Transcript Document 7368056

Physics 2120
University Physics
Dr. Bill Robertson
Middle Tennessee State University
Course Overview
• Text: Fundamentals of Physics, 8th Edition by
Walker, Halliday, and Resnick.
• Web site: www.mtsu.edu/~wroberts
• Office: WPS 207
• Office Hours: Monday & Wednesday mornings
• Grading:
– 3 Exams (In-class 3 worth 20%, 20%, 20%)
60%
– Homework
– Final Exam
10%
30%
Electrostatics and charge
• How was the concept of electric charge
discovered? History: Earliest experiments
on charge involved simple static attraction
and repulsion effects with dielectric rods
rubbed on cloth or fur. From these
experiments a number of basic terms and
a law of attraction (Coulomb’s law) was
derived.
Charge
• The electrostatic effect were attributed to
the fact that objects could become
charged.
• There were two types of charge: positive
and negative.
• Like charges repel and opposite charges
attract.
• Charge is measured in Coulombs
• Smallest unit of charge, e =1.6 x 10-19 C
Conductors and insulators
• In general terms materials fall into the
category of
– Insulators: materials in which charge cannot
move freely (rubber, plastic, pure water..)
– Conductors: materials in which charge can
move freely (metals, water with dissolved
impurities, bunnies…)
• We now understand that conduction
occurs because of the movement of
electrons (negative charges).
Example
• Induced charge: A charged insulating rod held
close to one end of a neutral metal rod attracts
the metal rod. Why?
Metal rod
--++++++++++++
Charged glass rod
+++
Save the bunny
• The physics is all wrong, but …
• http://www.youtube.com/watch?v=3K1Yzn
zxQmw
Coulomb’s law
• Charges were observed to attract or repel,
Coulomb’s law quantified the size and direction
of the force between charges.
q1q2
F  k 2 rˆ
r
• Similar to Newton’s gravitational law, inverse
square dependence on distance.
• k = 8.99x109 Nm2/C2 = 1/4peo
Example
• What is the force on an electron situated 1
cm away from a 0.1 C positive charge?
Example (leave the math!)
• Two otherwise identical metal spheres carry
charges Q1 and Q2. When their centers are a
distance 1.19 m apart, the spheres attract one
another with a force of magnitude 0.0853
N. The spheres are then briefly put into contact
with one another, and again separated to a
distance of 1.19 m. The spheres are then found
to repel one another with a force of magnitude
0.0196 N. What were the original charges on
the spheres if sphere 1 was initially positive?
Example
• Three charges Q1, Q2, and Q3 are
attached to the corners of an equilateral
triangle of side s = 0.50 m. Find the net
electrical force acting on the charge Q2 if
Q1 = 2.0 mC, Q2 = 7.0 mC, and Q3 = – 4.0
mC.
Circuits
• The practical side of the electromagnetism.
• We will cover some basic concepts that are
required for the lab work. We will define
these terms and then reconcile more
completely with the Coulomb force and
electric field material as the semester
progresses.
Current
• Current is the rate of flow of charge past a point in space
(e.g. in a wire or a beam of charged particles).
• Typically given the symbol i or I.
• Mathematically:
dQ
I
dt
• Unit of current is the Ampere (amp) 1 Amp = 1 C/s
• Current can flow in a conductor but not an insulator
• If the current is constant and 5 C passes through a wire in
10 seconds what is I?
EMF and Voltage
• In order to get charges to move around a circuit
there needs to be some “force” that causes them
to move. (EMF stands for Electromotive Force,
an olde worlde term no longer popular).
• An EMF source maintains a potential energy
difference between its terminals.
• The EMF source of initial interest is the battery.
More generally called a dc-power supply.
• dc [direct current] as opposed to ac [alternating
current]
EMF and Voltage
• EMF is defined in terms of the work done on a
charge that travels from one terminal to the
other.
• EMF is defined as the work done per unit charge.
dW
ΕMF 
dQ
• Units of EMF are work/charge i.e.
Joules/Coulomb = Volt (1 Volt = 1 Joule per
Coulomb)
Examples
• How much work is done when a steady
current of 5 mA flows for 3 minutes out of
a 9 Volt battery?
• What is the instantaneous power
expended by the battery during this time?
– Review: Power = Work done per unit time and
is measured in Watts.
Power from a battery
• Power from battery=Voltage x Current
Energy measured in electron volts
• The unit of the Volt is energy per coulomb;
thus we can express energy as charge
times voltage.
• When a an electron moves through a
voltage of 1 volt it acquires an increase in
energy of 1 electron volt.
Charge accelerator
• Electrons are accelerated (in vacuum) by
a voltage of 20 kV. What is the energy of
an electron in electron volts and in joules?
How fast is the electron traveling?
Electrons emitted
from heated wire
20 kV
Resistance
• We have said that there are two classes of
materials: conductors and insulators.
• A perfect conductor lets charge flow freely
whereas an insulator will not let charge
flow at all.
• In practice materials, and their geometry
(e.g. thin wire versus thick wire), offer
some impediment to the flow of charge.
This property is known as resistance.
Resistance and Ohm’s Law
• A perfect conductor would have zero resistance
• A perfect insulator would have infinite resistance
• How is resistance defined?
V
R
I
• This relation is known as Ohm’s law. (I tend to remember
it as V=IR)
• The unit of resistance is the Ohm (W) (1 volt per amp
equal 1 Ohm)
Circuit Symbols
• The resistor
• The battery (positive terminal long
line, negative short line). Current
always flows from positive to
negative.
• The ground connection
Example
• A 9 V battery is attached to a 250 Ohm resistor. What
current flows in the circuit?
9V
R=250W
• Circuit diagrams: Which is the positive terminal of the
battery? Which direction does current flow?
Example
• A 15V battery is connected “end-to-end” with a 1.2 kW
resistor. The negative terminal of the battery is grounded.
• (a) Draw the circuit diagram, label the circuit elements, and
draw in and label the current.
• (b) Wires used to connect circuit elements have a negligible
resistance within a circuit (compared to other resistances in
that circuit). Discuss the consequences of this fact when
connecting a circuit and analyzing its behavior.
• (c) Find the value of the current in the circuit.
• (d) Plot a graph of Voltage vs. Position around the circuit
• (e) Plot a graph of Current vs. Position around the circuit
Solving single loop circuits
• Rule 1: The sum of the potential around a closed
loop of any circuit must equal zero.
– Across a battery the potential change is positive from
the negative to the positive terminal; negative
otherwise.
– The potential drop across a resistor is negative if the
path goes in the same direction as the current across
the resistor; positive otherwise.
• Rule 2: The sum of the currents into a junction
equals the sum of currents out of a junction.
Example
• Solve for the current in the circuit below.
R = 50 W
2V
9V
R = 100 W
Resistors in parallel and in series
• What is the one resistance that could replace the
two resistors in the following circuits?
• What is common (current or voltage) for the (a)
series and (b) parallel resistors?
R1
V
V
R2
R1
R2
Combining resistors summary
• Resistors in parallel
1
1
1
 
RT R1 R2
• Resistors in series
RT  R1  R2
Solving circuits
• If possible see if the resistors can be
combined using the parallel and series
combination rules. Simplifying the circuit
by replacing parallel and series resistors
by their equivalent values can allow the
circuit parameters to be determined.
Example
• Solve the total current drawn from the
battery and the current in R3. R1 = R3 =
30W, R2  40 W R4 60 W. V = 12 V.
R1
V
R2
R4
R3
Real batteries—internal resistance
• Ideally a battery has a given EMF value.
When current is drawn from a real battery
the voltage between the terminals falls
from the ideal EMF value. The larger the
current the more the reduction from the
ideal EMF value. This effect is modeled as
an internal series resistance within the
battery.
Real batteries
• A 9 V battery gives a voltage of 8.7 V
when a current of 250 mA is drawn from it.
What is the internal resistance of the
battery? What voltage would there be if 25
mA were drawn from the battery?
r
Internal resistance
V=E-Ir
E
Multiple loop circuits
• Apply the loop potential rule around all the
loops of the circuit. Each loop will give one
equation relating the potentials and the
loop currents. Use simultaneous equations
to solve for the quantity of interest.
Example
• Find the value of the current and its direction for
each resistor in the circuit shown below.
• (b) What would be the reading on a voltmeter if
its black lead were connected to point A and its
red lead were connected to point B?
V1 = 34 V
V2 = 20 V
R1 = 4.0 W
R2 = 7.0W
R3 = 10W
R4 = 9.0 W
Example for you to ponder
• What is the diagonal resistance of a cube
of resistors each with resistance of 1 W?
Power dissipation in a resistor
• When a current I flows through a resistor what is
the power generated?
• Every charge, q, changes its energy as it
crosses the resistor by a value of q DV where DV
is the voltage drop across the resistor. By Ohm’s
law DV=IR, and the charge that flows per
second is I. Thus the energy change per second
is I2R.
• Power dissipated= I2R
• How else can we write this using Ohm’s Law?
Capacitance
• A capacitor is a device that stores charge.
• A capacitor consists of two conductors
separated by an insulator.
• The simplest capacitor consists of a pair of
parallel metal plates.
• When charge is moved from one plate to
the other a voltage develops between the
plates. Why?
Capacitance
• The voltage difference between the plates, V,
and the charge on the plates,Q, are related by
Q = CV
• The units of capacitance are Coulombs per Volt
which is named the Farad (1 F = 1 C/V).
• Capacitance is determined by the geometry of
the conductors (e.g. area of plates and their
separation).
RC discharge
• Imagine a charged capacitor connected in
series with a switch and a resistor. If the
capacitor initially hold a charge Q0 ,what is
capacitor charge Q as a function of time
after the switch is closed?
Q  Q0 e

t
RC
C
R
Charging RC circuit
• We could similarly derive
the equation for the
charging of a capacitor as
Q  Q0 (1  e

t
RC
)
• The quantity RC is called
the time constant often
given the symbol t. Note
Ohms times Farads equals
seconds
R
C
V
Energy stored in a capacitor
• A capacitor, C, that holds a charge Q
stores potential energy U
2
Q
U
2C
• How else can this expression be written
(remember Q=CV)
R1 = 100 kW, R2 = 50 kW,
C = 20 mF, Vin = 50 V
Consider the circuit shown in the diagram. At
t=0 the switch is thrown to position A; the
capacitor is initially uncharged. (a) What is
the current in the circuit immediately after the
switch is closed? (b) At what time, t0, is the
capacitor charge 90% of its maximum value?
(c) What is the voltage across the capacitor
at t0?(d) At what rate is energy being
delivered by the voltage source at t0?
R1 = 100 kW, R2 = 50 kW,
C = 20 mF, Vin = 50 V
At time to, the switch is thrown to position B.
(e) How long does it take after the switch is
thrown to position B for the charge on the
capacitor to be reduced by 90%? Call this
time t1. (f) At what rate is energy being lost
due to Joule heating at time t1?(g) At what rate
is energy being delivered by the capacitor at
time t1? (Does this result make sense?)
Electric Field
• The electric field is the “force field” that a
single charge or a collection of charges
creates in the surrounding space.
• The Electric field is defined as the force
per unit charge; i.e. the E-field at a point in
space is equal to the force that a 1
Coulomb charge (a unit charge) would feel
at that point.
E-field due to a point charge
• From Coulomb’s law the force between
two charges is
q1q2
F  k 2 rˆ
r
• If one charge (say q2) is a unit charge then
E due to q1 is defined as the force per unit
charge
q
q
E k
r
1
2
rˆ 
1
4peo r
2
rˆ
E-field is a vector field
• The electric field is a vector field—it has a
magnitude and a direction at every point.
• The E-field vector points away from a
positive charge and towards a negative
charge.
• The electric field at a point due to a
collection of charges is the vector sum of
the E-field created by every charge.
• A charge does not see its own E-field.
Field Line representation
• Maxwell came up with the field line
representation as a way to visualize the electric
field.
• The field lines show the direction of the E-field
and the density of lines is related to the strength
of the field.
• Field lines cannot cross!
• The field line model is a useful representation
rather than a quantitative method
E-Field due to a Point Charge
• Drawing not so
good!
• Lines should be
uniformly
spaced and
emanate from
the center of
the charge
+
E-field due to a sheet of charge
• What will the E-field due to an infinite
sheet of charge look like?
• We won’t get a quantitative answer (yet)
but we will figure out the direction of the
field.
Example
Find the field at a distance r along the x-axis
of two opposite charges spaced by a
distance d (see figure). Show that in the limit
that r>>d the field is
2Qd
E
3
4peo r
-Q
+Q
+
x=-d/2
x=0
x=+d/2
x=r
Charge density distributions
• In many cases the charge is distributed
along a line, over a surface, or in a volume.
• The charge is defined as a linear, areal, or
volume density
• E.g. A 1 m2 surface has 0.01 C of charge
evenly distributed. s=0.01/1 C m-2
• The charge in an infinitesimal area dA =
sdA
• A line of charge of length L has a nonuniform
linear charge density given by l = Ax, where A is
a constant. The line is situated along the positive
x-axis, as shown in the diagram below.
• (a) What are the SI units of the constant A?
• (b) Find an expression for the total line charge, Q.
• (c) Find an expression for the x-component of the
electric field at the point P, a distance d from the
end of the line.
Example
• What is the linear Uniform linear charge, Q, in a circle
charge density?
r
• Calculate the E-field
at a distance L
L
perpendicular to the
circle of charge, Q.
• In the limit of very
large L does the
result make sense?
Example
• Find the surface charge density and the E-field
at the point A. Examine the case x→∞. What is
E as R→ ∞? (Infinite plane of charge).
Uniform circular disk of charge.
Total charge Q.
R
x
A
Electric Field and Gauss’s Law
• Gauss’s law relates the flux of the electric
field through a closed surface to the total
charge enclosed.
 E .dA 
Qenc
eo
• Define flux
• Define closed surface and surface integral
• Dielectric permittivity eo
Flux
• A measure of the “flow” of electric field
through a surface.
• Analogy: imagine the flow of water through
wire loop. Can you see how the flow
depends on the orientation of the loop and
the direction of the flow of water? When is
the flow the greatest?
Flux
• The electric field flux, fe, through a surface
is given by the surface integral
f   E.dA
• What is the flux through a 1 cm2 loop in a
uniform E-field is the area of the loop is (a)
parallel to E, (b) 45 degrees to E and (c)
90 degrees to E?
• Note the direction of dA is outward from a
closed surface
Example
• What is the flux through the shaded area
in the figure due to an E-field given by
E  3e
x y
ˆx  17 y 3 ln( ay) yˆ  cx zˆ
y
y=h
0
x=w
x
Gauss’s law for a point charge
• We know the E-field (direction and
magnitude) for a point charge.
• Let’s examine the E-field flux through a
closed spherical surface that has the point
charge Q at its center.
• What does Gauss’s law give for the value
of E?
E-field due to a non-conducting
sheet of charge
• Imagine a large non-conducting sheet of
charge with surface charge density of s C
m-2. Use Gauss’s law to find an expression
for the E-field above and below the charged
sheet.
• Consider a pill box Gaussian surface
E-field due to capacitor plates
• Capacitor plates (when charged) can be
represented by two parallel sheets of
opposite charge. Using the result of the Efield due to a single sheet of charge figure
out the field above below and in between
Charge density +s
two capacitor plates.
d
Charge density -s
E-field due to a charged conductor
• Inside a conductor the E-field must be
zero. Why? Charges in a conductor are
free to move, if there is an E-field they will
move! They keep moving until E=0 inside
the conductor.
• This argument means that any excess
charge on a conductor resides on the
surface and that E is perpendicular to the
surface of the conductor.
E-field due to charged conductor
• A sphere of radius a has a charge Q
distributed uniformly throughout its
volume. The sphere is surrounded by a
conducting spherical shell of inner radius b
and outer radius c. (See picture on the
following slide). The conducting shell
carries charge +5Q
• Find an expression for the charge density
of the inner sphere.
E-field
• Find E for r<a,
a<r<b, b<r<c,
r>c
• What is the
charge on the
inner and outer
surface of the
conducting
shell?
Conducting shell
with charge +5Q
b
a
+Q
c
Example
A solid, non-conducting sphere of radius R
has a non-uniform volume charge density
a
given by

r
where a is a constant. Find E(r) in the
range 0 < r < R.
Spherical co-ordinates
• Volume element
• dV=r2sinqdqdf
Cylindrical symmetry
• A long straight wire of radius a carries a
charge per unit length l. What is the E
field around the wire for r<a and for r>a?
Example
A long, non-conducting cylindrical shell of
inner radius a and outer radius b has a nonuniform charge density given by
or

2pa
where o is a constant.
(a) Find the electric field in the region r < a.
(b) Find the electric field in the region a < r < b.
(c) Find the electric field in the region b < r.
Electrostatic potential
• The potential approach to electrostatics
uses work and energy instead of force and
acceleration. Compare this approach to
the two techniques used in mechanics
forces/accelerations etc versus potential
and kinetic energy.
• Which approach is more fundamental?
• Why two approaches?
Definition of potential
• In any volume of space where there is an
E-field we can define an electric potential
between two points as the negative of the
work done per unit charge by the E-field to
move from one point to the other.
b
DVab  Vb  Va    E .dl
a
The point charge
• Where is a
good choice for
zero potential?
• What is the
potential at a
radius R?
E-field
Equipotential
surfaces
+
Potential due to charge distributions
• Single point charge
V
Q
4pe0 r
• Charge distributions
V

all dQ
dQ
4pe0 r
Example
• Find the potential in the middle of the
square of charges.
+Q
-Q
d
+Q
+Q
d
Example
• A line of charge of length L is situated
along the x-axis as shown. The line has a
linear charge density given by l(x) = ax,
where a is a constant. What is the
electrostatic potential at point P, a
distance D from the end of the line?
x=D
x=0
x=L
Example
• A solid sphere of radius R has a total
charge Q distributed uniformly throughout
its volume. What is the electrostatic
potential at the center of the sphere
(relative to infinity)?
Finding E from V
• The electric field at a point in space can be
found from the gradient of the potential V.
• Gradient is given by the partial
differentiation
V
Ex  
x
V
Ey  
y
V
Ez  
z
• In words, the component of E in any
direction is the rate at which V changes in
that direction.
Surface of a conductor
• The surface of a conductor must be an
equipotential. Why? If V changes then
there is a gradient of V leading to an
electric field E, leading to a force on the
surface charges of the conductor.
Example
• V as a function of position in Cartesian
coordinates is given by
V ( x, y, z )  4 xz  5 y  3z
• What is E at the point (2,-1,3)
2
Capacitance
• We have already defined capacitance
earlier as the ratio of charge to voltage for
a pair of conductors carrying opposite
charges Q.
C=Q/V
• The voltage difference between
conductors can be foundb by
DVab  Vb  Va    E .dl
a
which means we can determine
how C
depends on conductor geometry.
Finding C of pairs of conductors
• Assume each conductor has equal
magnitude charge but opposite sign.
• Find E (Gauss’s law usually) in the region
between the conductors
• Find the potential difference between the
conductors, DV, by integrating E.dl along a
path joining the conductors
• Use C=Q/DV to find an expression for C.
Example
• Find an expression for the capacitance of
a pair of parallel conducting plates of area,
A, and separation distance, d.
• How does this result guide you in
determining how to combine capacitors in
parallel?
C1
C2
Combining capacitors in series
• What quantity is the same for the
capacitors in series? (Hint: rhymes with
barge.)
• What single capacitor CT could replace C1
and C2 in series?
C1
V
C2
Example
• Find an expression for the capacitance of
two concentric spheres of radii a and b (b
> a)
• What is the capacitance of an isolated
metal sphere?
Energy in a capacitor
• The stored energy in a capacitor can
easily be shown to be
q'
dW  dq 'V  dq '
C
q
q'
q2
W 
dq' 
C
2C
0
• Energy density in a parallel plate capacitor
U
CV 2 e 0 AV 2 1 V 
1
2
u



e

e
E
0
0
2

Ad 2 Ad
2 Ad
2 d 
2
2
Dielectrics
• Insulating materials (also called
dielectrics) in a capacitor alter the E field.
• In insulators the charges are not
completely free to move but they can
distort; i.e. positive tends to move in one
direction whereas the negative moves in
the opposite. Polarization.
+
-
E
Free and bound charges
• Use Gauss’s law to find E in the dielectric
q  q'
q
E

Ae 0
Ake 0
• What is q’ in term so k and q?
Free charge, q
++++++++++++++++
- - - - - - - - Dielectric
+ + + + + + + + +
- - - - - - - - - - - - - - - - -
Metal plates
of area A
Bound charge, q'
Example
• A coaxial cable consists of a wire of
radius a surrounded by a metal cylinder of
radius b. The intervening space is filled
with a dielectric with k.
• What is the capacitance of 1 m of the
coaxial cable?
• What is C if a=2 mm, b=5 mm & k=9?
• What is the free charge if V=20 V?
• What is the bound charge?
Magnetic fields
• Electric fields are created by charges
• Magnetic fields are created by currents (i.e.
moving charges)
– Electromagnets i.e. due to real currents in
wires
– Spin and angular momentum of electrons in
atoms—magnetic materials
• We define E due to the force it exerts on
another charge. Same method for
magnetic field.
Magnetism
• Magnetic field representation—similar to
E-field a vector field based on the force
exerted on charges (but only moving
charges).
• No magnetic monopoles—i.e. no isolated
sources of magnetic field
Force on a moving charge
• The magnetic field is given the symbol B.
• The force on a moving charge in a
magnetic field B is given by
F q v x B
• Can a magnetic field accelerate a particle?
(yes) Can it change a particles speed?
(no).
Magnetic field, B
• Measured in unit of Teslas
• 1 T = 1 Newton second/Coulomb meter =
1 Newton/Amp meter
Right Hand Rule
Example
• An electron moves with a velocity of 3
x105 m/s in the y direction in a region of
magnetic field given by
B  1.4 xˆ  2.1yˆ Teslas
• What is the magnetic force acting on the
electron?
Example
• An ion of mass 3.2 x 10-26 kg and charge
+e is accelerated through a voltage of 833
V. The charge then enters a uniform
magnetic field 0f 0.92 T as shown in the
figure. What is the radius of the resulting
circular motion? What is the period and
linear frequency?
Moving charged
particle
B field into page
Magnetic force on a wire
• Consider the force on each of the charges
that makes up the current in a wire of
length L and cross-sectional area A.
• Force on a wire
F iL xB
Example
• A conducting rod is supported
horizontally by two conducting
springs in a uniform horizontal x
magnetic field. The linear
x
mass density of the rod is
x
0.040 kg/m, and the magnetic
field strength is 3.6 T. Find
the current through the rod
that results in zero tension in
the support springs.
Support springs
x
x
x
x
x
x
x
x
x
x
x
B (into page)
x
Force on a wire loop in a B field
• Operating basis of many electric motors
Loop viewed from above
x
x i x
x
x
x
x
x
x
x
a
x
x
x
x
x
x
x
x
x
x
n
i
B (into page)
b
b
• N turns of wire in loop. What is the torque t?
t  NBiab sin q  NiAB sin q
Magnetic dipole moment
• A current loop behaves like a little bar
magnet aligning with a magnetic field.
• The magnitude of the dipole moment is
m  NiA
• The direction of the dipole moment vector
is given by right hand rule
m
i
Torque on a dipole
• From original definition
t  NBiab sin q  NiAB sin q
• Now with m=NiA
t m xB
Biot-Savart Law
• Moving charges are affected by magnetic
fields; similarly moving charges (currents)
create magnetic fields.
• Biot-Savart law
m0 I
B
4p

dl x r̂
2
r
Example
• A wire is configured as shown in the
diagram. What is the magnetic field at P
due to a current I =75 ma. a = 1.2 cm b =
3.5 cm, q=p/3 radians.
Test 2 topics
• You need to be able to:
– derive the E-field for planar, cylindrical, and spherical
geometries using Gauss’s law (with and without
dielectrics)
– Integrate E to get the voltage between conductors and
hence find capacitance
– Find the potential, V, due to a collection of charges
– Find the force (magnitude & direction) on charges in a
magnetic field
– Determine the magnetic field due to a current using the
Biot-Savart law (HW questions Ch 29: 6,24)
Ampere’s Law
• Ampere’s law is to magnetism what
Gauss’s law is to electrostatics.
B
.
d
l

m
I
0
enc

• This method works in cases with high
symmetry where the properties of the B
field can be inferred, figured out, whatever.
Current enclosed
• Current density J is the current per unit
area through a wire.
I enc   J .dA
• Example: What is J for a 6 mm diameter
wire carrying a uniformly distributed
current of 5 Amps?
Example
• Use the Biot-Savart law to determine the
direction of the magnetic field around a
current in a long-straight wire.
• Use Ampere’s law to find the magnitude of
the magnetic field a distance r from the
center of the wire.
• Find the value of the magnetic field
magnitude a distance of 10 cm from a wire
carrying a current of 1.0 A.
Example
• A current of 2.5 A flows through a solid
wire cable of radius 2.5 cm. Assuming
that the current is distributed uniformly
across the cross section of the wire, find
the magnetic field magnitude inside the
wire at a distance of one-half the wire’s
radius from its central axis.
Example
• A solenoid consists of loops of wire
carrying a current I. Let the solenoid be
wrapped with n turns-per-unit-length of
wire. Find an expression for the magnetic
field inside the solenoid.
Faraday’s law and Lenz’s law
• Faraday’s law describes how magnetic
fields can create voltages i.e. we are now
connecting magnetic and electric
phenomena
• In words a time varying magnetic flux
through a circuit loop creates a voltage
difference between the ends of the loop.
• Lenz’s law indicates the polarity of the
voltage
Faraday’s law
• Vind is the induced voltage between the
ends of the loop, fM is the magnetic flux
through the loop
df M
Vind 
dt
• Flux through 1 loop is given by
fM 
B
.
d
A

1loop Area
• Flux through N loops is N fM
Lenz’s law
• The voltage induced is always such as to
keep the flux through the circuit constant.
The direction of the voltage is oriented to
create a current in the loop such that the
flux remains the same.
• For example, what is the current direction
in the loop below as B increases into the
page?
X
X B into page
increasing in time
Solving Faraday law problems
• Draw a picture of the configuration
• Calculate an expression for the flux
through the circuit
• Determine the voltage across the loop
from
df M
Vind 
dt
• Determine the polarity of the voltage from
Lenz’s law.
Example
A single loop of wire has a rectangular cross
section of area A = 8.0 x 10–4 m2 and contains a
resistor of resistance R = 2.0 W. The loop is in a
region of uniform magnetic field that points
perpendicularly into the coil from your
perspective. The magnetic field magnitude
decreases linearly from Bi = 2.5 T to Bf = 0.50 T
over a time interval of 1.0 s. Find the magnitude
and direction of the current that is induced in the
resistor while the magnetic field is changing.
Example
A small loop of 250 turns and radius r = 6.0 cm
is inside a large solenoid with 400 turns/m. The
axis of the small loop is coincident with the axis
of the solenoid. The current in the solenoid
windings varies with time according to
I (t )  I 0 (1  e at )
where Io = 30 A and a = 1.61 s-1. Find the
voltage induced around the small loop at t = 1.0
s.
Motional EMF
• A metal rod moving through a uniform B
field as shown. The motion creates a force
on the charges in the rod that causes them
to move in the rod. What is the voltage
between the ends of the rod? (2 methods)
x
L
x
x
x
v
x
x
x
x
x
x
x
x
Example
• A metal rod of length L has
one end a distance a from a
long straight wire carrying a
current I. The rod has a
velocity that is parallel to the
current in the wire. Find the
induced voltage across the
rod and specify the polarity.
v
a
L
Faraday’s law: big picture
• The fundamental point of Faraday’s law is
that a time-varying magnetic flux, fM, leads
to an induced voltage and thus an E-field.
• In the briefest of terms
– A changing magnetic field produces an
electric field.
Inductance
• Earlier we defined capacitance as a ratio of
stored charge to voltage for 2 conductors.
• Here we define a similar quantity,
inductance, that relates magnetic flux
through circuit, NfM, to the current, i.
Nf M
L
i
Inductance of a solenoid
• A solenoid has a length L, cross-sectional
area, A, and a total of N turns. What is the
inductance?
• Note that inductance only depends on the
geometrical properties of the object.
Inductors in circuits
• The definition of inductance says
NfM  Li
• Faraday’s law says
d ( Nf M )
Vind  
dt
• Thus, in terms of inductance
di
VL   L
dt
• Sign of VL depends on i increasing or decreasing
RL circuit
• When the switch is closed what does the
circuit do? What is the direction of VL?
Switch
R
V
L
RL behavior
• With voltage source
i  iMAX (1  e

t
tL
• Without voltage source
i ie
'

t
tL
)
L
tL 
R
• What do these equations look like graphically?
Energy stored in an inductor
• Consider the circuit below. By the loop rule
di
V  L  iR
dt
• Multiply both sides by i; interpret each term
Vi  Li
• Conclusion
dU B
di
 Li
dt
dt
di 2
i R
dt
Switch
R
dU B  Li di
1 2
U B  Li
2
V
L
Energy density in a B field
• Consider a solenoid. Inductance per unit
length is L=mon2A
• Now use the energy stored in an inductor
formula UB and rearrange to find the
energy density uB (i.e. energy per unit
volume).
1
UB 
2
Li 2
2
2
1 Li 2 1
i
B
uB 
 ( m0 n 2 A) 
2 lA 2
A 2m0
Energy density in E and B fields
• Compare the energy density in E and B
fields
1
2
uE  e 0 E
2
2
1B
uB 
2 m0
• Remember energy density is equivalent to
pressure.
A circuit contains 2 resistors, a power source, inductor and
switch as shown. At t = 0, the switch is thrown from
neutral to A. Data: R1 = 4.0 W R2 = 7.0 W L = 8.0 mH V = 6.0 V
(a) What is the time constant of this circuit?
(b) What is the maximum current that can flow in this circuit?
(c) What is the current in the circuit at to = 250 ms?
(d) What is the rate at which energy is being stored in the
magnetic field of the inductor coils at time to?
(e) At time to the switch is set to B. How much later has the
rate at which energy loss due to Joule heating in the
resistors is down by a factor of 10 from its value when the
switch was set to position B? Switch
A
B
R2
V
R1
L
VL
LC oscillations
• LC circuit: Apply loop rule to get
Q
di
L 0
C
dt
Q  Qmax cos( t   )

1
LC
Example
• Show that the total energy in an LC circuit
(that is, the energy in the electric field
between the capacitor plates and the
magnetic energy in the magnetic field
around the inductor coils) is a constant.
Example
An LC circuit with inductor L=80 mH and capacitor C=5.0 mF.
The initial charge on the capacitor and current through the
inductor at t=0 are Qo=1.2 mC and Io= –6.7 mA. Find:
(a) the natural frequency of oscillation of the circuit.
(b) the phase constant.
(c) the charge amplitude.
(d) the current amplitude.
(e) the period of the current oscillations.
(f) the charge on the capacitor plates when t = 2.0 ms.
(g) the current through the inductor coils when t = 2.0 ms.
(h) the time at which the energy stored in the magnetic field
around the inductor coils first goes to zero after t = 0.
(i) the rate at which the energy being stored in the electric
field between the capacitor plates is changing at t = 2.0
ms.
Maxwell’s Equations
• Maxwell added one
item to the effects we
have studied so far—
a changing E field can
act like a current and
create a B field.
Compare this with a
changing B creating
an E field (Faraday’s
law).
• The final four
Q
 E .dA  e
o
 B .dA  0
df B
 E .dl   dt
dfE
 B .dl  moe o dt  mo I
The wave equation
• Maxwell combined the 4 equations to show that
in a charge free and current free region (Q=0 and
I=0) the equations combined to predict a coupled
electromagnetic wave. The E and B of this wave
in vacuum are described by
E  EM cos(kx  t )
B  BM cos(kx  t )
Wave properties
• E and B are perpendicular and they are in phase
• The direction of the wave is described by
v̂  Eˆ  Bˆ
• The wave speed v (c in vacuum) is given by
v
• E=cB
1
m oe o
 3 x 108 m / s
Poynting Vector
• The flow of energy in an EM wave is described
by the Poynting vector S defined by
S
1
mo
E B
• We can show that the wave intensity, I, is given
by
2
RMS
E
I
cm o
Energy density & pressure
• Imagine an EM wave carrying an energy U in a
time Dt onto an absorbing surface. What is the
force on the surface?
Absorbing wall
Energy, U
Area, A
cDt
Example
A laser of average power output 10 mW produces
a coherent beam of radius r = 0.80 mm.
(a) What is the average output intensity of the
laser?
(b) What is the average energy density in the
beam?
(c) What is the electric-field amplitude of the light
wave produced by the laser?
(d) What is the magnetic-field amplitude in the
output beam?
Example
The intensity of solar radiation at the earth’s
position is I = 1000 W/m2. The radius of the
earth is RE = 6.37 x 106 m, the masses of the
earth and sun are ME = 5.98 x 1024 kg and MS =
1.99 x 1030 kg, the distance from the earth to the
sun is D = 1.496 x 1011 m, and the universal
gravitational constant is G = 6.67 x 10–11 N.
m2/kg2.
(a) What is the average power radiated by the sun?
Example (continued)
The albedo of a planet is defined to be the
ratio of the total light reflected from it to the
total light incident on it. The approximate
albedo of Earth= 0.34.
(b) What approximate total force acts on the
earth due to radiation pressure?
(c) What is the ratio of the gravitational force
exerted on the earth by the sun to the total
force due to radiation pressure?
Introduction to Optics
• Optics is generally the study of a narrow
part of the entire EM spectrum—primarily
the part we can see (wavelengths from
700 nm to 400 nm)
Wave basics
• In optics the wavelength (rather than the
frequency) is generally used to describe
light of a fixed “color”.
• Of course wavelength and frequency are
related by
c  fl
• Red light 650 nm, green light 540 nm, blue
470 nm.
Geometrical Optics
• Light is a wave and waves diffract.
• Diffraction is negligible if the apertures and
objects that the radiation interacts with are
much larger than a wavelength.
• Geometrical optics describes light
propagation where diffraction is not
important.
• Light travels as rays. (Also called ray
optics).
Index of refraction
• Light in vacuum travels at c (=3 x 108 m/s)
• In transparent materials (dielectrics) the
speed of light slows. The extent of slowing
depends on the materials and it is described
by the refractive index, n, of the material.
• Glass n=1.5 (approximately) i.e. vglass=c/n
• Water n=1.33
• Picture of a wave in a material—what
changes?
Chromatic dispersion
• The refractive index in a material changes
with the wavelength of the radiation—this
effect is called dispersion.
• Normal dispersion means that the index
rises with decreasing wavelength.
• Dispersion causes a prism to split white
light into its colors and causes different
colors to focus at different distances from
a lens.
Example
Orange light of wavelength 600 nm in air enters
water having an index of refraction of 1.33.
(a) What is the frequency of this light?
(b) What is the speed of the light in the water?
(c) What is the wavelength of the light in the
water?
(d) If a person with normal vision were swimming
under water and looked at this light,
approximately what color would the person
see?
Refraction
• When a ray of light moves from one
material to another the change in velocity
causes the ray to deviate. Snell’s law
ni sin qi  nr sin q r
qi
Air
Glass
qr
Total-internal-reflection
• Consider a ray traveling from a high
refractive index material into a low
refractive index material. Above what
angle of incidence will all of the light be
reflected? Use the example of glass
(ng=1.5) and air (n=1)
Example
A small fish is swimming 1.2 m below the
surface of a calm pond. You are standing
on a small walking bridge over the pond
looking directly down at the fish. How far
beneath the water’s surface does the fish
seem to be to you, given that the index of
refraction of the pond water is 1.35?
Example
• A narrow beam of white light is incident at
an angle of 50o from the normal of a 60o
prism made of fused quartz. Find the
angular dispersion of the light emerging
from the far side of the prism.
• n(656nm)=1.4564
• n(588nm)=1.4585
• n(486nm)=1.4631
Example
• A small dot is placed at the center of a
piece of paper. Glass of thickness 1.26 cm
and index of refraction 1.48 is placed on
top of the paper. What is the smallest
radius of a coin that, when placed on top
of the glass and centered over the dot,
prevents anyone from being able to see
the dot from above the glass?
Lenses and image formation
• Lenses are created by making one or both
surfaces of a glass plates spherically
curved.
• By applying Snell’s law to rays hitting the
lens parallel rays of light can be brought to
a focus.
• http://www.fhsu.edu/~ktrantha/JavaOptics/j
avalens.html
Light from objects
• Light from illuminated objects diverges
from every point of the object. Every point
of the object is a source of scattered light
rays traveling in all directions.
• Light rays from distant objects is
essentially parallel. Thus, direct sunlight
consists of parallel rays of light.
Lens fundamentals
• Convex Lens—converging lens
do
di
Object
Image
ho
F1
F2
hi
Lens Fundamentals
• Concave lens—diverging lens (note F1 and F2
reversed)
do
Object
ho
hi
F2 Image
F1
di
Lenses and image formation
• A lens intercepts a selection of the scattered light from
an object and focuses the rays. In order to determine the
position of an image formed by a lens we consider only 3
easy to calculate rays.
– Ray 1: A ray leaving the tip of the object traveling parallel to the
optical axis will pass through the focal point F2 after passing
through the lens.
– Ray 2: A ray leaving the tip of the object and passing through the
focal point F1 will emerge from the lens traveling parallel to the
optical axis.
– Ray 3: The ray leaving the tip of the object and passing through
the center of the lens will emerge from the lens undeviated.
Real and virtual images
• If the rays after the lens converge to a
point then the image is real. Real means
that it can be displayed on a screen placed
at the point where the rays converge.
• If the rays after the lens diverge then the
image is said to be virtual. A virtual image
cannot be displayed on a screen but they
can be observed by eye. The lens in the
eye makes a real image on the retina.
The lens formula
• The positions of the image and object positions
are given by
1
1 1


f do di
• f is the focal length (positive for converging lens
negative for diverging lens)
• di is positive for a real image, negative for a
virtual image
• do and di are positive on left and right of lens
respectively
Magnification
• The magnification of the lens is given by how
much bigger the image is compared to the
object
hi
| m |
h0
• By examining the center ray it should be clear
that
di
m
d0
Example
An object is placed 10 cm from a 15-cm
focal length converging lens.
(a) Find and describe the image analytically.
(b) Find and describe the image using a ray
diagram.
Example
A diverging lens has a focal length of
magnitude 15.2 cm. You view an object of
height 3-cm through the lens when it is 10
cm from the center of the lens. Locate
and describe the image using analytical
means, and then draw a rough ray
diagram to help support your calculations
and conclusions.
Mirrors
• The ray reflection
from a mirror: the
angle of incidence is
equal to the angle of
reflection. qi = qr
qi
qr
Example
• You are looking at your image in a plane
mirror mounted flat on the wall 1 meter
away directly in front of you. The mirror
does not come all the way down to floor
level yet you can just see your shoes in
the image from the mirror. How high off the
floor is the mirror (assume you are 1.6 m
tall).
Spherical mirrors
• Curved mirrors (concave and convex) can form
images just like lenses
• For a spherical mirror with radius of curvature
10 cm, at what point is the focal length?
Ray parallel to
the optical axis
f
c
The 3 rays of happiness for mirrors
• Ray 1: The ray leaving the tip of the object
traveling parallel to the optical axis will pass
through the focal point after reflecting from the
mirror.
• Ray 2: The ray leaving the tip of the object and
passing through the focal point will reflect from
the mirror traveling parallel to the optical axis.
• Ray 3: The ray leaving the tip of the object and
passing through the center point of the mirror
will be reflected from the mirror undeflected.
Ray diagram
do
Ray parallel to
the optical axis
1
1 1


f do di
Object
f
Image
di
c
di and do are positive
on the reflecting side
of the mirror.
Example
A 1.00 cm-high object is placed 10.0 cm
from a concave mirror whose radius of
curvature is 30.0 cm.
(a) Draw a ray diagram to approximately
locate and describe the image.
(b) Locate and describe the image using
analytical methods.
Multi-lens systems
• Apply the lens formula to the lens nearest
the object—find the image. Use that image
as the object for the second lens to find
the image of the compound lens system.
• The magnification of a two lens system is
just the product of the magnifications of
successive lenses
mtotal = m1 m2 m3 …
Example
A compound microscope consists of two converging
lenses. The objective lens is a short focal-length lens that
is closest to the object being studied. The eyepiece is the
lens through which the viewer looks to see the enlarged
image of the object. The distance between the two lenses
should be substantially greater than the sum of the two
focal lengths of the two lenses, and the object is typically
placed just outside the focal point of the objective lens.
In this problem we will study the magnification of an ant
using a compound microscope. The objective lens has a
focal length of 1.0 cm, and the eyepiece has a focal length
of 2.5 cm. The ant being studied is placed 1.1 cm in front
of the objective lens, which is 13.4 cm from the
eyepiece. The ant is 0.2 mm wide.
Locate and describe the image of the ant as viewed
through this microscope.
Pinhole
• How does the pinhole allow imaging?
Wave nature of light
• EM radiation, including light, is a wave
phenomenon.
• The wave nature of light (or any wave
phenomenon) does not become apparent
until the light interacts with openings,
barrier, etc that are on the order of size of
the wavelength of the radiation.
• Wave effects: diffraction & interference
Properties of waves
• Principle of superposition: 2 parts
– Two or more waves will pass through one
another without their direction or speed being
changed. Waves don’t collide off each other
like particles do.
– Where two waves overlap the amplitude at a
point in space is just the sum of the individual
wave amplitudes at that point.
• The second point leads to wave interfere.
Interference
• When two single frequency waves
combine the relative phase between them
determines the extent of interference.
• Two extreme cases are when the waves
are exactly in phase—amplitudes add
leading to constructive interference.
• When waves are exactly out of phase the
amplitudes cancel—destructive
interference.
Huygens Principle
• Huygens, developed a
method of predicting the
propagation of waves. Each
point of a wave front becomes
a secondary source of new
waves. This method illustrates
(but does not really explain)
effects like diffraction.
Demos
• http://www.ngsir.netfirms.com/englishhtm/
Diffraction.htm
• http://physics.uwstout.edu/PhysApplets/acity/physengl/huygensengl.htm
Path length difference & phase
• If two waves travel different length paths to
between two points the path difference
can be expressed as a phase difference.
Path & phase
• If the path difference is Dx then the phase
difference (for waves of wavelength l) is
2p
Df 
Dx
l
• Constructive interference occurs if Dx is 0,
l, 2l… ie Df=0, 2p, 4p ..
• Destructive interference occurs if Dx is l/2,
3l/2, 5l/2… ie Df=p, 3p, 5p ..
Young’s double slit
• Demonstrated light was a wave
Incident
light
q
d
Bright fringe
Dark fringe
d sin(q)
ml  d sin q
1
(m  )l  d sin q
2
Intensity of the two slit diffraction
• Phasor concept
• At any point on the screen the time varying
E-field from the two slits is given by
E1  E cos(t )
E2  E cos(t  f )
• The phase difference f is found from the
path difference at angle q.
2p
f
d sin q
l
Phasor sum
• Individual phasors (left) and summed
phasors (right)
E actual
E actual
0
0
Etotal
f
0
 t+f
t

t
0
f 
Etotal  2 E0 cos   2 E0 cos  
2
Intensity
• The intensity of light is given by the square of
the total E-field at any point on the screen
• The E-field varies across the screen because
of interference
• Phasor picture and a little bit of mathematical
jiggery-pokery…
f 
I  4 E cos  
2
2
0
2
Example
• Light of wavelength 587.5 nm is incident
normally on a set of two slits separated by
a distance of 0.20 mm, and then projected
onto a distant screen. The second-order
maximum of the interference pattern thus
produced is found to be at the position y2
= 2.0 mm. Find the distance from the slits
to the screen.
Single slit diffraction pattern
• Derivation of the single slit diffraction equation
• Minima given by ml  a sin q m
First minimum
q1
a
Central maximum
a/2
l/2
Intensity for single slit
 f  Eq
sin   
2  2R

R
• Phasors, but with a
f/2 f/2
whole bunch of
sources side by side
EM
f
across the opening.
R
• EM is the sum of all
E
phasors end to end
f
• Eq is the amplitude

we seek.
EM
f 
Eq 
sin  
• I(q)=Eq2
f
2
q

2
Phase, f
• The phase, f, in the diagram is the
difference in phase between the paths to
the screen from opposite ends of the slit
opening.
• Path difference = a sinq
2
p
• Phase difference =
a sin q
l
Intensity for a single slit
• Final result
 sin a 
I (q )  I M 

a


2
f
pa
a   sin q
2 l
• This is the sinc function. What is its value
for a=0?
Double slit / slit width
1
0.9
Red is double slit, blue is single slit
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
-20
-15
-10
-5
0
Position on screen
5
10
15
20
Double slit
1
0.9
Diffraction of double slit, width accounted for
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
-20
-15
-10
-5
0
Position on screen
5
10
15
20
Full double slit formula
• Double slit with slit width, a, and slit
separation, d.
I (q )  I M
pd

sin q
l
 sin a 
cos  

a



2

a
ap
l
sin q
2
Example
Light from a green laser (540 nm) shines
through a set of double slits and is incident on
the board at the front of the room.
(a) By inspection of the pattern of laser light
formed on the board, but without performing any
distance measurements, estimate the ratio of the
slit separation to the slit width.
(b) Use distance measurements to help estimate
the separation of the slits and the width of each
slit.
Diffraction from a Grating
Diffraction grating
d
q
ml  d sin q
Each slit has path difference d sin q with
respect to the next slit
Gratings
• Grating are usually specified by the number of
rulings or grooves per unit length. For example,
300 lines per mm. You can get d easily from
this definition.
• Diffraction orders are labeled m=0, ±1, ±2…
• Grating diffraction leads to much narrower welldefined bright diffraction spots because of the
larger illuminated aperture.
• Resolution
l
Dq hw 
Nd
Example
Light from a green laser (540 nm) shines through
a diffraction grating of width 4.5 cm and is
incident on the board at the front of the room.
(a) What is the separation between adjacent
grooves?
(b) What is the linear groove density in the
grating?
(c) What is the total number of grooves in the
grating?
Thin film interference
• An everyday example of thin film interference is
the colorful pattern created by gasoline films on
water in parking lot puddles.
• The effect is due to interference between light
reflected from the top and bottom surfaces of the
thin gas layer. Cancellation of light occurs when
the phase difference between the 2 reflected
rays differs by p, 3p, 5p …
• The effect also occurs in any situation where
there are two closely spaced reflecting surfaces.
Phase change on reflection
• Whenever light hits an interface between 2 media
with different refractive indices some light is
reflected and some transmitted.
• If the light is reflected from a low-index to highindex transition then the reflected light is inverted
on reflection (i.e. it suffers a phase change of p
radians).
• Zero phase changes occurs for a high n to low n
transition.
• No phase change for transmitted light.
Path difference
• In a thin film interference problems:
– figure out the path difference between the two
rays and
– account for phase changes on reflection, if
any.
– If the total phase change is 0, 2p, 3p.. the
interference is constructive (bright fringe). If
total phase is p, 3p/2, 5p/2 … the phase
change is destructive (dark fringe).
Example
Thin films of material are often used as “nonreflective” coatings on camera lenses. A thin film
of magnesium fluoride having an index of
refraction of 1.38 coats a glass lens having an
index of refraction of 1.55. The film has a
thickness of 0.73 mm.
What wavelengths in the visible region of the
electromagnetic spectrum can be clearly seen
upon reflection from the thin film?
Final Exam
•
•
•
•
•
7 problems
Gauss’s law
Circuits, R, RL, RC
Mirrors/lenses and imaging
Covers everything up to thin films but not
polarization
Example
• One glass plate is positioned on top of
another with the end of a hair between the
plates at one end. The plates are
illuminated from above with light of
wavelength 589.0 nm from a sodium
vapor lamp. An interference pattern is
observed due to the variations in the
thickness of the air layer between the
glass plates. Use the interference pattern
to estimate the diameter of the hair.
Polarization
• Polarization refers to the orientation of the
E-field vector in an electromagnetic wave.
• Most light is unpolarized, i.e. the E-field of
the light is equally distributed over all
angles.
• A polarizer is a filter that allows one
direction of the E-field pass through.
Polarizer
• E-field vector component parallel to the
polarizer direction is allowed through.
Polarizer oriented vertically
Incident light
q
E0
Transmitted light
E0 cosq
Intensity
• We measure intensity not E-field by eye
and using light meters.
• What intensity of unpolarized light is
transmitted through a polarizer? What is
its polarization angle after transmission?
– What is the total incident intensity?
– What intensity is transmitted?
Polarized light transmission
• If linearly-polarized light is incident on a
polarizer what is the intensity of
transmission as a function of angle, q,
between the E-field and the polarizer?
– Law of Malus
I  I 0 cos 2 q
Example
• If unpolarized light of intensity 30 mW/m2
falls on a horizontally oriented polarizer,
what intensity is transmitted and what is
the orientation of the transmitted light’s Efield?
• If a second polarizer is added at 45
degrees to the horizontal, what intensity is
transmitted through the combination and
what is the orientation of the E-field?
Polarization on reflection
• Brewster’s angle: red p-pol.; blue s-pol.
1
0.9
0.8
Reflectivity
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
10
20
30
40
50
Angle (degrees)
60
70
80
90
Brewster’s angle
• The angle of minimum reflection of ppolarized light is given by
n2
q B  tan
n1
1
Example
• Discuss with as much detail as possible
the Polaroid sunglasses commercial on TV
in which swimmers beneath the water’s
surface in a swimming pool cannot be
seen due to glare from the sun’s light
reflected off of the water surface, but when
Polaroid sunglasses are placed over the
TV-camera lens, the swimmers can be
seen clearly.