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Math Help - Path of the light

  1. #1
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    Path of the light

    If there is a density function or refractive index function N(y)(continous) , can i find out the path of the light just solving Euler's equation
     F_{y} - \frac{d}{dx}( F_{y'}) = 0 ?

    where  \int_0^a F(y,y') dx is defined as the total time for the light travelling from O to (a , y(a))



     v(y) = \frac{ v_0 N_0 }{N(y)} = \frac{ds}{dt} = \frac{\sqrt{1+(y')^2} dx}{dt}



     Total~Time = \int_0^{t(a)} dt = \int_0^a \frac{ \sqrt{1+(y')^2} N(y) }{ N_0 v_0} dx

     F = \frac{ \sqrt{1+(y')^2} N(y) }{ N_0}{v_0}

    Since F is independent of x the equation leads to another equation :
     F - y' F_{y'} = C_1

    then

     \frac{N(y)}{\sqrt{1+(y')^2}} = C_1 v_0 N_0
    Last edited by simplependulum; July 28th 2009 at 03:53 AM.
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  2. #2
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    Quote Originally Posted by simplependulum View Post
    If there is a density function or refractive index function N(y)(continous) , can i find out the path of the light just solving Euler's equation
     F_{y} - \frac{d}{dx}( F_{y'}) = 0 ?

    where  \int_0^a F(y,y') dx is defined as the total time for the light travelling from O to (a , y(a))



     v(y) = \frac{ v_0 N_0 }{N(y)} = \frac{ds}{dt} = \frac{\sqrt{1+(y')^2} dx}{dt}



     Total~Time = \int_0^{t(a)} dt = \int_0^a \frac{ \sqrt{1+(y')^2} N(y) }{ N_0 v_0} dx

     F = \frac{ \sqrt{1+(y')^2} N(y) }{ N_0}{v_0}

    Since F is independent of x the equation leads to another equation :
     F - y' F_{y'} = C_1

    then

     \frac{N(y)}{\sqrt{1+(y')^2}} = C_1 v_0 N_0

    Oh , I remember that if the refractive index only varies with y ,

    there is a relation :

     N_0 \sin{\theta_0 } =  N_1 \sin{\theta_1 } = ... =  N(y) \sin{\theta_y } =  ~ constant

    therefore  \frac{N(y)}{\sqrt{1+(y')^2}} = C , It is very easy to get it !

    But if N is a fucntion of  x and  y ,
    i think using this calculus of variations is better and i want to ask one more question : How can we generalize it in polar coordinates ???
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