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AP Physics Notes Volume #3
Chapter 17 Current and Resistance
17.1 Electric Current
I = ΔQ / Δt= Coulombs/seconds = Amps
1 Amp = 1 Coulomb / 1 second
Thus the electric current (I) = The amount of charge (∆Q) that passes through an area in the given
time interval (Δt) time
It is conventional to give the current the same direction as the flow of the positive charge. In
some cases, gases and electrolytes, the current is the result of the flow of both positive and
negative charges. In a conductor such as Copper the current is a result of the motion of electrons,
thus the direction of the current is opposite the flow of charge carriers. But, in a beam of positive
charged protons the flow of charge is in the same direction as the current. It is common to refer to
a moving charge, ( positive or negative) as a mobile charge carrier. Either one is called a “charge
carrier”.
Ex. The amount of charge that passes through the filament of a light bulb in 2 seconds is 1.67C.
Find the current in the bulb and the number of electrons that pass through the filament each
second.
a. I = ∆Q/∆t = 1.67C/2.0s = .835A
b. Nq = N(1.6 x 10-19 C/electrons) = .835A
N = 5.22 x 1018 electrons
17.2 Current and Drift Speed
The volume of length x is A ΔX
Area multiplied by the change in x position.
A = Area = the cross section of a conductor that the current is being carried (flow) through.
Let N = number of charge carriers (mobile) per unit volume.
Thus the number of carriers per unit volume is NAΔX.
Also the charge Q of this element is ΔQ = (number of carriers)(charge per carrier) = (NAΔX)(q)
If the carriers move at speed Vd = drift speed for a period (Δt) then ΔX = VdΔt
So ΔQ = N A Vd Δt q
I = ΔQ/Δt = N A Vd Δt q/Δt
When a potential difference is applied to a conductor (like hooking up a battery). This creates an
electric force that accelerates the electrons in a zig-zag pattern in the opposite direction of the
current. The electrons collide with the atoms in the metal causing friction-induced heat. The
direction of the electric current movement will be denoted with E and a vector arrow. Because the
work done by the moving electrons exceeds the heat given off, a steady current (I) in produced.
In the classical model of electronic conduction in a metal, we treat electrons like molecules in a gas
(in the absence of an electric field) and say they have (0) zero average velocity V = O
Example:
A copper wire with cross section area 3x10-6 m2 carries a current of 10 amps. Assume each copper
atom contributes one free electron to the metal and find the drift speed. Use ρ copper = 8.95 g/cm3
Given: I= 10 amps A=3x10-6 m2 q= 1.6x10-19 C
We must find Vd and N.
N= mole
volume
volume V= M = 63.5g (mass of 1 mole) = 7.09 cm3
ρ 8.95 g/cm3
N= 6.02x1023 = 8.48x1022 electrons X 106 cm3 = 8.48x1028 electrons
7.09 cm3
cm3
m3
m3

Using I= NqVdA
Solving for Vd
Vd = I = 10 c/s
NqA (8.48x1023 electrons/m3)( 1.6x10-19e)(3x10-6m2)
= 2.46x10-6 m/s

Notice: Drift speed is very slow, at this velocity it would take 68 mins. to move 1 meter.
This shows that electrons are all in the light line before you switch on the light. (like water
in a hose)
Assign Pg. 575-576 1, 2, 3, 5, 6, 7, 9 (like example)
17.3 Resistance and Ohm’s Law
R= ΔV = Volts = Ω
I Amps

R = Resistance
I = Current
ΔV = Change in Voltage
Resistance is in Ohms Ω
Ex. An electric appliance connected to 120V carries a current of 6A
R = 120V = 20Ω
6A

Resistance occurs due to collisions between electrons caring the current and fixed charges
inside the conductor. (This works like friction)
Ohm’s Law
Resistance is constant over a wide range of applies voltage.
Example: A 3V battery produces twice the current on identical system as a 1.5V battery.
Rewrite R = ∆V/I as ΔV = IR *this is the way we will usually use it.
Materials that obey ohm’s law are called ohmic
Materials that don’t obey ohm’s law they are nonohmic
Ex. if an appliance states it carries a current of 6.4A at 120V
I = ΔV = 120V= 19Ω
R 6.4A
Ex. if an appliance has R = 48Ω, what is its amperage at 120V
I = ΔV = 120V= 2.5 Amps
R 48Ω
17.4 Resistivity
R=pL
L = length R = Resistivity p = constant of proportionality of the material
A
A = Cross section of material
p is found on a table on page 563
17.4 Resistivity
For ohmic materials, the ratio of the current density and the electric field, which gives rise to the
current is equal to a constant, σ, which is the conductivity of the material. The reciprocal of the
conductivity is called the resistivity, p. Each ohmic material has a characteristic resistivity that
depends only on the properties of the specific material and is a function of temperature.
The above equation shows that the resistivity of a cylindrical conductor is proportional to its length
and inversely proportional to its cross sectional area. (Like the flow through a hose or wire.) If cross
sectional area increases, its resistivity goes down and more water (electrons) can flow through it.
This is why light bulbs dim when they age. Their filament is tungsten, as it is used some of the wire
burns off, reducing its cross-section area, thus reducing electrons flow, and not as much light.
Ex. A 22 gauge nichrome wire with a radius of 321mm. Find its resistance per units length.
A = πR2 = π(.321x10-3)2=3.24x10-7m2
ρ=1.5x10-6Ω (from table page 563)
R= ρ L/A and R/L = ρ/A = 1.5x10-6Ωm = 4.6 Ω/m
3.24x10-7m2
If one meter of wire has a resistance of 4.6Ω, the current through the wire with a potential
difference of 10 Amps would be,
Ohms Law
I = ∆V = 10v = 2.2 Amps
R 4.6Ω
ASSIGN PAGE 576 10, 12, 14, 16
Section 17.5 Temperature Variation of Resistance
The resistivity of a substance depends upon a number of factors but the most important is heat.
For most metals, resistivity increases in a linear fashion with temperature (up to a certain point).
Remember some metals melt at temps that are not considered high.
T = Temperature in Celsius
ρo = Resistivity at a Reference Temp.
α = Temperature Coefficient of Resistivity from chart (pg 563, second column)
ρ = ρ0[ 1 + α ( T-To)]
Because resistance of a conductor is proportional with resistivity;
R = Ro [ 1 + α (T-To)]
ex. In a resistance thermometer (a device that measures temp by the change in conductivity of a
metal) is made of platinum. Its resistance is 50Ω @ 20oC when immersed in a vessel containing
melting indium, its resistivity increases to 76.8Ω. Find the melting point (in Co) if indium.
Solving: R = Ro [ 1+ α (T-To)]
 for T-To
T-To = R-Ro = 76.8Ω - 50Ω = 137oC
α Ro
(3.92x10-3)( 50Ω)
To = 20oC + 137o(ΔT)  TF = 157oC
17.6 Super Conductors
A super conductor is a class of materials whose resistance falls to virtually zero below a certain
temperature (called the critical temp) usually marked as Tc.
In super conductors a current (once applied) will persist indefinitely with out any in additional
applied voltage (because R=0) there has been laboratory tests run on super conductors where
current has persisted for years (in a closed loop) without any additional voltage.
ASSIGNMENT #1Page 576 21, 23, 26.
#2 22, 24, 29
17.7 Electrical Energy and Power
In a circuit consisting of a battery and a resister (R) positive charge flows clockwise from the
positive to the negative terminal of the battery.
I
B
C
A
D
+
_
.
This is because the positive terminal is always at a higher potential than the negative.
From the sketch note that point A is a ground (with potential = 0). So as the “charge” moves
through the battery it picks up potential (the battery loses an equal amount) and it (the charge) is
carried to the resister where it loses potential due to collisions (heat) in the resistor. No loss
occurs from BC and DA.
Power = P = ΔQ ΔV = I ΔV
ΔT
SO… p = I ΔV = I(IR) = I2R
but ΔV = IR
The loss of electrical potential due to heat is called joule heating often referred to as I 2R loss.
The unit of electrical energy we use at home is called a kilowatt-hour.
1 kwh = (103w)(3600 sec) = 3.60x106J
Have students turn to page 569 and explain Applying Physics 3 at top of page.
Then do Applying Physics at bottom of page.
ex. An electric heater applies potential difference of 50V to a nichrome wire with R=8Ω. Find the
current carried by the wire and the power rating of the heater.
I = ΔV = 50V = 6.25A
R 8Ω
w = watts
P = I2R = (6.25A) 2 (8Ω) = 131w
Assign page 576-577. 34, 36, 40, 43.
Ex. Find the cost to burn a 100w bulb for 24 hours, if electrical energy costs $0.080 per kilowatthour.
100w = 0.1kw
Energy = (0.1kw)(24h) = 2.4kwh
Costs = (2.4kwh)($0.08/kwh) = $0.19
Ex. How many watts will an electric oven use to cook a turkey for 5 hours using 20 amps @ 220
volts?
I = ρ = 20A = ρ  ρ = (220V)(20A) = 4400w
ΔV 220V
4400w = 4.4kw
E = (4.4kw)(5h) = 22kwh
Costs = (22kwh)($0.08/kwh) = $1.76 (Book Response was 1.80 == sig figs.)
Talk about 17.8
Voltage in medicine – show chest x-ray and print out of EKG’s. Abnormal EKG’s on page 72 show
irregular electrical wave beats. (Yes, electricity is produced naturally in the human body) There is
also a section about pacemakers.
Chapter 18.1
Direct Current Circuits
To start we will only use “steady state” currents – circuits with constant magnitude and direction.
Emf = E This was originally started as electro magnetic force hence the name Emf. But it was
found not to be a force so the long name is rarely used.
The potential drop across a battery (the terminal voltage) equals the Emf of the battery. But since
all batteries have internal resistance (R) thus terminal voltage of a battery equals the change in
voltage from a point prior to a point after (the battery) in a single circuit.
∆V = E – IR
E = Emf
I = Current of circuit
R = Resistance
So E = Terminal voltage in the real world only when IR = O ie no current flow. This is called an
open circuit voltage.
E = Emf
E = IR + IR
Big R = External Resistance
Little R = Internal Resistance
I = Current
Solving for I = E .
R+R
If R » R
» = Much greater than
We can neglect Little R (at all times you can assume this unless otherwise stated)
IE = Total Power Output at Source of Emf
multiply
E = IR + IR
by I
IE = I2R + I2R
18.2 Resistance in Series
*All currents in resisters in series are the same.
Req = R1 + R2 + R3 + …
ΔV = IReq  by ohms law and conservation of energy.
This tells us that the Req is always greater than any single RN.
EX:
ww
2
ww
4
ww
5
ww
7
ww
18
=
+ -
6V
+
-
6V
Find the Current in the Above Circuit
I = ΔV = 6V = 1 = .3333 A
Req 18Ω 3
Because the current in Req = 1/3 A
each RN =
ΔV2Ω = 2/3 V
ΔV4Ω = 4/3 V
ΔV5Ω = 5/3 V
ΔV7Ω = 7/3 V
Assign page 600-601. 2, 3,
the current in each
R N = 1/3 A
the voltage drop in
ΔVRN = (RN)(IReq)
18.3 Resisters in Parallel
*When resisters are connected in parallel, the potential differences (equal to the voltage of the
battery) across them are all equal.
*The currents generally are not equal.
Look at the sketch on page 584. Path of least resistance rule; which ever resister is least will
have more current flowing through it. Due number seven in class.
Assign page 600-601. Day #1 1, 4, 8, 9,13
DAY #2 Go over answers and assign, 12, 14.
18.4 Kirchoff’s Rules and Complex DC Circuits
4 rules of solving Kirchoff’s type circuits.
1. Draw and assign labels for each unknown and known quantities, assigning the direction of the
current in each part (if you assign it backwards don’t worry, your answer will have correct
magnitude just wrong sign). Do not change +/- directions once you start, stay consistent
throughout.
2. Apply the junction rule (the sum of current entering any junction must equal the sum leaving) to
any junction(s) in the circuit you may do this as many times needed as long as a new current
appears in the resulting equation.
3. Apply the loop rule to as many loops as needed to solve for all the unknown currents. loop
Rule: The sum of potential differences across all elements around any closed-circuit loop must be
zero.
4. Solve all equations simultaneously using algebraic rules for unknowns.
Go over example step by step starting on page 589 and continuing to the bottom of page 590.
Now go over example problem 18.5 page 591. Point out that the negative values for I2 and I3 are
okay just showing wrong direction for current flows.
Do problem #17 page 602 as an example.
E
a
I1
24V
I3
R
3Ω I2
6Ω
b
Given I1 = 3.0A. Find I2 and I3, if E and R are unknowns.
Solution for loop #1 from ab through 6Ω ;
∆Vab = +24V + (-I)R
= -24V + (-3A)(6Ω) = 24V – 18V = +6V
By the rules set fourth for parallel circuits –[the same change in potential (∆V) must go through all
parallel resisters in a circuit] lets us say that from ab through 3Ω resister yields.
∆Vab = -I2 (3Ω) = 6V
I2 = -2A
This (the negative on the amperage) tells us that the current flows from ba not ab as we
assumed.
From Kirchoff’s Junction Rule at point b
I3 = I2 + I1 = 3A + -2A = 1A
The positive sign on I3 tells us that the current ba through resister “R”.
Day #1 Assign page 602 15, 16, 18.
DAY #2 Assign page 602-603 22, 23, 25
Chapter 19
Magnetism
Do several demos with +/- poles as well as Stanley’s pipe demo using the magnet that seems to
defy gravity.
6 bar magnets – tape and get 6 bowls for water and 6 smaller ones to do Quick Lab on page 615 to
show students how easy it is to make a compass.
19.1 Magnets
They have a north and South Pole. If you cut it, the new ones will still have the same north and
south poles.
N
S

N
S N
S
Magnetic field lines run from the North Pole to the south pole (like electric field lines they never
cross).
POLES
Unlike attract and likes repel.
Hard magnetic material – hard to magnetize but it lasts a long time (cobalt, nickel).
Soft magnetic material – easy to magnetize but doesn’t last a long time (iron).
Magnetic field is “B” in this book.
19.2
Magnetic field of the Earth is just slightly off of straight up and down. North pole of Earth is 1300
miles from magnetic South Pole and South Pole of Earth is 1200 miles from magnetic north pole.
These distances vary slightly with time. While this may sound backwards, it makes sense in the
fact that the needle of your compass that points north is attracted to the south magnetic pole. But
at our place on Earth, your compass pointing north is really 4o west of north. (See page 615)
19.3 Magnetic Field
Right Hand Rule: Hold up right hand fingers pointing in direction of magnetic (B) field, thumb in
direction of motion (velocity) and out of the palm is the force of the magnetic field (F).
In Motion: Magnetic field force is
q = charge
F = qvB sinθ
B = definition of field
V = velocity
θ = angle between
Direction of V and B
F
B
+q
θ
V
The units on a magnetic field are in T = Tesla, sometimes called Wb = Weber/m2.
1 T = 1 Wb/m2
B = T = Wb = N = N .
m2 C·m/s A·M
But in practice we use Gauss = G.
1T = 104G
Conventional lab magnets have a field of 25000G or 2.5T.
Super conducting magnets are 30T = 3x105G
If we use F = qVB sin θ it has its max value at 90o = sin = 1
and min value at 270o = sin = -1
also note that if (B) field is parallel to the velocity (V) sin = 0 so F= 0.
Regardless of the θ between B and V force (F) is always perpendicular to both B and V.
Example 19.2 A Proton Moving in a Strong Magnetic Field
A proton moves at 8.0 X 10^6 m/s along the x-axis. It enters a region in which there is a magnetic
field of magnitude 2.5 T, directed at an angle of 60º with the x-axis and lying in the xy plane.
Calculate the initial force on and acceleration of the proton.
ex: 19.2 from page 618
Exercise uses F = qVBsinθ
F = (1.6X10-19)(8X106 m/s)(2.5T)(sin 60) = 2.8X10-12N
Using the right hand rule, the force is coming out of the hand (protons are positive)
Thus your force (F) is in the positive z direction.
Using the mass of a proton solve for acceleration.
a=F/m = 2.8X10-12N/1.6x10-27 kg= 1.7X1015m/s2
now try the problem above using an electron.
(Answer is a = 3X1018 m/s2 in the negative z direction )
Everyone try the review problem on page 641 now.
Hints: R=pL/A
n= L/2πr
L=length of the wire, solve for L
n=number of turns the wire of length L makes around a solenoid
B= μonI
solve for little n
l= N/n
l=length of solenoid
ANSWERS 1. 57.8M
2. 920 TURNS
3. 7.9X103 M
4. 11.6CM
ex. #7 page 642
An electron is accelerated through 2400V from rest and then enters a region in which there is a
uniform 1.70T magnetic field. What are the (a) maximum and (b) minimum magnitudes of the
magnetic force this charge can experience?
The gain in KE is equal to the loss in PE.
THUS ΔKE= - ΔPE= qΔV
ΔKE=1/2mv2 - 0 = eΔV solve for v = 2.9X107 m/s
using F = qvBsinθ
=(1.7 T)( 1.6 X10-19)( 2.9X107 m/s)sinθ
max at sin θ= 1 and min at sinθ = -1
a. 7.89X10-12 N
b. 0 N
because the velocity is parallel to the field
Assign 1-6, 10
19.4 Magnetic Force on a Current-Charging Conductor
*The total magnetic force on a closed current loop in a uniform field = 0
*For forces going into the page use X’s (tail of arrows)
*For forces coming out of the page use dots (tips of arrows)
The sketches below assume X’s (all forces going into the page)
I >0
I=0
I >0
XX
XXX
X
XX
XXXX
Wire flexes due to current (V) left ↑, (V) right ↓. This “flexing” is due to the sum of the
individual forces on the charge particles moving through the wire and colliding with the atoms in the
wire.
If we assume, for the drawing below that the magnetic field B is perpendicular to Vd
B x x x x
A
q
---------->
A= CROSS SECTION AREA CUT
We can find the magnitude of F = qVdB where Vd = drift speed of the charge q
The total force found by multiplying the force on one charge carrier by the number (N) of charge
carriers in each segment.
Because the volume of the segment is Al (A=area of cross section cut, l=wire length) the
magnitude of the total force is:
Fmax = (qVdB)(NAl)
where N = number of atoms
But, I = NqVdA (from chapter 17) so we simplify our equation into F = BIl
(This equation can only be used when the current and the magnetic
field are perpendicular to each other)
*If the field direction and velocity are not perpendicular F = BIl (sinσ) (use right hand rule).
*If the field direction and velocity are parallel then F = 0 (sin 0˚ and sin 180˚ both = 0)
Please read page 620-621 starting at application and ending at example 19.3.
EX: Page 642 #18
For a thin horizontal copper rod 1 m long with a mass of 50g, find the minimum current that will
cause it to float in a horizontal magnetic field of 2.0T.
To cause the rod to float its magnetic force must be equal and opposite of gravity.
Since F= mg we can say that
mg = BILsinσ
So
I = mg = (5x10-2kg)(9.8m/s2) = .245A
BLsinσ
(2.0T)(1.0m)
Why is sinσ = 1
(Hint: the min value will occur at sinσ=1)
Assignment: Problems 11-15.
19.8 Magnetic Field of a Long Straight Wire and Ampere’s Law
For a long straight wire wrap the fingers of your right hand around the wire pointing in the direction
of the magnetic field (B) and the current will go in the direction of your thumb.
This is given by the equation;
B = uo I
Where uo = 4(π)x 10-7Tm
2(π)r
A
uo = Permeability of free space
The above equation tells us that as the wire’s length increases its field strength decreases
(displaces).
19.9 Magnetic Force Between Two Parallel Conductors
Parallel conductors carrying currents in the same direction attract each other, where as if the
currents are moving in opposite directions they repel each other.
The magnitude of the magnetic force per unit length between parallel conductors is given by
F = uoI1I2 L
d = distance
2(π)d
L = unit length
uo = permeability of free space
I1 = current wire #1
I2 = current wire #2
Please look at example 19.8 page 631
Assign Page 644-645 36, 37, 42-44
19.10 Magnetic Field of a Current Loop
19.11 Magnetic Field of a Solenoid
Look at your textbook pages 632-633. All I am worried about is you can correctly sketch
the magnetic field lines.
QUIZ Chapter 19 tomorrow.
Chapter 22
Light and Optics
22.1 The Nature of Light
According to Newton, light is a particle (flowing out in a stream). But according to a Dutchman in
1678, Christian Hygens, it was a wave. (A wave bends around objects). They both were correct,
AKA Duality.
Vlight = 2.998 x 108m/s for this class
In 1905, Einstein said light was composed of corpuscles or discontinuous “quanta” of energy. Thus
he explained the photoelectric effect, ie photons. He stated that light was energy called photons &
E = hf
E = Energy
h = Planck’s Constant = 6.63 x 10-34 J∙s
f = Frequency of the electromagnetic wave
The photoelectric effect is the result of energy transfer from a single photon to an electron in metal.
The electron interacts with one photon of light as if it had been struck by a particle. Yet light also
displays wave characteristics because exhibits interference.
This leads us back to duality, sometimes it acts like a wave, sometimes like a particle.
22.2 Measurements of the Speed of Light
Light has been measured by several hundred scientists all garnering the same basic answer,
3 x 108m/s. The first accurate test was in 1849 by Armand Fizeau. He used a light source, gear,
and a mirror. He used the speed of rotation, distance, and which tooth the light passed through on
the way in and out, simple geometry to get 3.1 x 108m/s.
C = 2d
T
EX: A 360-tooth gear rotating at 27.5 rev/s and the light heads towards a mirror through the gap
between tooth A & B and its reflection is blocked by tooth B. If the distance from the gear to the
mirror is 7500m. Find the speed of light.
The following formula is from chapter 7
T= σ
ώ
σ = Theta
T = Time
ώ = omega – Rotational Velocity
T = 1/720 rev = 5.05 x 10-5sec
27.5 rev/s
C = 2d = 2(7500m) = 2.97 x 108m/s
T 5.05 x 10-5 sec
( Number 4 part A.)
The time the light beam would take to make it through both slots (ie time to travel distance S) must
equal the time for the disks to rotate through the angle σ, c = speed of light.
T = S = σ  C = Sώ
C ώ
σ
Now you guys do part b. tonight’s assignment page 746 1-4.
22.3 Light – The Ray Approximation in Geometrics Optics.
For this and all subsequent chapters we shall use the following premise from geometric optics.
*Light will travel in a straight line until it encounters a boundary between two different materials.
Note: The book will draw all light waves as straight lines with arrowheads (vectors) perpendicular to
their wave fronts.
22.4 Reflection & Refraction
Start with labs from Page 731 and 732
First Reflection – smooth surfaces – spectacular reflection – waves inbound and out bound are all
parallel.
mirror
Rough surface – diffused reflection – outbound waves scatter.
For all smooth surfaces
σ1 = σ1I
σ1
σ1I
Why are peoples eyes red in pictures??
Retro reflection : The light from a flash too close to a lens returns to the lens during the
photo imaging process displaying the light (red from the blood vessels in an eye, most light from
the eyes is red) on the picture. To stop this, move light source away from camera lens.
Example: Two mirrors make an angle of 120o.
M1 = Mirror 1
M2 = Mirror 2
Find the direction of the ray
after it leaves M2.
σ2
M2
65o
25o
M1
Since inbound = 65o from perpendicular out bound = 25o from parallel and m1 and m2 form 120o.
Thus 180o – 120o – 25o = 35o into m2 and 35o out of m2 thus 90o – 35o = 55o = σ2.
Notice incident ray is from the perpendicular line not from the parallel.
Refraction of Light : i.e. Light bends due to entering a different medium. (material it is passing
through must be transparent). Some of the ray is reflected away and part reflects into the medium.
Incident Ray
Reflected Ray
σ1
σ1
\
Examples of Refraction (use the overhead from physics)
light moving from air to glass
σ1
σ1 > σ2
σ2
light from glass into air
σ1
σ1 < σ2
σ2
Whether the light bends towards the normal line (dashed line) or away depends upon the
properties of the material the light is leaving and entering. To find the angle we use the velocities
of light in each medium (yes light has a different speed in various mediums). It’s velocity changes
but its frequency does not. To solve, use the following formula, aka, Snells Law.
sin θ2 = V2
sin θ1 V1
When light moves from a material with a high velocity to one with a lower velocity, the angle of
refraction θ2 will be less than θ1.
θ1 > θ2
this shows that light moves slower through glass than air
θ1
AIR
GLASS
θ2
The opposite will be true if the light is entering a medium with a higher velocity, it will bend away
from the normal line and θ2 > θ1.
22.5 Law of Refraction
A little N (n) will be used for the index of refraction.
n = speed of light in vacuum = C
speed of light in medium V
therefore, n = C/V
Recall the frequency doesn’t change,
only the velocity.
V = velocity
Also recall that v = fλ
f = frequency
λ = lambda (wavelength)
Using wavelength only we can still find the index of refraction.
n = λc
λm
λc = λ in vacuum
λm = λ in medium
Further, Snells Law can be written as
n1 sin θ1 = n2 sin θ2 normally seen version
Ex:
A light ray of wavelength 589nm (sodium lamp) travels through air is incident on a smooth
slab of crown glass at an angle of 30o to the normal.
AIR
GLASS
30o
θ2
find θ2, using n1 and n2 from page 730.
n1 sin θ1 = n2 sin θ2
(1.0)(sin 30o) = (1.52)(sin θ2)
.329 = sin θ2
19.2o = θ2 bent towards the normal if the light was moving through
the glass upward at 30o from normal. What angle from the normal
would it refract when it passed into the air?
1.0 sin θ1 = 1.52 sin 30o
θ1 = 49.5° (away from normal)
*If the light goes through more than one refraction do Snells Law for first refraction and use new
angle for Snells Law for second refraction.
*Remember T = d/Vm
T = time; d = thickness; Vm = speed in medium
Day One Assignment: 5-12 page 747
Day Two Assignment: 13-20 all
(Do lab page 736 before section)
22.6 Dispersion & Prisms
We proved in the last section that light of various wavelengths (λ) will be bent at different angles (θ)
when incident upon a refraction material.
Thus blue light (λ = 470nm) and red light (λ = 650nm) will bend at different angles. (Hint: the lower
the λ, the greater the bend).
For the colors of the spectrum ROYGBIV (red, orange, yellow, green, blue, indigo, violet), with red
as the longest and violet as the shortest, each color of light bends different through a spectrum.
The angle they bend is δ = the angle of deviation. This depends entirely on this wavelength λ the
longer the λ the less the deviation. Thus, red with longest λ deviates δ least and violet with the
shortest λ deviates δ most.
All hot-low pressure gases emit their own characteristic gasses. Scientists can identify the gas in
the air by using a spectrometer, which shows the colors.
Ex: Sodium emits two wavelengths both yellow, mercury emits 4 wavelengths orange, blue, green,
violet.
Assign Pg 749 27 and 28.
22.7 Rainbows
All you need to know about rainbows is that they are formed by sunlight passing through rain
droplets. The light enters the droplet and is reflected off the back (inside) back out the same side it
entered.
Since red has a lower angle of deviation δ, it appears at the top and violet with the highest δ at the
bottom.
22.8 Huygen’s Principle
It deals with wave points (wave fronts) being identified as points where secondary waves can form.
Huygen’s Principle states:
All points on a given wave front are taken as point sources for the production of spherical
secondary waves, called wavelets, which propagate outward with speed characteristics of
waves in that medium. After some time has elapsed, the new position of the wave front is
a surface tangent to the wavelets.
See Pages 738-739
*On page 740 it shows how Huygen’s Principle can be used to derive Snell’s Law.
22.9 Total Internal Reflection
When moving from a medium with high index of refraction to low, all light has a critical angle in
which light will not leave one medium for another, it will simply be reflected back down inward.
This angle can be found by using two simple facts;
1. Sin 90o for the second index of refraction is one so the light would be parallel to the
normal line.
2. Snell’s Law with Rule #1 inserted
n1 sinθ1 = n2 sin90o
sin θ1 = n2/n1
call θ#1 the critical angle and solve.
*for all n1>n2 because if n2>n1, sin θ>1 which is undefined
Hand out prisms and start to play while explaining and demonstrating fiber optics.
DEMO – laser, fiber optic cable
- laser/light and fiber optic tubes.
ASSIGN page 749 30-35
HOORAY! QUIZ on CHAPTER 22
Chapter 23
Mirrors and Lenses
Turn to page 753. Why is the reflection the same?
23.1 Flat Mirrors
Start with quick lab on page 754.
All mirrors have certain properties and recognizing which type of mirror one is working with will help
one to solve problems involving image placement, magnification, inverted or upright images, and
real or virtual images.
We start with flat mirrors. Their properties are as follows.
1. Image is as far behind the mirror as object is in front.
2.The image is:
a. Unmagnified
b. Virtual
c. Upright. (upright means, if an object arrow points upward, so does its image)
3.The image will exhibit right-left reversal (flipped).
*REAL images are ones in which light actually intersects or passes through the image point.
*VIRTUAL images are ones in which light does not pass through the image point but appears to
come (diverge) from that point.
Real images can be displayed on a screen but virtual images cannot. What are you looking at
now? A real image.
To find out where an image is formed it is necessary to trace two paths of light and use simple
geometry.
EX:
p
q
O
I
o = object
I = Image
p = distance
q = distance
*p = q for all flat mirrors
*all lines are straight lines (a
basic property of light).
Your Eye
Flat Mirror
To look at it geometrically;
p
Q
P
q
θ1
θ2
R
P1
ΔPQR ≈ ΔP1QR
θ1 = θ2
p = q for flat mirrors
Notice object (P) and image (Pi) are upright, and both P and P1 are identically sized. If we were
working with a mirror that was not flat we would notice lateral magnification, M.
M = Image Height
Object Height
Exercise page 782 1, 2 #2 uses the picture from page 754.
23.2 Images Formed By Spherical Mirrors
Spherical mirrors get their name from their shape. The light reflects from its silver inner liner
(concave surface) these are called concave mirrors.
They have a radius of curvature, R.
The center of the curve is C.
The center of the mirrors curve is V.
(make copies of pictures below & explain what each letter stands for!!)
h1 = image height
h = object height
m = h1/h = index of refraction = -q/p
q = image distance
C = center of curvature
Line through C & V = principle axis
f = focal length
p = object distance
o = object
I = image
V = center of mirror (spherical segment)
R = radius of circle (distance from center of
mirror to center of curvature)
Real Image: When reflected light actually passes through the point where the image is formed.
A Spherical Aberration: Rays that come from a point not on the principal axis form a blurred image
at some other point on the principal axis. This is present in all spherical mirrors but more prevalent
in some.
The inverse of the distance of the object from the mirror and the inverse of the image from
the mirror are equal to twice the inverse of the center of the curvatures distance.
However if the objects distance “P” is great enough compared with “R” then “P”
approaches infinity and 1/P≈0 and g≈R/2 will give the formation of our image.
*Thus, if the object is a long way away from the mirror, the image is exactly halfway
between the center of the curvature “C” and the center of the mirror “V”, and then we can assume
the incoming rays are parallel. In this special case we call the image a focal point (F) and the
distance a focal length (f) where f = R/2 changing our equation to
1+ 1 = 1
p q f
*Remember that rays at ∞ are always focused at the focal point.
23.3 Convex Mirrors and Sign Conventions
A convex mirror that is silvered so light reflects from the outer convex surface (in the last
section it was silvered on the inner convex surface) is called a diverging mirror. Because the light
rays diverge after reflection as through they were coming from a point behind the mirror. The
image is called virtual (rather than real). It is upright (erect) and smaller than the object.
Everyone needs to memorize the chart on page 760 table 23.1
If you have the position of the object and the location of the center of curvature you can locate the
image by using (3) rays. Start all three rays from the tip of the object for concave mirrors and draw
them as follows:
Ray #1:Draw parallel to the principle axis and back through the focal point F.
ray #2: Draw through the focal point F and back parallel to the principle axis.
Ray #3: Through the center of curvature C and back into itself.
*Any two rays that intersect form an image (but the 3rd should go through this point also).
*The image point (distance) can be checked by solving for q in the mirror formula.
NOTE:
For all concave mirrors; as the object moves closer to the real, invented image the object moves to
the left until it reaches the focal point. At the focal point an object will produce and image hat is a
distance to the left. Once the object is inside the focal point (between f and the mirror), the image
becomes virtual and upright.
NOTE:
With a convex mirror the image is always virtual and upright. As the distance of the object from the
mirror increases the virtual image shrinks as the object approaches the focal point “p” approaches
infinity.
Do lab on Page 761 with candles and concave, convex mirrors.
*Talk about which type of mirror you would use inside and outside a car.
*Ex 23.2 in book:
Using a concave spherical mirror with focal length 10cm, find
a. image of object with distance 25cm and describe the image.
1+1=1
p q f
1+1=1
25 q 10
q = 16.7cm
m = magnification
*image is smaller than object
inverted (m is negative)
Real (q is positive)
m = -q = -16.7 = -.667
p
25
b. object distance is 10cm
1+1=1
10 q 10
q=∞
c. object distance is 5 cm
1+1=1
5 q 10
q = -10cm
(light comes from focal point)
m = -q / (-10/5) = 2.0
p
Image is larger (magnified twice)
Positive (M) Image is Upright
-q Image is virtual (behind the mirror)
Day 1 Assign 3-9
Day 2 Assign 10-17
23.6 Thin Lenses
By definition, a piece of glass/plastic ground so each side is either sphere or plane is a thin lens.
Converging – If they are thicker in the center than on edges
Diverging – Thinner in the middle than the edges.
Just as with mirrors, we need to find the focal point (F) and the focal length (f) but a thin lens has
two focal points because rays that enter the lens parallel to the principle axis can come from both
sides.
(see figure 23.3 page 535 student solution guide, scan into a disk insert here!!)
*Always remember a thin lens within converging or diverging has two sides.
To find the focal points and images use the following rules and 3 rays of light.
Ray 1: Parallel to principle axis from the tip of image then through focal point on the other side.
Ray 2: Through center of lens, from tip of image.
Ray 1 and Ray 2 combine at tip of image.
Ray 3: Through focal point on same side of lens as object then parallel with principle axis from
lens to mage tip.
#1
#2
θ
θ
#3
F
f
p
q
Tan θ = PQ
f
1
M = h = -q
h
p
(PQ = h)
Tan θ =
-h1
q-f
Since both of the above = Tan θ
h = -h1
and
h1 = -q-f
f q-f
h
f
N = Index of refraction
R1 = curvature of front lens
R2 = curvature of back lens
But 1/f = (N-1)(1/R1 – 1/R2)
But in the last section we said h1/h = -q/p
Using both values of h1/h
You once again 1/p + 1/q = 1/f
So we use the same equations for thin lenses as we do for mirrors. (Note: It doesn’t matter if the
lens is convex or concave, use this formula).
You need to remember that a converging lens (see figure 23.22 page 770) has a positive focal
length and are thickest in the middle and a diverging lens has a focal length and are thinnest in the
middle.
Hand all students a copy of student solution guide pages 532, 533 and 535.
Do this before assigning problems!
#31 A diverging lens (w/N = 1.50) is shaped like figure 23.2 c (concave on both sides). The radius
of the first surface (R2) = 10cm (back). Find
a. focal length of the lens
1 = (n-1)( 1 – 1 ) = 1.5 – 10)( 1 – 1 ) = -12 cm
f
R1 R2
15 10
b. position of images for object distance ∞
1 + 1 = 1 IF p = ∞ THEN 1 = 1
p q f
q f
and since f = -12cm q = -12cm thus image is virtual and 12cm in front of lens.
c. position of image for 3|f|
p = 3|-12| = 36cm
IF 1 + 1 = 1 THEN q = pf = (36cm)(-12cm) = -9cm
p q f
p-f 36cm - (-12cm)
Thus, virtual image 9cm in front of lens.
d. position of image if p = |f| then p = 12cm
q = (12cm)(-12cm) = -6 cm
(12cm)-(-12cm)
Thus, virtual image 6cm in front of lens.
e. position of image if p = |f| = |-12cs| = 6cm
2
2
q = (6cm)(-12cm) = -4cm
(6cm)-(-12cm)
Thus, image is virtual and 4cm in front of lens.
Assign page 784 27-33
Day 2: Go over homework 27-33
Do #39
If you are going to use a diverging lens to produce a virtual image 1/3 as tall as the object, where
would you place the object?
Since a diverging lens always produces a virtual, upright image of any real object and we need a
magnification that is a positive 1/3 we use }m} = 1/3.
Recall magnification m = Image Height = +1 = -q
Object Height 3 p
Thus, q = -p
using thin lens equation
3
1 + 1 = 1 you get 1 - 3 = 1
p q f
p p f
So, -2 = 1 OR p = -2f
p f
Since the focal length of a divergent lens is always negative (see handout page 535 figure 23.3) we
can say that –f = |f|. Thus, the object distance should be p = 2|f|.
We can state that if you want an image that is 1/3 of the object height using a diverging lens you
have to place the object in front of the lens at a distance equal to twice it’s focal lengths absolute
value. (notice that you do not have to get a numerical value here, cool eh?)
Assign Page 784 34-43
That’s all folks!! Time to start reviewing for A.P. Test.
We will do chapters 27 and 28 after the test. Not in depth, but we will look at them.
Until that time, we will do A.P. Test every day.
27.1 Start with Maxwell’s Theory (Zit Book)
Electron Corral: Made by “STM”, Scanning Tunneling Microscope.
STM: can produce pictures of surface of metals, insulations, and strands of DNA.
Light and infrared radiation is produced by vibration of changed particles within atoms of a body
that is so hot it glows, i.e. incandescent.
Total power goes up as temperature increases. The amount of energy emitted every second by
electromagnetic waves is proportional to the temperature in K raised to the fourth power.
Sun surface 5800K – yellow color. Sun radiates 4 x 1026 Watts on average. Every square meter of
earths surface receives 1000J of energy each second.
E = nhf
E = energy
n = Integer (1, 2, 3…)
Energy is quanitized
h = Planck’s Constant 6.63 x 10-34 J/Hz
i.e. in integer multiples
f = Frequency of Vibration of the atom
only (hf, 2hf, 3hf…)
Further, only when Quanta changes i.e. from 3hf  5hf does an atom radiate.
Photoelectric Effect: The emission of electrons when electromagnetic radiation falls on an object.
*Not all radiation results in a current. Electrons are ejected only if the frequency of the radiation is
above a certain minimum value known as the threshold frequency fo. ( fo varies for each metal).
1905 Einstein explained the photoelectric effect. Light and other forms of radiation consist of
discrete bundles of energy (later named photons). The energy of each photon depends on the
frequency of the light. Recall a Hz = 1/s = s-1. Then Planck’s constant, 6.63 x 10-34 J/Hz = Js.
According to Einstein:
A photon must have minimum energy greater than or equal to hfo to eject an electron from metal. If
the photon has frequency that is below fo, the photon does not have the energy necessary to eject
an electron. Thus, no electrons will be ejected. If light has a frequency > fo, then is has the
necessary energy to eject a photon. The excess energy becomes the kinetic energy of the photon
using the formula,
KE = hf - hfo  Conservation of energy.
*For an incoming photon E = hf. An amount of energy, hfo is needed to free the electrons from the
metal. The reminder is the kinetic energy of the electron and is converted to that KE by the formula
above.
NOTES:
(1) An electron can not collect E < hfo until it gets enough to be released, it must all come at once.
Thus, only one photon can interact with one electron.
(2) Some electrons take more hfo than others to release.
(3) The expression, “kinetic energy of the ejected electrons” refers to the maximum Kinetic Energy
KE an electron can have. Some electrons will have less!
(4) Looking at the chart on the top of page 630, the maximum KE at the cathode is equal to the
work done by an electric field in stopping them.
Thus, W= KE= -qVo
-q = electron charge = -1.60 x 10-19 C
Vo = Magnitude of stopping potential in Volts (J/C)
THE WORK FUNCTION OF A METAL IS SIMPLY W= hfo
A photon that is from red light has the least energy of the color spectrum and a photon from violet
light has the most, this is a result of the differences in wavelengths.
Since a joule (J) is too large a unit for atomic system we will use the eV – electron Volt
1eV = (1.6 x 10-19C)(1V) = 1.6 x 10-19C∙V = 1.6 x 10-19J
Additionally, we can state (thanks to Einstein)
E= hc/λo = hfo hc is a constant 1240 eV nm ( the c is the speed of light)
so
KE = hf - hfo can be written as KE= E= hc/λ - hc/λo
Go over examples on page 631-633
Point out that h = Planck’s Constant.
h = ΔK slope for all metals on the graph is similar. (Look a graph on pg 632)
Δf
DAY 1. Assign page 633 page 1-4
Day 2 page 642 16-19 all
The Compton effect:
The photoelectric effect demonstrates that a photon even though it has no mass, has KE just as a
particle does.
In 1916, Einstein stated, “photons have other particle like properties and one can predict a
photons momentum”, it is given by hf/c.
Since f/c = 1/λ, thus, ρ = hf/c = h/λ
Recall, E = hf and f = c/λ  E = hc/λ
A Proton’s energy is inversely proportional to its wavelength.
When Compton shot x-rays at a target (graphite) some scattered at the same wavelength,
others at new wavelengths, meaning they lost or gained KE, as well as momentum. This shift in
energy of scattered photons is called the Compton Effect. It is only 10 -3nm and is measurable only
with x-rays having wavelength of 10-2nm.
Compton proved that photons have both energy and ρ. Also, they are a particle that has no mass.
Further, he showed they also move at the speed of light.
Assign page 642 22-23 all.
27.2 page 637
Recall p = mv = h/λ
 λ = h/p = h/mv
λ = h/mv = de Broglie wavelength of a particle. He also proved electrons and photons should
show wavelike particles. He showed in 1927 they both acted just like light.
ex: de Broglie wavelength of a 0.25kg baseball leaving the bat at 20m/s.
λ = h/mv = (6.63x10-34 j/s) = 1.3x10-34m very short wavelength.
(0.25kg)(20m/s)
ex. An electron accelerated by potential difference of 75V. Find de Broglie wavelength.
KE = -qV = 1/2mv2 = qV  v =
2qv / m
V= voltage
v= velocity
m=mass
q = charge of an electron/proton
ρ = mv = (4.6x10-24m/s) = 5.1x106m/s
ρ= momentum
assignment day #1 page 642 27 -30
day #2 page 642 31 -35