The Role of FIDO in a Cyber Secure Netherlands: FIDO Paris Seminar.pptx
Solution i ph o 17
1. Answers Question 1
(i) Vector Diagram
If the phase of the light from the first slit is zero, the phase from second slit is
θ
λ
π
φ sin
2
d=
Adding the two waves with phase difference φ where
−=
λ
πξ
x
ft2 ,
( )
( ){ }βξβξφξ
φξφξφξ
+=++
+=++
coscos2)cos()cos(
2/)2/cos(2)cos()cos(
aaa
aaa
This is a wave of amplitude βcos2aA = and phase β. From vector
diagram, in isosceles triangle OPQ,
θ
λ
π
φβ sin
2
1
d== )2( βφ =NB
and
.cos2 βaA =
Thus the sum of the two waves can be obtained by the addition of two vectors of
amplitude a and angular directions 0 and φ.
(ii) Each slit in diffraction grating produces a wave of amplitude a with phase 2β relative to
previous slit wave. The vector diagram consists of a 'regular' polygon with sides of
constant length a and with constant angles between adjacent sides.
Let O be the centre of circumscribing circle passing through the vertices of the
polygon. Then radial lines such as OS have length R and bisect the internal angles
of the polygon. Figure 1.2.
1
Figure 1.2
d
θ
θ
2. φ
φ
=
−==
SOT
OTSTSO
^
^^
and
)180(
2
1
In the triangle TOS, for example
βφ sin2)2/sin(2 RRa == as )2( βφ =
βsin2
a
R =∴ (1)
As the polygon has N faces then:
βφ NNZOTNZOT 2)(
^^
===
Therefore in isosceles triangle TOZ, the amplitude of the resultant wave, TZ, is given by
βNR sin2 .
Hence form (1) this amplitude is
β
β
sin
sin Na
Resultant phase is
( )
( )
β
φ
φ
φ
)1(
1
2
1
180
2
1
2
90
^^
^
−=
−−
−−
−
−=
=
N
N
N
ZTOSTO
STZ
(iii)
2
Intensity
β
β
2
22
sin
sin Na
I =
β
β
Na sin
sin
1
β
3. 37C
(iv) For the principle maxima ........210where ±±== ppπβ
'and0'
'
' 222
max βπββ
β
β
+===
= paN
N
aI
(v) Adjacent max. estimate I1 :
NN
pN
2
3
i.e
2
3
2,1sin 2 π
β
π
πββ ±===
±=
N
p
2
π
πβ does not give a maximum as can be observed from the graph.
23
2
3
1 22
2
2
1
Na
n
aI ==
π for N>>1
Adjacent zero intensity occurs for
N
π
πρβ ±= i.e.
N
π
δ ±=
For phase differences much greater than
sin
sin
aI, 22
a
N
=
=
β
β
δ .
(vi)
θ
λ
θ
λθθ
λ
πθ
λ
π
πβ
cos
cos
w.r.t,atingDifferenti
..........2,1,0sini.e.
maximumprincipleafor
d
n
nd
nnd
n
∆
=∆
∆=∆
±±==
=
Substituting m.102.1and2nnm.589.6nm,0.589 6−
×===∆+= dλλλ
2
1
−
∆
=∆
d
n
d
n
λ
λ
θ
as
2
1cosandsin
−==
d
n
d
n λ
θ
λ
θ
03
30.0or102.5 rads−
×=∆⇒ θ
3
I
0 π 2π 3π
β
4. v
R
tR
θ
θ
sin2
sin2EX =∴=
where v = vP for P waves and v = vS for S waves.
This is valid providing X is at an angular separation less than or equal to X', the tangential ray
to the liquid core. X' has an angular separation given by, from the diagram,
,cos22 1
= −
R
RC
φ
Thus
,cosfor,
sin2 1
≤= −
R
R
v
R
t C
θ
θ
where v = vP for P waves and v = vS for shear waves.
(ii) 3.831.0and5447.0 ==
P
CPC
v
v
R
R
From Figure 2.2
4
θ θ
R
O
R
φ
E X
X’
2.(i)
r
Figure2.1
Figure 2.2
α
O
X
E
BA
C
Rc
θ
i
Answers Q2
6. (ii) Continued
Snell’s Law gives:
.
sin
sin
CPv
v
r
i P
= (2)
From the triangle EAO, sine rule gives
.
sinsin i
R
x
RC
= (3)
Substituting (2) and (3) into (1)
−+
−= −−
i
R
R
ii
v
v C
P
CP
sinsinsinsin90 11
θ (4)
(iii)
Plot of θ against i.
Substituting into 4:
i = 0 gives θ = 90
i = 90° gives θ = 90.8°
6
For Information Only
For minimum 0, =
di
dθ
θ . 0
sin1
cos
sin1
cos
1 22
=
−
−
−
−⇒
i
R
R
i
R
R
i
v
v
i
v
v
C
C
P
CP
P
CP
Substituting i = 55.0o
gives LHS=0, this verifying the minimum occurs at this
value of i. Substituting i = 55.0o
into (4) gives θ = 75.80
.
0 550
900
900
75.80
7. Substituting numerical values for i = 0 → 90° one finds a minimum value at i = 55°
; the
minimum values of 0, θMIN = 75•8°.
Physical Consequence
As θ has a minimum value of 75•8° observers at position for which 2 θ <151•6° will not
observe the earthquake as seismic waves are not deviated by angles of less than 151•6°.
However for 2 θ ≤ 114° the direct, non-refracted, seismic waves will reach the observer.
(iv) Using the result
v
r
t
θsin2
=
the time delay Δt is given by
−=∆
PS vv
Rt
11
sin2 θ
Substituting the given data
θsin
85.10
1
31.6
1
)6370(2131
−=
Therefore the angular separation of E and X is
o
84.172 =θ
This result is less than
oC
R
R
114
6370
3470
cos2cos2 11
=
=
−−
7
0 cos min
1
θθ
= −
R
RC
C 900
9001800
Refracted
waves
θ
9. (v)
The observations are most likely due to reflections from the mantle-core interface. Using the
symbols given in the diagram, the time delay is given by
symmetrybyEXEDas
11
)ED(2'
11
)DXED('
=
−=∆
−+=∆
PS
PS
vv
t
vv
t
In the triangle EYD,
1cossincos2(ED)
)cos()sin((ED)
22222
222
=+−+=
−+=
θθθ
θθ
CC
C
RRRR
RRR
Therefore
−−+=∆
PS
CC
vv
RRRRt
11
cos22' 22
θ
Using (ii)
37s6mor7.396
cos2
sin
'
22
s
RRRR
R
t
t CC
⇒
−+
∆
=∆ θ
θ
Thus the subsequent time interval, produced by the reflection of seismic waves at the mantle core
interface, is consistent with angular separation of 17.840
.
9
E XY
R R
D
θ θ
O RC
10. Answer Q3
Equations of motion:
)()(
)()(
)()(
32312
3
2
21232
2
2
13122
1
2
uukuuk
dt
ud
m
uukuuk
dt
ud
m
uukuuk
dt
ud
m
−+−=
−+−=
−+−=
:andcos)0()(ngSubstituti
2
m
k
tutu onn == ωω
(c)0)0()2()0()0(
(b)0)0()0()2()0(
(a)0)0()0()0()2(
3
22
2
2
1
2
3
2
2
22
1
2
3
2
2
2
1
22
=−+−−
=−−+−
=−−−
uuu
uuu
uuu
ooo
ooo
ooo
ωωωω
ωωωω
ωωωω
Solving for u1(0) and u2(0) in terms of u3(0) using (a) and (b) and substituting into (c) gives the
equation equivalent to
0)3(
2222
=− ωωωo
22
3 oωω = ,
2
3 oω and 0
oo ωωω 3,3= and 0
(ii) Equation of motion of the n’th particle:
)()(
)()(
1
2
12
2
112
2
nnonn
n
nnnn
n
uuuuk
dt
ud
uukuuk
dt
ud
m
−+−=
−+−=
−+
−+
ω
Substituting t
N
nsutu snn ω
π
cos2sin)0()(
=
θθ
π
ωω
πππ
ω
π
ω
πππ
ω
π
ω
πππ
ω
π
ω
2cos1sin2As
),.....2,1(:2cos12
2sin2cos2sin22sin
)1(2sin
2
1
2sin)1(2sin
2
1
22sin
)1(2sin2sin2)1(2sin2sin
2
22
22
22
22
−=
=
−=∴
−
=
−
−−
+
+=
−
−+
−
+=
−
Ns
N
s
N
ns
N
s
N
ns
N
ns
N
sn
N
ns
N
sn
N
ns
N
sn
N
ns
N
sn
N
ns
os
os
os
os
This gives
),...2,1sin2 N(s
N
s
os =
=
π
ωω
.
2
to1rangetoingcorrespond;when22to0fromvalueshavecan
N
sN
m
k
os =∞→=ωω
10
n = 1,2……N
11. (iv) For s’th mode
+
=
+
N
sn
N
ns
u
u
n
n
π
π
)1(2sin
2sin
1
+
=
+
N
s
N
ns
N
s
N
ns
N
ns
u
u
n
n
ππππ
π
2sin2cos2cos2sin
2sin
1
(a) For small
1soand,02sinand12costhus,0,
1
≅≈
=≅
≈
+n
n
u
u
N
ns
N
ns
N
s ππ
ω .
(b) The highest mode, oωω 2max = , corresponds to s = N/2
as1
1
−=∴
+n
n
u
u ( )
( )
1
)1(2sin
2sin
−=
+ π
π
n
n
Case (a)
Case (b)
N odd
N even
11
12. (vi) If m' << m, one can consider the frequency associated with m' as due to vibration of m'
between two adjacent, much heavier, masses which can be considered stationary
relative to m'.
The normal mode frequency of m', in this approximation, is given by
'
2
'
2
'
2'
2
m
k
m
k
kxxm
=
=
−=
ω
ω
For small m', ω' will be much greater than ωmax,
DIATOMIC SYSTEM
More light masses, m', will increase the number of frequencies in region of ω' giving a band-
gap-band spectrum.
12
m
m m
0 2ωo ω ω
BAND
BAND GAP BAND
0 ω