Transcript Pre-Health Physics Review

Pre-Health Physics Review
Johnny B. Holmes, Ph.D.
Fundamentals
MKS system of units:
M for meters (distance)
K for kilograms (mass)
S for seconds (time)
Also add Q in Coulombs (electric charge)
Other units are combinations of these fundamental units:
Speed in m/s;
Acceleration in m/s2;
Force in Nt=kg*m/s2;
Energy in Joules = Nt*m (convert to calories, BTUs);
Power in Watts = Joules/sec
Pressure in Pascals = Nt/m2 (convert to psi, atm, mm of Hg, bars);
Current in amps = Coul/sec;
Voltage in volts = Joule/Coul.
Common Prefixes
milli = 10-3 (m)
micro = 10-6 (m)
nano = 10-9 (n)
pico = 10-12 (p)
femto = 10-15 (f)
Kilo = 103 (k)
Mega = 106 (M)
Giga = 109 (G)
Tera = 1012 (T)
Vectors
Space is three-dimensional, so many
quantities in physics have directions as
well as magnitudes. For 3-D, need three
numbers to specify the quantity (called a
vector).
Common forms for vectors are
rectangular (x,y,z)
in 2-D (x,y)
spherical (r,q,f)
in 2-D (r,q)
cylindrical (r,q,z)
To add vectors, must add in rectangular!
Motion
Velocity: v = dx/dt; for constant velocity,
have x = xo + vo*t (straight line)
v is the slope of the x vs t curve.
Acceleration: a = dv/dt; for constant acc,
have y = yo + vo*t + ½*a*t2 (parabola)
a is the slope of the v vs t curve.
For 2 & 3 dimensions, work in rectangular
components.
For trajectories, ax = 0 and ay = -g = -9.8 m/s2
Circular Motion
Work and think in rectangular (x,y)
Convert to polar (r,q) for easier equations:
r = constant
q changes in time
vr = 0
w = dq/dt = constant for uniform CM
vq = w*r = constant for uniform CM
ar = -w2*r directed towards the center due to direction of velocity
changing
aq = a*r where a = dw/dt is angular acc
= 0 for uniform CM
Also, w= 2*p*f and 1/f = T (f is frequency, T is period)
Newton’s Laws of Motion
SF = m*a
(vector equation so work in rectangular components)
The equation above is Newton’s 2nd law.
Newton’s 1st law is a special case of the 2nd :
if F=0, then a=0.
Newton’s 3rd law: for every action there is an equal
and oppositely directed action (or, you can’t
push yourself). The forces in Newton’s 2nd law
are the ones ON the object, not the ones BY the
object.
Forces
• Contact: Fc balances, perpendicular to surface
• Friction: Ff <= m*Fc parallel to surface
• Tension: T is same along the string; directed
parallel to the string
• Weight: near earth’s surface,
W = Fgravity = m*g directed down.
• Gravity far from earth’s surface:
Fgravity = G*M*m/r2 with G = 6.67x10-11Nt*m2/kg2
Example: Circular Orbits
For a satellite to orbit the moon at a height
of 100 km, how fast should it go and how
long will it take to make one orbit?
Radius of Moon is 1,700 km; mass of
moon is 7.2 x 1022 kg.
Satellite Problem – cont.
Recognize:
1. Circular orbit, so ar = w2*r, v = w*r, w=2p/T
2. Gravity is the only force: F = G*M*m/r2
3. Use SF = m*a
4. Recognize that r = Rmoon + h
= 1,700 km + 200 km = 1,900 km.
5. Therefore: G*M*m/r2 = m*w2*r
6. Recognize that m (mass of satellite)
doesn’t matter, solve for w, then use
v = w*r, w=2p/T to get both v and T.
Energy and
Conservation of Energy
Work = Force through a distance (a scalar)
W = F*s*cos(qFs)
Energy is the capacity to do work (in ideal situations)
Kinetic energy = KE = ½*m*v2
Gravitational Potential energy:
near the earth, PEgravity = m*g*h
in general, PEgravity = -G*M*m/r
Spring Potential energy: PEspring = ½k*(x-xo)2
Example: Escape Speed
S Energiesinitially = S Energiesfinally .
KEi + PEi =
KEf
+ PEf
(1/2)mvi2 - Gmearthm/ri = (1/2)mvf2 - Gmearthm/rf
(note that this is equivalent to saying DKE = DPE, or
½mvf2 – ½mvi2 = -Gmem/rf - -Gmem/ri)
We see that m is in each term, so we cancel it.
(1/2)*(vi)2 - (6.67x10-11 Nt*m2/kg2) *(6.0 x 1024 kg)/(6.4x106m) =
(1/2)*(0 m/s)2 - (6.67x10-11 Nt*m2/kg2) *(6.0 x 1024 kg)/(infinity)
We again have one equation in one unknown (vi).
Power
We now know what Force and Energy are,
but what is Power?
The definition of Power is that it is the rate of
change of Energy from one form into
another:
Power = DEnergy / Dt .
The units of power are: Joule/sec = Watt.
Another common unit is the horsepower, hp.
The conversion factor is: 1 hp = 746
Watts.
Review of Rotational Equations
Basically we replace F with t, m with I,
v with w, a with a and p with L (where
L is the angular momentum):
S F = ma
S t = Ia
Work = = F  ds
Work = t  dq
Power = F  v
Power = t w
KE = (1/2)mv2
KErotation = (1/2)Iw2
p = mv
L = Iw
S F = Dp/Dt
S t = DL/Dt .
Example: Your elbow
Let’s consider as an example of torque how
your muscles, bones and joints work.
Consider holding up a ball of weight 5 lb.
How does this work?
First we draw a diagram:
biceps
triceps
weight
= elbow
rw
r
b
Your elbow
From S F = 0 we have:
-Fc + Fb - W = 0
And from S t = 0 and measuring from the
elbow gives: Fc*rc + Fb*rb - W*rw = 0 .
We have two equations and we have two
unknowns (Fc and Fb).
Your elbow
• We can use the torque equation first, since
rc=0 eliminates one of the unknowns, Fc.
Fc*rc + Fb*rb - W*rw = 0 or Fb = W*rw/rb .
• Then we can use the force equation to find
Fc :
-Fc + Fb - W = 0, or Fc = Fb - W.
Example:
Object rolling down an incline
Conservation of Energy
(KEregular + KErotational + PEgravity)initial =
(KEregular + KErotational + PEgravity)initial + Elost
0 + 0 + mgh = (1/2)mv2 + (1/2)Iw2 + 0 + 0.
Substituting I=(2/5)mr2 and w=v/r gives:
mgh = (1/2)mv2 + (1/2)[(2/5)mr2][v2/r2] , or
mgh = (1/2)mv2 + (1/5)mv2 = (7/10)mv2 .
Momentum and
Conservation of Momentum
p = m*v (p and v are vectors!)
Conservation of Momentum
Fxext on 1 + Fxext on 2 = D(px1 + px2) / Dt .
If the external forces are small, or if the time of
the collision, Dt, is small, then we have:
D(px1 + px2) = 0. This can be re-written as:
(px1 + px2)i = (px1 + px2)f .
This is called Conservation of Momentum.
This is a vector law, so a similar equation
holds for each component of momentum.
1-D Collisions
In one dimensional collision cases, we can
apply two laws: Conservation of
Momentum and Conservation of Energy
(here we assume there are no PE’s that change):
(1/2)m1v1i2 + (1/2)m2v2i2 =
(1/2)m1v1f2 + (1/2)m2v2f2 + Elost
m1v1i + m2v2i = m1v1f + m2v2f
These are two equations with 7 quantities:
m1, m2, v1i, v2i, v1f, v2f, Elost . Hence if we
know five, we can solve for the other
two.
Explosions
Normally, in an explosion the initial object is
in one piece and at rest. After the
explosion, one piece goes forward.
Conservation of Momentum says the other
piece must then go backwards. (If we
brace ourselves, we can compensate for
that backwards push and not fall over.)
0 + 0 = m1v1f + m2v2f or v2f = - m1v1f / m2.
Pressure
P (pressure, not power or momentum)
P = Force/Area (definition)
(force is perpendicular to area, not parallel to it)
units of pressure:
– Nt/m2
– 1 atmosphere = 1.01 x 105 Nt/m2 = 14.7 lb/in2
– 1 bar = 1.00 x 105 Nt/m2
– 1 Torr = 1 mm of Hg, 760 Torr = 1 atmosphere
Ptop
Pressure
Pside
effects of gravity:
h
W=mg
– consider little cube of fluid
– consider forces on the fluid in y direction
Pbottom
1. weight acts down
2. pressure underneath pushes up
3. pressure on top pushes down
– SFy = -m*g + Pbottom*Abottom - Ptop*Atop = 0 ,
–
where m = *V = *A*h, and A = Abottom = Atop
so: Pbottom*Abottom - Ptop*Atop = m*g = *A*h*g ,
or: Pbottom - Ptop = *g*h .
Buoyant Force = P*A = *g*V
Fluid Flow
Conservation of Energy:
(1/2)mvi2 + mghi + Won = (1/2)mvf2 + mghf + Wby
divide each term by Volume, and note m/V=,
also note W = F*s, F=P*A, A*s=V, so Work = P*V:
(1/2)vi2 + ghi + Pi = (1/2)vf2 + ghf + Pf + Plost
examples:
• lift on wing of airplane
• coffee pot
• siphon
• oil well
Fluid Flow
Viscous Force: F = // dv/ds ;
for tubes (cylindrical hoses) with constant
velocity (Fapplied = Fresisted, F = DP*A )
DP * pr2 = 2prL) dv/dr
Q = pDP)R4 / (8L) wwhere DP = Plost
and Q is the volume per time flowing.
Power = Work/t = F*s/t = Pressure*A*s/t
= Pressure*V/t = Pressure*Q
Reynolds Number
Have laminar flow (previously assumed layer
over layer flow) as long as flow is slow
enough; otherwise have turbulent flow
Reynolds number:
R = 2vavgr/ = 2Q/(pr) (dimensionless!)
If R < 2,000, then laminar;
If R > 2,000, then turbulent.
Ideal Gas Law
P*V = N*k*T . We further define R = Na*k,
where Na = 6.02 x 1023 = 1 mole. Thus we
have: P*V = n*R*T , where
n = N/Na = number of moles in volume, V;
T must be in Kelvin, Not oF or oC !
k = experimental constant = 1.38 x 10-23 J/K ;
R = Na*k = 8.3 Joules/mole*Kelvin .
HEAT CAPACITY
The amount of energy necessary to heat a
material per temperature change is what
we call the heat capacity:
C (heat capacity) = Q/DT
where Q is the energy to raise temperature
of an amount of material by DT.
Usually we specify the heat capacity in one
of three ways: per object, per mole
(usually for gases), and per mass (usually
for liquids and solids).
Heat Capacity of Air
Cmolar-constant P = Cmolar-constant V + R
Air is made up mostly of N2 and O2. These
gases act approximately as diatomic ideal
gases. Usually, when we heat air it is
NOT in a contained volume but expands to
keep its pressure constant. This means
that most of the time, the heat capacity of
air is:
Cmolar - air - constant P = (5/2)R + R = (7/2)R .
Heat Capacity of Materials
By definition, a calorie is the energy
necessary to raise the temperature of 1
gram of water up 1oC.
Cwater = 1 cal/gm-oC = 4.186 J/gm-oC
Cethyl alcohol
= 2.400 J/gm-oC
Cwood
= 1.700 J/gm-oC
Cglass
= 0.837 J/gm-oC
Ccopper
= 0.387 J/gm-oC
Since liquids and solids don’t expand to fill the space like
gases do, we don’t usually distinguish between heat
capacities at constant pressure versus constant volume.
Latent Heat
For water, the latent heat of fusion (heat
needed to melt ice to water) is 79.7
cal/gm.
For water, the latent heat of vaporization
(heat needed to boil water) is 540 cal/gm.
For alcohol, the latent heat of vaporization
is less at 204 cal/gm.
Heat Transfer
There are four ways of moving heat:
•
Evaporation (using latent heat)
•
Convection (moving heat with a material)
•
Conduction (moving heat through a
material)
•
Radiation (light, usually mainly in the
infrared, both emitted and absorbed)
Heat Transfer: Conduction
Power = Q/t = k*A*DT/L
where k is a constant that depends on the
material, called the thermal conductivity;
where A is the cross sectional area;
LL
where L is the distance from the hot end to
the cold end;
A
hot
and DT is the temperature difference
k
Thi
between the hot and cold ends.
cold
Tlow
Blackbody Radiation:
Experimental Results
Ptotal = AT4
where  = 5.67 x 10-8 W/m2 *K4
peak = b/T where b = 2.9 x 10-3 m*K
Intensity
(log scale)
UV
blue
yellow
wavelength
red
IR
Thermodynamics
The First Law of Thermodynamics is a
fancy name for the Law of Conservation of
Energy applied to thermal systems. It says:
DU = Q - W
where DU indicates the change in the internal
energy of the system. This internal energy
is related to the temperature and heat
capacity of the system; Q is the heat energy
added to the system; and W is the work
done by the system.
Second Law of Thermodynamics
Entropy is a measure of the probability of
being in a state. Since things tend to go
to their most probable state, we can write
the 2nd Law of Thermodynamics as:
systems tend to have their entropy
increase.
Efficiency
Efficiency is a measure of how much you get out
versus how much you put in. For heat engines:
Efficiency = = Work done / Heat Added
By the first law, the work done is simply the difference
in the heat going into the engine minus the heat
coming out of the engine. The total heat added is
the heat going into the engine.
= (Qhot - Qcold) /Qhot . For the most efficient engine
possible: Carnot = (Thot - Tcold) / Thot
Oscillations
with a mass on a spring?
y = A sin(wt + qo)
The oscillation speed, w, describes how fast
the mass oscillates. But what does this
oscillation speed (ω=dqphase/dt ) depend on?
By putting in our solution for y into Newton’s
Second Law (the differential equation), we
can get a prediction: w = (k/m) .
For stiffer springs and lighter masses, the
frequency of the oscillation increases.
Note: the Amplitude does NOT affect the frequency!
Energy: Amplitude and frequency
Since Energy = (1/2) mw2A2 , as the frequency
goes up (ω), to keep the same energy the
amplitude (A) needs to go down. Can you make
sense of that relationship?
Since kinetic energy depends on velocity
(squared), and since v = dx/dt , a higher
frequency means that for the same distance
(amplitude) we have a smaller dt. To keep the
same v, we need a smaller distance (amplitude)
to go with the smaller dt (higher frequency).
Waves (in general)
y = A sin(q) where q is a phase angle
in a moving wave, q changes with both
– time (goes 2p radians in time T) and
– distance (goes 2p radians in distance )
so q = (2p/)*x +/- (2p/T)*t
– where 2p/T = w
and
– where 2p/ = k and so
phase speed: v = distance/time = /T = f =
w/k
Standing Waves
To create what are called standing waves (we will
play with these in the last lab), we need to create
constructive interference from both ends. This
leads to the following condition: #(/2) = L ,
which says: we need an integer number of half
wavelengths to “fit” on the Length of the string
for standing waves.
We can vary the wavelength by either varying the
frequency or the speed of the wave: recall that phase
speed: v = distance/time = /Tperiod = f . For a
wave on a string, recall that v = f = (Ttension)/m) where
m = m/L.
Standing Waves
For stringed instruments (piano, guitar, etc.), the
string vibrates with both ends fixed.
However, with wind instruments (trumpet,
trombone, etc.), we can have the situation
where both ends are free and a different
situation where one end is free and one end
is fixed.
1. If both ends are free, we get the same
resonance condition as for both ends fixed:
#(/2) = L.
2. If one end is free and the other end is fixed,
we get a different condition: #odd(/4) = L,
where #odd is an odd number (1, 3, 5, etc.).
Sound Intensity
I(dB) = 10*log10(I/Io) where Io = 10-12 W/m2
The weakest sound intensity we can hear is what
we define as Io. In decibels this becomes:
I(dB) = 10*log10(10-12 W/m2 / 10-12 W/m2) = 0 dB .
The loudest sound without damaging the ear is 1
W/m2, so in decibels this becomes:
I(dB) = 10*log10(1 W/m2 / 10-12 W/m2) = 120 dB .
Electric Force
To account for repulsive and attractive
forces, we find that like charges
repel, and unlike charges attract.
We also find that the force decreases with
distance between the charges just like
gravity, so we have Coulomb’s Law:
Felectricity = k q1 q2 / r122 where k, like G
in gravity, describes the strength of
the force in terms of the units used.
Electric Force
Charge is a fundamental quantity, like
length, mass and time. The unit of
charge in the MKS system is called the
Coulomb.
When charges are in Coulombs, forces in
Newtons, and distances in meters, the
Coulomb constant, k, has the value:
k = 9.0 x 109 Nt*m2 / Coul2 . (Compare this
to G which is 6.67 x 10-11 Nt*m2 / kg2 !)
Fundamental Charges
When we break matter up, we find there are
just a few fundamental particles: electron,
proton and neutron. (The proton and neutron are
now thought to be made up of more elementary particles
called quarks, while the electron remains elementary.)
electron: qe = -1.6 x 10-19 Coul; me = 9.1 x 10-31 kg
proton: qp = +1.6 x 10-19 Coul; mp = 1.67 x 10-27 kg
neutron: qn = 0;
mn = 1.67 x 10-27 kg
(note: despite what appears above, the mass of neutron and
proton are NOT exactly the same; the neutron is slightly
heavier; however, the charge of the proton and electron
ARE exactly the same - except for sign)
Electric Field for a point charge
If I have just one point charge setting up the
field, and a second point charge comes
into the field, I know (from Coulomb’s Law)
that
Fon 2 = k q1 q2 / r122
and
Fon 2 = q2 * Eat 2
which gives:
E at 2 due to 1 = k q1 / r122 for a point charge.
Electric Potential Energy
Since Coulomb’s Law has the same form as
Newton’s Law of Gravity, we will get a very
similar formula for electric potential energy:
PEel = k q1 q2 / r12
Recall for gravity, PEgr = - G m1 m2 / r12 .
Note that the PEelectric does NOT have a
minus sign. This is because two like
charges repel instead of attract as in gravity.
Voltage
Just like we did with forces on particles to
get fields in space,
(Eat 2 due to 1 = Fon 2/ q2)
we can define an electric voltage in
space (a scalar):
Vat 2 due to 1 = PEof 2 / q2 .
We often use this definition this way:
PEof 2 = q2 * Vat 2 .
Voltage and Field
DV = -E
Ds , or Ex = -DV / Dx .
Note also the minus sign means that electric field
goes from high voltage towards low voltage.
Note also that this means that
positive charges will tend to “fall” from high
voltage to low voltage (like masses tend to fall from
high places to low places) , but that
negative charges will tend to “rise” from low
voltage to high voltage (like bubbles tend to rise) !
Review
F1on2 = k q1 q2 / r122 PE12 = k q1 q2 / r12
Fon 2 = q2 Eat 2
PEof 2 = q2 Vat 2
Eat 2 = k q1 / r122
Vat 2 = k q1 / r12
use in
use in
S F = ma
KEi + PEi = KEf +PEf +Elost
VECTOR
scalar
Ex = -DV / Dx
Electric Circuits
In electricity, the concept of voltage will be like
pressure. Water flows from high pressure
to low pressure (this is consistent with our
previous analogy that Voltage is like height since
DP = gh for fluids) ; electricity flows from
high voltage to low voltage.
But what flows in electricity? Charges!
How do we measure this flow? By Current:
current = I = Dq / Dt
UNITS: Amp(ere) = Coulomb / second
Resistance
By experiment we find that if we increase
the voltage, we increase the current: V is
proportional to I. The constant of
proportionality we call the resistance, R:
V = I*R
Ohm’s Law
UNITS: R = V/I so Ohm = Volt / Amp.
The symbol for resistance is W (capital omega).
Resistance
The resistance depends on material and
geometry (shape). For a wire, we have:
R=L/A
where  is called the resistivity (in Ohmm) and measures how hard it is for
current to flow through the material, L is
the length of the wire, and A is the
cross-sectional area of the wire. The
second lab experiment deals with Ohm’s Law
and the above equation.
Electrical Power
The electrical potential energy of a charge is:
PE = q*V .
Power is the change in energy with respect to
time: Power = DPE / Dt .
Putting these two concepts together we have:
Power = D(qV) / Dt = V(Dq) / Dt = I*V.
Capacitance
We define capacitance as the amount of
charge stored per volt: C = Qstored / DV.
UNITS: Farad = Coulomb / Volt
Just as the capacity of a water tower depends
on the size and shape, so the capacitance
of a capacitor depends on its size and
shape. Just as a big water tower can
contain more water per foot (or per unit
pressure), so a big capacitor can store more
charge per volt.
Capacitance
While we normally define the capacity of a
water tank by the TOTAL AMOUNT of water
it can hold, we define the capacitance of an
electric capacitor as the AMOUNT OF
CHARGE PER VOLT instead.
There is a TOTAL AMOUNT of charge a
capacitor can hold, and this corresponds to a
MAXIMUM VOLTAGE that can be placed
across the capacitor. Each capacitor DOES
HAVE A MAXIMUM VOLTAGE.
Review:
Capacitors: C = Q/V
PE = ½CV2; C// = KA/[4pkd]
Series: 1/Ceff = 1/C1 + 1/C2
Parallel: Ceff = C1 + C2
series gives smallest Ceff , parallel gives largest Ceff .
Resistors: V = IR
Power = IV; R = L/A
Series: Reff = R1 + R2
Parallel: 1/Reff = 1/R1 + 1/R2
series gives largest Reff , parallel gives smallest Reff .
Magnetic Force Law
magnitude:
Fmagnetic = q v B sin(qvB)
direction: right hand rule:
thumb = hand  fingers
Point your right hand in the direction of v,
curl you fingers in the direction of B,
and the force will be in the direction of
your thumb; if the charge is negative,
the force direction is opposite that of
your thumb (or use you left hand).
Magnetic Force and Motion
Since the magnetic field is perpendicular to
the velocity, and if the magnetic force is the
only force acting on a moving charge, the
force will cause the charge to go in a circle:
SF = ma, Fmag = q v B, and a = w2r = v2/r
gives: q v B = mv2/r, or r = mv/qB .
This is the basis of the mass spectrometer and
the cyclotron.
Torque on
rectangular current loop
Recall that torque is: t = r F sin(qrF). For magnetic
force, F = qvB becomes F = ILB. In the figure
below we can see that r = w/2. Thus the Fleft
gives a torque of (w/2)ILB, and the Fright also
gives a torque of (w/2)ILB.
r
N
F
I
w
B
 F
I S L
Lenz’s Law
DV = D[ (N B A cos(qBA) ] / Dt
The above formula is for determining the
amount of voltage generated. But what
is the direction of that voltage (what
direction will it try to drive a current)?
The answer is Lenz’s Law: the direction
of the induced voltage will tend to
induce a current to oppose the change
in magnetic field through the area.
RMS Voltage and Current
In order to work with AC circuits just as we did with
DC circuits, we create a voltage and current
called rms (root mean square).
Vrms = Vo (1/2)1/2 and Irms = Io (1/2)1/2
so that we have
Pavg = Irms Vrms
and Vrms = Irms R .
Note that the power formula and Ohm’s Law are
the same for DC and for AC-rms, but NOT for
instantaneous AC.
Review of Circuit Elements
Resistor: VR = R I
where I = Dq/Dt
Capacitor: VC = (1/C)q (from C = q/V)
Inductor: VL = -L DI/Dt
We can make an analogy with mechanics:
q is like x;
V is like F;
t is like t;
L is like m;
I = Dq/Dt is like v = Dx/Dt; C is like 1/k (spring);
DI/Dt is like a = Dv/Dt;
R is like air resistance.
AC Circuits
A resistor obviously limits the current in a circuit. But,
as we just saw, a capacitor and an inductor also
limit the current in an AC circuit. However, the
reactances do not just add together. Using the
fundamental relations and the calculus, we come up
with the concept of impedance, Z: V = IZ
where Z takes into account all three
reactances: XR=R, XL=wL and XC= 1/wC:
Z = [R2 + (wL - 1/wC)2]1/2.
Power, however, is still: Pavg = I2R (not P=I2Z).
Property 1: Speed of Light
particle (photon):
no prediction
Maxwell’s Eqs.
in vacuum:
v = [1 / {o mo}]1/2
where
o = 1/{4pk} = 1 / {4p * 9x109 Nt-m2/Coul2}
mo = 4p * 1x10-7 T-s /Coul
v = [4p*9x109 / 4p*1x10-7 ]1/2 = 3 x 108 m/s = c
wave (E&M):
Property 2: Color
Experiment:
– invisible as well as visible
– total spectrum order:
•
•
•
•
•
•
radio
microwave
IR
visible
UV
x-ray and gamma ray
Property 2: Color
Experiment:
– visible order:
•
•
•
•
•
•
red
orange
yellow (yellow)
green
blue
violet
Property 3: Reflection
A white paper is rough on a microscopic level, and
so a beam of light is reflected in all directions:
Blue is
incoming,
red is
outgoing
rough paper
smooth mirror
A mirror is smooth on a microscopic level, and
so a beam of light is all reflected in just one
direction.
Property 4: Refraction
Snell’s Law: n1 sin(q1) = n2 sin(q2)
• NOTE: If n1 > n2 (v1 < v2), THEN q1 < q2 .
• NOTE: All q2 values (in the faster medium) between
0 & 90 degrees work fine.
• NOTE: Not all values of q1 (in the slower medium)
work!
Example: If n1 = 1.33, n2 = 1, and q1 = 75o, then
q2 = inv sin [n1 sin(q1) / n2] = inv sin [1.28] =
ERROR - this is called total internal reflection
Refraction and Thin Lenses
We break the THIN LENS equation:
(nglass – nair)
1
1
1 1
*{
+ } =
+
nair
R 1 R2
s s’
Into the LENS MAKERS equation
and the
LENS USERS equation:
(nglass – nair) * { 1 + 1 } = 1 & 1 = 1 1
+
nair
R 1 R2
f
f
s s’
where f is a distance called the focal length.
Refraction and the Lens-users Eq.
For s>f (lens used with camera or projector)
– Note that a real image is formed.
– Note that the image is up-side-down.
object
image
f
f
ray
1
ray 3
ray 2
Refraction and the Lens-users Eq.
For s<f (lens used as a magnifying glass)
Notice that: s’ is on the “wrong” side, which
means that s’ < 0 , and that |s’| > |s| so f > 0.
ray 1
h’
f
s
s’
f
ray 2
ray 3
Example: 1 / 5 cm = 1 / 4 cm + 1 / -20 cm
Refraction and the Lens-users Eq.
Notes on using a lens as a magnifying
glass:
• hold lens very near your eye
• want IMAGE at best viewing distance
which has the nominal value of 25 cm
so that s’ = -25 cm.
Microscope
M1 = -s1’/s1
M2 = -s2’/s2
Mtotal = M1 * M2 = (s2’*s1’) / (s2*s1)
s1
s1 ’
objective
lens
s2
s2 ’
L = s1 ’ + s2
eyepiece
1/s1 + 1/s1’ = 1/f1
1/s2 + 1/s2’ = 1/f2
Object 1
Image 2
Image 1
Object 2
NOTE: s2’ = -25 cm
so Mtotal < 0 !
Property 5: Shadows
Double Slit Experiment
n = d sin(qn)  d tan(qn) = d (xn / L)
bright
x
dim

d
bright
dim
L
SCREEN
bright
Diffraction: single slit
REVIEW:
-2 -1 0 1 2
• For double (and multiple) slits:
n = d sin(qn) for MAXIMUM
(for ALL n)
-2
-1 0
1
2
• For single slit:
n = w sin(qn) for MINIMUM
(for all n EXCEPT 0)
Diffraction: circular
opening
If instead of a single SLIT, we have
a CIRCULAR opening, the change
in geometry makes:
the single slit pattern into a series
of rings;
and
the formula to be: 1.22 n = D sin(qn) .
Rayleigh Criterion: a picture
In this case: qlimit = sin-1(1.22 /D)
= tan-1(x’/s’) = tan-1(x/s) .
D lens
x
x’
s
s’
Limits on Resolution:
• Imperfections in the eye (correctable with
glasses)
• Rayleigh Criterion due to wavelength of
visible light
• Graininess of retinal cells (Note that in low light
where only the rods are activated, we cannot resolve
very well because the rod cells are not packed as closely
as the cone cells are. Also in low light we only see in
black and white – not in color.)
Cone cells & Color Recognition
Cone cell sensitivity to different
wavelengths
400 450 500 550 600 650 700 (in nm)
If only the “blue” cone is activated, the color is violet.
If both the “blue” and “green” cones are activated, and the “blue” gives
a stronger signal, the color is blue.
If both the “blue” and “green” cones are activated, and the “green” gives
a stronger signal, the color is green.
Polarization: Wave Theory
Three polarizers in series:
Sailboat analogy:
North
wind
sail
force on
sail
boat goes along
direction of keel
Polarization: Wave Theory
#2 Polarization by reflection
– Brewster Angle: when qrefracted + qreflected =
90o
– Sunglasses and reflected glare
incident ray
vertical
horizontal
reflected ray
no problem with horizontal
almost no vertical since vertical
is essentially longitudinal now
surface
vertical can be transmitted
refracted ray
Interference: Thin Films
• Recall that the light is in the FILM, so the
wavelength is not that in AIR: f = a/nf.
reflected red interferes with
refracted/reflected/refracted blue.
air
film
water
t
Interference: Thin Films
• reflection: no difference if
180 degree difference if
• distance: no difference if
180 degree difference if
nf < nw;
nf > nw.
t = a/2nf
t = a/4nf
• Total phase difference is sum of the above
two effects.
Interference: Thin Films
• Total phase difference is sum of the two
effects of distance and reflection
• For minimum reflection, need total to be
180 degrees.
– anti-reflective coating on lens
• For maximum reflection, need total to be 0
degrees.
– colors on oil slick
Photons and Colors
• Electron volts are useful size units of energy
1 eV = 1.6 x 10-19 Coul * 1V = 1.6 x 10-19 J.
• radio photon: hf = 6.63 x 10-34 J*s * 1 x 106 /s
= 6.63 x 10-28 J = 4 x 10-9 eV = 4 neV
• red photon: f = c/ = 3 x 108 m/s / 7 x 10-7 m =
4.3 x 1014 Hz,
red photon energy = 1.78 eV
• blue:  = 400 nm; photon energy = 3.11 eV .
• X-ray photon with  = 1 nm; photon energy =
1,243 eV = 1.24 keV
Photoelectric Effect
Light hits a metal plate, and electrons are
ejected. These electrons are collected
in the circuit andlight
form a current.
ejected electron
A
- +
V
Photoelectric Effect
Put into a nice equation:
• hf = W + e*Vstop
– where f is the frequency of the light
– W is the “WORK FUNCTION”, or the amount
of energy needed to get the electron out of the
metal
– Vstop is the stopping potential
• When Vstop = 0, f = fcutoff , and hfcutoff = W.
Rutherford Scattering
The results of the scattering were
consistent with the alphas scattering off
a tiny positive massive nucleus rather
than the diffuse positive pudding.
The results indicated that the positive
charge and heavy mass were located in
a nucleus on the order of 10-14 m
(Recall the atom size is on the order of 10-10 m).
The Bohr Theory
Bohr Theory: angular momentum, radius and Energy are all
quantized (with quantum number, n)
r = n22/(meke2) = (5.3 x 10-11 m) * n2
(for n=1, this is just the right size radius for the atom)
and
E = [-mek2e4/22]*(1/n2) = -13.6 eV / n2
(where 1 eV = 1.6 x 10-19 Joules).
This says the electron energy is QUANTIZED.
The Bohr Theory - an example
DE = hf = [-13.6 eV]*[(1/nf2) - (1/ni2)]
Example:
In the case of ni = 3, and nf = 2,
DE = (-13.6 eV)*(1/4 - 1/9) = 1.89 eV
DE = hf = hc/ , so in this case,
emitted = hc/DE =
(6.63x10-34 J-sec)*(3x108 m/s)/(1.89 x 1.6x10-19 J)
= 658 nm (red light).
The Bohr Theory
• Note that we have quantized energy states
for the orbiting electron.
• Note that for all nfinal = 1, we only get UV
photons.
• Note that for all nfinal > 2, we only get IR
photons.
DeBroglie Hypothesis
Problem with Bohr Theory: WHY L = n ?
• have integers with standing waves:
n(/2) = Length
• consider circular path for standing wave:
n = 2pr , and so from Bohr theory:
L = mvr = n = nh/2p, get 2pr = nh/mv = n,
which means  = h/mv = h/p .
Quantum Theory
What we are now dealing with is the
Quantum Theory:
• atoms are quantized (you can have 2 or
3, but not 2.5 atoms)
• light is quantized (you can have 2 or 3
photons, but not 2.5)
• in addition, we have quantum numbers
(L = n , where n is an integer)
Heisenberg Uncertainty
Principle
A formal statement of this (from Fourier
analysis) is: Dx * Dk = /2
(where k = 2p/, and D indicates the
uncertainty in the value)
But from the DeBroglie Hypothesis,  =
h/p, this uncertainty relation becomes:
Dx * D(2p/) = Dx * D(2pp/h) = 1/2 , or
Dx * Dp = /2.
Heisenberg Uncertainty
Principle
A similar relation from Fourier analysis for
time and frequency: Dt * Dw = 1/2 leads to
another part of the Uncertainty Principle
(using E = hf = w):
Dt * DE > /2 .
There is a third part:
Dq * DL > /2
(where L is the angular momentum value).
All of this is a direct result of the
wave/particle duality of light and matter.
Nuclear Physics
Stability: see sheet detailing stable isotopes
Radiations:
1) a, b-, b+, g are all emitted;
2) protons and neutrons are NOT emitted,
except in the case of mass numbers 5 and 9;
3) alphas are emitted only for mass numbers
greater than 209, except in the case of mass
number 8.
Alpha (a) decay
234 + a4 + g
example: 92U238
Th
90
2
(it is not obvious whether there is a gamma
emitted; this must be looked up in each
case) Mass is reduced!
NOTE: 1. subscripts must be conserved
(conservation of charge) 92 = 90 + 2
2. superscripts must be conserved
(conservation of mass) 238 = 234 + 4
Beta minus b-) decay
14 + b0 + u0
example: 6C14
N
7
-1
0
(a neutron turned into a proton by emitting an
electron; however, one particle [the neutron]
turned into two [the proton and the electron].
Charge and mass numbers are
conserved, but since all three (neutron,
proton, and electron) are fermions [spin 1/2
particles], angular momentum, particle
number, and energy are not! Need the
anti-neutrino [0u0] to balance everything!
Positron (b+) decay
11 +
0 + u0
example: 6C11
B
b
5
+1
0
a proton turns into a neutron by emitting a
positron; however, one particle [the proton]
turned into two [the neutron and the positron].
Charge and mass numbers are conserved,
but since all three are fermions [spin 1/2
particles], angular momentum, particle
number, and energy are not! Need the
neutrino [0u0] to balance everything!
Nuclear Physics
General Rules:
1) a emitted to reduce mass, only emitted if
mass number is above 209
2) b- emitted to change neutron into proton,
happens when there are too many neutrons
3) b+ emitted (or electron captured) to change
proton into neutron, happens when there are
too few neutrons
4) g emitted to conserve energy in reaction,
may accompany a or b.
Mass Defect & Binding Energy
Similarly, the missing mass was converted into
energy (E=mc2) and emitted when the
carbon-12 atom was made from the six
protons and six neutrons:
Dm = 6*mproton + 6*mneutron - mC-12 =
6(1.00782 amu) + 6(1.008665 amu) - 12.00000 amu
= .099 amu;
BE = Dm*c2 =
(0.099 amu)*(1.66x10-27kg/amu)*(3x108m/s)2
= 1.478x10-11J*(1 eV/1.6x10-19J) = 92.37 MeV
Activity
N(t) = No e-t
A = N = Aoe-t
T(half life) = ln(2) /  .
If the half life is large,  is small. This means
that if the radioactive isotope will last a
long time, its activity will be small; if the
half life is small, the activity will be large
but only for a short time!
Review:
Radioactivity around us
an example
For 1 gram of carbon in a living plant,
Ao = 15.0/min . Also, carbon-14 has a half
life of 5,730 years.
If a 1 gram carbon sample from a dead plant
has an activity of 9.0/min, then using:
A = Aoe-t ,
we have 9.0/min = 15.0/min * e-(ln2/5730yrs)t ,
or -(ln2/5730 yrs)*t = ln(9/15) , or
t = 5730 years * ln(15/9) / ln(2) = 4,200 years.
Radioactivity around us
Another radioactive isotope found in the earth is
238 . Since it is well above the 209 mass limit, it
U
92
gives rise to a whole series of radioactive isotopes
with mass numbers 238, 234, 230, 226, 222, 218,
214, 210. The 226 isotope is 88Ra226, which is the
isotope that Marie Curie isolated from uranium
ore. The 222 isotope is 86Rn222 which is a noble
gas.
Other radioactive isotopes found in nature are
232 ,
235, and
40 . Both
232 and
235
Th
U
K
Th
U
90
92
19
90
92
have decay chains that lead down to 82Pb (lead).
X-rays
example
Eionization = 13.6 eV * (Z-1)2 where the -1
comes from the other inner shell electron.
If the electrons have this ionization energy,
then they can knock out this inner
electron, and we can see the characteristic
spectrum for this target material.
For iron with Z=26, the ionization energy is:
13.6 eV * (26-1)2 = 1e * 8,500 volts.
X and g ray penetration
I = Io e-mx where m depends on the material
the x-ray is going through and the energy
of the x-ray.
In a similar way to half lives, we can define a
half-value-layer, hvl, where hvl = ln(2)/m .
Measuring Radioactivity
• How do we measure radioactivity?
The Bq (dis/sec) and Curie (1 Ci = 3.7 x 1010 Bq)
measure how many decays happen per time.
However, different radioactive materials emit
different particles with different energies.
• What is the source of the health effects of
radiation?
Radiation (a, b, g) ionizes atoms. Ionized atoms
are important to biological function, and so
radiation may interfere with biological
functions.
• Can we devise a way to measure the health
effects of radiation?
Measuring Health Effects
Can we devise a way to measure the health
effects of radiation?
A unit that directly measures ionization is the
Roentgen (R) = (1/3) x 10-9 Coul created per
cc of air at STP. This uses air, since it is
relatively easy to collect the charges due to
ionization. It is harder to do in biological
material, so this method is best used as a
measure of EXPOSURE dose.
Measuring Health Effects
Can we devise a way to measure the health
effects of radiation?
In addition to measuring ionization ability in air, we
can also measure the energy that is absorbed
by a biological material: Rad = .01 J/kg
MKS: Gray (Gy) = 1 J/kg = 100 rads.
This is called an ABSORBED dose.
Generally, one Roentgen of exposure will give one
rad of absorption.
Measuring Health Effects
This difference in penetrating ability (and
localization of ionization) leads us to create an
RBE (radiation biological equivalent) factor and
a new unit: the rem. The more localized the
ionization, the higher the RBE.
# of rems = RBE * # of rads . This is called an
EFFECTIVE dose.
RBE for gammas = 1; RBE for betas = 1 to 2;
RBE for alphas = 10 to 20.
Levels of Radiation and
Measurable Health Effects
200 millirems/year: background
Here are some more benchmarks based on our
experience with acute (short time) doses:
20,000 millirems: measurable transient blood changes;
150,000 millirems: acute radiation sickness;
200,000 millirems: death in some people;
350,000 millirems: death in 50% of people.
Chain Reactions
In some cases, a very heavy nucleus, instead of
undergoing alpha decay, will spontaneously split in two.
Example:
238
U
92
129 +
106 + 3 n1 + energy
Sn
Mo
50
42
0
The amount of energy coming from this reaction is on the
order of 200 MeV, which is about 200 million times more
than a chemical reaction.
This fissioning of uranium does not always result in these
two resultant atoms - there is a whole range of resulting
atoms. But it always gives a few neutrons.
In some cases a neutron can stimulate a heavy nucleus to
split in two. This, if properly set up, can cause a chain
reaction. This chain reaction is the basis for both the
nuclear bomb and the nuclear power station.
Fusion
1 + H1
H
1
1
1 + D2
H
1
1
1 + T3
H
1
1
2 +
0 +  + energy
D
b
1
+1
3 +
0 +  + energy
T
b
1
+1
4 + energy
He
2
so we have four hydrogens becoming one
helium, with about 24 MeV of energy and
two neutrino’s produced plus two positrons
which will combine with the extra two
electrons from the 4 H’s to give another 2
MeV’s of energy.