1
Answers Question 1
(i)
Vector Diagram
θ
d
θ
If the phase of the light from the first slit is zero, the phase from second slit is

2

d sin 

x
Adding the two waves with phase difference  where   2  ft   ,


a cos(   )  a cos( )  2a cos( / 2)   / 2
a cos(   )  a cos( )  2a cos  cos   
This is a wave of amplitude A  2a cos  and phase β. From vector
diagram, in isosceles triangle OPQ,
1
2
 
and

d sin 

( NB   2 )
A  2a cos  .
Thus the sum of the two waves can be obtained by the addition of two vectors of
amplitude a and angular directions 0 and  .
(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.
Figure 1.2
2
^
^
O S T  OTS 
1
(180   )
2
^
and
T OS 
In the triangle TOS, for example
a  2R sin( / 2)  2R sin  as (  2 )
R 
a
2 sin 
(1)
As the polygon has N faces then:
^
^
T O Z  N (T O Z )  N  2 N
Therefore in isosceles triangle TOZ, the amplitude of the resultant wave, TZ, is given by
2R sin N .
Hence form (1) this amplitude is
a sin N
sin 
Resultant phase is
^
ZT S
^
^
 OT S OT Z
 1

 90    180  N 
2 2

1
 N  1
2
 ( N  1) 
(iii)
1
sin 
a sin N
Intensity I 
β
a 2 sin 2 N
sin 2 
3
I
β
37C
π
0
2π
3π
(iv) For the principle maxima   p where p  0  1  2........
 N ' 
  N 2 a 2
I max  a 2 

'


 '  0 and   p   '
(v) Adjacent max. estimate I1 :
sin 2 N  1,   2p 
3
3
i.e   
2N
2N
 

   p  2 N  does not give a maximum as can be observed from the graph.


I1  a 2
1
3 2
2n
Adjacent zero intensity occurs for    
For phase differences much greater than  ,
(vi)

N

a2N 2
for N>>1
23
i.e.   

N
 sin N
I  a 2 
 sin 

  a 2

.
  n for a principle maximum

i.e. d sin   n
n  0,1,2..........

Differenti ating w.r.t, 
d cos   n
n
 
d cos 
Substituting   589 .0nm,     589.6 nm. n  2 and d  1.2 10 6 m.
 
n
 n 
d 1 

 d 
2
n
 n 
as sin  
and cos   1  

d
 d 
   5.2 10 3 rads or 0.30 0
2
4
Answers Q2
2.(i)
E
X
Figure2.1
R
θ θ
O
φ
R
X’
EX  2 R sin   t 
2 R sin 
v
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,
R 
2  2 cos1  C ,
 R 
Thus
t
2R sin 
,
v
R 
for   cos1  C ,
 R 
where v = vP for P waves and v = vS for shear waves.
(ii)
RC
 0.5447
R
vCP
 0.831 .3
vP
and
Figure 2.2
A
i
E
B
r

C
Rc
X
θ
O
From Figure 2.2
^
^
  A O C  E O A    (90  r )  (1   )
(1)
5
(ii) Continued
Snell’s Law gives:
sin i v P

.
sin r v CP
(2)
From the triangle EAO, sine rule gives
RC
R

.
sin x sin i
(3)
Substituting (2) and (3) into (1)

 v CP

 vP

R

sin i   i  sin 1  C sin i 
 R


  90  sin 1 
(4)
(iii)
For Information Only
For minimum  ,
d
 0 .  1
di
 v CP

 vP

 cos i

v

1   CP sin i 
v
 P

2

 RC

 R

 cos i

R

1   C sin i 
R


2
0
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.
P l o t of θ a ga i n s t i .
900
75.80
550
0
900
Substituting into 4:
i=0
gives
θ = 90
i = 90°
gives
θ = 90.8°
Substituting numerical values for i = 0  90° one finds a minimum value at i = 55°; the
minimum values of 0, θMIN = 75•8°.
6
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.
Refracted
waves
θ
0
R
 C  cos 1  C
 R



 min
1800
(iv)
900
900
Using the result
t
2r sin 
v
the time delay Δt is given by
1
1 
t  2 R sin    
 vS vP 
Substituting the given data
1 
 1
131  2(6370 ) 

 sin 
 6.31 10 .85 
Therefore the angular separation of E and X is
2  17 .84 o
 RC
 R
This result is less than 2 cos 1 

 3470 
o
  2 cos 1 
  114
 6370 

And consequently the seismic wave is not refracted through the core.
7
(v)
E
Y
X
D
R
R
θ θ
O
RC
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
1
1 
t '  (ED  DX )   
 vS vP 
1
1 
t '  2(ED )    as ED  EX by symmetry
v
v
P 
 S
In the triangle EYD,
(ED) 2  ( R sin  ) 2  ( R cos   RC ) 2
(ED) 2  R 2  RC2  2 RRC cos 
sin 2  cos 2   1
Therefore
1
1 
t '  2 R 2  RC2  2 RRC cos    
v
v
P 
 S
Using (ii)
t
2
R 2  RC  2 RRC cos 
R sin 
 396 .7 s or 6m 37s
t ' 
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.
8
Answer Q3
Equations of motion:
m
m
m
d 2 u1
dt 2
d 2u2
dt 2
d 2u3
dt 2
Substituting un (t ) u n (0) cos t and o 
2
 k (u 2  u1 )  k (u 3  u1 )
 k (u 3  u 2 )  k (u1  u 2 )
 k (u1  u 3 )  k (u 2  u 3 )
k
:
m
(2 o   2 )u1 (0)  o u 2 (0)  o u 3 (0)  0
(a)
 o u1 (0)  (2 o   )u 2 (0)  o u 3 (0)  0
(b)
 o u1 (0)  o u 2 (0)  (2 o   )u 3 (0)  0
(c)
2
2
2
2
2
2
2
2
2
2
2
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
2
2
(3o   2 ) 2   0

2
 3 o , 3o and 0
2
  3 o ,
2
3 o and 0
(ii) Equation of motion of the n’th particle:
d 2un
 k (u1 n  u n )  k (u n 1  u n )
dt 2
d 2un
2
 k (u1 n  u n )   o (u n 1  u n )
2
dt
 

Substituting u n (t )  u n (0) sin 2ns  cos  s t
N

m
n = 1,2……N
 

 
 




2
2
  s  sin 2ns     o sin 2(n  1) s   2 sin 2ns   sin 2(n  1) s 
N 
N
N
N 


 
 
 
 
  1 
 



2
2 1
  s  sin 2ns    2 o  sin 2(n  1) s   sin 2ns   sin 2(n  1) s 
N 
N
N 2 
N 

 
2 
 
   
 



2
2
  s  sin 2ns    2 o sin 2ns  cos 2s   sin 2ns 
N 
N  N
N 

 
 
  
2
2
  s  2 o 1  cos 2s  : ( s  1,2,..... N )
 N 

As 2 sin 2   1  cos 2
This gives
 s 

 N 
 s  2 o sin
 s can have values from 0 to 2 o  2
(s  1,2,... N )
k
N
when N  ; corresponding to range s  1 to .
m
2
9
(iv) For s’th mode
un

u n 1
un
u n 1
 

sin  2ns 
N

 

sin  2(n  1) s 
N



sin 2ns 
N


  
  


sin 2ns  cos 2s   cos 2ns  sin 2s 
N
N
N  N

 


un
 
 
 s 


1.
  0, thus cos 2ns   1 and  sin 2ns   0, and so
N
N
u n 1
N


(b) The highest mode,  max  2 o , corresponds to s = N/2
(a) For small  , 

Case (a)
Case (b)
N odd
N even
un
sin 2n 
 1
 1 as
sin 2(n  1) 
u n 1
10
(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
m‘
m
m
m' x  2kx
'2 
2k
m
' 
2k
m'
For small m', ω' will be much greater than ωmax,
BAND
2ωo
0
ω
ω
DIATOMIC SYSTEM
More light masses, m', will increase the number of frequencies in region of ω' giving a bandgap-band spectrum.
BAND
GAP
BAND
DD
0
ω