Snell's Law
I.N. Galidakis
following problem, which is related to Snell's Law:
Given two foci B and C, with |BC| = RL, determine the shape of the curve y(x)
which is defined as follows:
http://misc.virtualcomposer2000.com/Lens.gif
If the line normal to the curve at A = (x,y(x)) is AE then theta1 and theta2
always satisfy Snell's Law, with sin(theta1)/sin(theta2) = v1/v2, where v1 and
v2 are known speeds, as x varies in some interval 0 < R1 <= x <= R2 or as ABD
varies as 0 < ABD < some theta0.
I was able to get some relations from the above shape, but I am unable to
determine the curve shape. Please refer to the figure on the link, above. Angles
are refered to as ABC with the apex being the middle letter. The focus B lies at
the origin for simplicity, so C = (RL, 0):
arctan(dy/dx) = ADE =>
arctan(dy/dx) = BAD + ABD =>
arctan(dy/dx) = Pi/2 - theta1 + arctan(y/x) (1)
The dual relation for the right circle is:
Pi - (arctan(dy/dx) + Pi/2) = theta2 + ACE =>
Pi/2 - arctan(dy/dx) = theta2 + arctan(y/(RL - x)) (2)
The third equation is of course Snell's Law:
sin(theta1)/sin(theta2) = v1/v2 (3)
Solving (1) and (2) for theta1 and theta2 resp., we get:
theta1 = -arctan(diff(y(x),x))+1/2*Pi+arctan(y(x)/x) (4)
theta2 = 1/2*Pi-arctan(diff(y(x),x))+arctan(y(x)/(-3+x)) (5)
Substituting (4) and (5) into Snell's Law (3), after lots of simplifications
with Maple 9, I get the equation:
>
eq:=(9-6*x+x^2+y(x)^2)^(1/2)*(x+diff(y(x),x)*y(x))/(-3+x+diff(y(x),x)*y(x))/(x^2
+y(x)^2)^(1/2) = 5/4 (for v1=100, v2=80, RL=3).
Now Maple is unable to solve the resultant differential equation:
> dsol:=dsolve({eq4, y(RL/3)=0},y(x));
Fails.
The following fails too:
> dsol:=dsolve({eq4,
y(RL/3)=0},y(x),type=numeric,range=0..RL,output=listprocedure);
The numeric solution tells me that there are no solutions left of 1.
Can anyone see a way out to get the shape of this &^%^%$#* curve?
Many thanks,
--
Ioannis
I.N. Galidakis
Never mind. I think I got it. For those interested, the analysis seems to be
easier with the shape being vertical:
http://misc.virtualcomposer2000.com/Lens2.gif
We have for the red circle:
arctan(dy/dx) = CEG =>
arctan(dy/dx) = Pi/2 + CFB =>
arctan(dy/dx) = Pi/2 + FAC + ACF =>
arctan(dy/dx) = Pi/2 + arctan(x/(RL - y)) + Pi/2 - theta1 (1)
The dual for the lower blue circle is:
arctan(dy/dx) - Pi/2 = CIB =>
arctan(dy/dx) - Pi/2 = arctan(y/x) - theta2 (2)
Snell's Law:
sin(theta1)/sin(theta2) = v1/v2 (3)
WIth RL = 3, v1 = 100, v2 = 80, solving (1) and (2) for theta1 and theta 2 and
substituting into (3), I get (after expanding and simplifying):
eq4 :=
((x^2+y(x)^2))^(1/2)*(-3*diff(y(x),x)+diff(y(x),x)*y(x)+x)/(x+diff(y(x),x)*y(x))
/(9-6*y(x)+y(x)^2+x^2)^(1/2) = 5/4;
Now THAT can be solved numerically with Maple 9:
> dsol:=dsolve({eq4,y(0)=1/3*RL},
y(x),range=0..RL,type=numeric,output=listprocedure);
> y:=eval(y(x),dsol);
> with (plots):
> plot(y(x),x=-1..1);
and the resultant shape:
http://misc.virtualcomposer2000.com/LensShape.gif
Which looks like a hyperbola. It must be, because if the initial condition
changes to y(0)=2/3*RL, I get the hyperbola's dual wing, opening upwards.
To conclude: The shape of a lens which focuses waves emanating from the focus F1
onto focus F2 based on Snell's Law, appears to be a hyperbola. This means that
the lens works basically as a consequence of the hyperbola's "reflection
property":
http://en.wikipedia.org/wiki/Hyperbola
(Section "Geometrical Properties")
Thanks all,
--
Ioannis
rancid moth
>
>
> eq:=(9-6*x+x^2+y(x)^2)^(1/2)*(x+diff(y(x),x)*y(x))/(-3+x+diff(y(x),x)*y(x))/(x^2
> +y(x)^2)^(1/2) = 5/4 (for v1=100, v2=80, RL=3).
[cut]
your equation is
(9-6*x+x^2+y(x)^2)^(1/2)*(x+diff(y(x),x)*y(x))/(-3+x+diff(y(x),x)*y(x))/(x^2
+y(x)^2)^(1/2) = 5/4
the reduced characteristic equation is
-(x^2-y^2-3x)/3y +
{4x*sqrt(9-6x+x^2+y^2)+(15-5x)*sqrt(x^2+y^2)}/(y*(4*sqrt(9-6x+x^2+y^2)-5*sqrt(x^2+y^2)))
= 0
Solving this we get the following solutions
y(x) = +/-i*x, +/-i*(x-3), +/- 1/3*sqrt(-9x^2+150x-225)
Don't know if that helps any.
I.N. Galidakis
Thanks. The third solution looks like one of the solutions (hyperbola?) for the
horizontal case.
In any case, I dropped the horizontal case because the differential equation has
a singularity at the tip of the hyperbolic solution. It's easier doing it
vertically.
If you wish, apply your reasoning to the vertical case and let me know what you
get, so I can check it against the solution found.
Please see my previous post (not the one you quoted, the next)
--
Ioannis
rancid moth
okay. doing the same thing for that other equation i get these as
solutions
y(x) = +/- ix, 3 +/- ix, -16/3 +/- 1/3 * sqrt{(20 - 3x)*(3x + 20)}
I.N. Galidakis
Your solution looks the same to Maple's modulo a constant. Your y (call it
y1(x)) has:
y1(0)=4/3. If I ask Maple to solve the differential equation numerically and
specify the initial condition y(0)=4/3, here's what I get:
> dsol:=dsolve({eq4,y(0)=4/3},
y(x),range=0..1,type=numeric,output=listprocedure);
> y:=eval(y(x),dsol);
> plot(y(x),x=-1..1);
Result:
http://misc.virtualcomposer2000.com/Sol1.gif
your solution is:
> y1:=x->-16/3+1/3 *sqrt((20 - 3*x)*(3*x + 20));
> plot(y1(x),x=-1..1);
result:
http://misc.virtualcomposer2000.com/Sol2.gif
They look the same to me.
How did you manage to solve those differential equations?
Maple 9 could not find a closed-form solution. It was grinding for half an hour
before I stopped it and tried the option "type=numeric". Do you mind elaborating
on this issue of "equation of reduced coefficients" a bit?
Thanks,
--
Ioannis
krokodyle
"I.N. Galidakis" <morp...@olympus.mons> wrote
> Further consideration of a problem I posted about the Sound Lens, leads to
> the following problem, which is related to Snell's Law:
>
> Given two foci B and C, with |BC| = RL, determine the shape of the curve
> y(x) which is defined as follows:
>
> http://misc.virtualcomposer2000.com/Lens.gif
Apparently you want all the rays emanating from B to end up focused at C
If this is the case you have to define the indices of refraction of the B
medium and the C medium.
You want to derive the equation of the curve limiting the two mediums.
This is a classical optics problem first solved in the 17th century by
Descartes.
The curve limiting the two nediums is called a Cartesian oval.
It is a well known and well studied curve.
Please look up any good book on geometrical optics in a library or even on
the internet.
The curve in question is of the fourth degree.
There is no need to engage in complicated derivations and solving DEs. :-)
For some reason this problem enthralls beginners and it keeps being
reinvestigated all the time by people who think they have come up
with something new.
rancid moth
consider a general first order ODE of the form
y'(x) = w(x,y) [1]
assume that this solution has symmetries. that is if y=f(x) solves [1],
then (Y,f(X)) also solve [1] where Y=Y(x,y),X=(x,y) are transformations of
y and x. this criteria, the fact that if f solves [1], then (Y,f(X)),
solves Y'=w(X,Y), yields an infinitesimal criteria for a one parameter
group of infintesimal transformations that map solutions curves into
solution curves, or indeed leaves [1] invarient.
the condition for this is in part given by noting that if
dy/dx = w(x,y), then
dY/dX = w(X,Y) when dy/dx = w(x,y). therefore
(Y_x+ y'Y_y)/(X_x+y'X_y) = w(X,Y) when y'=w(x,y).
(where underscore = partial diff)
So
(Y_x + w*Y_y)/(X_x + w*X_y) = w(X,Y) [2]
Assume, for now, that any such transformation that maps solutions into
solutions can be parameterised so we have
(X,Y) = (X(x,y;t),Y(x,y;t))
where
(x,y) = (X(x,y;0),Y(x,y;0))
these set of points are coordinates on what is termed the orbit of the
group through (x,y). The action of this transformation to points on the
orbit maps them to other points on the same orbit. at least i think
that's what they're called - check this!
expand the X,Y in a taylor series
X = x+t*E(x,y)+O(t^2) [3]
Y = y+t*N(x,y)+O(t^2)
where
E = dX/dt | t=0 [4]
N = dY/dt | t=0
substituting these into [2], gives you
w+t*(N_x+(N_y-E_x)*w-E_y*w^2) +O(t^2) = w + t*(E*w_x+N*w_y) + O(t^2)
Equating terms yeilds
N_x+(N_y-E_x)*w-E_y*w^2 = E*w_x+N*w_y [5]
which is a linear PDE involving two dependent variables. solving this
gives you E and N. the transformations then come from solving [3] with
[4].
one must make an ansatz on the form for E and N in order to solve [5].
usually a n-degree polynomial in y is enough (depending on w)...though
more complicated forms may be required.
I don't have maple, but i guess that if it is using symmetry methods for
solving ODE, then it would have algorithms that work their way through
various ansatz for first order ODE. There are plenty of programs out there
that can derive (E,N) for you (i use BigLIE by Alan Head...it's
free)...and the reason that most of them usually obtain solutions is that
[5] is linear and overdetermined. continuing...
We now have an expansion of an infinitesimal transformation that tells us
how to turn x,y into X,Y such that dY/dX = w(X,Y) when y'=w, i.e. turns
solutions into solutions. Solution points that are mapped to themselves by
every symmetry transformation admitted by the ODE are called invariant.
from [3] we see that the point (x,y) is only invariant if E(x,y)=N(x,y)=0.
Note also that the vector (E,N), given what E and N are, is the tangent
vector to the orbit at the point (X,Y).
Invariant points are then points where the tangent vector is zero.
Invariant curves are then curves that have their tangent parallel to
the tangent vector of the orbit. So take an arbitrary curve y=y(x), the
tangent to this curve at (x,y(x)) is the direction (1,y'), which is only
an invariant if it is parallel to (E,N) i.e. if N-y'E = 0 on the curve. If
it is invariant, then it is also a solution to [1]. The reason is
that the E,N are derived from this very requirement...i.e. they are
defined as the tangent vectors of transformations that map solutions to
solutions. N-y'E = 0 is known as the characteristic. substituting for y'
from [1] gives the reduced characteristic
N(x,y)-w(x,y)E(x,y) = 0 [6]
which is satisfied if y(x) is a solution of [1] (by the construction of E
&N).
Such solutions are known as "invariant solutions". if N/E = w
identically, then the reduced characteristic is trivialy zero and the
symmetry involved is a trivial symmetry and can be of no
use in determining invariant solutions.
Therefore if you find (E,N), and plug them into (6), solving the
algebraic equation that results will give you an invariant solution to [1].
That's all i did.
cheers
moth
I.N. Galidakis
> "I.N. Galidakis" <morp...@olympus.mons> wrote
>
>> Further consideration of a problem I posted about the Sound Lens, leads to
>> the following problem, which is related to Snell's Law:
>>
>> Given two foci B and C, with |BC| = RL, determine the shape of the curve
>> y(x) which is defined as follows:
>>
>> http://misc.virtualcomposer2000.com/Lens.gif
>
> Apparently you want all the rays emanating from B to end up focused at C
That's right. But before we proceed, please take a look at this thread on
sci.optics:
> If this is the case you have to define the indices of refraction of the B
> medium and the C medium.
I know the speeds of the explosives v1 and v2 in the two different mediums,
which is equivalent.
> You want to derive the equation of the curve limiting the two mediums.
That's right.
> This is a classical optics problem first solved in the 17th century by
> Descartes.
>
> The curve limiting the two nediums is called a Cartesian oval.
>
> It is a well known and well studied curve.
>
> Please look up any good book on geometrical optics in a library or even on
> the internet.
The Cartesian Oval was the first solution I got, using totally elementary
algebra. It was also suggested by Ken Pledger in another sci.math thread. The
problem is that the plane lens .pdf reference (for pattent US4729318) shown in
the sci.optics thread, specifies a hyperbola. So I am not sure the Cartesian
Oval works here, but for what it's worth, here is the derivation of the oval:
Let's assume that the detonator is located at (0,RL) and we want to focus the
waves at the pit, at (0,0). Then, if the speeds in the two mediums are v1 and f2
resp., a parametrization with respect to time of the emanating waves at the
detonator, is:
x(t)^2 + (y(t) - R_L)^2 = (v1*t)^2 (1)
The waves must transition from (1) to:
x(t)^2 +y(t)^2 = (R_2 - v1*(t-t_0))^2 (2)
Solving (1) and (2) for y(t) (and taking reverse sqrts to preserve orientation),
we get:
y_1(x,t) = RL - sqrt((v1*t)^2 - x(t)^2), for 0<=t<=t0 (3)
y_2(x,t) = sqrt((RL - v1*t)^2 - x(t)^2) for t0<t<=tc, (4)
where t0 = R1/vf, the subradius of the fast explosive, before it reaches the
slow Baratol explosive transition region, and R1 + R2 = RL.
It follows immediately that the transition curve has to satisfy the following
functional equation:
h(x,t)=y_2(x,t) - y_1(x,t) (5)
modulo a constant Rh, which is the height at which the Baratol explosive will be
placed. A quick calculation shows that:
Rh=R2*cos(theta1)-h(R2*sin(theta1),R1/v1) (6)
Hence, the shape of the delimiting surface between the fast and slow explosive
is:
H(x,t0) = h(x,t0) + R_h (7)
Now, if we set:
x = r*cos(theta)
y = r*sin(theta) (8)
and substitute (8) into (7) for RL=3, R1=1/3*RL, R2=2/3*RL, v1=100 (m/s, say)
we get:
> eq:=h(x,R1/v1)=y; #omitting the factor Rh for simplicity)
> eq:=subs({x=r*cos(theta),y=r*sin(theta)},eq);
> eq := (4-r^2*cos(theta)^2)^(1/2)-3+(1-r^2*cos(theta)^2)^(1/2) = r*sin(theta)
(9)
Solving the above for r, and graphing parametrically, we get:
> sol:=solve(eq,r):
>
plot(sol[2],theta=-2*Pi/3..2*Pi/3,coords=polar,scaling=CONSTRAINED,numpoints=100
0);
Result:
http://misc.virtualcomposer2000.com/CartesianOval.gif
which is, indeed, a Cartesian Oval plus something else, we don't care about.
In view of the US patent mentioned in sci.optics, which is a hyperbola, I am not
sure the Cartesian Oval is the right solution, also in view of the fact that the
transition curve seems independent of v2. Hence I set forth to derive an
alternate curve based on Snell's Law, which should hold for explosive waves as
well.
> The curve in question is of the fourth degree.
Solving equation (9) for r, indeed requires the solution of a fourth degree
polynomial. One solution gives the oval, the others, the rest of the shape.
> There is no need to engage in complicated derivations and solving DEs. :-)
You think, eh?
> For some reason this problem enthralls beginners and it keeps being
> reinvestigated all the time by people who think they have come up
> with something new.
I am not sure what gave you the idea that I am a "beginner", but whatever it is,
you can now take it out of your mind and store it safely in your cupboard.
--
Ioannis
I.N. Galidakis
Funny and powerful method. I don't recall any such method in my books. I am
surprised Maple 9 does not use it.
In any case, many thanks. The only thing I need now, is to find someone who
knows *for sure* whether the correct solution to my problem is a hyperbola or
that Cartesian Oval thing :-)
> cheers
> moth
--
Ioannis
Peter Webb
"I.N. Galidakis" <morp...@olympus.mons> wrote in message
news:1259503299.610738@athprx04...
No, it is refraction, not reflection.
And it is a parabola, certainly looks like one, and should be, unless you
used a relative index of refraction less than 1.
Peter Webb
> In any case, many thanks. The only thing I need now, is to find someone
> who
> knows *for sure* whether the correct solution to my problem is a hyperbola
> or
> that Cartesian Oval thing :-)
If you assume that Snell's Law applies to sound in the same manner as it
does to light (as seems likely), the shapes of the lenses will be identical
for sound and light.
The shape depends only upon 3 things: the distance from object to lens, lens
to focus, and the relative refractive index of the two materials, which is
about 1.4 for glass. It is very, very uncommon to see relative refractive
indexes less that 1, as that means the lens is made out of material less
dense than the sorrounding medium - which is usually air. While
theoretically possible, I have never seen such a lens.
For relative refractive indexes greater than 1, the answer given by Snell's
law is a ellipse rotated about its major axis. The further away the subject
is from the lens, the closer this approximates a parabola, and in practice
all single element lenses are shaped as parabolas as this allows them to
focus at infinity.
A lens with the unlikely characteristic of using a material with lower
refractive index than the surrounding medium (air, in almost all cases)
would be a rotated hyperbola, but that is of more mathematical interest than
of practical use.
>
>> cheers
>> moth
> --
> Ioannis
>