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Light ray equation: derivation?

  1. Aug 15, 2006 #1
    For a course in Experimental Aerdynamics I have to study a section on Optical Measurement Techniques. My oral exam is coming up and there is this little thing I don't understand. It annoys me expremely. It is the ray equation. It is about light ray path in a medium with changing refractive index. I will copy what's in the lecture notes, and maybe someone will know the derivation of this formulae.

    [tex]\vec{e}_r[/tex]: vector towards centre of curvature
    [tex]\vec{e}_s[/tex]: vector in path direction

    Deflection (1):
    [tex] n \frac{\partial \vec{e}_s}{\partial s} = \frac{\partial n}{\partial r} \vec{e}_r [/tex]

    Acceleration (2):
    [tex]\frac{\partial n}{\partial s} \vec{e}_s [/tex]

    Combining (adding (2) to both sides of (1))

    [tex]\frac {\partial (n \cdot \vec{e}_s)}{\partial s}=\frac{\partial}{\partial s} (n \cdot \frac{\partial \vec{x}}{\partial s})=\nabla n [/tex]

    with [tex]\vec{x}=[x,y,z]^T[/tex]

    Can somebody help me out by giving the derivation or a link to a page that contains it?

    Kind regards,

    Maciej

    PS: So you can't let the tex automatically span more lines?
     
    Last edited: Aug 15, 2006
  2. jcsd
  3. Aug 16, 2006 #2
    Does somebody know anything more about this set of equations?
     
  4. Aug 16, 2006 #3

    Physics Monkey

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    Hi mvvdsteen,

    You can easily obtain this equation from the standard variational principle for light. This principle, called Fermat's principle, states that light will take the path which extremizes the time of flight.

    Ok, so how does it work out? The time of flight is given by the integral [tex] T = \int \frac{ds}{v} ,[/tex] where [tex] ds [/tex] is the infinitesimal arc length and [tex] v [/tex] is the speed of light. You know that the speed of light in a medium of refractive index [tex] n [/tex] is just [tex] v = c/n [/tex], so you rewrite the time of flight as [tex] T = \frac{1}{c} \int n(\vec{x}(s)) ds, [/tex] where I have indicated the dependence on arc length [tex] s [/tex]. Following the usual variational approach, we try to find which path extremizes this quantity. Consider varying the path of the light ray from [tex] \vec{x}(s) [/tex] to [tex] \vec{x}(s) + \delta \vec{x}(s) [/tex] with end points fixed. The first order variation in the time of flight is given by [tex] \delta T = \frac{1}{c} \int \[ \delta n ds + n \delta ds \], [/tex] and note that it is important to remember that the infinitesimal arc length changes too! We can easily calculate the first term as [tex] \delta n = \nabla n \cdot \delta \vec{x}. [/tex] The variation of arc length is also easy to calculate: [tex] \delta ds = \sqrt{(d\vec{x} + d \delta \vec{x})^2} - \sqrt{(d\vec{x})^2} = ds \frac{d\vec{x}}{ds} \cdot \frac{d \delta \vec{x}}{ds}, [/tex] where I have kept only first order terms in [tex] \delta \vec{x} [/tex]. We can now write the variation of T as [tex] \delta T = \frac{1}{c} \int [ \nabla n \cdot \delta \vec{x} + n \frac{d\vec{x}}{ds} \cdot \frac{d \delta \vec{x}}{ds} ] ds = \frac{1}{c} \int [ \nabla n - \frac{d}{ds}\left(n \frac{d\vec{x}}{ds}\right) ]\cdot \delta \vec{x} ds. [/tex] The last equality there follows from an integration by parts and the fact that [tex] \delta \vec{x} [/tex] vanishes at the end points. The extremal path has [tex] \delta T = 0 [/tex] for arbitrary variations [tex] \delta \vec{x} [/tex]. This implies that the extremal path satisfies [tex] \nabla n - \frac{d}{ds}\left(n \frac{d\vec{x}}{ds}\right) = 0, [/tex] which is the desired equation.

    If you haven't seen it before, the variational approach is extremely powerful and elegant. I like it a lot. Hope this helps.
     
    Last edited: Aug 16, 2006
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