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

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Homework Help Overview

The problem involves atmospheric refraction and its effect on the visibility of a duck in a desert, with a refractive index given as a function of height. The observer is positioned 26.2 km away from the duck, and the task includes determining the apparent position of the duck and graphing the ray path.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply the eikonal equation and manipulate it to derive a relationship between y and x. There are questions regarding the correctness of their approach and how to express y as a function of x.
  • Some participants suggest using hyperbolic trigonometric substitutions and question the nature of the integral results, noting the logarithmic form.
  • There is a discussion about the difficulty of inverting the integral results and the implications of using different mathematical techniques.

Discussion Status

Contextual Notes

Participants note the complexity of the problem due to the refractive index's dependence on height and the implications for the ray path. There is an acknowledgment of the original poster's potential language barrier in understanding certain mathematical concepts.

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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!
 
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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 a lot, anyway.
 

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