WES C. ERBSEN
September 2010 - December 2011
Contents i 1 Electrodynamics II 1 1.1 Homework #1 . . . 1 1.2 Homework #2 . . . 14 1.3 Homework #3 . . . 25 1.4 Homework #4 . . . 34 1.5 Homework #5 . . . 40 1.6 Homework #6 . . . 46 1.7 Homework #7 . . . 54 1.8 Homework #8 . . . 63 1.9 Homework #9 . . . 78 2 Quantum Mechanics II 89 2.1 Homework #1 . . . 89 2.2 Homework #2 . . . 101 2.3 Homework #3 . . . 124 2.4 Homework #4 . . . 144 2.5 Homework #5 . . . 161 2.6 Homework #6 . . . 174 2.7 Homework #7 . . . 187 2.8 Homework #9 . . . 198 2.9 Homework #10 . . . 213 Appendix A* . . . 222 Appendix B* . . . 225 3 Statistical Mechanics 227 3.1 Homework #1 . . . 227 3.2 Homework #2 . . . 235 3.3 Homework #3 . . . 247 3.4 Homework #4 . . . 262 3.5 Homework #7 . . . 275 4 Mathematical Methods 283 4.1 Homework #3 . . . 283 4.2 Homework #4 . . . 288 4.3 Homework #5 . . . 294 4.4 Homework #6 . . . 303
4.5 Homework #7 . . . 310 4.6 Homework #8 . . . 315 4.7 Homework #9 . . . 324 4.8 Homework #10 . . . 332 4.9 Homework #11 . . . 342 4.10 Homework #12 . . . 355 5 Departmental Examinations 367 5.1 Quantum Mechanics . . . 367 5.2 Electrodynamics . . . 410 5.3 Modern Physics . . . 449 5.4 Statistical Mechanics . . . 497
Electrodynamics II
1.1
Homework #1
Problem 9.1
A copper ring of radius a is fixed at a distance d (with a d) directly above an identical copper ring. Each ring has a resistance R for circulating currents. An increasing current I = Iot/τ is applied in the lower ring. Neglect
the self-inductance of each ring, and make appropriate approximations. a) Find the dipole moment in the lower ring.
b) Find the magnetic flux through the upper ring.
c) Find the induced EMF and the current in the upper ring. d) Find the induced dipole moment of the upper ring.
e) Show that the force between the rings is∗
F = 12π
4a8I2 ot
c3Rd7τ2 (1.1.1)
Is the force repulsive or attractive?
Solution
a) An expression for the magnetic dipole moment for a circular loop can be realized by applying multipole expansion of the magnetic scalar potential (Φm) and noting that the coefficients of
P`(cos θ) /r`+1are in fact the magnetic dipole moments (from Franklin p. 209). ?
µ`= − 2πIa`+1 c −1/ 2 `+1/ 2 −→ µ1= Iπa2 c (1.1.2)
∗Errata indicates that c3
→ c4
Where µ`=1 represents the magnetic moment of a dipole. The elementary magnetic dipole µ is
found by taking the limit of (1.1.1) as a → 0 keeping Ia2 fixed. We subsequently arrive at
µ=πa
2I ot
cτ ˆz (1.1.3)
b) The magnetic flux is of course given by
ΦB =
I
s
B · dA
While the magnetic field of a magnetic dipole is given by (7.108) on p. 209 as B = −∇ µ · rr3
=3 (µ · ˆr) ˆr − µ
r3 (1.1.4)
The closed surface we choose to integrate over is the upper hemisphere closed on the bottom by a disk, as inspired by Griffiths in Ex. 8.2 on p. 353. ?
The general idea is that since the source of the field lies outside the surface, the total flux through the volume is zero: HvB · dA = 0. We then note that the flux entering through the disk is the same as that exiting through the top of the hemisphere: Φdisk= −Φhemisphere. It i easiest to
calculate the flux through the hemisphere.
We also recognize that since a d, we can use the following approximation: tan θ ≈ a/d −→ sin θ = a/d cos θ ≈ a/d, as per the small angle approximation. We can then proceed to find the flux through the hemisphere:
ΦB= Z θ 0 B · 2πr2sin θ dθ =2π Z θ 0 3 (µ · ˆr) ˆr − µ r3 r2sin θ dθ ˆz =2π r Z θ 0 Iπa2 c ˆz sin θ dθ ˆz =2π 2a2I cr Z θ 0 sin θ dθ ≈2π 2a2I cr Z θ 0 a d d (a/d) ≈2π 2a2I cr a2 d2 (1.1.5)
Since d a, we can approximate r = √d2+ a2 ≈ d2, and if we recall that I = I
ot/τ, (1.1.5) becomes ΦB≈ 2π 2a4I ot cd3τ (1.1.6)
c) The induced EMF in the upper ring is given by E= −1
c dφB
Substituting (1.1.6) into (1.1.7), we have E≈ −1 c d dt 2π2a4I ot cd3τ −→ E ≈ −2π 2a4I o c2d3τ (1.1.8)
And the current in the upper ring (I0) can be found from Ohm’s Law,
I0≈ E
R −→ I0 ≈ − 2π2a4I
o
c2d3τ R (1.1.9)
d) To find the induced dipole moment in the upper ring (µ0), we proceed in the same spirit that led
us to (1.1.3), but this time we use I0 from (1.1.9):
µ0≈ −2π 2a4I o c2d3τ R πa2 c ˆz −→ µ 0≈ −2π3a6Io c3d3τ R ˆz (1.1.10)
e) The force on a dipole µ in a magnetic field B is given by Franklin as (7.144) on p. 216:
F = ∇ (B · µ) (1.1.11)
Assuming that the charges are fixed and that the force of the lower loop on the upper loop is the same as that of the upper loop and the lower loop, we can say that the field is provided by the lower loop, and the opposing magnetic moment is that of the upper ring. Substituting the expression for B and (1.1.10) into (1.1.11), F ≈∇ πa2I ot cd3τ ˆz · −2π 3a6I o c3d3τ R ˆz ≈ − ∇ 2π4a8I2 ot c4d6τ2R (1.1.12) And in our case the gradient reduces to ∇ = (∂/∂r) ˆr where r → d, and (1.1.12) becomes
F ≈ 12π
4a8I2 ot
c4d7τ2R ˆz (1.1.13)
Which is identical to (1.1.1) (with correction). Since (1.1.13) is positive, the force is repulsive .
Problem 9.2
b) Use their mutual inductance to find the EMF induced in the upper ring for the current I = Iot/τ in the lower
ring.
c) Find the mutual interaction energy in this system.
Solution
a) We know from (9.18) on p. 245 of Franklin that
Φ2= cM21I (1.1.14)
Using (1.1.6) from the previous problem, we can solve (1.1.14) to find the mutual inductance:
M21≈ 1 cI 2π2a4I cd3 −→ M21≈ 2π2a4 c2d3 (1.1.15)
b) As before, the induced EMF is given by
E=1 c
dφB
dt (1.1.16)
The magnetic flux can be found from the mutual inductance by substituting (1.1.15) into (1.1.14), and the induced EMF can then be found by substituting the result into (1.1.16):
E≈ 1 c d dt cIot τ 2π2a4 c2d3 −→ E ≈ 2π 2a4I o c2d3τ (1.1.17) which is identical to (1.1.8).
c) The mutual interaction energy is most easily calculated from (5.152) from Jackson on p. 215: ? w = 1 2 N X i=1 LiIi2+ N X i=1 N X j6=i MijIiIj (1.1.18)
Ignoring the self inductance terms in (1.1.18), we find that w = 1
2[M12I1I2+ M21I2I1] (1.1.19) Where the mutual inductances are of course equal. Substituting (1.1.9) and (1.1.15) into (1.1.19),
w ≈ 2π2a4 c2d3 Io t τ −2π 2a4I o c2d3τ R −→ w ≈ −4π 4a8I2 ot c4d6τ2R
Problem 9.3
a) Find the electric and magnetic fields at the surface of the wire.
b) Integrate the Poynting power flux through the surface of a piece of the wire of length L to show that the power through the surface equals I2R.
c) Find the electromagnetic energy and momentum in this piece of wire.
Solution
a) Assuming that the wire can be approximated as infinite, the magnetic field can be found using Amp´ere’s Law: I c B · d` =4π c Ienc −→ B = 2I ca φˆ (1.1.20)
To find the electric field,
V = − Z f
i
E · d`
Where in our case E = |E|ˆz and d` = |d`|ˆz = dz ˆz. We also note that V = EL = IR −→ E = IR
L ˆz (1.1.21)
b) The “Poynting power flux” is most likely referring to the Poynting vector, which is really the “Poynting power flux density.” We must first find S using (9.48) on p. 249
S = c
4π(E × H) (1.1.22)
Where in our case H = B. First, we calculate E × H using (1.1.20) and (1.1.21), E × H = IR L ˆz × 2I ca φˆ =2I 2R caL ˆz × ˆφ = − 2I 2R caL ˆr (1.1.23)
The power loss is defined by P = HSS · dA. It is to our advantage to integrate in cylindrical coordinates, and we recall that in our case ˆr · ˆz = 0. Therefore, we can find the Poynting power flux using (1.1.22) and (1.1.23): P = c 4π I S (E × H) · dA = c 4π −2I 2R caL Z 2π 0 Z L 0 ˆr · aˆr dφdz = − I 2R 2πL(2π) (L) (1.1.24)
It is easy to see that (1.1.24) reduces to
P = I2R
c) The electromagnetic energy density is given by (9.47) on p. 249 as UEM= 1
8π(E · D + B · H) (1.1.25)
Where in our case D = E and H = B. Therefore, (1.1.25) becomes UEM=
1 8π E
2+ B2 (1.1.26)
Where E and B are the values within the conductor. If we assume that the electric field is uniform, then we can use our previous value. To find the magnetic field,
I C B · d` = 4π c πr2I πa2 −→ B = 2I ca2r ˆφ (1.1.27)
Substituting (1.1.21) and (1.1.27) into (1.1.26), UEM= 1 8π " IR L 2 + 2I ca2r 2# (1.1.28) To find the electromagnetic energy, we integrate the electromagnetic energy density (1.1.28) over the entire volume:
w = Z V UEMdτ = 1 8π Z V " IR L 2 + 2I ca2r 2# dτ = 1 8π I2R2 L2 Z r drdφdz + 4I 2 c2a4 Z r3drdφdz = 1 8π " I2R2 L2 Z a 0 r dr Z 2π 0 dφ Z L 0 dz + 4I 2 c2a4 Z a 0 r3dr Z 2π 0 dφ Z L 0 dz # = 1 8π I2R2 L2 a2 2 (2π) (L) + 4I 2 c2a4 a4 4 (2π) (L)
Simplifying this expression leads to:
w = I
2R2a2
8L + I2L
4c2
The electromagnetic momentum stored in the field is given by (9.80) on p. 255 in Franklin: PEM= 1 4πc Z V E × B dτ (1.1.29)
First calculating the cross product, E × B = IR L ˆz × 2I ca2r ˆφ = −2I 2R ca2Lr ˆr (1.1.30)
Substituting (1.1.30) into (1.1.29), PEM= 1 4πc Z V −2I 2R ca2Lr dτ = − I 2R 2πc2a2L Z L 0 Z 2π 0 Z a 0 r drdφdz −→ P EM= 0
Problem 9.4
The upper and lower curves of the hysteresis loop of a hard ferromagnetic are given by
B+=Bo[tanh (H/Ho+ .5) − .1] , (1.1.31a)
B−=Bo[tanh (H/Ho− .5) + .1] , (1.1.31b)
respectively, for −1.5 < H/Ho< +1.5.
a) Find the energy lost from the magnetic field in one cycle.
b) If Bo= 3 kilogauss, Ho= 1 gauss, and the frequency is 60 Hz, find the power loss in watts.
Solution
a) The work done per unit volume in each cycle of a hysteresis loop is given by the total area, w =
I
B · dH (1.1.32)
Where B = (B+− B−). Applying this to (1.1.32) over the specified intervals,
w = Z +1.5 −1.5 (B+− B−) dH = Z +1.5 −1.5 Bo tanh H Ho + .5 − .1 − Bo tanh H Ho − .5 + .1 dH (1.1.33) If we define α = H/Hoand dα = (1/Ho) dH, (1.1.33) becomes
w =Ho Z +1.5 −1.5 [Bo(tanh (α + .5) − .1) − Bo(tanh (α − .5) + .1)] dα =BoHo Z +1.5 −1.5 tanh (α + .5) dα − Z +1.5 −1.5 tanh (α − .5) dα − .2 Z +1.5 −1.5 dα =BoHo ( ln [cosh (α + .5)] +1.5 −1.5 − ln [cosh (α − .5)] +1.5 −1.5 − .2 α +1.5 −1.5 dα )
=BoHo{ln [cosh (2)] − ln [cosh (1)] − ln [cosh (1)] + ln [cosh (2)] − .6} (1.1.34)
Evaluating (1.1.34) numerically leads to
b) To find the energy loss according to the given parameters, w ≈ (1.18244)(1 gauss)(3 · 10 3gauss) 1/60 Hz −→ w ≈ 212.832 · 103erg · s−1 (CGS) w ≈ 212.832 · 10−4Watts (SI)
Problem 9.5
Two point charges, each of charge q, are a distance 2d apart.
a) Find the Maxwell stress tensor [T] on the plane surface midway between the charges.
b) Find the force on either charge by integrating [T] · dS over a plane surface, closing the surface with a large hemisphere of radius R. (Show that the integral over the hemisphere vanishes in the limit R → ∞.)
c) Repeat parts (a) and (b) if the charges have opposite signs.
d) What would the force on either charge be if they were immersed in a simple dielectric of infinite extent?
Solution
a) The Maxwell stress tensor is given in Griffiths as (8.19) on p. 352 as [T] = εo EiEj− 1 2δijE 2 + 1 µo BiBj− 1 2δijB 2 (1.1.35) For the case in question, our charges are stationary so that the second portion of (1.1.35) vanishes. We can then express [T] in matrix form as
[T] = TTxxyx TTxyyy TTxzyz Tzx Tzy Tzz = εo E 2 x−1/2E2 ExEy ExEz ExEy Ey2−1/2E2 EyEz EzEx EyEz Ez2−1/2E2
In order to resolve the individual components of [T], we must find the corresponding electric fields. We note that the charges are located on the x-axis, and the plane between the charges is the yz-plane. Furthermore, we recall that the electric field of an electric dipole is given by
E = 1 4πεo
p
r3 (1.1.36)
And in our case r = x2+ y2+ z21/2
⇒ d2+ y2+ z21/2
. With this in mind, we can now find the electric fields in the x, y, and z directions respectively:
Ex=0 Ey = 1 2πεo ey (d2+ y2+ z2)3/2 ˆy ⇒ 1 2πεo er cos θ (d2+ r2)3/2 ˆr Ez= 1 2πεo ez (d2+ y2+ z2)3/2 ˆz ⇒ 1 2πεo er sin θ (d2+ r2)3/2 ˆr
Where I shamelessly switched to polar coordinates. The total electric field is then E2=E2x+ E2y+ E2z = 1 2πεo er cos θ (d2+ r2)3/2 ˆr !2 + 1 2πεo er sin θ (d2+ r2)3/2 ˆr !2 = e 2πεo 2 r2 (d2+ r2)3 cos 2θ + sin2θ = e 2 4π2ε2 o r2 (d2+ r2)3
We now have all the tools required to calculate the components of [T]: Txx= − e2 8π2ε2 o r2 (d2+ r2)3 Tyy= e 2 4π2ε2 o r2 (d2+ r2)3 cos2θ − 12 Tyz =Tzy = e2 4π2ε2 o r2 (d2+ r2)3sin θ cos θ Tzz = e 2 4π2ε2 o r2 (d2+ r2)3 sin2θ − 12 Txy=Txz= Tyx = Tzx= 0 We now define β = e 2 4π2ε2 o r2 (d2+ r2)3 (SI) =4e2 r2 (d2+ r2)3 (CGS)
And the Maxwell stress tensor becomes
[T] = εo
−β/20 β cos20θ − 1/2 β sin θ cos θ0
0 β sin θ cos θ β sin2θ − 1/2
b) To integrate the Maxwell stress tensor, we take the route suggested in the prompt. We take a hemisphere of radius R with the base coinciding with the plane surface halfway between the charges (the yz-plane). If we let the hemisphere expand to very large values, then the field at the boundary of the hemisphere is seen as a dipole, ∝ R−3, and the surface integral then varies like R−4. If we
take R out to infinity, then this portion of the surface integral vanishes since the components of [T] go to zero. There remains then only the force across the plane boundary in the yz-plane.
To find the force on either charge, we integrate [T] · dS over the interstitial plane. We note that the force is in the x-direction, and the transversal forces are cancelled. Therefore, the only component of [T] that we need to integrate is Txx, as tabulated previously. We also recall that for
the equatorial disk dS = −r drdφ ˆx, and so the force is F =
Z
S
= − 2π Z ∞ 0 Txxr dr ˆx = − 2πεo Z ∞ 0 − e2 8π2ε2 o r2 (d2+ r2)3 ! r dr ˆx = e 2 4πεo Z ∞ 0 r3 (d2+ r2)3 dr ˆx ) partial fractions = e 2 4πεo Z ∞ 0 " r (d2+ r2)2 − d2r (d2+ r2)3 # dr ˆx = e 2 4πεo "Z ∞ 0 r (d2+ r2)2 dr − d 2 Z ∞ 0 r (d2+ r2)3 dr # ˆ x ) u = d2+ r2, du = 2rdr = e 2 4πεo 1 2 Z ∞ 0 1 u2 du − d2 2 Z ∞ 0 1 u3 dr ˆ x = e 2 4πεo 1 2 −(d2+ r1 2) ∞ 0 −d 2 2 −2 (d21+ r2) ∞ 0 ˆ x = e 2 4πεo 1 2 1 d2− d2 2 1 2d4 ˆ x (1.1.37)
From (1.1.37), it is only a short leap to our answer:
F = 1 4πεo e2 (2d)2 xˆ (SI), F = e2 (2d)2 xˆ (CGS)
Which is exactly what we expected; proceeding through Coulomb’s law yields the precisely same result.
c) If the charges are opposite, then the symmetry changes, however the process is very much the same. The electric fields in this case are
Ex= 1 2πεo ed (d2+ r2)3/2 ˆx Ey=Ez= 0
And the total electric field is
E2= Ex2= 1 4π2ε2 o e2d2 (d2+ r2)3
And now to calculate the components of [T], Txx= 1 4π2ε2 o e2d2 (d2+ r2)3 Tyy= − 1 4π2ε2 o e2d2 (d2+ r2)3 Tzz = − 1 4π2ε2 o e2d2 (d2+ r2)3 Txy=Txz= Tyx = Tyz= Tzx= Tzy = 0
And let’s go ahead and define β = 1 4π2ε2 o e2d2 (d2+ r2)3
Then the Maxwell stress tensor is
[T] = εo β0 −β0 00 0 0 −β To find the force on either charge, we follow the same logic as before,
F = Z S [T] d · S = − 2πεo Z ∞ 0 Txxr dr ˆx = − 2πεo Z ∞ 0 1 4π2ε2 o e2d2 (d2+ r2)3 ! r dr ˆx = −e 2d2 2πεo Z ∞ 0 r (d2+ r2)3 dr ˆx ) u = d2+ r2, du = 2rdr = − e 2d2 4πεo Z ∞ 0 1 u3 du ˆx = − e 2d2 4πεo " − 1 4 (d2+ r2)2 ∞ 0 ˆ x = − e 2d2 4πεo 1 4d4 ˆ x (1.1.38)
From (1.1.38) it is easy to see that the force is F = − 1 4πεo e2 (2d)2 xˆ (SI), F = − e2 (2d)2 ˆx (CGS) Which is what we would expect.
d) If the system is immersed into a linear dielectric of permittivity ε, then this will effect our answers only by including a factor of ε in the denominator of our expressions, which weakens the force of attraction or repulsion. vacuum dielectric equal charges F = e 2 (2d)2 xˆ −→ F = 1 ε e2 (2d)2 ˆx opposite charges F = − e 2 (2d)2 ˆx −→ F = − 1 ε e2 (2d)2 ˆx (CGS)
Problem 9.7
A magnetic dipole µ is located at the center of a uniform electric charge distribution of radius R, charge e, and mass m.
a) Find the electromagnetic angular momentum of this configuration.
b) Find the value of the radius R for which the g-factor of this configuration equals 2.
Solution
a) We first recall that the electromagnetic angular momentum is given by (9.113) on p. 260 as LEM= 1 4πc Z V r × (E × B) dτ (1.1.39) Now, we find the electric field by
E = e 4/ 3πr3 4/ 3πR3r2 ˆr = er R3 ˆr (1.1.40)
The magnetic field is given by (7.108) on p. 209 as
B = 3 (µ · ˆr) ˆr − µ
r3 (1.1.41)
We also recall that in our case µ = |µ| ˆz, and therefore (1.1.41) becomes
B = −rµ3 ˆz (1.1.42)
We now use (1.1.40) and (1.1.42) to examine the cross product in (1.1.39): E × B = er R3 ˆr × −rµ3 ˆz = −Reµ3r2(ˆr × ˆz) = − eµ R3r2 − sin θ ˆφ= eµ R3r2sin θ ˆφ (1.1.43)
Now substituting (1.1.43) into (1.1.39), LEM= 1 4πc Z V r × eµ R3r2sin θ ˆφ dτ = e 4πcR3 Z V sin θ r ˆr × ˆφdτ = −4πcRe 3 Z V sin θ r θrˆ 2sin θ drdθdφ = −4πcRe 3 Z R 0 rdr Z π 0 Z 2π 0 sin2θ ˆθdθdφ (1.1.44) We now recall from the back cover of Griffiths that
ˆ
θ= cos θ cos φ ˆx + cos θ sin φ ˆy − sin θ ˆz (1.1.45) Substituting (1.1.45) into (1.1.44), LEM= − µe 4πcR3 Z R 0 rdr Z π 0 Z 2π 0
= −8πcRµe Z π
0
Z 2π 0
[sin2θ cos θ cos φ ˆx
| {z }
I
+ sin2θ cos θ sin φ ˆy
| {z } II − sin3θ ˆz | {z } III ] dθdφ (1.1.46)
Taking the angular integrals of I, II and III separately yields I = Z π 0 sin2θ cos θ dθ Z 2π 0 cos φ dφ ˆ x = 0 (1.1.47a) II = Z π 0 sin2θ cos θ dθ Z 2π 0 sin φ dφ ˆ y = 0 (1.1.47b) III = Z π 0 sin3θ dθ Z 2π 0 dφ ˆz = 2π Z π 0 sin θ dθ − Z π 0 sin θ cos2θ dθ ˆz u = cos θ, du = − sin θ =2π Z π 0 sin θ dθ + Z −1 1 u2du ˆ z =2π " − cos θ π 0 + u3 3 −1 1 # ˆ z =2π 2 − 2 3 ˆ z =8π 3 ˆz (1.1.47c)
Substituting (1.1.47a)-(1.1.47c) into (1.1.46), LEM= − µe 8πcR 0 + 0 −8π 3 ˆz −→ LEM= µe 3cR ˆz (1.1.48) b) We recall from (7.134) on p. 214 that µ = −gµBJ, and also that LEM= ~J. Combining these, as
well as implementing (1.1.48), yields 3cRLEM e = gµBLEM ~ −→ 3cR e = gµB ~ −→ R = gµBe 3c~ (1.1.49)
Recall that the Bohr Magneton is given by
µB = e~
2me (1.1.50)
The appropriate values we require are
e =1.60217733 · 10−19C ~ =1.05457162 · 10−34Kg · m2
· s−1 c =2.99792458 · 108m · s−1
me=9.10938215 · 10−31Kg
Now substituting (1.1.50) into (1.1.49) as well as the appropriate constants, R = (2) 1.60217733 · 10
−19C2
Problem 9.8
a) Calculate e2/~c in Gaussian units to verify Eq. (9.121).
b) Calculate e2/(4π
o~c) in SI units.
Solution
a) The values needed in this calculation are
e =4.80320680 · 10−10cm3/2
· g1/2
· s−1 ~ =1.05457162 · 10−27erg · s
c =2.99792458 · 1010cm · s−1 And so, we can now calculate α:
e2 ~c = 4.80320680 · 10−10cm3/ 2g1/2 s−1 2 1.05457162 · 10−27erg s · 2.99792458 · 1010cm s−1 −→ e2 ~c = 0.00729736 ⇒ 1 137.036 b) The necessary values to make the calculation in SI units are
e =1.60217733 · 10−19C εo=8.85418781 · 10−12F · m−1 ~ =1.05457162 · 10−34J · s c =2.99792458 · 108m · s−1 And now, e2 4πo~c = 1.60217733 · 10 −19C2 4π (8.85418781 · 10−12 F · m−1) (1.05457162 · 10−34J) (s · 2.99792458 · 108m · s−1) −→ e 2 4πo~c = 0.00729736 ⇒ 1 137.036
1.2
Homework #2
Problem 10.1
The intensity of sunlight at the Earth’s surface is 12 · 105ergs/(cm2· sec).
a) Find the electric field of this radiation at the Earth’s surface. Express the answer in stavolts/cm and in volts/m (1 statvolt = 300 volts). [Answer: 960 V/m]
b) Find the radiation pressure (in dynes/cm2) if the sunlight is 100% reflected at normal incidence. Compare
this to atmospheric pressure. [Answer: 2 × 10−5 dynes/cm2]
c) Find the radiation pressure if
i) the sunlight is absorbed (no reflection).
ii) the sunlight is specularly reflected from white sand. (The reflected radiation has the same intensity at any angle.)
d) What would the intensity, electric field, and radiation pressure be at the surface of the sun?
Solution
a) The intensity is called the time averaged Poynting vector, and is given by (10.27) of Franklin: ? S = cˆk
8π rε
µ|E0|
2 (1.2.1)
In SI units, the intensity is given by (9.61) in Griffiths on p. 381: ? S = 1
2cε0|E0|
2ˆk (1.2.2)
Since we are given the intensity, it is a trivial task to find the electric field. We solve (1.2.2) for Eo;
S = 1 2cε0|Eo| 2ˆ k −→ |E0| = s 2S ε0c (1.2.3)
We recall that 1 erg = 10−7 J and ε
0≈ 8.8542 A2· s4/(Kg · m3), (1.2.3) becomes |E0| ≈ s (2)(1200 Kg · m2/s) 8.8542 A2· s4/(Kg · m3) · 3 · 108m/s −→ |E0| ≈ 950.54 V/m |E0| ≈ 3.1685 · 10−2statvolt/cm (1.2.4)
b) The radiation pressure is given by (10.33) in Franklin as prad=
2 c
õ S (1.2.5)
If we recall that 1 erg = 1 dyne· cm, then we can solve (1.2.5) directly: prad=(2)(12 · 10
5dyne/(cm · sec))
3 · 1010cm/sec −→ prad= 8 · 10−5dynes/cm
2 (1.2.6)
The atmospheric pressure is patm≈ 106dynes/cm2, which means that the radiation pressure
mul-tiplied by a factor of 1.25 · 1010is approximately equal to the atmospheric pressure at 1 atm.
c) The radiation pressure for the following unique cases is as follows:
i) If the radiation is completely absorbed, then the radiation pressure would be just half that as if it were completely reflected from (1.2.6):
prad=
1 c
√
ii) If the radiation is specularly reflected (evenly across all angles), then certainly the radiation pressure will be less than if it were completely reflected. In fact, for each angle, we are only interested in the 0oprojection, going back towards the sun. But wait! Things aren’t quite this
simple – we must integrate over all possible angles. Doing this yields the force: F = Z π 0 Z π/ 2 0 2S c cos θ sin θ dθdφ =2S c 2π Z π /2 0 cos θ sin θ dθ =2πS c (1.2.7)
And to find the force we must divide by the surface area (note that I have neglected the arbitrary radial term, as it vanishes anyway):
prad= F A = 2πS c · 1 2π −→ prad= 4 · 10 −5dynes/cm2
d) To find the intensity at the surface of the sun, we assume that the total radiation flux at an altitude of Earth is equal to that at the surface of the sun. We know the intensity on the surface of the Earth, and also the distance of Earth from the sun (R ≈ 1.5·1011m), so the total flux (as calculated
from Earth’s orbit) is
Stot = 4πR2SEarth
And we also know that the radius of the sun is r ≈ 7.0 · 108m, so the intensity at the surface can
be found as Ssun= R2 r2SEarth−→ Ssun≈ 5.5 · 10 10ergs/(cm2· sec)
Problem 10.2
A beam of elliptically polarized light, propagating in the z-direction, passes through a polarizer in the xy-plane. The maximum transmitted intensity is 9Iowhen the polarizer is set at 30oto the x-axis. The minimum transmitted
intensity is Io when the polarizer is set at 120oto the x-axis.
a) Find r, α, E+, and E−in the circular basis.
b) Find Ex and Ey in the plane basis.
c) What would the transmitted intensities be if the polarizer were set along the x-axis, and then along the y-axis?
Solution
I am going to solve this problem using Jones’ Calculus, starting in the linear basis. We note that the electric field of an (arbitrary) polarized wave can be expressed as
E(z, t) = |Eox|eiδxˆx + |Eoy|eiδyˆyei(k·z−ωt)
= Eoxˆx + Eoyeiδˆy
ei(k·z−ωt) (1.2.8) And now (1.2.8) can be expanded and put into matrix form, ditching the explicit time dependence:
E(z, t) = Eox Eoyeiδ = Eox Eoy(cos δ + i sin δ) (1.2.9) We also recall that the arbitrary Jones’ Matrix for a linear polarizer at some angle θ is given by
M =
cos2θ sin θ cos θ
sin θ cos θ sin2θ
We can apply the Jones’ Matrix to our incident elliptical wave given by (1.2.8), the result gives the exiting field.∗ Starting with the maximum field when the polarizer is oriented at θ = 30o,
|E300 |2= 3/4 √3 /4 √ 3 /4 1/4 Eox Eoy(cos δ + i sin δ) =1 4 3E2ox+ Eoy2 + 2 √ 3 EoxEoycos δ (1.2.10) And similarly if the polarizer is oriented at 120o:
|E0120|2= 1/4 −√3 /4 −√3 /4 3/4 Eox Eoy(cos δ + i sin δ) =1 4 Eox2 + 3E2oy− 2 √ 3 EoxEoycos δ (1.2.11) The intensity, of course can be rewritten to allow us to rewrite (1.2.10) and (1.2.11):
9Io= c 8π 1 4 3E2 ox+ E2oy+ 2 √ 3 EoxEoycos δ (1.2.12a) Io= c 8π 1 4 Eox2 + 3E 2 oy− 2 √ 3 EoxEoycos δ (1.2.12b) We introduce another equation, the derivation of which will not be shown here:
tan(2α) = 2EoxEoycos δ E2
ox− Eoy2
(1.2.13)
Solving (1.2.12a)-(1.2.13) allows us to solve for our unknowns. We also note that α = 2 · 30o, which is deducible from the fact that the maximum intensity is observed at 30o. The tabulated results follow in (a) and (b).
a) The phase angle is found to be δ = 40.89o , while E
+ and E− may be found from the results of
part (b):
E+= Ex+ iEy=
r Io
c (13.26 cos(ωt) + i8.86 cos (ωt − 40.89
o))
E−= Ex− iEy=
r Io
c (13.26 cos(ωt) − i8.86 cos (ωt − 40.89
o))
b) We found Eoxand Eoy directly after the prompt. The values for these are:
Eox= 13.26 r Io c , and Eoy= 8.86 r Io c From which we may calculate Exand Ey:
Ex= 13.26 r Io c cos(ωt), and Ey= 8.86 r Io c cos (ωt − 40.89 o)
c) If the polarizer is placed first precisely along the x-axis, then the intensity is: I0=
c
8π(13.26)
2
−→ I0= 7.53 Io
And if perfectly along the y-axis, I90= c
8π(8.86)
2
−→ I90= 3.12 Io
Problem 10.3
Consider a beam of partially plane polarized light with the same maximum and minimum intensities as in the previous problem.
a) What is the percent polarization of this light?
b) What would the transmitted intensities be if the polarizer were set along the x-axis, and then along the y-axis?
c) How could you tell whether an incident light beam were elliptically polarized or partially plane polarized?
Solution
a) The polarization Π can be found as follows
Π = Smax− Smin Smax+ Smin
(1.2.14) We know from the previous problem that the maximum intensity is Smax= 9Io while the minimum
intensity is Smin= Io, so (1.2.14) becomes
Π = 9Io− Io
9Io+ Io −→ Π =
8
b) If we imagine the partially-plane polarized light incident on the polarizer to be composed of orthog-onal linear components along the x and y-directions, then the total intensity is
S0 = S2x+ S2y+ 2
p
SxSy cos θ
Where θ is the angle between Sxand Sy, which is most definitely 90o. We can find Smax and Smin
by taking these limits: Smax= q S2 x+ Sy2 2 , and Smin= q S2 x− Sy2 2
And now using (1.2.14) we can say that Smax
Smin
=1 + Π
1 − Π (1.2.15)
From Malus’ Law, we can now say that
S0=1 − Π 1 + Πcos
2(θ)S max
With the help of (1.2.15). We can now say that the transmitted intensity when the polarizer is placed perfectly along the x and y-axis, respectively are
Sx0 = 0.75 Io, and Sy0 = 0.25 Io
c) To tell if incident radiation were elliptically polarized or partial plane polarized, I would place a linear polarizer in front of the incident beam followed by an energy meter. If the intensity changes monotonically with polarization angle then the radiation is said to be more linearly polarized than elliptically.
Problem 10.5
A horizontal light ray is incident on a 60o-60o-60o glass prism (n = 1.5) that is resting on a table. At what
angle with the horizontal does the ray leave the prism? (Assume that the ray does not strike the bottom of the prism before exiting.)
Solution
The initial ray is coming in completely horizontal, however the incident angle to the medium is propor-tional to the dimensions of the prism. In our case, the normal component of the front face of the prism makes and angle that is precisely 30o from the incident ray. We now apply Snell’s Law,
nasin θa= ngsin θg−→ θg= sin−1
sin (30)
1.5
≈ 19.47o (1.2.16) Now the ray is within the prism, and the next interface is that of the opposite surface of the prism. The normal component of the exiting face is 60oin the positive direction to the first, so that the angle between
the incident beam calculated in (1.2.16) and this normal angle is 60o− 19.47o= 40.53o. Applying Snell’s
ngsin θg= nasin θa−→ θa = sin−1[1.5 sin (40.53)] −→ θa≈ 77.10o
Problem 10.6
For the prism in the preceding problem, what is the smallest angle of incidence for which a light ray will pass directly through the prism without total internal reflection?
Solution
Total internal reflection occurs at θc= sin−1(1/1.5) ≈ 41.81o. This is, of course, relative to the normal
component of the interior of the exiting interface. The simplest way to find the incident angle is to work backwards from what we did in the preceding problem.
We then only must apply Snell’s Law once, for the first interface. The critical angle calculated for the outer face is oriented to the normal of the first face as 60o− 41.81o = 18.19o. We can now apply
Snell’s Law:
ngsin θg= nasin θa−→ θa= sin−1[1.5 sin (18.19o)] −→ θa≈ 27.92o
Problem 10.7
You are standing in front of a rectangular fish tank filled with water (n = 1.33).
a) Show that you can always see through the back of the tank without total internal reflection as you look through the front face.
b) Show that if you look through the front face toward the right side of the tank, there is a maximum angle of incidence for which there is total internal reflection from the right face. What is this angle?
c) What is the minimum index of refraction of the liquid in the tank such there that would always be total internal reflection from the right face?
d) Show that the fact that there is a thickness of glass (n = 1.5) between the water and the air does not affect this problem.
Solution
a) To show that you can always see through the back of the fish tank when viewed through the front, it is sufficient to show that when viewing the fish tank at the most extreme angle, that the exiting angle is less than the critical angle. In our case, the critical angle for glass/air interface is θc= sin−1(1/1.5) ≈ 41.81o.
There are 5 distinct zones in the problem with 4 interfaces. Starting from the front of the fish tank, the air is region 1, then the glass is region 2 and so on. The critical angle refers to the glass/air interface between regions 4 and 5, so to satisfy the requirement that light entering from the front of the fish tank (region 1) will always exit through the back (region 2) we must show that θ4< θc.
We imagine the incident angle to be as wide as possible, say 89.99o, which corresponds to θ 1:
n1sin θ1 =n2 sin θ2−→ θ2= sin−1
1 1.5sin (89.99 o) ≈ 41.81o n2sin θ2 =n3 sin θ3−→ θ3= sin−1
1.5 1.33sin (41.81 o) ≈ 48.75o n3sin θ3 =n4 sin θ4−→ θ4= sin−1
1.33 1.5 sin (48.75 o) ≈ 41.80o (1.2.17) From (1.2.17), we can see that θ4 < θc (41.80o< 41.81o) so that total internal reflection is never
achieved, and you can always see through the back of the fish tank .
b) The process here is very much the same as in the preceding problem, except that we need to work backwards. The critical angle has not changed, θ4 = θc≈ 41.81o. So, working backwards from θ4,
we have
n3sin θ03=n4sin θ4 −→ θ30 = sin−1
1.5 1.33sin (41.81 o) ≈ 48.75o n2sin θ2=n3sin θ3 −→ θ2 = sin−1
1.33 1.5 sin (90 o − 48.75o) ≈ 35.77o
n1sin θ1=n2sin θ2 −→ θ1 = sin−1[1.5 sin (35.77o)] ≈ 61.27o (1.2.18)
Hence from (1.2.19) we have shown that the maximum angle of incidence to allow for total internal reflection on the right face of the fish tank is θ1 = 61.27o.
c) To find the minimum index of refraction of the liquid such that light incident to the front of the fish tank will always satisfy total internal reflection, we once again precede using Snell’s Law. The only difference is that this time we will leave the index of refraction for the liquid to be arbitrary.
n1sin θ1=n2sin θ2−→ θ2= sin−1
1
1.5sin (θ1)
n2sin θ2=n3sin θ3−→ θ3= sin−1
1.5 n sin sin−1 1 1.5sin (θ1) = sin−1 1 nsin (θ1)
n3sin θ3=n4sin θ4−→ sin (θ4) =
n 1.5sin 90o− sin−1 1 nsin (θ1) (1.2.19) It is possible to solve (1.2.19) for n, using the appropriate trig substitutions:
1.5 sin (θ4) =n sin 90o− sin−1 1 nsin (θ1) =n sin (90o) cos sin−1 1 nsin (θ1) − cos (90o) sin sin−1 1 nsin (θ1) =n 1 −sin 2(θ 1) n2 1/ 2 =n2− sin2(θ 1) 1/ 2 (1.2.20)
If we recall that θ4= θcand imagine our incident angle to the most extreme position of θ1= 89.99o,
(1.2.20) can be solved to find the minimum index of refraction: n =h(1.5)2sin2(41.81o) + sin2(89.99o)i
1
/2
−→ n ≈ 1.41 (1.2.21) d) The fact that the width of the glass is negligible can be shown by recomputing the results of parts (a) and (b) with assuming that there is no glass media. For part (a), where we are asked to show that total internal reflection is never satisfied if viewing through the front of the tank, we start with the critical angle at the second interface, which is θc= sin−1(1/1.33) ≈ 48.75o, and applying Snell’s
Law:
n1sin θ1= n2sin θc−→ θ1= sin−1[1.33 sin (48.75o)] = 90o
Since you can never see through the fish tank at exactly 90o, we have shown that the critical
condition cannot be satisfied, as previously shown in (1.2.17).
For part (b), we can similarly work backwards from the critical angle:
n1sin θ1= n2sin θc−→ θ1= sin−1[1.33 sin (90o− 48.75o)] ≈ 61.27o (1.2.22)
As can be seen, (1.2.22) matches (1.2.19), and therefore from these two instances we are forced to admit that the width of the glass is irrelevant .
Problem 10.9
a) Solve Eqs. (10.117)-(10.120), to get the transmitted and reflected electric fields given by Eqs. (10.121) and (10.122).
b) Use these fields to get the transmission and reflection coefficients for the coated surface.
c) For an original air (n = 1) to glass (n = 1.5) interface, find the index of refraction and thickness of a coating that would have no reflection at the incident wavelength of 5, 000˚A.
Solution
a) Equations (10.117)-(10.120) read E1+ E01=E2+ E20eik2d (1.2.23a) n1(E1− E10) =n2(E2− E02eik2d) (1.2.23b) E2eik2d+ E02=E3 (1.2.23c) n2(E2eik2d− E20) =n3E3 (1.2.23d)We start with (1.2.23a) and rearrange it:
We now substitute (1.2.24) into (1.2.23b): n1 E1− E2+ E20eik2d− E1 =n2 E2− E02eik2d n1 E1− E2− E20eik2d+ E2 =n2 E2− E02eik2d 2n1E1− n1E2− n1E02eik2d =n2E2− E20n2eik2d n2E20eik2d− n1E02eik2d =n2E2+ n1E2− 2n1E1 −→ E0 2= e−ik2d E2(n2+ n1) − 2n1E1 n2− n1 (1.2.25) We now take (1.2.25) and substitute it into (1.2.23c):
E2eik2d+ e−ik2d E2(n2+ n1) − 2n1E1 n2− n1 = E3 E3(n2− n1) =E2(n2− n1)eik2d+ E2(n2+ n1)e−ik2d− 2n1E1e−ik2d 2E1n1e−ik2d+ E3(n2− n1) =E2(n2− n1)eik2d+ E2(n2+ n1)e−ik2d =E2 n2eik2d− n1eik2d+ n2e−ik2d+ n1e−ik2d =E2n2 eik2d+ e−ik2d− n1 eik2d− e−ik2d
=E2[2n2cos(k2d) − 2in1sin(k2d)]
=2E2[n2cos(k2d) − in1sin(k2d)]
−→ E2=2E1n1e
−ik2d+ E
3(n2− n1)
2 [n2cos(k2d) − in1sin(k2d)]
(1.2.26) We perform a similar exercise by substituting (1.2.25) into (1.2.23d):
n3E3=n2 E2eik2d− e−ik2d E2(n2+ n1) − 2n1E1 n2− n1 n3E3=n2E2eik2d−E2(n2+ n1)n2e −ik2d+ 2n 1n2E1e−ik2d n2− n1 n3(n2− n1)E3=n2(n2− n1)E2eik2d− n2E2(n2+ n1)e−ik2d+ 2n1n2E1e−ik2d
n3(n2− n1)E3− 2n1n2E1e−ik2d=n2E2n2eik2d− n1eik2d− n2e−ik2d− n1e−ik2d
=n2E2 n2 eik2d− e−ik2d − n1 eik2d− e−ik2d =n2E2[2in2sin(k2d) − 2n1cos(k2)d)]
−→ E2=
n3(n2− n1)E3− 2n1n2E1e−ik2d
2n2[in2sin(k2d) − n1cos(k2d)]
(1.2.27) We can now set (1.2.26) and (1.2.27) equal to one another to eliminate E2:
2E1n1e−ik2d+ E3(n2− n1)
2 [n2cos(k2d) − in1sin(k2d)]
=n3(n2− n1)E3− 2n1n2E1e−ik
2d
2n2[in2sin(k2d) − n1cos(k2d)]
(1.2.28) If we let α = sin k2d and β = cos k2d, then (1.2.28) becomes
2E1n1e−ik2d+ E3(n2− n1) n2β − in1α = n3(n2− n1)E3− 2n1n2E1e−ik2d in2 2α − n1n2β 2E1n1e−ik2d n2β − in1α +2n1n2E1e−ik 2d in2 2α − n1n2β | {z } I =n3(n2− n1)E3 in2 2α − n1n2β − E3(n2− n1) n2β − in1α | {z } II
Where I have separated our expression for simplicity. The components are: I =2E1n1e−ik 2d n2β − in1α · n2β + in1α n2β + in1α +2n1n2E1e−ik 2d in2 2α − n1n2β · in2 2α + n1n2β in2 2α + n1n2β =2E1n1e−ik 2d(n 2β + in1α) n2 2β2+ n21α2 − 2n1n2E1e−ik2d in22α + n1n2β n2 1n22β2+ n42α2 =2E1n1(n2− n1)E1(iα + β) (n2α + in1β)(n1α + in2β) = 2E1n1(n2− n1)E1(i sin(k2d) + cos(k2d)) (n2sin(k2d) + in1cos(k2d))(n1sin(k2d) + in2cos(k2d))
= 4in1(n1− n2)E1
2in1n2cos(2k2d) + (n21+ n22) sin(2kdd)
(1.2.29) And now II =n3(n2− n1)E3 in2 2α − n1n2β · in2 2α + n1n2β in2 2α + n1n2β − E3(n2− n1) n2β − in1α · n2β + in1α n2β + in1α = −n3(n2− n1)E3 in 2 2α + n1n2β n2 1n22β2+ n42α2 − E3(n2− n1) (n2β + in1α) n2 2β2+ n21α2 = − E3(n2− n1) n3 n2(n1β − in2α)+ in1n2αβ n2 1α2+ n22β2 = − E3(n2− n1) n 3 n2(n1cos(k2d) − in2sin(k2d)) + in1n2sin(k2d) cos(k2d) n2 1sin(k2d)2+ n22cos(k2d)2 (1.2.30) Setting (1.2.29) equal to (1.2.30) and shifting to one side, we see that
(n1− n2) E3(n22+ n1n3) sin(k2d) + iE3n2(n1+ n3) cos(k2d) − 2iE1n1n2
n2(2in1n2cos(2k2d) + (n21+ n22) sin(2k2d)) = 0 (1.2.31)
Solving (1.2.31) for E3yields
E3= 2n1n2
n2(n1+ n3) cos(k2d) − i(n22+ n1n3) sin(k2d)
E1
Using the results from part (b), it is also easily seen that E10 =
n
2(n1− n3) cos(k2d) − i(n1n3− n22)2sin2(k2d)
n2(n1+ n3) cos(k2d) − i(n1n3+ n22)2sin2(k2d)
E1
b) The transmission coefficient can be found by T = ˆn · S2 ˆ n · S1 =n3 n1 EE31 2 = n3 n1 n2(n1+ n3) cos(k2d) − i(n2n1n2 2 2+ n1n3) sin(k2d) 2 (1.2.32) From (1.2.32) it can be seen that the transmission coefficient is finally given by
T = 4n1n3n 2 2 n2 2(n1+ n3)2cos2(k2d) + (n1n3+ n22)2sin2(k2d) (1.2.33) Which matches (10.123). The reflection coefficient may be calculated via similar means, however the easiest way is to recognize that R + T = 1, so that from (1.2.33), we calculate
R =n 2 2(n1+ n3)2cos2(k2d) + (n1n3+ n22)2sin2(k2d) n2 2(n1+ n3)2cos2(k2d) + (n1n3+ n22)2sin2(k2d)− 4n1n3n22 n2 2(n1+ n3)2cos2(k2d) + (n1n3+ n22)2sin2(k2d) =n 2 2(n21+ n23+ 2n1n3) cos2(k2d) + (n21n23+ n42+ 2n1n3n22) sin2(k2d) − 4n1n3n22 n2 2(n1+ n3)2cos2(k2d) + (n1n3+ n22)2sin2(k2d) =n 2
2(n21+ n23+ 2n1n3) cos2(k2d) + (n21n23+ n42+ 2n1n3n22) sin2(k2d) − 4n1n3n22(sin2(k2d) + cos2(k2d))
n2
2(n1+ n3)2cos2(k2d) + (n1n3+ n22)2sin2(k2d)
Combining terms yields R = n 2 2(n1− n3)2cos2(k2d) + (n1n3− n22)2sin2(k2d) n2 2(n1+ n3)2cos2(k2d) + (n1n3+ n22)2sin 2(k 2d)
c) To find the thickness of the coating, we use (10.127), which reads d = λ1
4 r
n1
n3
Applying the provided values,
d =5000 · 10−10m 4 r 1 1.5 −→ d ≈ 1.02 · 10 −7m
And by looking at our equation for the reflection coefficient, we can see that it will be zero if n2=√n1n3, so the index of refraction of the coating is n2≈ 1.22 .
1.3
Homework #3
Problem 11.1
A plane electromagnetic wave is incident at an angle θ from vacuum onto the flat surface of a perfect conductor. a) Use conservation of momentum to find the radiation pressure on the surface of the conductor.
b) For a wave that is polarized perpendicular to the plane of incidence, find the radiation pressure by calculating the magnetic force on the surface current.
c) For a wave that is polarized parallel to the plane of incidence, find the radiation pressure by calculating the magnetic and electric forces on the surface and surface charge.
Solution
a) The equation for radiation pressure was derived using conservation of momentum by Franklin as (10.33):
prad=
4π|E0|
This is for the case that the incident light is completely reflected (no absorption) and is normal to the reflecting surface. Since the radiation is at an angle θ, we must take the normal component of (1.3.1) in order to discount the radiation not contributing momentum pressure. Furthermore, we must also recognize that we must only recognize the component of the actual momentum normal to the conductor (since we are recognizing both incoming and outgoing rays, the other component cancels). Therefore,
prad(θ) =
4π|E0|
2
cos θ · cos θ −→ prad=
1 4π|E0|
2cos2θ (1.3.2)
Since we are coming from vacuum, ⇒ 0⇒ 1 in CGS units.
b) &c) It is clear from (1.3.2) that the radiation pressure is independent of polarization, so the result will once again be:
prad= 1
4π|E0|
2cos2θ
Problem 11.2
Use the physical properties of copper to estimate the frequency and wave length for which its conductivity develops an appreciable (∼ 10%) imaginary part. Assume two conduction electrons per atom.
Solution
We are primarily motivated by the availability of (11.62) in Franklin: σ(ω) = zcN e
2
m (γ − iω) (1.3.3)
We can separate (1.3.3) into it’s real and imaginary parts, Re {σ(ω)} =zcN e 2 m γ γ2+ ω2 (1.3.4a) Im {σ(ω)} =zcN e 2 m ω γ2+ ω2 (1.3.4b)
We are told that zc is the number of conduction of electrons, which in the case of Copper is just 2.
Furthermore, the combined quantity zcN is the number of free electrons per unit volume, while γ is the
damping constant. According to Jackson on p. 312, the relevant quantities can be given numerically as: ?
N ' 8 · 1028atoms/m3 σ ' 5.9 · 107(Ω · m)−1
The damping constant can now be calculated using (1.3.3): γ ' zcN e 2 mσ '(2)(8 · 10 28atoms/m2 )(1.602 · 10−19 C)2 (9.109 · 10−31kg)(5.9 · 107(Ω · m)−1) ' 7.641 · 1013Hz (1.3.5)
We are interested at the frequencies at which the imaginary part of (1.3.3) develops an imaginary part that is equal to ∼ 10% of the real part:
Im {σ(ω)} = (0.10)Re {σ(ω)} −→ ω = (0.10)γ (1.3.6) Where I used the real and imaginary components of (1.3.3) as shown in (1.3.4a) and (1.3.4b). According to (1.3.6), the approximate frequency where this is possible is:
ω ' 7.641 · 1012Hz (1.3.7)
If we recall that λ = 2πc/ω, and using (1.3.7) we can find the corresponding wavelength: λ = 2π(3 · 10
8m/s)
(7.641 · 1012Hz)−→ λ ' 246.7 µm
Problem 11.3
a) Find the real and imaginary parts of , as given in (11.52).
b) Show that is analytic in the upper half of the complex ω plane. In finding any particular pole position, the contributions from other modes can be ignored since they are analytic in that region.
Solution
a) Equation (11.52) reads
(ω) = 1 + 4πχ = 1 + 4πN Σiziηi 1 − 4π3 Σiziηi
(1.3.8) And we also recall that the molecular polarizability is from (11.52) in Franklin
ηi=
e2
m [ω2
i − ω2− iωγi] (1.3.9)
We also note that the frequency-dependent susceptibility takes the form χ = N Σiziηi 1 − 4π 3 Σiziηi = N e 2 m 1 ω2 i − ω2− iωγi2 (1.3.10)
Then the real and imaginary parts of from (1.3.8) are Re {(ω)} = 1 +4πN e 2 m ω2 i − ω (ω2 i − ω2)2+ ω2γ2 Im {(ω)} =4πN e 2 m ωγ (ω2 i − ω2)2+ ω2γ2
b) To show that is analytic in the upper half of the complex plane, we first substitute (1.3.10) into (1.3.8): (ω) = 1 +4πN e 2 m 1 ω2 i − ω2− iωγi2 (1.3.11) Taking the complex integral of (1.3.11) leads to
Z ∞ −∞ (ω) dω = Z ∞ −∞ 1 + 4πN e 2 m 1 ω2 i − ω2− iωγi2 dω =4πN e 2 m Z ∞ −∞ 1 ω2 i − ω2− iωγ2i dω (1.3.12)
It is clear that (1.3.12) has poles in the lower half plane at ω1,2= − iγi 2 ± ω2i − γ2 i 4
Since there are no poles at the boundary or in the upper half plane, and if we assume that the kernel is finite, then we can say that (ω) is analytic in the upper half plane .
Problem 11.4
Find the Fourier transform of the function in (11.74) to verify (11.75). (Hint : Complete the square in the exponent in doing the integral.)
Solution
We first recall that the functional forms of the integral Fourier transform is f(x, 0) =
Z ∞ −∞
f(ω, 0)eikx dk (1.3.13a)
A(k, 0) = 1 2π
Z ∞ −∞
f(x, 0)e−ikxdx (1.3.13b) And the wave packet function (11.74) is given by
f(x, 0) = e−x2/2L2eik0x (1.3.14)
A(k, 0) = 1 2π Z ∞ −∞ exp − x 2 2L2 · exp [ik0x] exp [−ikx] dx =1 2π Z ∞ −∞ exp − x 2 2L2 · exp [i (k0− k) x] dx =1 2π Z ∞ −∞ exp − x 2 2L2 − iαx dx (let α = k − k0) (1.3.15)
At this point, per the suggestion in the prompt, we wish to complete the square in the exponent in the integrand in (1.3.15): − x 2 2L2 − iαx = − 1 2L2 x 2+ 2iL2αx ⇒ −2L12 h x + iL2α2+ L2α2i (1.3.16) We now substitute (1.3.16) into (1.3.15),
A(k, 0) = 1 2π Z ∞ −∞ exp − 1 2L2 h x + iL2α2+ L2α2idx = 1 2πexp −L 2α2 2 Z ∞ −∞ exp −2L12 h x + iL2α2idx (1.3.17) We wish to employ u0du0 substitution (where the primes are present to avoid confusion later on), where
u0=√1
2L2 x + iL
2α, and du0=√1
2L2 dx
With this substitution, (1.3.17) becomes A(k, 0) = 1 2πexp −L 2α2 2 √ 2L2 Z ∞ −∞ e−u02du0 =√L 2 πexp −L 2α2 2 Z ∞ −∞ e−u02du0 (1.3.18) The integral in (1.3.18) is just the standard Gaussian integral, whose solution is simply √π , and in substituting our value for α, (1.3.18) becomes
A(k, 0) = √L 2π exp " −L 2(k − k 0)2 2 # (1.3.19)
We can see that (1.3.19) is identical to (1.3.13b), which is also the same as (11.75), which is what we were asked to show.
Problem 11.5
A wave packet is given by
and f(x, 0) is zero elsewhere. a) Plot |f(x, 0)|2.
b) Find the Fourier transform, A(k), of f(x, 0). Plot |A|2 for m = 1 and m = 2.
c) Define a reasonable ∆x and ∆k for this wave packet. Calculate ∆x, ∆k, and ∆x∆k.
Solution
We first note that due to boundary conditions, we can rewrite (1.3.20) as f(x, 0) = i sin mπx
L
(1.3.21) a) The item to be plotted is (please see last page for printout of plots):
|f(x, 0)|2= sin2mπx L
(1.3.22) b) The Fourier transform can be found the same way as we did in the last problem:
A(k, 0) = i 2π Z L 0 sin mπx L e−ikxdx −→ A(k, 0) = iπL2 e
−ikLmπ cos(mπ) + ikL sin(mπ) − mπeikL
k2L2− m2π2 (1.3.23)
c) A reasonable value for ∆x is L, and likewise for ∆k is 1/L, so that ∆x∆k = 1: ∆x = L, ∆k = 1
L, ∆x∆k = 1
Problem 11.6
a) Derive (11.81) and (11.82).
b) The spectral line in the decay of an excited state of sodium has a wave length (λ = 5, 893 ˚A), and a lifetime of 2 · 10−8 seconds. Calculate its natural line width.
Solution
a) In order to derive (11.81), we must take the Fourier Transform of (11.80), which reads E(x, t) = ( E0exp [i (k0x − ωt)] · exp x−ct 2cτ , x < ct, 0 x > ct
E(x, 0) = E0exp i k0− i 2cτ x (1.3.24) We take the Fourier Transform of (1.3.24) by substituting it into (1.3.13b),
E(k) =E0 2π Z ∞ −∞ exp i k0− i 2cτ x · exp [−ikx] dx =E0 2π Z ∞ −∞ exp i (k0− k) + 1 2cτ x dx =E0 2π 1 i (k0− k) + 1/(2cτ )exp i (k0− k) + 1 2cτ x ∞ −∞ −→ E(k) = E0 2π [i(k0− k) + 1/(2cτ )] (1.3.25)
It is clear that (1.3.25) is identical to (11.81). In order to derive (11.82), we simply take the modulus squared of (1.3.25) |E(k)|2= E0 2π [−i(k0− k) + 1/(2cτ )] E0 2π [i(k0− k) + 1/(2cτ )] (1.3.26) At this point we make the following assignments: α = (k0− k), and β = 1/(2cτ ), since otherwise
we will run out of paper. With these new definitions, (1.3.26) becomes |E(k)|2= E0 2π [−iα + β] E0 2π [iα + β] =E 2 0 4π2 1 (−iα + β) (iα + β) =E 2 0 4π2 1 α2+ β2 (1.3.27)
Backsubstituting our previous definitions for α and β, (1.3.27) becomes
|E(k)|2= E
2 0
4π2[(k − k
0)2+ 1/(2cτ )2] (1.3.28)
And it is easy to see that (1.3.28) is identical to (11.82). b) According to (11.84)∗, the natural line width is given by
Γ = λ
2
2πcτ (1.3.29)
With the data provided, this becomes
Γ = 5893 · 10
−10m2
2π · (3.00 · 108m/s) · (2 · 10−8s)−→ Γ ≈ 9.212 · 10−15m
Problem 11.7
A glass block has an index of refraction given by
n = 1.350 −100 ˚λA (1.3.30)
in the visible region.
a) What is the angular difference between red light (Use λ = 6, 500 ˚A) and blue light (λ = 4, 500 ˚A) in the glass if the light enters with an angle of incidence of θ = 30o?
b) What are the phase and group velocities for each color light in the glass?
Solution
a) This problem requires a simple application of Snell’s Law; so solving for the exiting angle θ2,
n1sin(θ1) = n2sin(θ2) −→ θ2= sin−1
n1 n2 sin(θ1) (1.3.31) So the goal here would be to calculate the exiting angle for both red and blue light, and take the difference. We first address incident light in the case that it is red, and find the index of refraction from (1.3.30): ∗
nr= 1.350 −
100 ˚A
6, 500 ˚A ≈ 1.3346 (1.3.32) We now take (1.3.32) and substitute it into (1.3.31):
θr= sin−1 1 1.3346 sin(30 o) ≈ 22.002o (1.3.33)
We now do the same for the shorter wavelength, starting with calculating the index of refraction from (1.3.30):
nb= 1.350 −
100 ˚A
4, 500 ˚A≈ 1.3278 (1.3.34) And now we can calculate the angle from (1.3.31):
θb= sin−1 1 1.3278 sin(30 o) ≈ 22.121o (1.3.35)
And therefore the angular difference is given by
∆θ = θb− θr= 22.121o− 22.002o−→ ∆θ = 0.11927o
b) The phase velocities for both wavelengths in the glass are given simply by (11.93): vp=
c
n (1.3.36)
∗Two assumption have been made. First, I have assumed that all the constant in (1.3.30), as well as all given wavelengths, are exact,
so that we can take the result out to arbitrary precision. Otherwise, we would be bound to the rules of significant figures, which would limit the transparency of the answer. Second, I have taken the incident index of refraction to be that of a vacuum, i.e. n1 = 1.00.
Completing this equation using first the index of refraction of the glass for the longer wavelength form (1.3.32), we have vp,r =2.9979 · 10 8m/s 1.3346 −→ vp,r ≈ 2.2463 · 10 8m/s
And now finding the phase velocity for the shorter wavelength we do the same calculation using (1.3.34),
vp,b =2.9979 · 10 8m/s
1.3278 −→ vp,b ≈ 2.2578 · 10
8m/s
To find the group velocity, we first recall that vg = dω(k)/dk, and also that ω(k) = ck/n(k).
With this information, as well as the expression given (1.3.30), the generic expression for the group velocity is vg= d dk ck 1.350 − 100 ˚A k/(2π) = d dk 2πck 2π1.350 − k100 ˚A =2πc 1 2π1.350 − k100 ˚A d dk(k) + k d dk 1 2π1.350 − k100 ˚A =2πc " 1 2π1.350 − k100 ˚A+ k100 ˚A 2π1.350 − k100 ˚A2 # = 2πc 2π1.350 − k100 ˚A 1 + k100 ˚A 2π1.350 − k100 ˚A = 2πc 2π1.350 − 100 ˚A (2π/λ) 1 + 100 ˚A (2π/λ) 2π1.350 − 100 ˚A (2π/λ) = c 1.350 − 100 ˚A (λ)−1 " 1 + 100 ˚A (λ) −1 1.350 − 100 ˚A (λ)−1 # (1.3.37) And at this point we can now calculate the group velocity given our two frequency components. Starting with the longer wavelength,
vr,g= 2.9979 · 10 18˚A/s 1.350 − 100 ˚A 6500 ˚A−1 " 1 + 100 ˚A 6500 ˚A −1 1.350 − 100 ˚A 6500 ˚A−1 # −→ vr,g≈ 1.7719 · 10 m/s
And now for the shorter wavelength, vb,g= 2.9979 · 10 18˚A/s 1.350 − 100 ˚A 4500 ˚A−1 " 1 + 100 ˚A 4500 ˚A −1 1.350 − 100 ˚A 4500 ˚A−1 # −→ vb,g≈ 1.7902 · 10 m/s
1.4
Homework #4
Problem 12.1
A coaxial wave guide consists of two concentric copper cylinders (conductivity σ = 0.50 × 1018 sec−1), an
in-ner cylinder of radius A = 0.10 cm, and an outer clinder of radius B = 0.40 cm. The dielectric between the clinders has a permittivity = 2.0, and permeability µ = 1.0. The outer cylinder is grounded, and a potential V0= 120 volts (convert this to esu), oscillating at a frequency ν = 12 MHz, is applied to the inner cylinder.
a) Find the fields E(r, φ, z, t) and H(r, φ, z, t) for a wave traveling between the cylinders in the positive z-direction.
b) Find the power (convert it to Watts) transmitted by this wave. c) Find the attenuation length for this wave.
Solution
a) We start by solving Laplace’s Equation in cylindrical coordinates: ∇2 Tϕ = 1 r ∂ ∂r r∂ϕ ∂r + 1 r2 ∂2ϕ ∂φ2 = 0 (1.4.1)
Where we note that due to symmetry, the potential is independent of the angle φ, and so (1.4.1) becomes 1 r ∂ ∂r r∂ϕ ∂r = 0 −→ ϕ(r) = C1ln r + C2 (1.4.2)
Our boundary conditions are that the outer conductor is grounded, ϕ(b) = 0, and the inner con-ductor is at some potential (120 V in our case), ϕ(a) = V0. Applying the first boundary condition,
ϕ(B) = C1ln B + C2= 0 −→ C2= −C1ln B (1.4.3)
Substituting (1.4.3) into (1.4.2) and simplifying, we have
ϕ(r) = C1ln (r/B) (1.4.4)
We now apply the second boundary condition to (1.4.4),
ϕ(A) = C1ln A/B= V0−→ C1= V0
ln (A/ B)
(1.4.5) And now substitute (1.4.5) into (1.4.4),
ϕ(r) = V0 ln (A/
B)
ln (r/
B) (1.4.6)
If we take the negative derivative of (1.4.6), we can find the field: E = −∇ϕ −→ E = ln (VB0/
A)
1
rˆr (1.4.7)
H = ±√µ ˆz × E Using (8.28), the magnetic field becomes
H =√µ V0 ln (B/ A) 1 r ˆz × ˆr −→ H = √µ V0 ln (B/ A) 1 r φˆ (1.4.8)
We can include the additional space and time dependence of the electric and magnetic fields from (1.4.7) and (1.4.8), respectively: E(r, φ, z, t) = V0 ln (B/ A) 1 re i(kz−ωt)ˆr, H(r, φ, z, t) =√µ V0 ln (B/ A) 1 re i(kz−ωt)φˆ
b) The transmitted power can be found by integrating the component of the Poynting vector pointing in the z-direction over the cross-sectional area:
P = c 8π
Z
(E∗×H) · ˆz dA (1.4.9) Substituting the expressions calculated for the fields in the previous section into (1.4.10), we obtain
P = c 8π Z V 0 ln (B/ A) 1 re−i(kz−ωt)ˆr × √µ V0 ln (B/ A) 1 rei(kz−ωt)φˆ ·ˆz · r drdφ = c 8π V0 ln (B/ A) 2 √µ Z 2π 0 dφ Z B A 1 r dr =c 4 V0 ln (B/ A) 2 √µ ln A/ B =c√µ 4 V2 0 ln (B/ A) (1.4.10) And substituting the appropriate constants into (1.4.10) yields∗
P = 3 × 10 8m/sp(1)(2) 4 (0.4 esu)2 ln (.004/.001)−→ P = 1.224 × 10 7esu2 · m/s −→ P = 1.224 × 10−2Watts· m
Problem 12.3
A rectangular wave guide with copper walls has cross section dimensions 4.0 × 6.0 mm. It is filled with air. a) Find the first five TM cutoff frequencies (in Hz) for this wave guide.
b) Sketch nodal planes of Ezfor each of these TM modes.
c) Find the first five TE cutoff frequencies for this wave guide. d) Sketch nodal planes of Hz for each of these TE modes.
Solution
a) The relation for the TM cutoff frequencies in Hz is given by (12.57) in Franklin: fm` = c 2õ `2 A2 + m2 B2 1/ 2 (1.4.11) Where we note that neither ` nor m can equal zero. Furthermore, õ = n = 1, and (1.4.11) becomes fm` = c 2 `2 A2 + m2 B2 1 /2 (1.4.12)
Table 1.1: TM`,m cutoff frequencies
m ` fm` (Ghz) 1 1 45.07 2 1 62.50 1 2 79.06 3 1 83.85 2 2 90.14
b) See attached plots
c) The expression for the TE cutoff frequencies is identical to (1.4.11) except that either ` or m can equal zero:
Table 1.2: TE`,m cutoff frequencies
m ` fm` (Ghz) 1 0 25.00 0 1 37.50 1 1 45.07 2 0 50.00 2 1 62.50
d) See attached plots
Problem 12.4
a) The magnitude of Ez at the center of the wave guide in the preceding problem is 6, 000 volts/meter (convert
to esu). Find the power transmitted by this wave guide for a TM wave with a frequency halfway between the lowest TM cutoff frequencies.
b) Find the attenuation length for this wave.
Solution
a) We know that 1 esu = 300 volt, so 6, 000 volts/meter = 20 esu/meter. Furthermore, the frequency halfway between the two lowest TM cutoff frequencies is f = (62.46 − 45.04)/2 + 45.04 = 53.75 GHz. We know from (12.75) in Franklin that
P = ωkE
2 0AB
32πγ2 (1.4.13)
And we also know from class that ωk γ2 ∝ f fc c √µ s f fc 2 − 1 (1.4.14)
Substituting (1.4.14) into (1.4.13), we have P = E 2 0AB 32π f fc r µc s f fc 2 − 1 (1.4.15)
And now substituting the appropriate constants into (1.4.15) with = 0= 1 and µ = µ0= 1,
P = (20 esu/m) 2 (.004 m)(.006 m) 32π · 53.75 × 109Hz 45.04 × 109Hz · 3 × 10 8m/ss53.75 × 109Hz 45.04 × 109Hz 2 − 1 −→ P = 2.227 × 104esu · m/s
b) The attenuation is given by (12.86) and (12.87) in Franklin as Latten= πµσ c 2µc 1/ 2"AB `2B2+ m2A2 `2B3+ m2A3 # r 1 − ω2 c/ω2 ω (1.4.16)
Jiggiling (1.4.16) around a little bit and noting that ` = m = 1, Latten= hσc 4 i1/ 2 " AB B2+ A2 B3+ A3 # s 1 − f2 c/f2 f (1.4.17)
Substituting the appropriate values into (1.4.17) yields Latten= 0.50 × 1018sec−1 4 1/ 2" (.004 m)(.006 m)(.006 m)2+ (.004 m)2 (.006 m)3+ .(004 m)3 # · r 1 − (45.04)2/(53.75)2 53.75 × 109Hz −→ Latten= 3.709 m
Problem 12.5
a) The magnitude of Hzat the corner of the wave guide in the preceding problem is 0.20 gauss. Find the power
transmitted by this wave guide for a TE wave with a frequency halfway between the two lowest TE cutoff frequencies.
b) Find the attenuation length for this wave.
Solution
a) As in the previous problem, we find the cutoff frequency halfway between the two lowest TE frequencies to be (37.47 − 23.98)/2 + 23.98 = 30.73 GHz. The power is given by (12.76) in Franklin as
P = µωkH
2 0AB
16πγ2 (1.4.18)
If we substitute (1.4.14) into (1.4.18) and simplify, P = H 2 0AB 16π f fc c sf fc 2 − 1 (1.4.19)
And substituting the appropriate values into (1.4.19) yields P = (.20 gauss) 2(.004 m)(.006 m) 16π · 30.73 × 109Hz 24.98 × 109Hz· 3 × 10 8m/s s 30.73 × 109Hz 24.98 × 109Hz 2 − 1 −→ P = 5.047 gauss · m/s
b) The attenuation length can be found from (12.86) and (12.88): Latten= πµσc 2µc 1 /2" AB `2B2+ m2A2 ηAB (`2B + m2A) + (ω c/ω)2(`2B3+ m2A3) # r 1 − ω2 c/ω2 ω (1.4.20) Getting rid of the appropriate constants and recalling that for the lowest cutoff frequency ` = 1 and m = 0, (1.4.21) becomes Latten= hσc 4 i1/ 2 AB A/2 + B(fc/f)2 s 1 − f2 c/f2 f (1.4.21)
And substituting the appropriate constants into (1.4.21), Latten= 0.50 × 1018sec−1 4 1 /2 (.004 m)(.006 m) (.004 m)/2 + (.006 m)(24.98/30.73)2 r 1 − 24.982/30.732 30.73 × 109GHz −→ Latten= 7.085 m
Problem 12.7
A circular wave guide with copper walls has a radius R = 4.0 mm. It is filled with air. a) Find the first five TM cutoff frequencies (in Hz) for this wave guide.
b) Sketch nodal surfaces of Ez for each of these TM modes.
c) Find the first five TE cutoff frequencies for this wave guide. d) Sketch nodal surfaces of Hz for each of these TE modes.
Solution
a) An expression for the TM cutoff frequency can be found from (12.68) in Franklin: fm`= c
2π jm`
R (1.4.22)
Substituting the values from Table 12.2 from Franklin into (1.4.22) yields
Table 1.3: TM`,m cutoff frequencies
m ` fm` (Ghz) 0 1 28.71 1 1 45.74 2 2 61.30 0 2 65.89 1 2 83.74
b) See attached plots
c) Using (12.69) we can find the cutoff frequency: fm`= c
2π j0
m`
R (1.4.23)
And using the values from Table 12.2 we can evaluate (1.4.23) for the desired modes:
Table 1.4: TE`,m cutoff frequencies
m ` fm` (Ghz) 1 1 21.98 2 1 36.46 0 2 45.74 1 2 63.64 2 2 80.00
d) See attached plots
Problem 12.8
The magnitude of Ez at the center of the wave guide in the preceding problem is 6, 000 volts/meter (convert
to esu). Find the power transmitted by this wave guide for a TM wave with frequencies halfway between the two lowest TM cutoff frequencies.
Solution
As before, 6, 000 volts/meter = 20 esu/meter. Additionally, the frequency halfway between the two lowest TM cutoff frequencies is f = (45.74 − 28.71)/2 + 28.71 = 37.23 GHz. We know from (12.79) in Franklin that P = ωkR 2E2 0 16γ2 m` [Jm0 (γm`R)]2 (1.4.24)
To find the derivative of the mth Bessel function we can use a recursion relation. Using Mathematica,
we find that J0 m(γm`R) = −0.5192, and (1.4.24) becomes P = π 2(37.23 GHz)2(.004 m)2(20 esu/m)2 4(3 × 108m/s)(2.4048/.004 m)2 [−0.5192] 2 −→ P = 5.439 × 104esu · m/s
1.5
Homework #5
Problem 12.10
A rectangular cavity (with and µ = 1) has dimensions 2.0 × 3.0 × 4.0 cm.
a) Find the first six resonant frequencies of this cavity, and identify any degenerate modes. b) Write down the E and H fields for the lowest frequency resonant mode.
Solution
We first recall that for a TM wave, Ez must vanish at the walls, and according to (12.107) in Franklin,
Ez= E0sin lπx A sin mπy B cos nπz C [TM mode] (1.5.1)
From which we can see that only n may be zero for a TM wave. Similarly, for a TE wave Hzmust vanish
at the walls, and from (12.108), Hz= H0cos lπx A cosmπy B sin nπz C [TE mode] (1.5.2) We can gather from this that both l and m can be zero for a TE wave. For both TM and TE waves, the resonant frequency is given by (12.109),
flmn = c 2õ "l A 2 + m B 2 + n C 2# 1/ 2 (1.5.3) a) We can find the first six resonant frequencies for the cavity by evaluating (1.5.3) for a variety of
different modes and finding the lowest six:
Resonant frequencies l m n flmn(Ghz) TM/TE d 0 1 1 6.250 TE 1 1 0 1 8.350 TE 1 1 1 0 9.014 TM 2 0 1 2 9.014 TE 2 1 1 1 9.768 TE 1 1 0 2 10.61 TE 1 0 2 1 10.68 TE 1
We note that both l and m cannot both be zero, otherwise all transverse components of E and H will vanish, and we will not have a standing wave.
b) The field components are just given by (12.63)-(12.65), if we allow A > B: Hz(x, y) =H0cos πx A Hz(x, y) = −ikA π H0sin πx A Ey(x, y) = iµωA cπ H0sin πx A
Problem 12.11
A cubical cavity of dimensions 2.0 × 2.0 × 2.0 cm has copper walls. Find the fundamental frequency and Q value for that frequency.
