PHY4324/Fall 09: EM II
HOMEWORK ASSIGNMENT #6: SOLUTIONS due by 11:45 p.m. Wednesday 10/15
Instructor: D. L. Maslov maslov@phys.ufl.edu 392-0513 Rm. 2114
Please help your instructor by doing your work neatly.
Every (algebraic) final result must be supplemented by a check of units. Without such a check, no more than 75% of the credit will be given even for an otherwise correct solution.
1. G9.10 [We did this problem in class but the problem is so good that it’s worth doing it again. Here is a link to the original 1901 paper by P. N. Lebedev describing his experiments on measuring the light pressure http://web.ihep.su/dbserv/compas/src/lebedev01/eng.pdf] [20 points]
Solution: The momentum density of the electromagnetic wave is P = I/c 2 .
Consider a slab of area A and height cdt.
Over time dt, all the momentum in this slab will fall on the surface. For a perfectly absorbing surface, the change in the momentum is P cdtA = ( I/c ) dtA.
Pressure is the force per unit are
P = I/c =
1 .
3 × 10 3
3 × 10 8
= 4 .
3 × 10
− 6
N/m
2
.
Atmospheric pressure is about 10 5 N/m 2 , so the pressure of light is about 4 .
3 × 10 − 11 reflecting surface, the pressure is twice bigger=8 .
6 × 10 − 6 N/m 2 atm. For a perfectly
2. G9.14 [25 points]
Solution: Let
E i
E r
E t
= E
0 i
= E
0 r
(cos α ˆ + sin α y )
= E
0 t
(cos β ˆ + sin β y ) .
From Faraday’s law, ∇ × E = − ∂ B /∂t → B = ( k × E ) /ω →
B
B i r
E t
=
= − ( k
1
× E ) /ω = −
E
0 r v
1
= (
E v
1 k
0 i
2
× E ) /ω =
E
0 t v
2
( −
( − cos β cos
ˆ
α ˆ + sin
+ sin β x ) .
α x )
The boundary conditions E 1
||
= E 2
|| coefficients of ˆ in the B field gives and B 1
||
/µ
1
= B 2
||
/µ
2
.
Equating the coefficients of ˆ in the E field and the
E
0 r sin α = E
0 t sin β
E
0 r
µ
1 sin v
1
α
= −
E
0 t
µ
2 sin β v
2
→ E
0 r sin α = −
µ
1
µ
2 v
1 v
2
E
0 t sin β.
These two equations are compatible only if α = β = 0 .
Q. E.D.
3. G9.33 [30 points]
Solution a)
1. Gauss’s law
∇ · E =
1 r sin θ
∂E
φ
∂φ
= 0 .
X

2. Use the Faraday’s law to find the magnetic field.
−
∂ B
∂t
= ∇ × E =
1 r sin θ
∂
∂θ
(sin θE
φ
) ˆ −
1 r
∂
∂r
( rE
φ
Denote u ≡ kr − ωt and observe that
∂
∂r
∂ cos u = − k sin
∂r sin u = k cos u.
u
Then,
−
∂ B
∂t
=
1 r sin θ
E
0 r
2 sin θ cos θ cos u −
1 kr
× ( − k sin u + sin u kr 2
− cos u r sin u ˆ −
1 r
E
0 sin θ
Integrate over t using R dt cos u = − 1
ω sin u and R dt sin u = 1
ω cos u
B =
2 E
0 cos
ωr 2
θ 1 sin u + kr cos u ˆ +
E
0 sin θ
ωr
− k cos u + cos u kr 2
+ sin u r
3. Having found B , we can now check if the ”no monopoles” law is satisfied.
∇ · B = 0
∇ · B = r
1
2
= r
1
2
∂
∂r
∂
∂r
= r
2
B r
+
2 E
0 cos θ
1 r 2
2 E
0 cos θ
ω
ω k
∂ r sin θ ∂θ cos
1 sin u u
−
+
(sin θB
θ
)
1 kr cos u kr 2 cos u +
− sin r u
+
1
2 cos θE
0 r 2 ω
∂ r sin θ ∂θ
E
0 sin
2
θ
ωr
− k cos u + cos u kr 2
+ sin u r
− k cos u + cos u kr 2
+ sin u r
= 0 .
X
4. Check the Amp`ere’s law
(
∇ × B =
∇ × B =
∇ ×
1 c 2
B
∂
)
E
∂t
φ
=
=
=
=
=
1 c 2
1 r
∂ E
∂t
∂
∂r
( rB
θ
) −
∂B r
∂θ
1 r
∂
∂r
−
1 r
∂
∂θ
E
0 sin θ
ω
2 E
0 cos θ
ωr 2
− k cos u + cos u kr 2
+ sin u r sin u + cos kr u
1 E
0 sin θ r ω
+
2 E
0 sin θ
ωr 3 kE
0 sin θ
ωr k 2 sin u −
2 sin u r 2 k sin u + cos u sin u +
1 r kr cos u
− 2 cos u kr 3
+ k cos u r
1 c 2
E
0 sin θ r k 2
ω 2
ωE
0 sin θ kr
ω sin u +
ω kr k sin u + cos u =
1 r cos u =
1 c 2
ωE
0 sin θ kr kE
0 sin
ωr
θ k sin u +
1 r cos u k sin u +
1 r cos u = ( ∇ × B )
φ
.
X
2

3 b) Poynting vector
Recall that
S =
1
µ
0
E × B =
E
0 sin θ
µ
0 r
×
2 E
0 cos θ
ωr 2 sin u + cos u − sin kr u
1 kr cos u ˆ +
E
0 sin θ
ωr
− k cos u + cos u kr 2
+ sin u r
ˆ
ˆ form a right-handed triad so that ˆ ×
ˆ
= ˆ and ˆ × r = ˆ Then,
S =
=
E 2
0 sin 2
µ
0
ωr 2
θ
−
E 2
0 sin
µ
0
ωr
2
2
θ cos u − sin u kr sin u +
1 kr cos u cos u − sin u
− k cos u + kr cos u kr 2
+ sin u r
E 2
0 sin 2
µ
0
ωr 2
θ cos u sin u 1 −
1
( kr )
2
!
+
1 kr cos
2 u − sin
2 u
−
E 2
0 sin
µ
0
ωr
2
2
θ
!
− k cos 2 u +
1 r sin u cos u + cos 2 u kr 2
− sin u cos u k 2 r 3
+ cos u sin u
− r sin
2 u kr 2
.
Average over the period using h cos u sin u i = 0 , h cos
2 the ˆ component becomes equal to u i = h sin
2 u i = 1 / 2 .
The ˆ component of S vanishes while h S i = kE 2
0
2 µ
0 sin
2
ωr 2
θ
ˆ =
E 2
0 sin
2
2 µ
0 cr 2
θ c) Power
Z h S i· d a =
=
=
E 2
0
2 µ
0 c
πE
µ
0 c
2
0
Z
Z
1
− 1
Z Z sin
2
θ r 2 sin θdrdθdφ = r 2
Z
1
1 − x
2 dx =
2 πE 2
0
µ
0 c
0
πE
µ
0 c
2
0
1 − x
2
2 πE 2
0
µ
0 c
1 −
1
3
=
4 πE 2
0
.
X
3 µ
0 c
Z
π sin
3
θdθ
0 dx
4. G9.36 [25 points]
Solution
Substituting numbers into Eq. (9.199),
5.
T
− 1
=
=
1
4 (4 / 3) (1)
(
4
3
+ 1
2
+
49
48
+
85
(48) (36) sin
2
3 ωd
2 c
16
9
−
9
4
1 −
9
4
9
4
.
sin
2
3 ωd
2 c
)
Obviously, if T − 1 sin
2 3 ωd
2 c
= 1 .
So, is maximal, then T is minimal. The maximum of T − 1
T min
=
49
48
+
1
85
(48)(36)
≈ 0 .
935 .
is at
3 ωd
2 c
= π
2
(2 n + 1) , where
Likewise, when sin
2 3 ωd
2 c
= 0 , T − 1 reaches its minimum and T reaches its maximum
T max
=
48
49
≈ 0 .
980 .
Since Eq. (9.199) is unchanged if n
1 is replaced by n
3 and n
3 is replaced by the reverse direction is the same and the fish sees you as well as you see it.
n
1
, the transmission coefficient in

4
6.
Bonus: G9.34 [25 points]
Let r
12
= n
1
− n
2 n
1
+ n
2 be the (amplitude) reflection coefficient for the wave coming from medium 1 to medium 2. Similarly r
21
= − r
12
= n
2
− n
1 n
1
+ n
2 is the reflection coefficient for the wave coming from medium 2 to medium 1, and r
23
= n
2
− n
3 n
2
+ n
3 is the same for the wave coming from medium 2 to medium 3. Also, t
12
= n
1
2 n
1
+ n
2 is the (amplitude) transmission coefficient for a wave coming from medium 2 to medium 3, and t
23
= n
2
2 n
2
+ n
3 is the same for the wave coming from medium 2 to medium 3. Choose the amplitude of the incident wave to be equal E
0
.
Follow the ”trajectory” of the wave after a couple of reflections. The first transmitted wave has the amplitude t
12 t
23 and it has traveled through glass once, so it accumulated the phase factor e ik
2 d , where k
2
= ωn
2
/c.
The wave reflected from the right inner surface of the glass has an amplitude t a phase factor e ik
2 d
12 r
23 and
.
After reflection from the inner left surface of glass, the amplitude of the wave changes to t
12
P r
23
.
The net amplitude of the second wave coming from glass is t
12 is t
12 t e 5 ik
2 d
.
The next transmitted wave
, etc. Summing up transmitted waves with the help of the formula for geometric series n =0 r
21
, and one more passage through the right inner glass surface adds a factor of t
23 t
23 r
21 r
23 and a phase factor of e 3 ik
2 d
23 x n r 2
21 r 2
23
= 1 / (1 − x ), we get
E t
= E
0 t
12 t
23 e ik
2 d
= E
0 t
12 t
23 e ik
2 d
+ t
12 t
23 r
21 r
23 e
3 ik
2 d
+ t
12 t
23 r
2
21 r
2
23 e
5 ik
2 d
1 + r
21 r
23 e 2 ik
2 d + r 2
21 r 2
23 e 4 ik
2 d + . . .
+ . . .
t
12 t
23 e ik
2 d
= E
0
1 − r
21 r
23 e 2 ik
2 d
.
The intensity of the wave is | E | 2 /vµ
0
, so the energy transmission coefficient is
T =
| E t
| 2 v
3
:
| E
1
|
2 v
1
= n
3 n
1 t
12 t
23 e ik
2 d
1 − r
21 r
23 e 2 ik
2 d
2
= n
3 n
1
1 + r 2
21 r 2
23 t 2
12 t 2
23
− 2 r
21 r
23 cos (2 k
2 d )
.
The rest is just some algebra.
T = 16 n n
3
1 n 2
1 n 2
2
( n
1
+ n
2
)
2
( n
2
+ n
3
)
2
1 + n
2 n
1
− n
1
+ n
2
2 n
2
− n
3 n
2
+ n
3
1
2
− 2 n
2 n
1
− n
1
+ n
2 n
2
− n
3 n
2
+ n
3 cos (2 k
2 d )
= 16 n
=
1 n
3 n
2
2
( n
1
+ n
2
)
2
( n
2
+ n
3
)
2
1
+ ( n
2
− n
1
)
2
( n
2
− n
3
)
2
− 2 ( n 2
2
− n 2
1
) ( n 2
2
( n
1
+ n
2
)
2
( n
2
+ n
3
)
2
+ ( n
2
− n
1
)
2
(
16 n
1 n
3 n 2
2 n
2
− n
3
)
2
− 2 ( n 2
2
− n 2
1
) ( n 2
2
− n 2
3
− n
) 1 − 2 sin
2
2
3
) cos (2 k
2 d k
2 d )
=
( n
1
+ n
2
)
2
( n
2
+ n
3
)
2
16 n
1 n
3 n 2
2
+ ( n
2
− n
1
)
2
( n
2
− n
3
)
2
− 2 ( n 2
2
− n 2
1
) ( n 2
2
− n 2
3
) + 4 ( n 2
2
− n 2
1
) ( n 2
2
− n 2
3
) sin
2 k
2 d

FIG. 1: Problem 5. Notice that the directions of the wave propagation are drawn at an angle to the slab only to avoid cluttering the plot. In reality, the incidence is normal.
1 2 3
5 t12 t12t23exp(ik2d) t12r23 t12r23r21 t12t23r23r21exp(3ik2d) t12t23(r23r21)^2exp(5ik2d)
Denoting α = n
1
+ n
2
, β = n
2
+ n
3
, γ = n
2
− n
1
, δ = n
2
− n
3
, we have
( n
1
+ n
2
)
2
( n
2
+ n
3
)
2
+ ( n
2
− n
1
)
2
( n
2
+ n
3
)
2
− 2 n
2
2
− n
2
1 n
2
2
− n
2
3
= α 2 β 2
= n
1 n
2
+ γ 2 δ 2
+ n
2
2
+
− n
2
1
αγβδ n
3
+ n
= (
2 n
3
αβ
−
− n
2
2
γδ ) 2
+ n
1
= [( n
1 n
2
+ n
+ n
2
2 n
3
) ( n
2
+ n
3
) − ( n
2
2
− n
1
) ( n
2
− n
1 n
3
= 4 [ n
1 n
2
+ n
2 n
3
]
2
− n
3
)]
2
= 4 n
2
2
( n
1
+ n
3
)
2
.
Substituting this back into T, we get
T =
( n
1
+ n
3
)
2
+
4 n
1 n
3
( n 2
2
− n 2
1
)( n 2
2 n 2
2
− n 2
3
) sin
2 k
2 d
.
X