Download PHYS 110B - HW #5

PHYS 110B - HW #5
Spring 2004, Solutions by David Pace
Any referenced equations are from Griffiths
Problem statements are paraphrased
[1.] Problem 9.10 from Griffiths
Sunlight hitting the earth has an intensity of approximately 1300 W/m2 . What pressure
does sunlight exert on a perfect absorber? What pressure does it exert on a perfect reflector?
How does this compare to atmospheric pressure on the surface of the earth?
Solution
The radiation pressure on a perfect absorber is given in Griffiths in terms of the intensity
of the light wave in question.
I
(1)
Prad =
c
where c is the speed of light.
In this problem the intensity is a given and the pressure on a perfect absorber is,
Pabs
1300
3 × 108
=
=
4.3 × 10−6
Pa
(2)
where the SI unit of pressure, Pascal, is used.
As stated on page 382 of Griffiths, the pressure exerted on a perfect reflector is twice that
given in (1). Momentum must be conserved, and in the case of reflection the momentum
of the light is reversed. Such a reversal means that δp in equation 9.64 of Griffiths is twice
as large. For a perfect reflector then,
Pref
=
2Pabs
=
8.6 × 10−6
Pa
(3)
Atmospheric pressure at sea level is approximately Patm = 1.03 × 105 Pa. Comparing to the
radiation pressure on the perfect reflector above gives,
Pref
Patm
=
8.6 × 10−6
1.03 × 105
=
8.3 × 10−11
(4)
so the radiation pressure due to sunlight is entirely insignificant compared to the atmospheric pressure. The relation between the radiation pressure on a perfect absorber and
the atmospheric pressure is half of (4).
[2.] Problem 9.13 from Griffiths
Determine the reflection and transmission coefficients without letting µ1 = µ2 = µo . Verify
that R + T = 1.
Solution
Begin with the definition of the coefficients,
R =
IR
II
Eq. 9.86
(5)
T =
IT
II
Eq. 9.87
(6)
1
The intensities mentioned above are,
1
I = vEo2
2
(7)
Define region 1 to contain the incident and reflected waves (the left side of the interface
with the incident wave moving to the right) and region 2 to contain the transmitted wave.
The reflection coefficient becomes,
R =
=
1
v E2
2 1 1 oR
1
v E2
2 1 1 oI
(8)
2
EoR
2
EoI
(9)
Now consider the following general relation between the reflected and incident amplitudes,
∼
1−β ∼
Eq. 9.82
(10)
E oI
E oR =
1+β
µ1 v 1
µ2 v 2
β ≡
(11)
If we divide the incident amplitude from both sides in (10) and then square the result, the
complex factors will become unity. Therefore,
∼
E oR
!2
=
∼
E oI
2
EoR
=
2
EoI
1−β
1+β
2
1−β
1+β
2
(12)
Returning to (9) we have solved for the reflection coefficient,
2
1−β
R =
1+β
(13)
(14)
Following the same method to solve for T ,
T =
1
v E2
2 2 2 oT
1
v E2
2 1 1 oI
2 v2
=
1 v1
EoT
EoI
(15)
2
(16)
The amplitude ratio is found from,
∼
E oT =
2
2
1+β
∼
E oI
(17)
∼
E oT
!2
=
∼
E oI
EoT
EoI
2
=
2
1+β
2
2
1+β
2
(18)
(19)
Insert this result into (16).
2 v2
T =
1 v1
2
1+β
2
(20)
The velocities are related to and µ according to,
1
v=√
µ
Eq. 9.68
(21)
Use this to simplify the expression for T .
T = β
2
1+β
2
(22)
Making use of,
µ1 µ2 2 v 2
·
µ1 µ2 1 v 1
=
v12 µ1 v2
·
v22 µ2 v1
µ1 v 1
µ2 v 2
=
=
β
(23)
Expressing everything in terms of β allows us to easily show R + T = 1.
R+T =
1−β
1+β
2
+β
2
1+β
2
(24)
(1 − β)2 + 4β
=
(1 + β)2
=
1 − 2β + β 2 + 4β
1 + 2β + β 2
(25)
=
1
(26)
[3.] Problem 9.14 from Griffiths
Prove that the polarization of reflected and transmitted waves have the same polarization
as the incident wave. Griffiths Hint: Define the polarization vectors of the reflected and
transmitted waves as,
n̂T = cos θT x̂ + sin θT ŷ
n̂R = cos θR x̂ + sin θR ŷ
and then use boundary conditions to prove that θT = θR = 0.
Solution
3
(27)
Regardless of the changes in polarization, this is still a problem of reflection and transmission at normal incidence. The components of the electric and magnetic fields are everywhere parallel to the interface and we use the following boundary conditions (which are
expressed throughout the text and lectures),
~ 1k = E
~ 2k
E
(28)
1 ~k
1 ~k
B1 =
B
µ1
µ2 2
(29)
As usual, set the direction of the incident wave propagation to be in the +ẑ direction.
The electric fields for all of the waves may be written in the form,
∼
∼
~ (z, t) =E o ei(kz−ωt) n̂
E
(30)
where n̂ is the polarization vector.
In this problem the incident wave is,
∼
∼
~ I (z, t) = E oI ei(k1 z−ωt) x̂
E
∼
1 ∼
i(k z−ωt)
ŷ
E oI e 1
v1
~ I (z, t) =
B
(31)
(32)
While the reflected wave is written,
∼
∼
~ R (z, t) = E oR ei(−k1 z−ωt) n̂R
E
∼
= E oR ei(−k1 z−ωt) (cos θR x̂ + sin θR ŷ)
∼
~ R (z, t) =
B
=
(33)
(34)
1 ∼
i(−k z−ωt)
(−ẑ × n̂R )
E oR e 1
v1
(35)
1 ∼
i(−k z−ωt)
(sin θR x̂ − cos θR ŷ)
E oR e 1
v1
(36)
where the cross product is used because the magnetic field must be perpendicular to both
the electric field and the direction of propagation.
The transmitted wave is,
∼
∼
~ T (z, t) = E oT ei(k2 z−ωt) n̂T
E
∼
= E oT ei(k2 z−ωt) (cos θT x̂ + sin θT ŷ)
∼
~ T (z, t) =
B
=
(37)
(38)
1 ∼
i(k z−ωt)
(ẑ × n̂T )
E oT e 2
v2
(39)
1 ∼
i(k z−ωt)
(− sin θT x̂ + cos θT ŷ)
E oT e 2
v2
(40)
4
Condition (28) provides the following (the interface is placed at z = 0 so the exponential
dependence factors out of all the terms),
∼
∼
∼
E oI x̂+ E oR (cos θR x̂ + sin θR ŷ) = E oT (cos θT x̂ + sin θT ŷ)
∼
∼
E oR sin θR = E oT sin θT
(41)
(42)
where (42) is the y component of (41), the usefulness of which will be shown soon.
Condition (29) gives,
1
1 ∼
1 ∼
1
1 ∼
=
E oI ŷ +
E oR (sin θR x̂ − cos θR ŷ)
E oT (− sin θT x̂ + cos θT ŷ)
µ1 v 1
v1
µ2 v 2
This time keep the x component terms.
1 ∼
1 ∼
E oR sin θR = −
E oT sin θT
µ1 v 1
µ2 v 2
∼
∼
E oR sin θR = −β E oT sin θT
(43)
(44)
where β is given in (11).
One way to proceed from this point is to subtract (44) from (42). The result is,
0 = (1 + β) sin θT
(45)
This is satisfied when either β = −1 or the sine term is zero. Using the definition of β,
where the propagation speeds are positive definite, this condition could be met if one of
the permeabilities is negative. Such novel materials are studied and have unique properties, but you are not supposed to be concerned with them here. We generally assume all
permeabilities are close to that of free space and certainly not negative.
Our solution is sin θT = 0. For transmission we limit the angle to 0 ≤ θT ≥ π/2 (if the
angle was greater than π/2 then the wave would be propagating back through the glass
instead of being transmitted). This means that we have shown θT = 0. Now, going back to
either (42) or (44) and considering that the amplitude terms and β are non-zero we see that
sin θR = sin θT = 0 and therefore θR = θT = 0. Replacing this in the equations for all of the
original wave equations shows that they all have the same polarization.
[4.] Problem 9.18 from Griffiths
(a) Let there be an amount of free charge in a piece of glass. Roughly how long will it take
for this charge to entirely migrate to the surface?
(b) How thick should you make the silver coatings to protect yourself from 1010 Hz microwaves? Bear in mind that silver is expensive so you want to balance the protection it
affords with a minimal cost.
(c) What are the wavelength and propagation speed of 1 MHz radio waves in copper?
Compare this to the wavelength and propagation speed of these waves in air (which you
may approximate as vacuum).
5
Solution
(a) The characteristic time, τ , for free charge to get to the surface of its conductor is given
in,
ρf (t) = e−(σ/)t ρf (0)
Eq. 9.120
(46)
where τ ≡ /σ.
Table 7.1 in Griffiths gives the resistivity of glass, and the conductivity σ is the inverse of
the resistivity. Therefore, σ ≈ 10−12 Ω/m.
The permittivity of glass is = o r , where r is the relative permeability. Glass is nonmagnetic so we can write its index of refraction as,
√
n∼
Eq. 9.70
(47)
= r
From lectures during the week of 10-5-2004 and page 386 in Griffiths we know that nglass =
1.5. The permittivity of glass is,
=
o (1.5)2 ≈ 2 × 10−11
(48)
τ≈
2 × 10−11
≈ 20 s
10−12
(49)
The characteristic time we seek,
(b) Since silver is expensive we want to make these coatings as thin as possible while still
blocking the microwaves. The skin depth for microwaves in silver is a measure of how far
these waves can penetrate into the metal. Find this distance and then make the coating
slightly thicker.
Skin depth is given by d ≡ 1/κ where,
#1/2
r "r
σ 2
µ
1+
−1
κ≡ω
2
ω
Eq. 9.126
(50)
ω is the angular frequency of the wave and σ is the conductivity of the material into which
the wave is propagating.
Take the resistivity of silver from Table 7.1 in Griffiths to find that σ = 6.29 × 107 . The
angular frequency is ω = 2πf = 2π × 1010 . We can use ≈ o = 8.85 × 10−12 .
ω = 0.56
(51)
σ
= 1.12 × 108 ≫ 1
ω
(52)
Due to (52) we can simplify κ as,
r
κ = ω
r
= ω
r
=
µ
2
#1/2
σ 2
−1
ω
"r
µ h σ i1/2
2 ω
ωµσ
2
6
(53)
(54)
(55)
The permeability of conductors is approximately that of vacuum. Check Table 6.1 in Griffiths to verify this.
The skin depth is,
r
2
= 6.4 × 10−7 m
ωµσ
so the coatings should be made a little thicker than this.
d
=
(56)
(c) Wavelength, λ, can be written in terms of wave vector as,
λ=
2π
k
(57)
In a conductor the wave vector is given by,
#1/2
r "r
σ 2
µ
+1
k≡ω
1+
2
ω
Eq. 9.126
(58)
As in part (b) we get the conductivity of copper using Table 7.1. Also, the permittivity may
be approximated as o .
σ = 6 × 107
(59)
ω = o (2π × 106 )
=
5.56 × 10−5
σ
= 1.08 × 1012 ≫ 1
ω
We can skip to the same simplifications made in part (a) due to (61).
r
ωµσ
k =
2
= 15390.60
λ = 4.08 × 10−4 m
(60)
(61)
(62)
(63)
(64)
The propagation speed is,
v =
ω
k
= 408.25
Eq. 9.129
(65)
m
s
(66)
In vacuum the speed of and electromagnetic wave is always c = 3.0 × 108 m/s. The corresponding wavelength is,
λ
=
2π
2πc
=
k
ω
= 300 m
7
(67)
(68)
[5.] Problem 9.19 from Griffiths
(a) Show that in a poor conductor the skin depth is,
r
2 d=
σ µ
(69)
and then find the skin depth for water. Your condition of poor conductivity is σ ω.
(b) Show that for a good conductor,
λ
(70)
2π
where the condition for good conducting is σ ω. The wavelength represents that of
the wave while inside the conductor. Calculate the skin depth for a typical metal using
σ = 107 , ω = 1015 , = o , and µ = µo . Why are metals not transparent to this light, which is
in the visible range.
d=
(c) Using the typical metal described in part (b), show that in this situation the magnetic
field lags the electric field by 45◦ . What is the ratio of the magnetic field amplitude to the
electric field amplitude?
Solution
(a) Based on the condition of poor conductivity we can employ the binomial expansion,
(1 + x)n ≈ 1 + nx
(71)
for small x.
Use this to simplify the expression for κ, (50).
1/2
µ
1 σ 2
κ ≈ ω
−1
1+
2
2 ω
r µ
σ
√
= ω
2
2 ω
r
σ µ
=
2 r
(72)
(73)
(74)
The skin depth is the inverse of (74),
2
d=
σ
r
µ
(75)
To find the skin depth for water use parameters given in the text,
Table 4.2
= 80.1o
µ ≈ µo
Table 6.1
σ = 4.0 × 10−6
8
Table 7.1
(76)
(77)
(78)
Using these values the skin depth of water is d ≈ 1.19 × 104 m.
(b) With the condition of good conductivity the expressions for k and κ simplify as in
problem 9.18, (55) and (62). Notice that when this condition is met k = κ. Therefore,
d
=
1
κ
=
1
k
λ
2π
=
(79)
recalling (57).
Leaving the skin depth in terms of κ and using (55) the value for the numbers given is,
r
2
d =
= 1.26 × 10−8 m
(80)
ωµσ
which is approximately 13 nanometers.
Metals are opaque because the skin depth for visible light is very short. If you could obtain
a 5 nm thick sheet of metal you would be able to see through it. Interestingly enough, many
metals are transparent to ultraviolet light. This is a result of the frequency dependence of
the metal’s permittivity; a subject we are not able to cover this quarter.
(c) The magnetic field lags behind the electric field by φ,
φ = δB − δE
Eq. 9.136
(81)
κ
k
Eq. 9.134
(82)
φ ≡ tan−1
Part (b) showed that for a good conductor k = κ, so the lag of the magnetic field is,
φ
=
tan−1 (1)
The ratio of the amplitudes is given by,
s r
σ 2
Bo
= µ 1 +
Eo
ω
=
45◦
Eq. 9.137
(83)
(84)
Using the numerical values from part (b) allows for the following simplification (once again
the 1 has no noticeable effect on the term it is added to),
s r
σ 2
Bo
=
µ
(85)
Eo
ω
r
σ
=
µ
(86)
ω
r
µσ
=
(87)
ω
≈ 10−7
(88)
This ratio actually has units of s/m, which makes sense when one considers that we write
the magnetic components of electromagnetic waves as the electric field amplitude divided
by the wave speed.
9
[6.] Problem 9.21 from Griffiths
Solve for the reflection coefficient of light striking an air-silver interface. Use µ1 = µ2 = µo ,
1 = o , σ = 6 × 107 , and ω = 4 × 1015 .
Solution
The reflection coefficient is defined according to (9). A relationship between the complex
amplitudes is given in,


∼
∼
1− β  ∼
Eq. 9.147
(89)
E oR = 
E oI
∼
1+ β
∼
where β is now a complex quantity.
∼
β =
µ1 v 1 ∼
k2
µ2 ω
∼
∼
E oI
EoR
EoI
2
∼

(90)
Eq. 9.125
k 2 = k + iκ
Arrange this into a useful form,
!2
∼
E oR
Eq. 9.146
(91)
2
1− β 
= 
∼
1+ β



∼
∼
∗
1− β   1− β 
= 
∼
∼
1+ β
1+ β ∗
(92)
=
R
(93)
where the starred variable means complex conjugate.
From previous work in problems 9.18 and 9.19 we know that silver is a good conductor
and has the property k ≈ κ. Using (55) and (62) we can write,
r
ωµσ
k2 = κ2 =
(94)
2
r
∼
µ1 v1 ωµσ
β =
(1 + i)
(95)
µ2 ω
2
r
σ
= µ1 v 1
(1 + i)
(96)
2ωµ2
Now make the numeric substitutions given in the problem to find the prefactor in (96),
noting that v1 = vair ≈ c.
s
r
σ
6 × 107
µ1 v 1
= µo c
(97)
2ωµ2
2(4 × 1015 )µo
= 29
(98)
∼
β = 29(1 + i)
10
(99)
The reflection coefficient is,

∼
∼
∗


1− β   1− β 
R = 
∼
∼
1+ β
1+ β ∗
1 − 29(1 − i)
1 − 29(1 + i)
=
1 + 29(1 + i)
1 + 29(1 − i)
=
−57 + 292 (2)
59 + 292 (2)
(100)
(101)
(102)
(103)
= 0.93
This shows that 93% of the light incident on the air-silver interface is reflected.
[7.] Problem 9.34 from Griffiths
Reference figure 9.27 in Griffiths.
Light of angular frequency ω passes through three linear and homogeneous media. It
comes from the left in medium 1, then passes through a thickness d of medium 2 before
entering medium 3. Show that under normal incidence the transmission coefficient is,
1
(n21 − n22 )(n23 − n22 ) 2 n2 ωd
−1
2
sin
T =
(n1 + n3 ) +
Eq. 9.199
(104)
4n1 n3
n22
c
Griffiths Hint: Treat this as a five wave problem. The central medium will have a right
going and a left going wave inside of it. Write out all of the waves in the problem and then
match boundary conditions at the two interfaces to solve for T .
Solution
Griffiths’ suggested method:
The transmission coefficient is still given as the ratio of the final transmitted intensity over
the incident intensity.
T =
=
1
v E2
2 3 3 oT
1
v E2
2 1 1 oI
(105)
2
3 v3 EoT
2
1 v1 EoI
(106)
Write out all of the waves in a general manner. Beginning with the incident and reflected
waves in medium 1,
∼
∼
∼
~ I (z, t) = 1 E oI ei(k1 z−ωt) ŷ
B
v1
∼
~ I (z, t) = E oI ei(k1 z−ωt) x̂
E
∼
∼
~ R (z, t) = − 1 E oR ei(−k1 z−ωt) ŷ
B
v1
∼
~ R (z, t) = E oR ei(−k1 z−ωt) x̂
E
11
∼
(107)
(108)
In medium 2 I define a right-going and left-going wave,
∼
∼
~ r (z, t) = E or ei(k2 z−ωt) x̂
E
∼
~ r (z, t) = 1 E or ei(k2 z−ωt) ŷ
B
v2
∼
∼
∼
~ l (z, t) = − 1 E ol ei(−k2 z−ωt) ŷ
B
v2
∼
~ l (z, t) = E ol ei(−k2 z−ωt) x̂
E
∼
(109)
(110)
In medium 3 there is only the final transmitted wave.
∼
∼
∼
~ T (z, t) = 1 E oT ei(k3 z−ωt) ŷ
B
v3
∼
~ T (z, t) = E oT ei(k3 z−ωt) x̂
E
(111)
This is a normal incidence problem so we apply the boundary conditions of (28) and (29).
Since we are given that all of the permeabilities are equal, (29) becomes,
~ 1k = B
~ 2k
B
at z = 0
(112)
~ 2k = B
~ 3k
B
at z = d
(113)
Applying the boundary conditions at z = 0 provides (the exponential dependence factors
out as in other similar problems),
∼
∼
∼
∼
E oI + E oR = E or + E ol
∼
∼ 1 ∼
1 ∼
E oI − E oR =
E or − E ol
v1
v2
∼ ∼
∼
v1 ∼
−
−
=
E or E ol
E oI E oR
v2
(114)
(115)
(116)
where the reason for writing (116) (and (119) below) will become apparent soon.
At z = d only the angular frequency parts of the exponentials factor out when the boundary
conditions are applied,
∼
∼
∼
ik d
−ik d
ik d
E or e 2 + E ol e 2 = E oT e 3
(117)
∼
1 ∼
1 ∼
ik2 d
−ik2 d
ik d
e
−
e
=
E or
E ol
E oT e 3
v2
v3
(118)
v2 ∼
ik d
E oT e 3
v3
(119)
∼
∼
ik d
−ik d
E or e 2 − E ol e 2 =
The expressions in (114), (116), (117), and (119) relate the five amplitudes in this problem.
The final answer is a ratio of the transmitted versus incident intensity, so there are really
only four variables and these four equations with which to determine them.
Add (114) and (116) to remove the reflected amplitude.
∼
v1 ∼
v1 ∼
2 E oI =
1+
E or + 1 −
E ol
v2
v2
12
(120)
Add (117) and (119) to remove the left-going wave.
∼
v2 ∼
ik2 d
ik d
2 E or e
=
1+
E oT e 3
v3
∼
1
v2 ∼
id(k −k )
1+
E or =
E oT e 3 2
2
v3
Subtract (119) from (117) to remove the right-going wave.
∼
v2 ∼
−ik2 d
ik d
2 E ol e
=
1−
E oT e 3
v3
∼
1
v2 ∼
id(k +k )
1−
E ol =
E oT e 3 2
2
v3
(121)
(122)
(123)
(124)
By putting (122) and (124) into (120) we have an expression for the incident (complex)
amplitude in terms of the transmitted amplitude. Notice that the problem gives the inverse
of the transmission coefficient. At this step is should be observed that solving for T −1 is
much tidier algebraically.
∼
v1 1
v2 ∼
v2 ∼
v1 1
id(k3 −k2 )
id(k +k )
+ 1−
1+
1−
2 E oI =
1+
E oT e
E oT e 3 2
v2 2
v3
v2 2
v3
The rest is algebra. Some of the in-between steps are shown here. Just multiply everything
out and pull out factors as appropriate.
∼
∼
v1 v2 v1 v2 −ik2 d
v1 v2 v1 v2 ik2 d
E oT eik3 d
2 E oI =
(125)
1+
+
+
e
+ 1−
−
+
e
2
v2 v3 v2 v3
v2 v3 v2 v3
∼
E oT eik3 d
=
2
v1
v1 v2
−ik2 d
ik2 d
−ik2 d
ik2 d
1+
e
+e
+
e
−e
+
v3
v2 v3
(126)
Identities:
e−iθ − eiθ = −2i sin θ
∼
E oI
∼
E oT
2
EoI
EoT
e−iθ + eiθ = 2 cos θ
eik3 d
v1
v1 v2
=
1+
2 cos(k2 d) +
+
(−2i sin(k2 d))
4
v3
v2 v3
1
v1
v1 v2
2
2
=
1+
cos (k2 d) −
+
sin (k2 d)
4
v3
v2 v3
(127)
(128)
(129)
Use sin2 θ + cos2 θ = 1 to place the above equation completely in terms of the sine function
since that is requested in the final form. The transmission coefficient is simplified as,
2
µo 3 v3 EoT
2
µo 1 v1 EoI
2
v1 EoT
=
v3 EoI
T =
13
(130)
(131)
All of the velocities are related to their respective medium according to,
vj =
c
nj
Eq. 9.68
(132)
where c is the speed of light.
Using the relations derived and following the final steps prescribed will produce (104) as
the inverse transmission coefficient.
14