How Does Atmospheric Refraction Affect the Visibility of a Duck in a Desert?

  • Thread starter Thread starter avisha03
  • Start date Start date
avisha03
Messages
6
Reaction score
0

Homework Statement



A duck is walking in a flat desert (yes, the proff. is creative...). According to atmospheric conditions the Refractive index is n(y)=1.0006-0.0003y. y in KM.
An observer is on the ground, 26.2 KM far. Where will he see the duck? Show the gragh of the ray path.


Homework Equations


d(ne)/ds= grad n
e- the ray path direction

The Attempt at a Solution


I used the eikonal equation, and some matematical manipulations to get the following equation:
dy/dx= sqrt((a-by)^2/c^2-1)
(a=1.0006, b=0.0003, c=1.0006*cos(theta_0))
In order to try and find y as a function of x, I tried to solve the latter eq. by seperating it, and using the variable replacment: m=a-by
I got:
ln(m+sqrt(m^2-c^2))=exp(-bx/c)
now I don't know wheather my way till now is correct, and how to display y as y(x).
Thanks for your help!
 
Physics news on Phys.org
Welcome to PF!

Hi avisha03! Welcome to PF! :smile:

(have a square-root: √and a theta: θ and try using the X2 tag just above the Reply box :wink:)
avisha03 said:
… I used the eikonal equation, and some matematical manipulations to get the following equation:
dy/dx= sqrt((a-by)^2/c^2-1)
(a=1.0006, b=0.0003, c=1.0006*cos(theta_0))
In order to try and find y as a function of x, I tried to solve the latter eq. by seperating it, and using the variable replacment: m=a-by
I got:
ln(m+sqrt(m^2-c^2))=exp(-bx/c)
now I don't know wheather my way till now is correct, and how to display y as y(x).
Thanks for your help!

You mean dy/dx= √((a-by)/c)2 - 1) ?

(I haven't checked it, but assuming that it's correct …)

Easier to put n = (a-by)/c, dn = -bdy/c

dn/√(n2 - 1) = -(b/c)dx

and then use a hyperbolic trig substitution. :wink:
 
Thanks for your help & advices. :)
What do you mean by hyperbolic trig substitution?
The integral result is logarithmic, isn't it?
 
avisha03 said:
Thanks for your help & advices. :)
What do you mean by hyperbolic trig substitution?

Hi avisha03! :smile:

Haven't you come across cosh and sinh? if not, see http://en.wikipedia.org/wiki/Sinh#Standard_Integrals"

(See also standard integrals in the PF Library)

There are tables in which you can look up, for example, cosh(x) and cosh-1(x).
The integral result is logarithmic, isn't it?

Yes, but, as you've noticed, it's difficult to invert in that form (ie ln(x + √(x2 + a2))) :wink:
 
Last edited by a moderator:
I do know cosh & sinh, but since I'm not a native english speaker- I wanted to make sure. Thanks alot, anyway.
 
To solve this, I first used the units to work out that a= m* a/m, i.e. t=z/λ. This would allow you to determine the time duration within an interval section by section and then add this to the previous ones to obtain the age of the respective layer. However, this would require a constant thickness per year for each interval. However, since this is most likely not the case, my next consideration was that the age must be the integral of a 1/λ(z) function, which I cannot model.
Back
Top