```Physics 218 Practice Final Exam: Solutions
Spring 2004
Problems 2 (a and b), 3, 4, and 5 (a and c) were on the real final exam last time I taught
PHY 218. I already used the remaining problems on that exam for the practice midterm.
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Problem 1 (50 points)
a.
A spherical shell (inside radius a, outside radius b) has charge Q cos ωt spread on its
outer surface, and charge −Q cos ωt spread uniformly on its inner surface. What is the
electric field in the far-field zone, a distance r λ b from the center of the sphere?
Both spheres act as if their charge is concentrated at their centers. The total
charge is zero; so are all of the multipole moments, because the positive and
negative charges are located in exactly the same place. Because of the latter, there
is no radiation; E = 0 everywhere in the far field.
b.
Another shell is the same in all respect to the first one, except that the centers of the inner
and outer surfaces are displaced with respect to one another by a distance d. What is the
electric field in the far-field zone, a distance r λ b from the center of the sphere?
Now there’s a finite dipole moment:
p = zˆ Qd cos ωt ,
where ẑ points along the line between the centers, from minus to plus. The
system thus radiates like a dipole, and in the far field domain the electric field is
given as usual by
2
Qdω sin θ
E = −θˆ
cos ω ( t − r c )
rc 2
in cgs units. (Swap the 1 c 2 for µ0 4π to get the answer in MKS units.)
______________________________________________________________________________
Problem 2 (50 points)
a.
Calculate the distance y0 from the center of a rain drop, at which the “rainbow ray” is
incident: light that corresponds exactly to the scattering-angle extremum that in turn
corresponds to the primary rainbow. Refer to the diagram at right for the geometry.
In the coordinate system of the diagram, in
which z is the initial direction of the light,
y
dz
=
= tan θ
dy
r2 − y2
θ
y
,
θ′
r
so
sin θ =
cos θ = −
θ′
θ′
(r
y
2
−y
2
)+y
2
=
r2 − y2
(r2 − y2 ) + y2
y
r
,
= − 1−
θ′
θ
y2
r2
∆θ
.
Snell’s Law implies that
sin θ ′ =
y
1
sin θ =
n
nr
.
Thus the scattering angle is, as a function of y,
⎛y⎞
⎛ y ⎞
∆θ = 2θ − 4θ ′ + π = 2 arcsin ⎜ ⎟ − 4 arcsin ⎜ ⎟ + π
⎝r⎠
⎝ nr ⎠
.
Find the minimum as usual:
d
d
d
⎛y⎞
⎛ y ⎞
∆θ = 2 arcsin ⎜ ⎟ − 4 arcsin ⎜ ⎟
dy
dy
dy
⎝r⎠
⎝ nr ⎠
=
2
r 2 − y2
−
4
= 0 at y = y1 ;
n2 r 2 − y 2
(
n2 r 2 − y12 = 4 r 2 − y12
y1 =
b.
r
12 − 3n2
3
)
.
Repeat the calculation of part a, but for the secondary rainbow. Again, refer to the
diagram at right for the geometry.
For the secondary rainbow,
- 2 -
⎛y⎞
⎛ y ⎞
∆θ = 2 arcsin ⎜ ⎟ − 6 arcsin ⎜ ⎟ ,
⎝r⎠
⎝ nr ⎠
θ′θ′
and so
d
2
6
∆θ =
−
=0
2
2
2 2
2
dy
r −y
n r −y
θ
θ′
θ′
θ′
θ
(
n2 r 2 − y 22 = 9 r 2 − y 22
c.
r
y
at y = y 2 ;
y 2 = −r
θ′
9 − n2
8
)
∆θ
.
In the Bible (Genesis 9:13-15), it is written that shortly after the Flood subsided, God
said to Noah and his family,
I have set my bow in the clouds, and it shall be a sign of the covenant between me and the earth.
Henceforth when I bring clouds over the Earth and the bow is seen in the clouds, I will remember
my covenant that is between me and you and every living creature of all flesh; and the waters
shall never again become a flood to destroy all flesh.
So before the Flood, raindrops did not produce rainbows, but afterward they did.
Describe how the refractive index of water would have had to change during this
conversation, in order for the optical properties of raindrops to change like this.
I think this is the earliest literary reference to rainbows; had to work it in to the
course somehow…
Arguing from the results of parts a and b (which would certainly be good
enough to get full credit): There would be no rainbows, of at least the primary
and secondary sort, if n were large enough that y1 and y 2 came out imaginary,
which is the case for n > 3. So the index had to start off greater than 3, and drop
suddenly to its present value near 4/3.
If you want to be really clean about it, note that the N th rainbow lies at
∆θ = 2θ − 2 ( N + 1 )θ ′ + Nπ
( N = 1, 2,...)
⎛y⎞
⎛ y ⎞
= 2 arcsin ⎜ ⎟ − 2 ( N + 1 ) arcsin ⎜ ⎟ + Nπ
⎝r⎠
⎝ nr ⎠
such that
- 3 -
,
2 ( N + 1)
d
2
∆θ =
−
= 0 at y = y N ;
dy
r2 − y2
n2 r 2 − y 2
2
n2 r 2 − yN
= ( N + 1)
2
( r 2 − yN2 )
2
2
2 ⎡
yN
N + 1 ) − 1 ⎤ = r 2 ⎡( N + 1 ) − n 2 ⎤
(
⎣
⎦
⎣
⎦
yN = ±r
( N + 1 )2 − n2
.
( N + 1 )2 − 1
Thus there is no N th rainbow if n ≥ N + 1. Since in principle one could consider
N as high as one likes, n would formally have to be infinite in order not to have
rainbows of any order. In practice, though, it would suffice to have n simply a
good deal larger than 1 (say, 10), in order that the high-order rainbows still
allowed would be very faint, owing to the large reflectivity of the drop at high
index.
______________________________________________________________________________
Problem 3 (50 points)
Diffraction of a Gaussian beam. The
electric field in a laser beam, as the beam leaves
the end of the laser, is linearly polarized
vertically, is axially symmetric, and has a
magnitude which depends upon distance from
the laser's axis as follows:
E = E0 e −s
2
s02
= E0 e
(
− x′2 + y ′2
)
s02
Laser
.
Screen
x
y
z
The laser beam is pointed perpendicular to a
screen which lies a very long distance r ( s0 )
away from the laser. What is the electric field on this screen, as a function of distance
q = x 2 + y 2 from the point on the screen at which the laser is aimed?
Make a rough plot of the electric field amplitude as a function of q.
Hint: work in Cartesian coordinates initially, and complete the square in the exponent of the
integrand, to carry out the integral.
- 4 -
EN = E0 e
EF =
(
− x′2 + y ′2
ei( kr −ωt )
λr
)
s02 − iωt
e
∞ ∞
∫ ∫
E0 e
(a Gaussian),
(
− x′ 2 + y ′ 2
)
(
s02 − i kx x′+ ky y′
e
) dx′dy′
.
−∞ −∞
Note that
−
2
2
⎛ x ′2
x ′2
⎡ ikx s0 ⎤ ⎞ ⎡ ikx s0 ⎤
′
′
ik
x
ik
x
−
=
−
+
+
+
⎜
⎟
x
x
⎢⎣ 2 ⎥⎦ ⎟ ⎢⎣ 2 ⎥⎦
⎜ s2
s02
⎝ 0
⎠
2
⎛ x′ ik s ⎞
k 2 s2
= −⎜ + x 0 ⎟ − x 0
2 ⎠
4
⎝ s0
.
Similarly for y ′. The integral becomes
EF =
E0 e
i( kr −ωt )
λr
e
−
(
kx2 + ky2
)
s02
4
∞ ∞
∫ ∫
⎛ x′ ik s ⎞
−⎜ + x 0 ⎟
s
2 ⎠
e ⎝ 0
2
⎛ y′ iky s0 ⎞
−⎜ +
⎟
s
2 ⎠
e ⎝ 0
2
dx′dy ′ .
−∞ −∞
Now change variables:
x′ ikx s0
+
s0
2
y′ iky s0
v= +
2
s0
u=
EF =
du =
dx′
s0
−∞ < u < ∞
dv =
dy′
s0
−∞ < v < ∞
s02 E0 ei( kr −ωt ) −( kx2 + ky2 )s02
e
λr
4
∞ ∞
∫ ∫
e
(
− u2 + v 2
) dudv
,
−∞ −∞
and change them again:
ρ 2 = u2 + v 2 ⎫⎪
⎬ dudv = ρ d ρ dφ , 0 < ρ < ∞ , 0 < φ < 2π ;
tan φ = v u ⎪⎭
s02 E0 ei( kr −ωt ) −( kx2 + ky2 )s02
EF =
e
λr
4
2π
∫
0
∞
2
dφ ∫ ρ e − ρ d ρ
,
0
and change one of them yet again, and the integrals become trivial:
w = ρ2
, dw = 2 ρ d ρ
- 5 -
, 0<w<∞ ,
EF =
=
s02 E0 ei( kr −ωt ) −( kx2 + ky2 )s02
e
2λ r
4
π s02 E0 ei( kr −ωt ) −( kx2 + ky2 )s02
e
λr
2π
∫
0
∞
dφ ∫ e − w dw
0
4
.
(
)
We are asked to note that kx2 + ky2 = k 2 x 2 r 2 + y 2 r 2 = k 2 q 2 r 2 , which makes it
π s02 E0 ei( kr −ωt ) − k 2q 2 s02
EF =
e
λr
4r 2
.
This is also a Gaussian. Gaussian beams stay Gaussian as they propagate,
because the Fourier transform of a Gaussian is another Gaussian.
Near field
Far field
π s02 E0
λr
E0
2s0
2r ks0
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Problem 4 (50 points)
electric dipoles, separated by a distance d, are
oriented as shown in the figure at right. Using what
you know about the potentials for individual
dipoles, calculate the scalar potential V in the far p0 cos ωt
field ( r λ d ) .
θ+
θ
Hints: Keep only terms that are first order in d.
Note that neither dipole lies at the origin.
Superpose the retarded potentials from the two
dipoles, which are
z
r
r−
d
y
θ−
p ω cos θ ±
⎛ r ⎞
V± = ∓ 0
sin ω ⎜ t − ± ⎟ .
c
c ⎠
r±
⎝
− p0 cos ωt
- 6 -
P
r+
We can use the law of cosines and the binomial theorem to obtain approximations for
r± :
2
d
d
d
⎛d⎞
⎛
⎞
r± = r + ⎜ ⎟ ∓ 2r cos θ ≅ r 1 ∓ cos θ ≅ r ⎜ 1 ∓ cos θ ⎟ ,
r
2
2r
⎝2⎠
⎝
⎠
d
1 1⎛
⎞
≅ ⎜ 1 ± cos θ ⎟ ,
r± r ⎝
2r
⎠
2
to first order in d. Note also from the diagram – specifically, the triangle sides that lie
along the z axis – that
r± cos θ ± ±
d
= r cos θ
2
,
or
r cos θ ∓ d 2
d ⎞1⎛
d
⎛
⎞
≅ r ⎜ cos θ ∓ ⎟ ⎜ 1 ± cos θ ⎟
r±
2
2
r
r
r
⎝
⎠ ⎝
⎠
d
d
≅ cos θ ± cos 2 θ ∓
ignoring terms in d 2 !
2r
2r
d
d
= cos θ ∓
1 − cos 2 θ = cos θ ∓ sin 2 θ ,
2r
2r
cos θ ± =
(
(
)
)
also to first order in d. Thus the retarded-time factor becomes
ωd
d
⎡ r⎛
⎛ r ⎞
⎞⎤
⎛
⎞
sin ω ⎜ t − ± ⎟ = sin ω ⎢t − ⎜ 1 ∓ cos θ ⎟ ⎥ = sin ⎜ ωt0 ±
cos θ ⎟
2r
2c
c ⎠
⎝
⎠⎦
⎝
⎠
⎣ c⎝
⎛ ωd
⎞
⎛ ωd
⎞
= sin ωt0 cos ⎜
cos θ ⎟ ± cos ωt0 sin ⎜
cos θ ⎟
⎝ 2c
⎠
⎝ 2c
⎠
ωd
≅ sin ωt0 ±
cos θ cos ωt0 ,
2c
where we have abbreviated t0 = t − r c , and have used the small-angle approximation
for the terms in ω d 2c , to first order in d.
Putting all this together, we get, for the retarded potentials of the two dipoles,
p0ω cos θ ±
⎛ r ⎞
sin ω ⎜ t − ± ⎟
r±
c
c ⎠
⎝
p ω⎛
d
d
ωd
⎞1⎛
⎞⎛
⎞
cos θ cos ωt0 ⎟ .
= ∓ 0 ⎜ cos θ ∓ sin 2 θ ⎟ ⎜ 1 ± cos θ ⎟⎜ sin ωt0 ±
c ⎝
2r
2r
2c
⎠r⎝
⎠⎝
⎠
V± = ∓
- 7 -
When multiplying it out, remember to drop terms of order d 2 or higher:
p0ω ⎛
d
d
ωd
⎞
2
2 ⎞⎛
cos θ cos ωt0 ⎟
⎜ cos θ ∓ sin θ ± cos θ ⎟⎜ sin ωt0 ±
cr ⎝
2r
2r
2c
⎠⎝
⎠
p ω⎛
d
ωd
⎞
cos 2 θ cos ωt0 ±
cos 2 θ − sin 2 θ sin ωt0 ⎟ .
= ∓ 0 ⎜ cos θ sin ωt0 ±
cr ⎝
2c
2r
⎠
V± = ∓
(
)
Now we can finally do the superposition:
V = V+ + V−
p0ω ⎛
ωd
d
⎞
cos 2 θ cos ωt0 +
cos 2 θ − sin 2 θ sin ωt0 ⎟
⎜ cos θ sin ωt0 +
2c
2r
cr ⎝
⎠
p ω⎛
ωd
d
⎞
+ 0 ⎜ cos θ sin ωt0 −
cos 2 θ cos ωt0 −
cos 2 θ − sin 2 θ sin ωt0 ⎟
cr ⎝
2c
2r
⎠
p ω ⎛ ωd
d
⎞
=− 0 ⎜
cos2 θ cos ωt0 + cos 2 θ − sin 2 θ sin ωt0 ⎟
cr ⎝ c
r
⎠
(
=−
)
(
(
=−
)
)
p0ω 2 d ⎛
c
⎞
2
+
t
θ
ω
cos
cos
cos 2 θ − sin 2 θ sin ωt0 ⎟ .
⎜
0
2
ωr
⎠
c r ⎝
(
)
Since in the far field we have r c ω , we can neglect the second term in the brackets
compared to the first one, and obtain a rather simple expression:
V =−
p0ω 2 d cos 2 θ
p0ω 2 d cos 2 θ
⎛ r⎞
=
−
cos
t
cos ω ⎜ t − ⎟ .
ω
0
2
2
r
r
⎝ c⎠
c
c
Multiply the right side by 1 4πε 0 to get the MKS version of the answer.
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Problem 5 (50 points)
a.
Using the electromagnetic field tensor F µν and the dual tensor G µν , show that
E2 − B2
(cgs units) or E2 − c 2 B2
and
E◊B
are invariant under Lorentz transformations.
- 8 -
(MKS units)
There are three invariants one can make by the inner products of F µν and G µν .
Note while doing the sums that a zero covariant index brings in an extra minus
sign, and that the diagonal elements are zero:
Fµν F µν = −F 01F 01 − F 02 F 02 − F 03F 03 − F 10 F 10 + F 12 F 12 + F 13F 13
− F 20 F 20 + F 21F 21 + F 23F 23 − F 30 F 30 + F 31F 31 + F 32 F 32
= −Ex2 − Ey2 − Ez2 − Ex2 + Bz2 + By2 − Ey2 + Bz2 + Bx2 − Ex2 + By2 + Bx2
(
= 2 E 2 − B2
)
;
Gµν G µν = −G 01G 01 − G 02G 02 − G 03G 03 − G 10G 10 + G 12G 12 + G 13G 13
− G 20G 20 + G 21G 21 + G 23G 23 − G 30G 30 + G 31G 31 + G 32 G 32
= − Bx2 − By2 − Bz2 − Bx2 + Ez2 + Ey2 − By2 + Ez2 + Ex2 − Bz2 + Ey2 + Ex2
(
= 2 B2 − E 2
)
;
Fµν G µν = −F 01G 01 − F 02G 02 − F 03G 03 − F 10G 10 + F 12G 12 + F 13G 13
− F 20G 20 + F 21G 21 + F 23G 23 − F 30G 30 + F 31G 31 + F 32 G 32
= −Ex Bx − Ey By − Ez Bz − Ex Bx − Bz Ez − By Ey
− Ey By − Bz Ez − Bx Ex − Ez Bz − By Ey − Bx Ex
= −4 E ◊ B .
The first two results each show that E2 − B2 is Lorentz invariant. The third shows
that E ◊ B is invariant. (Q.E.D.) It works similarly in MKS units except for the
stray factors of c.
b.
Suppose that, in a certain inertial frame S, the electric field E and the magnetic field B
are neither parallel nor perpendicular. Show that in a different inertial frame S , moving
relative to S at velocity v given by
v=
c
γ
2
E¥B
B2 + E 2
(cgs units) or
1
γ
2
E¥B
B + E2 c 2
2
(MKS units), ,
the fields E and B are parallel.
Suppose the two frames move relative to one another along the x direction, so we
can use our usual forms of the Lorentz transformation. If x is also the direction of
E ¥ B , as we are asked to show, then E and B lie in the y-z plane. Let ψ be the
angle between the two fields in S, and for simplicity let one of them, say E, point
along z; then Bz = B cosψ and By = B sinψ . Thus the relativistic transformations
of the fields give us
- 9 -
Ey = γ ( − β B cosψ ) , Ez = γ ( E + β B sinψ ) ,
By = γ ( B sinψ + β E ) , Bz = γ ( B cosψ ) .
The x components of the field are still zero; this component is left unchanged by
the transformation. E and B are parallel if the ratios of their y and z components
are equal:
Ey
Ez
=−
By γ ( B sinψ + β E )
γβ B cosψ
.
=
=
γ ( E + β B sinψ ) Bz
γ B cosψ
Multiply it out:
− β B2 cos 2 ψ = ( E + β B sinψ )( B sinψ + β E )
(
)
= β E2 + β B2 sin 2 ψ + 1 + β 2 EB sinψ
(
) (
;
)
β E2 + β B2 cos2 ψ + sin 2 ψ + 1 + β 2 EB sinψ = 0
1 + β 2 ) EB sinψ
(
1
β =−
=−
2
E +B
2
E¥B
2
γ E 2 + B2
.
The way we’ve set up the coordinate axes, though, the cross product of the fields
is
xˆ
yˆ
zˆ
E¥B= 0
0
E
0 B sinψ
B cosψ
= xˆ
0
E
B sinψ
B cosψ
= − xˆ EB sinψ
(I guess that means I should have pointed B along the z axis), so the minus signs
cancel:
v=−
c.
E¥B
γ 2 E 2 + B2
c
, q.e.d.
Is there a frame in which the electric and magnetic fields are perpendicular?
No. Since the fields are not perpendicular in the original frame, E ◊ B ≠ 0, and
since this quantity is Lorentz-invariant, they can’t be perpendicular in any other
frame either.
- 10 -
______________________________________________________________________________
Problem 6 (50 points)
Unpolarized light with angular frequency ω is incident from
vacuum, at angle θ , on a planar conducting surface.
Calculate the degree of polarization,
δ=
I⊥ − I
I⊥ + I
x
Vacuum Conductor
,
of reflected light, and describe in a few words the state of the
light’s polarization: nature, magnitude and direction. (Here
⊥ and mean perpendicular and parallel to the plane of
incidence.)
θ
z
We need to handle the TE ( ⊥ ) and TM ( ) parts of the
incident light separately. TE first: the boundary
conditions on the components of E and H parallel to the
surface are
E0 I + E0 R = E0T
1
µ1
(−
,
)
µ1ε 1 E0 I cos θ I + µ1ε 1 E0 R cos θ I = −
1 ck2
µ2 ω
E0T cos θT
,
which we can rearrange in a familiar way:
E0 I + E0 R = E0T
⎛ cos θT
E0 I − E0 R = ⎜
⎝ cos θ I
,
⎞ ⎛ µ1 ck2 ⎞
⎟⎟ E0T = αβ E0T
⎟ ⎜⎜
⎠ ⎝ ε 1 µ 2ω ⎠
.
Multiply the first condition by αβ and subtract:
(αβ − 1) E0 I + (αβ + 1) E0 R = 0
⎛ E0 R ⎞
1 − αβ
⎜
⎟ =
⎝ E0 I ⎠ ⊥ 1 + αβ
;
.
And now for TM, for which the boundary conditions on the components of E and H
parallel to the surface are
- 11 -
E0 I cos θ I + E0 R cos θ I = E0T cos θT
1
µ1
(
)
µ1ε 1 E0 I − µ1ε 1 E0 R =
1 ck2
,
E0T
µ2 ω
,
which can also be rearranged in a familiar way, using the same α and β , and solved
just as before:
E0 I + E0 R =
E0 I − E0 R =
cos θT
E0T = α E0T
cos θ I
,
µ1 ck2
E0T = β E0T
ε 1 µ 2ω
,
⎛β
⎞
⎛β
⎞
⎜ − 1 ⎟ E0 I + ⎜ + 1 ⎟ E0 R = 0 ;
⎝α
⎠
⎝α
⎠
⎛ E0 R ⎞ α − β
.
⎜
⎟ =
⎝ E0 I ⎠ α + β
Now,
⎛ I R ⎞ ⎛ I R ⎞ 1 − αβ 1 − αβ * α − β α − β *
±
⎜ ⎟ ±⎜ ⎟ =
⎝ I I ⎠ ⊥ ⎝ I I ⎠ 1 + αβ 1 + αβ * α + β α + β *
=
1 − αβ − αβ * +α 2 β 2 α 2 − αβ − αβ * + β 2
±
1 + αβ + αβ * +α 2 β 2 α 2 + αβ + αβ * + β 2
;
1 − αβ − αβ * +α 2 β 2 α 2 − αβ − αβ * + β 2
−
I⊥ − I
1 + αβ + αβ * +α 2 β 2 α 2 + αβ + αβ * + β 2
δ=
=
I⊥ + I
1 − αβ − αβ * +α 2 β 2 α 2 − αβ − αβ * + β 2
+
1 + αβ + αβ * +α 2 β 2 α 2 + αβ + αβ * + β 2
.
This is an ugly expression, but at normal incidence (α = 1 ) it simplifies considerably:
1 − β − β * +β 2 1 − β − β * +β 2
−
1 + β + β * +β 2 1 + β + β * +β 2
δ=
=0 ;
1 − β − β * +β 2 1 − β − β * +β 2
+
1 + β + β * +β 2 1 + β + β * +β 2
unpolarized light reflected at normal incidence is still unpolarized. At oblique incidence
it obviously doesn’t vanish, so the reflected light is polarized.
- 12 -
I wouldn’t expect you to go any farther than this, on a real test. (I wish I had written the
problem just in terms of the fields; I didn’t think the result was going to be this ugly.) It
turns out that the polarization is slight ( a few percent at most), and perpendicular to
the plane of incidence; that is, δ turns out to be small and positive.
- 13 -
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