Therefore:
2
i.e. the wave profile remains unchanged.
If y= f2(ct+x), the expression for y at a time t+Δt and a position x+Δx, where c
x t=−Δ
Δ , is given by:
© 2008 John Wiley & Sons, Ltd
x t x
x t t
y x ct f
x x x ct f
x x c x t c f
x x t t c f y
, 1
1 1 1 ,
] [
] [
)]
( ) (
[
)]
( ) ( [
= +
=
Δ + + Δ
−
=
Δ + + Δ
−
=
Δ + + Δ +
Δ =
+ Δ +
i.e. the wave profile also remains unchanged.
5.3
5.4
The pulse shape before reflection is given by the graph below:
The pulse shapes after of a length of Δl of the pulse being reflected are shown below:
(a) Δl =l 4 y
x c y t y
∂
= ∂
∂
∂
x
x
l
4l 3 2l 1
(b) Δl=l 2
(c) Δl =3l 4
(d) Δl=l
5.5
The boundary condition yi+yr = yt gives:
) ( 2 ) ( 1 ) ( 1
kx t i kx
t i kx t
i Be Ae
e
A ω− + ω+ = ω− At x=0, this equation gives:
2 1
1 B A
A + = (5.5.1) Z1 Z2 =∞
4l 3
2l 1
Z1 Z2 =∞
l
Z1 Z2 =∞
The boundary condition t (yi yr) T x
x y T
Ma +
∂
− ∂
∂
= ∂ gives:
) ( 1 )
( 1 )
( 2
kx t i kx
t i kx
t
i ikTAe ikTBe
e ikTA
Ma=− ω− + ω− − ω+
At x=0, a= y&&t = y&&i+y&&r, so the above equation becomes:
1 1
2 2
2 B
c i T c A i T c A i T
MA ω ω ω
ω =− + −
−
i.e. 1 1 A2 c iT M c B
iT c A
iT ⎟
⎠
⎜ ⎞
⎝
⎛− +
=
− ω
Noting that T c=ρc, the above equation becomes:
( )
21
1 i cB M i c A
cA
iρ − ρ = −ω + ρ (5.5.2) By substitution of (5.5.1) into (5.5.2), we have:
( )
( 1 1)1
1 i cB M i c A B
cA
iρ − ρ = −ω + ρ + i.e.
iq iq A
B +
= −
1 1
1
where q=ωM 2ρc
By substitution of the above equation into (5.5.1), we have:
2 1
1 1 A A
iq
A iq =
− +
i.e.
iq A
A
= + 1
1
1 2
5.6
Writing q=tanθ, we have:
θ θ
θ θ
θ θ
e i
i i
iq A
A = −
= +
= +
= + cos
sin cos
cos tan
1 1 1
1
1 2
and
) 2 ( 1
1 sin
sin cos
sin tan
1 tan 1
π
θ θ
θ θ
θ θ
θ = − +
+
= − +
= − +
= − e i
i i i
i iq iq A
B
which show that A2 lags A1 by θ and that B1 lags A1 by (π 2+θ) for 0<θ <π 2
The reflected energy coefficients are given by:
and the transmitted energy coefficients are given by:
θ
Suppose T is the tension of the string, the average rate of working by the force over one period of oscillation on one-wavelength-long string is given by:
∫ ∫
− ∂∂ ∂∂kx dxdt t
which equals the rate of energy transfer along the string.
5.8
π π
ω
3 . 0 5 2 1 . 0
3 .
max 0 =
×
= ×
= A F c T
Noting that ρc=T c, the rate of energy transfer along the string is given by:
] 20[ 1 3 . 0 ) 5 2 3 ( . 0 2 1 2
1 2
2 2 2
2 2
2
W c A
T A
P c π π
ω π ω
ρ = = × × × × =
=
so the velocity of the wave c is given by:
] 30[ 1 . 0 ) 5 2 ( 01 . 0
20 3 2
2 1
2 2 2
2
= −
×
×
×
= ×
= ms
A c P
π π
π ρω
5.9
This problem is not viable in its present form and it will be revised in the next printing. The first part in the zero reflected amplitude may be solved by replacing Z3 by Z1, which then equates r with R′ because each is a reflection at a Z1Z2 boundary. We then have the total reflected amplitude as:
2 4
2
) 1 1
( R
R R tT R
R R tT
R − ′
+ ′
=
′ +
′ +
′ +
+ L
Stokes’ relations show that the incident amplitude may be reconstructed by reversing the paths of the transmitted and reflected amplitudes.
T is transmitted back along the incident direction as tT in Z1 and is reflected as T ′R in
Z2.
R is reflected in Z1 as (R)R=R2 back along the incident direction and is refracted as TR
in the T ′R direction in Z2.
We therefore have tT + R2 =1 in Z1, i.e. tT =1 R− 2 and T(R+ R′)=0 in Z2 giving
R
R=− ′, ∴tT =1−R2 =1−R′2 giving the total reflected amplitude in Z1 as R+ R′=0 with R=−R′.
Fig Q.5.9(a) Fig Q.5.9(b)
θ1 θ1 θ2
1 R
T
θ1 θ1 θ2
tT R
T θ2
R T ′
TR R2
Z1
Z2
Z1
Note that for zero total reflection in medium Z1, the first reflection R is cancelled by the sum of all subsequent reflections.
5.10
The impedance of the anti-reflection coating Zcoat should have a relation to the impedance of air
Zair and the impedance of the lens Zlens given by:
lens air lens
air
coat Z Z n n
Z = = 1
So the reflective index of the coating is given by:
22 . 1 5 .
1 = = 1 =
= air lens
coat
coat n n
n Z
and the thickness of the coating d should be a quarter of light wavelength in the coating, i.e.
] [ 10 12 . 22 1 . 1 4
10 5 . 5 4
7 7
n m d
coat
− −
×
× =
= ×
= λ
5.11
By substitution of equation (5.10) into
x y
∂
∂ , we have:
c t t
B t c A
x
y n
n n n n
n ω ω ω
ω ( cos + sin )cos
∂ =
∂
so:
c y c
t t B
t c A
x
y n n
n n n n n
2 2 2
2 2
2
sin ) sin cos
( ω ω ω ω
ω + =−
−
∂ =
∂
Noting that
k ωcn
= , we have:
2 0
2 2
2 2
2 2
= +
−
=
∂ +
∂ y
y c y c
x k
y ωn ωn
5.12
By substitution of the expression of (yn2)max into the integral, we have:
⎟⎟⎠
Noting that
l
, the above equation becomes:
)
which gives the expected result.
5.13
which is the superposition of standing waves.
5.14
The wave group has a modulation envelope of:
⎟⎠ difference. At a certain time t, the distance between two successive zeros of the modulation envelope Δx satisfies: equation becomes:
λ π
which shows that the number of wavelengths λ contained between two successive zeros of the modulating envelop is ≈λ Δλ
5.15
The expression for group velocity is given by:
dk
By substitution of the expression of v into the above equation, we have:
cos 2
At long wavelengths, i.e. k →0, the limiting value of group velocity is the phase velocity c.
5.16
Noting that the group velocity of light in gas is given on page 131 as:
⎟⎟⎠
The relation
2
2 2 2 2 =ωe +c k
ω (5.17.1) As ω →ωe, we have:
1 1
2
2
2 ⎟ <
⎠
⎜ ⎞
⎝
−⎛
= ω
ωe v
c
i.e. v>c, which means the phase velocity exceeds that of light c. From equation (5.17.1), we have:
) (
)
( 2 d 2 c2k2 d ω = ωe +
i.e. 2ωdω =2kc2dk
which shows the group velocity vg is given by:
c vc c v c c k
dk
vg = d = 2 = 2 = <
ω ω
i.e. the group velocity is always less than c.
5.18
From equation (5.17.1), we know that only electromagnetic waves of ω >ωe can propagate through the electron plasma media.
For an electron number density ne ~1020, the electron plasma frequency is given by:
] [
10 65 . 10 5
8 . 8 10 1 . 9 10 10
6 .
1 31 12 11 1
20 19
0
−
−
−
− = × ⋅
×
×
× ×
×
=
= rad s
m e n
e e
e ε
ω
Now consider the wavelength of the wave in the media given by:
] [ 10 10 3
65 . 5
10 3 2 2 2
2 3
11 8
c m v
v f
v
e e
× −
× =
×
= ×
<
<
=
= π
ω π ω
π ω λ π
which shows the wavelength has an upper limit of 3×10−3m.
5.19
The dispersion relation ω2 c2 =k2+m2c2 h2 gives
) (
)
( 2 c2 d k2 m2c2 h2
d ω = +
i.e. d kdk
c 2
2
2 ω =
ω
i.e. c2
The series in the problem is that at the bottom of page 132. The frequency components can be expressed as:
t t
which is a symmetric function to the average frequency ω0. It shows that at
ω
The frequency of infrared absorption of NaCl is given by:
]
The corresponding wavelength is given by:
]
which is close to the experimental value: 61μm
The frequency of infrared absorption of KCl is given by:
]
The corresponding wavelength is given by:
]
which is close to the experimental value: 71μm
5.22
Before the source passes by the observer, the source has a velocity of u, the frequency noted by the observer is given by:
ν ν c u
c
= −
1
After the source passes by the observer, the source has a velocity of −u, the frequency noted by the observer is given by:
ν ν c u
c
= +
2
So the change of frequency noted by the observer is given by:
) (
2
2 1 2
2 c u
cu u
c c u c
c
= −
⎟⎠
⎜ ⎞
⎝
⎛
− +
= −
−
=
Δν ν ν ν ν
5.23
By superimposing a velocity of −v on the system, the observer becomes stationary and the source has a velocity of u−v and the wave has a velocity of c−v. So the frequency registered by the observer is given by:
ν
ν c u
v c v u v c
v c
−
= −
−
−
−
= −
′′′ ( )
5.24
The relation between wavelength λ and frequency ν of light is given by:
ν = λc
So the Doppler Effect
u c
c
= −
′ ν
ν can be written in the format of wavelength as:
) (
2
u c
c c
= −
′ λ λ
i.e. λ λ c
u c−
′=
Noting that wavelength shift is towards red, i.e. λ′>λ, so we have:
λ λ λ λ c
−u
=
′−
= Δ
i.e. 5[ ]
10 6
10 10
3 1
7 11
8 −
−
− =−
×
×
− × Δ =
−
= c Kms
u λ
λ
which shows the earth and the star are separating at a velocity of 5Kms−1.
5.25
Suppose the aircraft is flying at a speed of u, and the signal is being transmitted from the aircraft at a frequency of ν and registered at the distant point at a frequency of ν′. Then, the Doppler Effect gives:
u c
c
= −
′ ν ν
Now, let the distant point be the source, reflecting a frequency of ν′ and the flying aircraft be the receiver, registering a frequency of ν′′. By superimposing a velocity of −u on the flying aircraft, the distant point and signal waves, we bring the aircraft to rest; the distant point now has a velocity of −u and signal waves a velocity of −c−u. Then, the Doppler Effect gives:
u c
u c c
u c u
u c
u c
−
= +
′ +
− =
−
−
−
−
′ −
′′=ν ν ν
ν ( )
which gives:
] [ 750 10
10 3 3 2
10 15 2
1 8
9
3 × × = −
×
×
= × Δ +
= Δ
′′+
′′−
= c c ms
u ν ν
ν ν
ν ν ν
i.e. the aircraft is flying at a speed of 750m s
5.26
Problem 5.24 shows the Doppler Effect in the format of wavelength is given by:
λ
λ c
u c−
′=
where u is the velocity of gas atom. So we have:
λ λ λ
λ c
= u
′−
= Δ
i.e.
] [ 10 1 10 10 3
6 10
2 8 3 1
7
12 −
−
− × × = ×
×
= ×
= Δ
′−
= c ms
u λ
λ λ λ
The thermal energy of sodium gas is given by:
kT u
mNa 2 3 2
1 2 =
where k =1.38×10−23[JK−1] is Boltzmann’s constant, so the gas temperature is given by:
] [ 10 900
38 . 1 3
1000 10
66 . 1 23
3 23
2 27
2
k K u
T mNa ≈
×
×
×
×
= ×
= − −
5.27
A point source radiates spherical waves equally in all directions.
⎟⎠
⎜ ⎞
⎝
⎛
− ′
′=
u c
v vc : Observer is at rest with a moving source.
⎟⎠
⎜ ⎞
⎝
⎛ − ′
′′= c
v
v c : Source at rest with a moving observer.
⎟⎠
⎜ ⎞
⎝
⎛
− ′
− ′
′′′=
u c
v
v c : Source and observer both moving.
5.28
By substitution of equation (2) into (3) and eliminating x′, we can find the expression of t′
given by:
⎥⎦⎤
⎢⎣⎡ − −
= ′
′ 1 ( )
vt x k k
x t v
Now we can eliminate x′ and t′ by substituting the above equation and the equation (2) into equation (1), i.e.
2
2 2 2 2
2 2
2 ( ) ( )⎥⎦⎤
⎢⎣⎡ − −
− ′
−
=
− k x vt
k x v vt c x k t c x
i.e.
0 1 1
1 2
1 2 2 2
2 2 2 2
2 2
2
2 2
2 ⎥ =
⎦
⎢ ⎤
⎣
⎡ ⎟⎟⎠−
⎜⎜ ⎞
⎝
⎛ −
⎥ +
⎦
⎢ ⎤
⎣
⎡ ⎟
⎠
⎜ ⎞
⎝⎛ − + ′
⎥ +
⎥⎦
⎤
⎢⎢
⎣
⎡ ⎟
⎠
⎜ ⎞
⎝⎛ − + ′
− c t
v v c k xt k k
v k c kv x k k
v k c
which is true for all x and t if and only if the coefficients of all terms are zeros, so we have:
2
2 2
2 1
1 ⎟
⎠
⎜ ⎞
⎝
⎛
− ′
=
− k k
v k c
2 2 2
2
1 v
k c v k
c ⎟⎟⎠ ′=
⎜⎜ ⎞
⎝
⎛ −
α o′
v s′
u
θ
θ cos u
u′= v′=vcosθ s′
o′
θ
u
θ cos u u′=
θ o′
v
θ cos v v′= s′
2 2 2
2(c v ) c
k − =
The solution to the above equations gives:
1 2
1 β
= −
= k′ k
where, β =v c
5.29
Source at rest at x1 in O frame gives signals at intervals measured by O as Δt=t2−t1
where t2 is later than t1. O′ moving with velocity v with respect to O measures these intervals as:
)
( 2
1
2 x
c t v k t t
t′ − ′=Δ ′= Δ − Δ with Δx=0 t
k t′= Δ Δ
∴ )
(x2 x1
l = − as seen by O, O′ sees it as (x2′ −x1′)=k[(x2−x1)−v(t2−t1)].
Measuring l′ puts t2′ =t1′ or Δt′=0
0 ) ( 2 1
2 =
⎥⎦⎤
⎢⎣⎡Δ − −
′= Δ
∴ x x
c t v k
t i.e. 2(x2 x1) t2 t1
c
t = v − = −
Δ
k x x x
c x x v x k t v x x k x x
l 2 2 1 2 1
2 1 2 1
2 1
2 [( ) ( )] ( ) ( ) −
⎥=
⎦
⎢ ⎤
⎣
⎡ − − −
= Δ
−
−
′=
′ −
′=
∴
k l l′=
∴
5.30
Two events are simultaneous (t1 =t2) at x1 and x2 in O frame. They are not simultaneous in O′ frame because:
⎟⎠
⎜ ⎞
⎝⎛ −
′ =
⎟≠
⎠
⎜ ⎞
⎝⎛ −
′ = 1 2 1 2 2 2 2
1 x
c t v k t c x
t v k
t i.e. x1 ≠ x2
5.31
The order of cause followed by effect can never be reversed.
2 events x1,t1 and x2,t2 in O frame with t2 >t1 i.e. t2− t1 >0(t2 is later).
⎥⎦⎤
⎢⎣⎡ − − −
′ =
′ − 1 ( 2 1) 2 ( 2 1)
2 x x
c t v t k t
t i.e.
⎥⎦⎤
⎢⎣⎡Δ − Δ
′=
Δ x
c t v k
t 2 in O′ frame.
Δt′ real requires k real that is v<c, Δt′ is +ve if ⎟
⎠
⎜ ⎞
⎝
> ⎛ Δ
Δ c
x c
t v where
c
v is +ve
but <1 and
c Δx
is shortest possible time for signal to traverse Δx.