Electrostatic forces are inverse square law forces (proportional to. 1/r. 2. ) •
Electrostatic force is proportional to the product of the amount of charge on each
...
R. D. Field
PHY 2049
Electrostatic Force and Electric Charge Electrostatic Force (charges at rest):
• Electrostatic force can be attractive q1 q2 • Electrostatic force can be repulsive • Electrostatic force acts through empty r space • Electrostatic force much stronger than gravity • Electrostatic forces are inverse square law forces (proportional to 1/r2) • Electrostatic force is proportional to the product of the amount of charge on each interacting object
Magnitude of the Electrostatic Force is given by Coulomb's Law: F = K q1q2/r2
(Coulomb's Law)
where K depends on the system of units K = 8.99x109 Nm2/C2 (in MKS system) K = 1/(4πε πε0) where ε0 = 8.85x10-12 C2/(Nm2)
Electric Charge: electron charge = -e proton charge = e
e = 1.6x10-19 C C = Coulomb
Electric charge is a conserved quantity (net electric charge is never created or destroyed!)
Chapter 22
chp22_1.doc
R. D. Field
PHY 2049
Units MKS System (meters-kilograms-seconds): also Amperes, Volts, Ohms, Watts Force: F = ma Newtons = kg m / s2 = 1 N Work: W = Fd Joule = Nm = kg m2 / s2 = 1 J Electric Charge: Q Coulomb = 1 C F = K q1q2/r2 K = 8.99x109 Nm2/C2 (in MKS system)
CGS System (centimeter-grams-seconds): Force: Work: Electric Charge: F = q1q2/r2
F = ma W = Fd Q K=1
1 dyne = g cm / s2 1 erg = dyne-cm = g cm2 / s2 esu (electrostatic unit) (in CGS system)
Conversions (MKS - CGS): Force: Work: Electric Charge:
1 N = 105 dynes 1 J = 107 ergs 1 C = 2.99x109 esu
Fine Structure Constant (dimensionless):
α = K 2π πe2/hc h = Plank's Constant
Chapter 22
(same in all systems of units) c = speed of light in vacuum
chp22_2.doc
R. D. Field
PHY 2049
Electrostatic Force versus Gravity Electrostatic Force : Fe = K q1q2/r2
(Coulomb's Law)
K = 8.99x109 Nm2/C2
(in MKS system)
Gravitational Force : Fg = G m1m2/r2
(Newton's Law)
G = 6.67x10-11 Nm2/kg2
(in MKS system)
Ratio of forces for two electrons : e = 1.6x10-19 C
m = 9.11x10-31 kg
e, m
e, m r
Fe / Fg = K e2 / G m2 = 4.16x1042
Chapter 22
(Huge number !!!)
chp22_3.doc
R. D. Field
PHY 2049
Vector Forces ^
r
q
Q
The Electrostatic Force is a vector: The force on q due to Q points along the direction r and is given by
r KqQ F = r$ 2 r
q1
F3 F2 Q
q2
F1 q3
Vector Superposition of Electric Forces: If several point charges q1, q2, q3, … simultaneously exert electric forces on a charge Q then
F = F1 + F2 +F3 + …
Chapter 22
chp22_4.doc
R. D. Field
PHY 2049
Vectors & Vector Addition The Components of a vector: y-axis
Ay =A sin θ
A θ Ax =A cos θ
x-axis
Vector Addition: y-axis C
B A
x-axis
To add vectors you add the components of the vectors as follows:
r A = Ax x$ + Ay y$ + Az z$ r B = Bx x$ + By y$ + Bz z$
r r r C = A + B = ( Ax + Bx ) x$ + ( Ay + By ) y$ + ( Az + Bz )z$
Chapter 22
chp22_5.doc
R. D. Field
PHY 2049
The Electric Dipole +Q
-Q d
An electric "dipole" is two equal and opposite point charges separated by a distance d. It is an electrically neutral system. The "dipole moment" is defined to be the charge times the separation (dipole moment = Qd). Example Problem: +Q d x
q
-Q A dipole with charge Q and separation d is located on the y-axis with its midpoint at the origin. A charge q is on the x-axis a distance x from the midpoint of the dipole. What is the electric force on q due to the dipole and how does this force behave in the limit x >>d (dipole approximation)?
Example Problem: -Q
d
+Q x
A dipole with charge Q and separation d is located on the x-axis with its midpoint at the origin. A charge q is on the x-axis a distance x from the midpoint of the dipole. What is the electric force on q due to the dipole and how does this force behave in the limit x >>d (dipole approximation)?
Chapter 22
chp22_6.doc
R. D. Field
PHY 2049
The Electric Field +Q
q E
The charge Q produces an electric field which in turn produces a force on the charge q. The force on q is expressed as two terms:
F = K qQ/r2 = q (KQ/r2) = q E The electric field at the point q due to Q is simply the force per unit positive charge at the point q:
E = F/q
E = KQ/r2
The units of E are Newtons per Coulomb (units = N/C). The electric field is a physical object which can carry both momentum and energy. It is the mediator (or carrier) of the electric force. The electric field is massless.
The Electric Field is a Vector Field:
r KQ E = 2 r$ r
Chapter 23
chp23_1.doc
R. D. Field
PHY 2049
Electric Field Lines
+Q
-Q
Electric field line diverge from (i.e. start) on positive charge and end on negative charge. The direction of the line is the direction of the electric field.
The number of lines penetrating a unit area that is perpendicular to the line represents the strength of the electric field.
+Q
Chapter 23
+2Q
chp23_2.doc
R. D. Field
PHY 2049
Electric Field due to a Distribution of Charge dE = K dQ/r2 r
dQ r
The electric field from a continuous distribution of charge is the superposition (i.e. integral) of all the (infinite) contributions from each infinitesimal dQ as follows:
r E =
∫
K $ 2 rdQ r
and
Charge Distributions:
• Linear charge density λ : length
Q = ∫ dQ λ(x) λ( ) = charge/unit
L
dQ = λ dx
For a straight line dQ = λ(x) dx and
Q = ∫ dQ = ∫ λ( x)dx
If λ(x) = λ is constant then dQ = λ dx and Q = λL, where L is the length.
Chapter 23
chp23_3.doc
R. D. Field
PHY 2049
Charge Distributions Charge Distributions:
• Linear charge density λ : length
λ(θ) = charge/unit arc
dQ = λ ds = λ R dθ R
For a circular arc dQ = λ(θ) ds = λ(θ) Rdθ and
Q = ∫ dQ = ∫ λ (θ ) ds = ∫ λ (θ ) Rdθ
If λ(θ θ) = λ is constant then dQ = λ ds and Q = λs, where s is the arc length.
• Surface charge density σ:
σ(x,y) σ( ) = charge/unit area
dQ = σ dA
For a surface dQ = σ(x,y) dA and
Q = ∫ dQ = ∫ σ( x, y)dA
If σ(x,y) = σ is constant then dQ = σ dA and Q = σA, is the area.
where A
• Volume charge density ρ: ρ(x,y,z) ρ( ) = charge/unit volume dQ = ρ dV
For a surface dQ = ρ(x,y,z) dV and
Q = ∫ dQ = ∫ ρ( x, y, z)dV
If ρ(x,y,z) = ρ is constant then dQ = ρ dV and Q = ρV, where V is the volume. Chapter 23
chp23_4.doc
R. D. Field
PHY 2049
Calculating the Electric Field Example: P
L x
A total amount of charge Q is uniformily distributed along a thin straight rod of length L. What is the electric field a a point P on the x-axis a distance x from the end of the rod?
r E=
Answer:
KQ x$ x ( x + L)
Example: P
A total amount of charge Q is uniformily distributed along a thin straight rod of length L. What is the electric field a a point P on the y-axis a distance y from the midpoint of the rod?
Answer:
y
L
r E=
KQ y y 2 + ( L / 2) 2
y$
Example: A infinitely long straight rod has a uniform charge density λ. What is the electric field a point P a perpendicular distance r from the rod?
P r λ
Answer:
Chapter 23
r 2 Kλ E= r$ r
chp23_5.doc
R. D. Field
PHY 2049
Some Useful Math Approximations:
(1 + ε ) ≈ 1 + pε p
ε b = r$ 2 r r E a 0 (increasing with time), B is positive which means that it points in the direction of my chosen orientation.
Chapter 32
chp32_5.doc
R. D. Field
PHY 2049
Simple Harmonic Motion Hooke's Law Spring: For a Hooke's Law spring the restoring force is linearly proportional to the distance from equilibrium, Fx = -kx, where k is the spring constant. Since, Fx = max we have
d 2x − kx = m 2 dt
or
d2x k + x = 0 , where x = x(t). dt 2 m
General Form of SHM Differential Equation: The general for of the simple harmonic motion (SHM) differential equation is
d 2 x (t ) + Cx (t ) = 0 , dt 2 where C is a positive constant (for the Hooke's Law spring C=k/m). The most general solution of this 2nd order differential equation can be written in the following four ways:
x ( t ) = Ae iωt + Be − iωt x ( t ) = A cos(ω t ) + B sin(ω t ) x ( t ) = A sin(ω t + φ ) x ( t ) = A cos(ω t + φ ) where A, B, and φ are arbitrary constants and
ω = C . In the chart, A is the amplitude of the oscillations and T is the period. The linear frequency f = 1/T is measured in cycles per second (1 Hz = 1/sec). The angular frequency ω = 2π πf and is measured in radians/second. For the
x(t) = Acos(ω ωt+φ φ) 1.0
T
0.5
A
0.0 0
1
2
3
4
5
6
-0.5 -1.0
ωt+φ (radians)
Hooke's Law Spring C = k/m and thus ω = C = k / m . Chapter 33
chp33_1.doc
7
8
R. D. Field
PHY 2049
SHM Differential Equation The general for of the simple harmonic motion (SHM) differential equation is
d 2 x (t ) + Cx (t ) = 0 , dt 2 where C is a constant. One way to solve this equation is to turn it into an algebraic equation by looking for a solution of the form
x ( t ) = Ae at . Substituting this into the differential equation yields, 2 a 2 Ae at + CAe at = 0 or a = − C .
Case I (C > 0, oscillatory solution): For positive C, a = ± i C = ± iω , where ω = C . In this case the most general solution of this 2nd order differential equation can be written in the following four ways:
x ( t ) = Ae iωt + Be − iωt x ( t ) = A cos(ω t ) + B sin(ω t ) x ( t ) = A sin(ω t + φ ) x ( t ) = A cos(ω t + φ ) where A, B, and φ are arbitrary constants (two arbitrary constants for a 2nd order differential equation). Remember that e where i =
± iθ
= cos θ ± i sin θ
−1.
Case II (C < 0, exponential solution): For negative C, a = ± − C = ±γ , where γ =
− C . In this case,
the most general solution of this 2nd order differential equation can be written as follows:
x ( t ) = A e γt + B e − γt
,
where A and B arbitrary constants.
Chapter 33
chp33_2.doc
R. D. Field
PHY 2049
Capacitors and Inductors Capacitors Store Electric Potential Energy:
UE
Q C
E
Q = C ∆ VC uE =
1 ε0 E 2 2
Q2 = 2C ∆ VC = Q / C (E-field energy density)
Inductors Store Magnetic Potential Energy: 1 2 U = LI B B 2 I Φ B = LI L = ΦB / I L dI εL = − L dt 1 uB = B 2 (B-field energy density) 2 µ0
Chapter 33
chp33_3.doc
R. D. Field
PHY 2049
An LC Circuit
Switch
Q +++ +++
L
C --- ---
At t = 0 the switch is closed and a capacitor with initial charge Q0 is connected in series across a inductor (assume there is no resistance). The initial conditions are Q(0) = Q0 and I(0) = 0. Moving around the circuit in the direction of the current flow yields
Q dI − L = 0. C dt
Since I is flowing out of the capacitor, I = − dQ / dt , so that
1 d 2Q Q = 0. 2 + dt LC This differential equation for Q(t) is the SHM differential equation we studied earlier with ω = 1 / LC and solution Q ( t ) = A cos ω t + B sin ω t . The current is thus,
I (t ) = −
dQ = A ω sin ω t − B ω cos ω t . dt
Applying the initial conditions yields
Q ( t ) = Q 0 cos ω t I ( t ) = Q 0ω sin ω t
Thus, Q(t) and I(t) oscillate with SHM with angular frequency ω = 1 / LC . The stored energy oscillates between electric and magnetic according to
Q 2 (t ) Q 02 U E (t ) = = cos 2 ω t 2C 2C 1 1 U B ( t ) = LI 2 ( t ) = L Q 02 ω 2 sin 2 ω t 2 2 Energy is conserved since Utot(t) = UE(t) + UB(t) = Q02/2C is constant.
Chapter 33
chp33_4.doc
R. D. Field
PHY 2049
LC Oscillations I
I t=0
Q
B
t = T/4
+++ +++
E
L
C
L
C
--- ---
1.0
Q ( t ) = Q 0 cos ω t I ( t ) = Q 0ω sin ω t
Q(t) 0.5
I(t) 0.0 0
1
2
3
4
5
6
7
-0.5
-1.0
ωt (radians)
Utot = UE + UB
1.0
UE(t)
Q 02 U E (t ) = cos 2 ω t 2C Q 02 U B (t ) = sin 2 ω t 2C
UB(t)
0.5
0.0 0
1
2
3
4
5
6
7
8
ωt (radians)
Chapter 33
chp33_5.doc
8
R. D. Field
PHY 2049
Mechanical Analogy I
t=0
t=0
Q +++ +++
E
k
L
C
--- ---
m x-axis
x0
At t = 0:
At t = 0: 1 Q 02 2C I =0
1 2 kx 2 0 v = 0
U =
E =
At Later t: dx v= dt x ( t ) = x 0 cos ω t k ω = m 1 1 E = mv 2 + kx 2 2 2
At Later t: dQ I =− dt Q ( t ) = Q0 cos ω t 1 ω = LC 1 1 E = LI 2 + Q2 2 2C
Constant
Correspondence: x (t ) ↔ Q (t ) v (t ) ↔ I (t ) m↔ L k ↔ 1/ C
Chapter 33
chp33_6.doc
R. D. Field
PHY 2049
Another Differential Equation Consider the 2nd order differential equation
dx ( t ) d 2 x (t ) + + Cx ( t ) = 0 , D dt 2 dt where C and D are constants. We solve this equation by turning it into an algebraic equation by looking for a solution of the form x ( t ) = Ae . Substituting this into the differential equation yields, at
2
a 2 + Da + C = 0
D D a = − ± −C . 2 2
or
Case I (C > (D/2)2, damped oscillations): For C > (D/2)2, a = − D / 2 ± i C − ( D / 2 )
ω′ =
= − D / 2 ± iω ′ , where
2
C − ( D / 2 ) 2 , and the most general solution has the form:
x ( t ) = e − Dt / 2 ( Ae iω ′t + Be − iω ′t ) x ( t ) = e − Dt / 2 ( A cos(ω ′t ) + B sin(ω ′t ) ) x ( t ) = Ae − Dt / 2 sin(ω ′t + φ ) x ( t ) = Ae − Dt / 2 cos(ω ′t + φ )
where A, B, and φ are arbitrary constants.
Case II (C < (D/2)2, over damped): For C < (D/2)2, a = − D / 2 ±
γ =
( D / 2 ) 2 − C = − D / 2 ± γ , where
( D / 2) 2 − C . In this case,
x ( t ) = e − D t / 2 ( A e γt + B e − γt ) .
Case III (C = (D/2)2, critically damped):
For C = (D/2)2, a = − D / 2 , and
x (t ) = A e − Dt /2
Chapter 33
.
chp33_7.doc
R. D. Field
PHY 2049
An LRC Circuit
Switch
Q +++ +++
L
C --- --R
At t = 0 the switch is closed and a capacitor with initial charge Q0 is connected in series across an inductor and a resistor. The initial conditions are Q(0) = Q0 and I(0) = 0. Moving around the circuit in the direction of the current flow yields
Q dI − L − IR = 0 . C dt
Since I is flowing out of the capacitor, I = − dQ / dt , so that
1 d 2Q R dQ + Q = 0. 2 + dt L dt LC This differential equation for Q(t) is the differential equation we studied earlier. If we take the case where R2 < 4L/C (damped oscillations) then
Q(t ) = Q0 e− Rt / 2 L cosω ′t , with ω ′ =
ω 2 − ( R / 2 L ) 2 and ω =
1 / LC .
1.0
Damped Oscillations
0.5
0.0 0
1
2
3
4
5
6
7
8
9
10 11 12 13 14 15 16 17 18
-0.5
Q(t) -1.0
Time
Chapter 33
chp33_8.doc
R. D. Field
PHY 2049
Traveling Waves A "wave" is a traveling disturbance that transports energy but not matter. Constructing Traveling Waves: To construct a wave with shape y = f(x) at time t = 0 traveling to the right with speed v simply make the replacement x → x − vt .
y = f(x) at time t=0
y = f(x-vt) v
x=0
x = vt
Traveling Harmonic Waves: Harmonic waves have the form y = A sin(kx) or y=Asin(kx) λ y = Acos(kx) at time t = 0, 1.0 where k is the "wave number" 0.5 (k = 2π π/λ λ where λ is the "wave length") and A is the 0.0 "amplitude". To construct an -0.5 harmonic wave traveling to the -1.0 right with speed v, replace x by kx (radians) x-vt as follows: y = Asin(k(x-vt) = Asin(kx-ω ωt) where ω = kv (v = ω/k). The period of the oscillation, T = 2π π/ω ω = 1/f, where f is the linear frequency (measured in Hertz where 1Hz = 1/sec) and ω is the angular frequency (ω ω = 2π πf). The speed of propagation is given by v = ω/k = λ f .
y = y(x,t) = Asin(kx-ω ω t) right moving harmonic wave y = y(x,t) = Asin(kx+ω ω t) left moving harmonic wave
Chapter 34
chp34_1.doc
A
R. D. Field
PHY 2049
The Wave Equation
∂ 2 y( x, t ) 1 ∂ 2 y( x, t ) − 2 =0 2 2 ∂x v ∂t Whenever analysis of a system results in an equation of the form given above then we know that the system supports traveling waves propagating at speed v. General Proof: If y = y(x,t) = f(x-vt) then
∂y = f ′ ∂x ∂y = − vf ′ ∂t
∂2y = f ′′ ∂x 2 ∂2y 2 = v f ′′ ∂t 2
and
∂ 2 y(x,t) 1 ∂ 2 y( x,t) − 2 = f ′′ − f ′′ = 0 . ∂x 2 v ∂t 2 Proof for Harmonic Wave: If y = y(x,t) = Asin(kx-ω ω t) then ∂2y 2 2 = − k A sin( kx − ωt ) ∂x
∂2y 2 2 = −ω A sin( kx − ωt ) ∂t
and
∂ 2 y( x,t ) 1 ∂ 2 y( x, t ) ω2 2 − 2 = − k + 2 A sin( kx − ω t ) = 0 , ∂x 2 v ∂t 2 v since ω = kv.
Chapter 34
chp34_2.doc
R. D. Field
PHY 2049
Light Propagating in Empty Space Since there are no charges and no current in empty space, Faraday's Law and Ampere's Law take the form r r dΦ B ⋅ = − E dl ∫ dt
y-axis
E
r r dΦ E ⋅ = B dl µ ε . 0 0 ∫ dt
x-axis
B
z-axis
Look for a solution of the form r E ( x , t ) = E y ( x , t ) y$ r B ( x , t ) = B z ( x , t ) z$ Faraday's Law: Computing the left and right hand side of Faraday's Law using a rectangle (in the xy-plane) with width dx and height h (counterclockwise) gives
y-axis
E(x+dx,t)
E(x,t) dx
h
∂Bz E y ( x + dx , t ) h − E y ( x , t ) h = − hdx ∂t
x-axis
or
∂E y
∂B =− z ∂x ∂t
B
z-axis
Ampere's Law:
y-axis
E
Computing the left and right hand side of Ampere's Law using a rectangle (in the xz-plane) with width dx and height h (counterclockwise) gives B z ( x , t ) h − B z ( x + dx , t ) h = µ 0 ε 0
or
∂E y ∂Bz − = µ0 ε 0 ∂x ∂t
Chapter 34
∂E y ∂t
h
hdx B(x,t) z-axis
x-axis
dx
B(x+dx,t)
chp34_3.doc
R. D. Field
PHY 2049
Electromagnetic Plane Waves (1) y-axis
We have the following two differential equations for Ey(x,t) and Bz(x,t):
∂E y ∂Bz = − ∂t ∂x
and
E
(1)
x-axis
∂E y
1 ∂Bz =− ∂t µ0 ε 0 ∂ x
z-axis
B
(2)
Taking the time derivative of (2) and using (1) gives 2 ∂ 2Ey 1 ∂ ∂ Bz 1 ∂ ∂Bz 1 ∂ Ey =− =− = ∂t 2 µ0 ε 0 ∂ t ∂ x µ0 ε 0 ∂ x ∂ t µ0 ε 0 ∂ x 2
which implies
∂ 2Ey ∂x 2
− µ0 ε 0
∂ 2Ey ∂t 2
= 0.
Thus Ey(x,t) satisfies the wave equation with speed v = 1 / ε 0 µ 0 and has a solution in the form of traveling waves as follows: Ey(x,t) = E0sin(kx-ω ω t), where E0 is the amplitude of the electric field oscillations and where the wave has a unique speed ω 1 v=c= = λf = = 2.99792 × 10 8 m / s (speed of light). k ε0 µ0 From (1) we see that
∂E y ∂Bz =− = − E 0 k cos( kx − ω t ) , ∂t ∂x
which has a solution given by
Bz ( x , t ) = E 0
k E sin( kx − ω t ) = 0 sin( kx − ω t ) , c ω
so that
Bz(x,t) = B0sin(kx-ω ω t), where B0 = E0/c is the amplitude of the magnetic field oscillations. Chapter 34
chp34_4.doc
R. D. Field
PHY 2049
Electromagnetic Plane Waves (2) The plane harmonic wave solution for light with frequency f and wavelength λ and speed c = fλ λ is given by
r E ( x , t ) = E 0 sin( kx − ω t ) y$ r B ( x , t ) = B 0 sin( kx − ω t ) z$
where k = 2π π/λ λ , ω = 2π πf, and E0 = cB0.
y-axis
E
Direction of Propagation
x-axis
z-axis
B
Properties of the Electromagnetic Plane Wave: • Wave travels at speed c ( c =1/ µ0ε0 ). r r E • E and B are perpendicular ( ⋅ B = 0).r r • The wave travels in the direction of E × B . • At any point and time E = cB. Electromagnetic Radiation:
Chapter 33
chp34_5.doc
R. D. Field
PHY 2049
Energy Transport - Poynting Vector Electric and Magnetic Energy Density: For an electromagnetic plane wave Ey(x,t) = E0sin(kx-ω ω t),
y-axis
E
Bz(x,t) = B0sin(kx-ω ω t),
x-axis
where B0 = E0/c. The electric energy density is given by
uE =
B
z-axis
1 1 ε0 E 2 = ε0 E02 sin 2 ( kx − ωt ) and the magnetic energy density is 2 2 1 2 1 1 2 uB = B = E = ε0 E 2 = uE , 2 2 µ0 2 µ0 c 2
where I used E = cB. Thus, for light the electric and magnetic field energy densities are equal and the total energy density is
1 2 B = ε0 E02 sin 2 (kx − ωt ) . µ0 r 1 r r S Poynting Vector ( = µ E × B ): 0 utot = uE + uB = ε0 E 2 =
The direction of the Poynting Vector is the direction of energy flow and the magnitude 1 E2 1 dU S = EB = = µ0 µ0 c A dt is the energy per unit time per unit area (units of Watts/m2). Proof: dU tot = utotV = ε0 E 2 Acdt so
Intensity
y-axis
E A Energy Flow x-axis
z-axis
B cdt
1 dU E2 E 02 2 S = = ε0c E = = sin 2 ( kx − ω t ) . A dt µ0 c µ0c of the Radiation (Watts/m2):
The intensity, I, is the average of S as follows: 1 dU E 02 E 02 2 I =S = = sin ( kx − ω t ) = A dt µ0 c 2 µ0 c .
Chapter 34
chp34_6.doc
R. D. Field
PHY 2049
Momentum Transport - Radiation Pressure Relativistic Energy and Momentum:
E2 = (cp)2 + (m0c2)2 energy
momentum rest mass
For light m0 =0 and
E = cp
(for light)
For light the average momentum per unit time per unit area is equal to the intensity of the light, I, divided by speed of light, c, as follows:
1 dp 1 1 dU 1 = = I. A dt c A dt c Total Absorption: dp 1 dU 1 F = = = IA dt c dt c F 1 P= = I (radiation pressure) A c Total Reflection: dp 2 dU 2 F = = = IA . dt c dt c F 2 P= = I (radiation pressure) A c
Chapter 34
Light
Total Absorption
Light
Total Reflection
chp34_7.doc
R. D. Field
PHY 2049
The Radiation Power of the Sun Problem:
P = 3.9 x 1026 W
The radiation power of the sun is 3.9x1026 W and the Sun distance from the Earth to d = 1.5 x 1011 m Earth the sun is 1.5x1011 m. (a) What is the intensity of the electromagnetic radiation from the sun at the surface of the Earth (outside the atmosphere)? (answer: 1.4 kW/m2) (b) What is the maximum value of the electric field in the light coming from the sun? (answer: 1,020 V/m) (c) What is the maximum energy density of the electric field in the light coming from the sun? (answer: 4.6x10-6 J/m3) (d) What is the maximum value of the magnetic field in the light coming from the sun? (answer: 3.4 µT) (e) What is the maximum energy density of the magnetic field in the light coming from the sun? (answer: 4.6x10-6 J/m3) (f) Assuming complete absorption what is the radiation pressure on the Earth from the light coming from the sun? (answer: 4.7x10-6 N/m2) (g) Assuming complete absorption what is the radiation force on the Earth from the light coming from the sun? The radius of the Earth is about 6.4x106 m. (answer: 6x108 N) (h) What is the gravitational force on the Earth due to the sun. The mass of the Earth and the sun are 5.98x1024 kg and 1.99x1030 kg, respectively, and G = 6.67x10-11 Nm2/kg2. (answer: 3.5x1022 N)
Chapter 34
chp34_7e.doc
R. D. Field
PHY 2049
Geometric Optics Fermat's Principle: In traveling from one point to another, light follows the path that requires minimal time compared to the times from the other possible paths.
Theory of Reflection: Let tAB be the time for light to go from the point A to the point B reflecting off the point P. Thus,
t AB
1 1 = L1 + L2 , c c
B
A
L2
L1 a
θi
where
b
θr d-x
x
L1 = x 2 + a 2 . L2 = ( d − x ) 2 + b 2
P d
To find the path of minimal time we set the derivative of tAB equal to zero as follows:
dt AB 1 dL1 1 dL2 = + = 0, dx c dx c dx which implies
dL1 dL =− 2, dx dx
B
but
dL1 x = = sin θi dx L1 dL2 − ( d − x ) = = − sin θr dx L2
A
L2
L1 a
θi
Chapter 34
d-x
x
so that the condition for minimal time becomes
sin θi = sin θr
θr
P d
θi = θr
.
chp34_8.doc
b
R. D. Field
PHY 2049
Law of Refraction Index of Refraction: Light travels at speed c in a vacuum. It travels at a speed v < c in a medium. The index for refraction, n, is the ratio of the speed of light in a vacuum to its speed in the medium, n = c/v, n1 A where n is greater than or equal to L1 one. a
θ1
Theory of Refraction:
t AB
d-x
x
Let tAB be the time for light to go from the point A to the point B refracting at the point P. Thus,
P θ2
1 1 = L1 + L v1 v2 2 ,
where
L2
b
n2
L1 = x 2 + a 2 . L2 = ( d − x ) 2 + b 2
d
B
To find the path of minimal time we set the derivative of tAB equal to zero as follows:
dt AB 1 dL1 1 dL2 1 dL1 1 dL2 = + = 0 , which implies =− dx v1 dx v2 dx v1 dx v2 dx , but dL1 x = = sin θ1 dx L1 dL2 − ( d − x ) = = − sin θ2 dx L2 so that the condition for minimal time becomes
1 1 sin θ1 = sin θ2 v1 v2
Chapter 34
n1 sin θ1 = n 2 sin θ2 Snell's Law
.
chp34_9.doc
R. D. Field
PHY 2049
Total Internal Reflection Total internal refection occurs when light travels from medium n1 to medium n2 (n1 > n2) if θ 1 is greater than or equal to the critical angle, θ c, where
n sinθc = 2 n1
θ2=90
θc
o
n2 n1
.
Problem: n 2 =1
R θc
n 1 =1.3
Chapter 34
A point source of light is located 10 meters below the surface of a large lake (n=1.3). What is the area (in m2) of the largest circle on the pool's surface through which light coming directly from the source can emerge? (answer: 455)
chp34_10.doc
R. D. Field
PHY 2049
Refraction Examples Problem: A scuba diver 20 meters beneath the smooth surface of a clear lake looks upward and judges the sun θ2 n2=1 o to be 40 from directly overhead. At the same n1=4/3 time, a fisherman is in a θ1 boat directly above the 20 m diver. (a) At what angle from the vertical would the fisherman measure the sun? (answer: 59o) (b) If the fisherman looks downward, at what depth below the surface would he judge the diver to be? (answer: 15 meters)
Chapter 34
chp34_11.doc
R. D. Field
PHY 2049
Spherical Mirrors Vertex and Center of Curvature: The vertex, V, is the point where the principal axis crosses the mirror and the center of curvature is the center of the spherical mirror with radius of curvature R.
Concave Mirror
R = Radius of Curvature C = Center of Curvature F = Focal Point V = Vertex
Light Ray Enters
R R-side
Real and Virtual Sides:
C
Principal Axis
F
V
V-side
Light Ray Exits
The "R" or real side of a spherical mirror is the side of the mirror that the light exits and the other side is the "V" or virtual side. If the center of curvature lies on the R-side then the radius of curvature, R, is taken to be positive and if the center of curvature lies on the V-side then the radius of curvature, R, is taken to be negative. Light Ray Exits R = Radius of Curvature C = Center of Curvature F = Focal Point V = Vertex
Light Ray Enters
R
Principal Axis R-side
Convex Mirror
V
F
C
V-side
Focal Point: A light ray parallel to the principal axis will pass through the focal point, F, where F lies a distance f (focal length) from the vertex of the mirror. For spherical mirrors a good approximation is f = R/2.
Concave and Convex Mirrors: A concave mirror is one where the center of curvature lies on the R-side so the R > 0 and f > 0 and a convex mirror is one where the center of curvature lies on the V-side so that R < 0 and f < 0. concave f>0 convex f 0 then converging lense if the image is not inverted
if f < 0 then diverging lense if the image is inverted
Example (converging lense):
R1 = R R2 = − R
Example (diverging lense): R1 = − R R2 = R
Chapter 35
R >0 2(n − 1) −R f = > d then to a good approximation the path length difference is,
θ
S1 d
θ
S2
L2
∆L = L2 − L1 = d sinθ , and thus
d sinθ
Maximal Constructive Interference: The condition for maximal constructive interference is
sin θ = m
Double Slit
λ m = 0,1,2,K d
P
y
(Bright Fringes - max constructive) Order of the Bright Fringe
Maximal Destructive Interference: The condition for maximal destructive interference is
S1 d S2
θ L
y = L tanθ
1 λ sin θ = m + m = 0,1,2,K 2 d (Dark Fringes - max destructive)
Chapter 36
chp36_3.doc
R. D. Field
PHY 2049
Thin Film Interference Thin film interference occurs when a thin layer of material (thickness T) with index of refraction n2 (the "film" layer) is sandwiched between two other mediums n1 and n2.
Incident Light
1
2
n1 "film"
∆1
T
n2 ∆2 n3
The overall phase shift between Phase Shift Condition Value the reflected waves 1 and 2 is n1 > n2 0 ∆1 given by, n1 < n2 ∆1 λ film/2 ∆ overall = 2T + ∆1 + ∆ 2 , n2 > n3 0 ∆2 where it is assumed that the n2 < n3 ∆2 λ film/2 incident light ray is nearly perpendicular to the surface and the phase shifts ∆1 and ∆2 are given the table.
Maximal Constructive Interference: The condition for maximal constructive interference is
∆ overall = 2T + ∆1 + ∆ 2 = mλ film m = 0,±1,±2,K (max constructive)
where λ film = λ 0/n2, with λ 0 the vacuum wavelength.
Maximal Destructive Interference: The condition for maximal destructive interference is
1 ∆ overall = 2T + ∆1 + ∆ 2 = m + λ film m = 0,±1,±2,K (max destructive) 2
Chapter 36
chp36_4.doc
R. D. Field
PHY 2049
Interference Problems
Double Slit Example: Red light (λ λ = 664 nm) is used with slits separated by d = 1.2x10-4 m. The screen is located a distance from the slits given by L = 2.75 m. Find the distance y on the screen between the central bright fringe and the third-order bright fringe. Answer: y = 0.0456 m
Thin Film Example: A thin film of gasoline floats on a puddle of water. Sunlight falls almost perpendicularly on the film and reflects into your eyes. Although the sunlight is white, since it contains all colors, the film has a yellow hue, because destructive interference has occurred eliminating the color of blue (λ λ 0 = 469 nm) from the reflected light. If ngas = 1.4 and nwater = 1.33, determine the minimum thickness of the film. Answer: Tmin = 168 nm
Chapter 36
chp36_5.doc
R. D. Field
PHY 2049
Diffraction Summary Single Slit
Single Slit-Diffraction:
P
Angular position of the dark fringes:
λ sinθ = m W
y
m = 1,2,K
θ
W
L
Width of the Slit y = L tanθ θ
(Dark Fringes - max destructive)
Round Hole-Diffraction: Angular position of the first dark ring:
Diameter of the Hole
sinθ = 1.22
λ D
(Dark Ring - max destructive)
Diffraction Grating:
Diffraction Grating
Angular position of the bright fringes:
sinθ = m
λ d
m = 0,1,2, K
P
y d
θ L
Slit
(Bright Fringes - max constructive)
y = L tanθ θ
Chapter 37
chp37_1.doc
R. D. Field
PHY 2049
Diffraction Problems Single Slit Example: Light passes through a slit and shines on a flat screen that is located L = 0.4 m away. The width of the slit is W = 4x10-6 m. The distance between the middle of the central bright spot and the first dark fringe is y. Determine the width 2y of the central bright spot when the wavelength of light is λ = 690 nm. Answer: 2y = 0.14 m
Chapter 37
chp37_2.doc
