Transcript Invisible Fishing Line?

Mathematical Theory of Fishing
Line Visibility
Is there such a thing as an “invisible”
fishing line?
Jeff Thomson
Invisible Fishing Line?
 2 types of line - monofilament &
fluorocarbon (FC)
– FC has index of refraction near water’s
– Touted by manufacturers as “invisible”
– Much more expensive than mono
 How much less visible is FC than mono?
– We will analyze scattering using Mie theory
– Limit in which diameter of line is much larger
than wavelength of light.
Outline of Presentation
 Simple cases suggest a strong dependence
on difference of refraction index:
– Reflection from plane
– Reflection from slab
 Mie theory analysis does not show this
 Tabletop comparison of mono and FC line
in water shows no observable differences
 Approximate refraction theory agrees with
Mie results gives insight into reason
Simple Case - Reflection
from Plane
In region 1:
Ee
ik1 x
k2
k1
 ik1 x
 re
In region 2:
E  te
ik2 x
Match E & B across x=0:
k2
1  r  t,1  r  t
k1
E
z
k
B=-(k/w)E
y
x
Plane Reflection continued
 Solve for reflection coefficient
k2
1 
  ,r
k1
1 
R
1   2

R  rr*  


1   
Obviously a strong dependence on ratio of
indexes of refraction

More Complex - Reflection from
a Slab
In region 1:
Ee
ik1 x
k1
 re
E  ae
k1
ik1 x
In region 2:
ik2 x
k2
 be
 ik2 x
X=0
X=L
In region 3:
E  te
ik1 x
Now we apply BC at x=0 and
X=L
Slab - continued
At x=0:
1  r  a  b,1 r   (a  b)
R
At x=L:
ae
ik2 L
 be
 (ae
ik2 L
ik2 L
ik 2 L
 be
 te
ik1 L
)  te
ik1 L
k1L=10

Solving for r:
(  1)  (   1)e i2k 2 L
r 2
2
2 i 2k2 L  1
(  1)  (   1) e
Also strong
dependence on 
Reflection from Cylinder
 We will use Mie scattering theory and
obtain an exact answer
– Incident plane wave is expressed in cylindrical
wave functions
– Scattered and internal fields are also expressed
in cylindrical wave functions
– Boundary conditions are applied at edge of
cylinder
• I.e. tangential fields (Ez, Bq) are continuous
Definition of Some Terms
 J and Y are Bessel functions of the first kind
 H(1) is the Hankel function of the first kind
 The Wronskian is W(J,Y) = J(z)Y(z)/ -
Y(z)J(z)/ = 2/pz
 Fields - E electric, B magnetic
Definition of fields
Incident field
y
z
Ez  eik1 x  e ik1 r cosq   i n Jn k1 rein q
x
n
Br  
Bq 
r
q
z
1  Ez
1

i n nJ n k1 re inq

iw r q
wr n
1  Ez k1

i n Jn ' k1r ein q

iw  r i w n
Ez   i nn Hn(1) k1r ein q
Scattered field
n
Bq 
Internal field
k1
iw
 i  H ' k re
n
n
1
n
Ez   i n n Jn k2 r ein q
n
n
inq
Bq 
k2
iw
 i  J k re
n
n
n
n
2
inq
Continuity of Tangential
Components
At r=a:
Jn (k1 a)  n Hn (k1a)  n Jn (k2 a)
Jn ' (k1 a)  n Hn ' (k1a)  n Jn ' (k 2 a)
Solving for n
n 
Jn ' ()Jn ()  J n ' ( )Jn ()
Jn ' ()Hn ()  Hn ' ( )Jn ()
Using H(1)=J+iY, we can write this as
n 
1
 J ' ( )Yn ( )  Yn ' ( )J n ( )
,a  n
1  ia
J n ' (  )J n ( )  Jn ' ( )Jn (  )
Scattering Coefficient
Differential intensity scattered
dI  dzR  dqr •

0
Integrating over q
2p
ExB
R
dz
* 
 k


n
in q
n
in q
1



dI  dzR  dq Re 
i

H

k
r

e
i

H
'

k
r

e




n
n
1
n
n
1
iw n
 n
 


0
2p
dI 2p k1 R
2
*


Im
H

k
R

H
'

k
R

 n
n 1
n
1
dz
w




Im Hn k1 RHn ' k1 R   Jn k1 R Yn ' k1 R  Yn k1 R J n ' k1 R  W(J n ,Yn ) 
*
dI
4

dz w

2
n
2
pk1 R
Scattering Coefficient - continued
The intensity of the incident plane wave is
dI0  2adz
k1
w
The scattering coefficient is obtained by dividing the total
Scattered light by the incident intensity on the cylinder:

2
2

(

,
k
a)
 n 1
k1a
Why you can believe this
Formal solution is the same as given in van de Hulst:
Light Scattering by Small Particles
Numerical solutions (using
Mathematica) are identical with
Van de Hulst’s

  1.25


  1.5

Calculate  as function of 
For large

diameter/wavelength
Scattering comes up
Quickly, oscillates around
Mean value
 = 10

For visible light, need to integrate
Over wavelength span,

Oscillations will average out
Unless the radius is very small
There will be no observable
Difference between =1.05
and 1.2
 = 100

Why is This?
 Cylinders do not collapse, in any limit, to
slabs
– All normally incident rays on slab will
propagate with some reduction in intensity
– Only one ray incident on a cylinder will do so,
all others are bent, thus scattered
– Even if -1 is small, the path length is many
wavelengths, so that the wave front is strongly
perturbed
Approximate Refraction Theory
 Since  is near unity and cylinder is very
large, can develop approximate theory
– Ray is not deflected much entering and leaving
– Phase change during passage is (-1)L/l
– Calculate scattered wave using
L
Huygen’s Principle
 Result agrees numerically
with Mie theory
 Criterion for “invisibility”: (-1)L/l<1
Effect of Salinity on Index of
Refraction
 For pure water, n = 1.33
 For FC, n=1.4, so =1.05, i.e. 5% difference
 Water n changes with wavelength,
temperature, and salinity
– Temperature effect very small
– Over visible range, n changes by about .7%
– Over salinity change of 0 - 40 ppt (fresh water
to very salty) n changes by about .5%
– At most  goes to 1.04, which is still too large
Conclusion
 Fluorocarbon does not live up to its
advertising hype - it is not “invisible”