PHys ].ooB
HU * A. Prolrlem * a
Kurl
Tho*pson
rlarf
,) ^) €*t"*
rs
6(*\?=O
up
"r',
.,- -, { *'o
'lp
/(-e?
L_
\, =
b #;
b
tr= il.
Lfr, L f #t
".
id,
.iad,
"?6r
r
eb
o"
{ i gizt**atfl6. *
\t 6, eis' ei*r.!
T"= ( ?.0:'* \0," * if lJ') t + (?.d,' +'r6,'r'tfL:\:y
r"= ti'P;* \/i' r \ €$r'tt -Li(,O;f * tof)" * t'(6i)^ * fr4'0,' +abiil" rcd,'oilt
S,^.b.tit^le \.b *\re 4i46*sltial eguc,ticrn, o^d o[rviJe \,,J ] ;
g.r?-,+
+Q=O
to |hei. por^rer o$ €-',
t
:.fr;"d,'k* (',fr", * (ri)* -rd; dr'le
cr.((or6r.q
(o - (d:)-\ * (i6J' -
iils gedera\--*va,'irh
(u\
\
6l'
(l)
-
\d','
-
I
|)
- to'rlt - zt(ol)" - ft(d'')* 'xr.6nt6,t - te'd,'dr'* rc' 6,'aL
iedj' + \ex6 ", * \{fi{
G*tli^.c terms
6,t=o
res.r\t
i\ in +\e
{tll--it.+--tou4L.( enueii
eq,uo*ionr.-
=o
(o',)t - ),do'6,t = o
l^6J d,'
,
--t
I
0
F1o*ti
dl
i\
tt) \f,
tt
e eikora"i e6u.*roo &" S.
"(
4
So\ve
t,1 Lei6nit i"tetro\
xoe lR
ls*-os&r
j" ..'\vo (e\ r-e nee,L-rlit
-Ql: [aP'l
l ta-fl -
=
Solr" e 16.r** io ,r t *)
:
\6n" _ I oo'd,, = o
o'oi + lql d,' -- o .\
IJQLQ,,= f, i q'q'a
t!
il,'
Qta
T\.a
Y *^
^l
aggoxic""*ion
e
=
L
-Y
ea'
,
L
T
6u
,'rx
LA,
f 1",4t )ot+) -! t^1ot- +-*-**"f
e
I\!.r".*eOco:j11q,{_i66
''-Tl- *-'" -- -- - *
i:_
J
cll*\
not
{ ck)-}O- }ler *,bs
-,1_#_ Q _(il
-aol
lO
_Lhea .*hs
-l
lq\C.\ +
fi' t or\
:a ., d! u-t{,'*
-f
,,-rlc.
*,$.
A
/,
,
Ae (-
is 3,ver
{,J
-ij;a*G-14 -it^tot + 16,
-"-*"- *- | ?r /X
-a{
j
+- e;e-_ u. i *, dtlqq+1
c,rc. € C
vo\-r"L*--Sr- (l.(d = O=---(-ful$-fe!e) *s"pl'.{ -rr :: 3:,r-rs zn *ffgrtkd-b
-&[!ttea-
t \qs--- omg&.J:rLkrt-- -go---as* ex4nmila\s
'-'Y{-s$er'
- '-- bl.,
5 -Js
-st.uJ,ior
possi
-gEn -.... lh-err{uce
-r
-
d
e,
a rliv+jer+ 5gier ,
in,e- rh.- serier .X en-'finLr\ is i,\
$envir\
\i^e- <eries nQe,!{ ,hr L€ tcrntc,,+ce$ ?.f*,1 lb poJua o. .{1*t Afi)r"xirr^e{ ion
S, Y, T,- qo,aera'^\ cor-Jlt\on= nee.l +. f,e rtret i'r otJer 4r *.
-6
t*J (')c-"yp"txi^r*ioo'- ((
tn-'ln(r\
L^ 6n*,t")
t-nilnl (*) << 1 6.s r -]o
Ita- -ii.s| 6enstiti on L&icrr+e r t*Loct e acl. {<,rr" i,\ +L.{- srj ie s rnr.r,sf be .,.uql.,
snncr\\er *ho,r {Le tfcviou{ t.orm. 1\r sccrJ cor.c}rt\on s'la&s {do.t
t\r
ter* cr4{er \-l'e" \c*ncc.*oC €u* *ruE\ \e- mueh \ nnct|lc. *han 1*For this
grublan \ \.e Cor.chtio,rs betonne
a^,[ (r)
:
3.)
Question: Consider an acoustic wave propagating in a 3D medium with index
of refraction n = n(x, y, z)
a.)
For a short wavelength, constant frequency excitation, so that the ray theory is
applicable, discuss under what circumstances you will be able to solve the ray
equations.
We begin with the wave equation:
1 ∂2ψ
∂2ψ
+ 2 2 =0
2
∂x
c ∂t
(1)
If we assume an oscillatory solution to Eq (1), and using the relation n = c0 /c,
then we can write:
∂ 2 ψ ω2
+ 2ψ=0
∂x2
c
∂ 2 ψ ω2 2
+ 2 n (x, y, z)ψ = 0
∂x2
c0
(2)
Now, the problem states that the light we are concerned with has a short wavelength. Thus the wave can be treated as a ray and we can approximate ψ(x)
as:
" ∞
#
i X m
ψ(x) = A(x) exp
φm (x)
m=0
i
(3)
ψ(x) ≈ A(x) exp φ0 (x)
We now plug Eq (3) into the wave equation, Eq (2). I suppress the arguments
of the functions for simplicity:
(∇φ0 )2
i 2
2i
ω2 2
2
−
A
+
∇
φ
A
+
(∇φ
·
∇A)
+
∇
A
+
n A=0
(4)
0
0
2
c20
We now take the limit→∞ . If we also assume that A(x) is slowly varying (so
that ∇2 A ≈ 0) the dominate term in Eq (4) becomes:
−(∇φ0 )2 A = −
(∇φ0 )2 =
1
ω2 2
n A
c20
ω2 2
n
c20
(5)
Which is just the Eikonal equation as expected. Expanding the previous equation out:
2 2 2
∂ψ0
∂ψ0
∂ψ0
ω2
(6)
+
+
= 2 n2 (x, y, z)
∂x
∂y
∂z
c0
Thus, we will only be able to solve the ray equations if:
φ0 (x) = φx (x) + φy (y) + φz (z)
We furthermore need n2 (x, y, z) to respect this structure as well, or
n2 (x, y, z) = a(x) + b(y) + c(z)
b.)
Question: In the case of a.), calculate the Eikonal phase φ(x)
Assuming the condition of part a.) are true:
2 2
∂φy
∂φz
ω2
+
+
= 2 [a(x) + b(y) + c(z)]
∂y
∂z
c0
) (
) (
)
(
2
2
2
ω2
∂φy
ω2
∂φz
ω2
∂φx
− 2 a(x) +
− 2 b(y) +
− 2 c(z) = 0
∂x
c0
∂y
c0
∂z
c0
∂φx
∂x
2
For the above to be true at all points in space, each of the bracketed terms must
be equal to some constant. Let’s define:
(
)
2
∂φx
ω2
− 2 a(x) = f1
∂x
c0
(
)
2
∂φy
ω2
− 2 b(y) = f2
∂y
c0
)
(
2
∂φz
ω2
− 2 c(z) = f3
∂z
c0
Where fi is a constant for i = 1, 2, 3. We can thus write:
s
∂φx
ω2
= ± f1 + 2 a(x)
∂x
c0
s
∂φy
ω2
= ± f2 + 2 b(y)
∂y
c0
s
∂φz
ω2
= ± −(f1 + f2 ) + 2 c(z)
∂z
c0
2
We can now integrate these equations to arrive at the final form for the Eikonal
phase, φ(x):
Z
φ(x) = ±
s
dx
ω2
f1 + 2 a(x) ±
c0
s
Z
dy
3
ω2
f2 + 2 b(y)dy ±
c0
s
Z
dz
−(f1 + f2 ) +
ω2
c(z)
c20