# Thread: Path of the light

1. ## 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

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

2. Originally Posted by simplependulum
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 ???