
polarized in the direction of the dipole
• radiation pattern is doughnut shaped & outward traveling
– zero amplitude directly above and below dipole
– maximum amplitude in-plane
Receiving E-M Radiation
receiving antenna
y
Speaker
x
z
One way to receive an EM signal is to use the same sort
of antenna.
• Receiving antenna has charges which are
accelerated by the E field of the EM wave.
• The acceleration of charges is the same thing as an
EMF. Thus a voltage signal is created.
Lecture 27, ACT 1
• Consider an EM wave with the E field
POLARIZED to lie perpendicular to the ground.
y
x
z
In which orientation should you turn your receiving
dipole antenna in order to best receive this signal?
a) Along S
b) Along B
C) Along E
Loop Antennas
Magnetic Dipole Antennas
• The electric dipole antenna makes use of the
basic electric force on a charged particle
• Note that you can calculate the related
magnetic field using Ampere’s Law.
• We can also make an antenna that produces
magnetic fields that look like a magnetic dipole,
i.e. a loop of wire.
• This loop can receive signals by exploiting
Faraday’s Law.
For a changing B field
through a fixed loop
Lecture 27, ACT 2
• Consider an EM wave with the E field
POLARIZED to lie perpendicular to the ground.
y
x
z
In which orientation should you turn your receiving
loop antenna in order to best receive this signal?
a) â Along S
b) â Along B
C) â Along E
Waves, Wavefronts, and Rays
• Consider a light wave (not necessarily visible) whose E
field is described by,
• This wave travels in the +x direction and has no
dependence on y or z, i.e. it is a plane wave.
3-D Representation
RAYS
Wave Fronts
EM wave at an interface
• What happens when light hits a surface of a material?
• Three Possibilities
– Reflected
– Refracted (transmitted)
– Absorbed
incident
ray
reflected
ray
MATERIAL 1
MATERIAL 2
refracted
ray
Geometric Optics
• What happens to EM waves (usually light) in
different materials?
– index of refraction, n.
• Restriction: waves whose wavelength is much
shorter than the objects with which it interacts.
• Assume that light propagates in straight lines,
called rays.
• Our primary focus will be on the REFLECTION and
REFRACTION of these rays at the interface of two
materials.
incident
ray
reflected
ray
MATERIAL 1
MATERIAL 2
refracted
ray
Reflection
• The angle of incidence equals the angle of reflection
 qi = qr , where both angles are measured from the normal:
· Note also, that all rays lie in the “plane of incidence”
qi qr
· Why?
» This law is quite general; we supply a limited
justification when surface is a good conductor,
• Electric field lines are perpendicular to the
conducting surface.
• The components of E parallel to the surface of the
incident and reflected wave must cancel!!
Ei
Er
qi
qr
qi qr
x
Index of Refraction
• The wave incident on an interface can not only
reflect, but it can also propagate into the second
material.
• Claim the speed of an electromagnetic wave is
different in matter than it is in vacuum.
– Recall, from Maxwell’s eqns in vacuum:
– How are Maxwell’s eqns in matter different?
 e0 e , m0 m
· Therefore, the speed of light in matter is related to the
speed of light in vacuum by:
where n = index of refraction of the material:
· The index of refraction is frequency dependent: For example
nblue > nred
Refraction
•
How is the angle of refraction related to the angle of
incidence?
– Unlike reflection, q1 cannot equal q2 !!
q1
n1
n2
» n1  n2  v1  v2
but, the frequencies (f1,f2) must be the same  the
wavelengths must be different!
Therefore, q2 must be different from q1 !!
q2
Snell’s Law
• From the last slide:
q2
q1 L
q2
q1
q1
q2
q2
The two triangles above each have hypotenuse L
\
But,
n1
n2
1
Lecture 27, ACT 3
• Which of the following ray diagrams could represent the passage of
light from air through glass and back to air?
(a)
(b)
(c)
air
air
air
glass
glass
glass
air
air
air
Lecture 27, ACT 3
• Which of the following ray diagrams could represent the passage of
light from air through glass and back to air?
(a)
(b)
(c)
air
air
air
glass
glass
glass
air
air
air
Lecture 27, ACT 4
• Which of the following ray diagrams could represent the passage of
light from air through glass and back to air?
(a)
(b)
(c)
(d)
A prism does two
things,
1. Bends light the
same way at both
entrance and exit
interfaces.
2. Splits colours due to
dispersion.
Index of refraction
Prisms
1.54
ultraviolet
absorption
bands
1.52
1.50
frequency
white light
prism
Prisms
Entering
q1
Exiting
q3
q2
For air/glass interface, we
use n(air)=1, n(glass)=n
q4
Prisms
Overall Deflection
f
q1
q3
q4
q2
• At both deflections the amount of downward deflection
depends on n (and the prism apex angle, f).
• The overall downward deflection goes like,
g ~ A(f) + B n
• Different colours will bend different amounts !
Lecture 27, ACT 5
White light is passed through a
prism as shown. Since n(blue)
> n(red) , which colour will end
up higher on the screen ?
A) BLUE
B) RED
?
?
LIKE SO!
In second rainbow
pattern is reversed
Total Internal Reflection
– Consider light moving from glass (n1=1.5) to air (n2=1.0)
n1
incident
ray
q1 qr
reflected
ray
GLASS
q2
refracted
ray
n2
AIR
ie light is bent away from the normal.
as q1 gets bigger, q2 gets bigger, but
q2 can never get bigger than 90 !!
2
In general, if sin q1  sin qC  (n2 / n1), we have NO refracted ray;
we have TOTAL INTERNAL REFLECTION.
For example, light in water which is incident on an air surface with
angle q1 > qc = sin-1(1.0/1.5) = 41.8 will be totally reflected. This
property is the basis for the optical fibers used in communication.
ACT 6: Critical Angle...
An optical fiber is
cladded by another
dielectric. In case I this
is water, with an index
of refraction of 1.33,
while in case II this is
air with an index of
refraction of 1.00.
Compare the critical
angles for total internal
reflection in these two
cases
a) qcI>qcII
b) qcI=qcII
c) qcI<qcII
water n =1.33
Case I
qc
glass n =1.5
water n =1.33
air n =1.00
Case II
qc
glass n =1.5
air n =1.00
ACT 7: Fiber Optics
The same two fibers are
used to transmit light
from a laser in one
Case I
room to an experiment
in another. Which
makes a better fiber,
the one in water (I) or
the one in air (II) ?
a) IWater
b) IIAir
Case II
water n =1.33
qc
glass n =1.5
water n =1.33
air n =1.00
qc
glass n =1.5
air n =1.00
Problem
You have a prism that from the side forms a
triangle of sides 2cm x 2cm x 22cm, and
has an index of refraction of 1.5. It is
arranged (in air) so that one 2cm side is
parallel to the ground, and the other to the
left. You direct a laser beam into the prism
from the left. At the first interaction with the
prism surface, all of the ray is transmitted
into the prism.
a) Draw a diagram indicating what happens to
the ray at the second and third interaction
with the prism surface. Include all reflected
and transmitted rays. Indicate the relevant
angles.
b) Repeat the problem for a prism that is
arranged identically but submerged in water.
A) Prism in air
Solution
• At the first interface q=0o, no deflection of initial light direction.
• At 2nd interface q=45o, from glas to air ?
• Critical angle: sin(qc)=1.0/1.5 => qc= 41.8o < 45o
• Thus, at 2nd interface light undergoes total internal reflection
• At 3rd interface q=0o, again no deflection of the light beam
B) Prism in water (n=1.33)
• At the first interface q=0o, the same situation.
• At 2nd interface now the critical angle: sin(qc)=1.33/1.5 => qc= 62o > 45o
• Now at 2nd interface some light is refracted out the prism
•
n1 sin(q1) = n2 sin(q2)
=> at q2 = 52.9o
• Some light is still reflected, as in A) !
• At 3rd interface q=0o, the same as A)