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whenever two media and two velocities are involved, one must follow Snell's law if one wants to take the shortest time.

Why snells law must be followed to travel diffrent media in shortest time? Does this mean that the path followed by light in travelling through different media will give a straight line if different media are transformed into like first media? I think so because one would get shortest time if one travel in a straight line.

So how does this apply to particles? Does this statement points wave particle duality?

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  • $\begingroup$ Can you please elaborate on this ? Does this mean that the path followed by light in travelling through different media will give a straight line if different media are transformed into like first media? $\endgroup$ – SamD97 May 21 '16 at 12:29
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    $\begingroup$ I think the OP means that if you have two different media, the shortest travel time path between two points for light is not necessarily a straight line. But if you change one of the media so it has the same index of refraction as the first medium, then the shortest travel time path between the two points is a straight line. And the answer is yes, that is the case, as you can see if you try shining a laser between two points through a homogeneous medium. I think the first question is a request for a derivation of Snell's law starting from a travel time minimization requirement. $\endgroup$ – Brionius May 21 '16 at 12:38
  • $\begingroup$ I am pretty sure wave particle duality is not involved in any way, it's the principle of least action instead. $\endgroup$ – user108787 May 21 '16 at 12:41
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    $\begingroup$ Related: physics.stackexchange.com/q/127037/2451 and links therein. $\endgroup$ – Qmechanic May 21 '16 at 12:59
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    $\begingroup$ To emphasize that there is nothing intrinsic to light in Snell's Law, I ask mechanics students to solve for the optimal path of a lifeguard running and swimming to rescue a swimmer who is both down the beach and out to sea if the lifeguard moves faster on land than on water. That is satisfy Fermat's principle for a human locomotion scenario. The answer (under an all or nothing assumption) is that given by Snell's Law. $\endgroup$ – dmckee May 21 '16 at 16:00
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(the following answer is included essentially in "The Feynman LECTURES ON PHYSICS-Mechanics, Radiation & Heat ,Vol. 1, 26-3 Fermat's principle of least time.)

Suppose you are at point A in the land and a screaming girl is at point B in the sea. You can run with a speed $\:v_{1}\:$ on the land greater than the speed $\:v_{2}\:$ you can swim in the sea. At a moment you decide to follow the path ACB spending time $\: t_{1}=AC/v_{1}\:$ running on the land and time $\: t_{2}=CB/ v_{2}\:$ swimming in the sea, that is total time

\begin{equation} t_{tot}=t_{1}+t_{2}=\dfrac{AC}{v_{1}}+\dfrac{CB}{v_{2}} \tag{01} \end{equation}

But after a while you change your mind and decide to displace the point C on shoreline a little to the right to the point D. But then you are wondering if by such a displacement you shorten the total time or not.
For infinitesimally small displacement $\:CD\equiv \Delta x\:$ you can do the following approximations :

\begin{equation} AE \approx AC \qquad \theta_{1}^{\prime} \approx \theta_{1} \qquad BZ \approx BD \qquad \theta_{2}^{\prime} \approx \theta_{2} \tag{02} \end{equation}

You realize that on one hand you decrease the swimming distance by $$ BC-BD \approx CZ=\Delta x \cdot \sin\theta_{2} \tag{03} $$ so changing (decreasing) the swimming time by $$\Delta t_{2}=t_{2}^{\prime}-t_{2}=-\Delta x \cdot \dfrac{\sin\theta_{2}}{v_{2}} \tag{04} $$ On the other hand you increase the running distance by $$ AD-AC \approx DE=\Delta x \cdot \sin\theta_{1} \tag{05} $$ so changing (increasing) the running time by $$\Delta t_{1}=t_{1}^{\prime}-t_{1}=+\Delta x \cdot \dfrac{\sin\theta_{1}}{v_{1}} \tag{06} $$

So, balancing, the total time change is $$ \Delta t_{tot}=t_{tot}^{\prime}-t_{tot}=\left(t_{2}^{\prime}+t_{1}^{\prime} \right)-\left(t_{2}+t_{1}\right)=\Delta t_{2}+\Delta t_{1}= \Delta x \cdot\left(\dfrac{\sin\theta_{1}}{v_{1}}- \dfrac{\sin\theta_{2}}{v_{2}}\right) \tag{07} $$ This means that if $$ \left(\dfrac{\sin\theta_{1}}{v_{1}}- \dfrac{\sin\theta_{2}}{v_{2}}\right) > 0 \tag{08} $$ then moving infinitesimally to the right, $\: \Delta x >0 \:$, we increase the time while moving to the left, $\: \Delta x <0 \:$, we decrease the time. So in case that the condition (08) is valid then in order to find a shorter time we must search to the left of point C.

If $$ \left(\dfrac{\sin\theta_{1}}{v_{1}}- \dfrac{\sin\theta_{2}}{v_{2}}\right) < 0 \tag{09} $$ then moving infinitesimally to the right, $\: \Delta x >0 \:$, we decrease the time while moving to the left, $\: \Delta x <0 \:$, we increase the time. So in case that the condition (09) is valid then in order to find a shorter time we must search to the right of point C.
But finally, if $$ \left(\dfrac{\sin\theta_{1}}{v_{1}}- \dfrac{\sin\theta_{2}}{v_{2}}\right) = 0 \tag{10} $$

then either moving to the right,$\: \Delta x >0 \:$, or moving to the left , $\: \Delta x <0 \:$, the change is infinitesimally zero. This is the definition of the extreme points of a function. So, condition (10) is the one of the shortest time and if you are a light ray then in terms of refraction indices

$$ v_{1}=c_{1}=\dfrac{c_{0}}{n_{1}}, \quad v_{2}=c_{2}=\dfrac{c_{0}}{n_{2}} \tag{11} $$ and (10) is Snell's Law

$$ \bbox[#FFFF88,12px]{n_{1}\sin\theta_{1}=n_{2}\sin\theta_{2} \qquad \textbf{(Snell's Law)}} \tag{12} $$


EDIT enter image description here

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So how does this apply to particles? Does this statement [point to] wave particle duality?

No, the fact of Snell's law cannot be thought of as being any evidence for the wave particle duality of light, even though Frobenius' Answer / Feynman's derivation superficially considers a particle's path.

This is because rays of light can equally well be interpreted wholly in wave terms, namely, as the unit normals to the phase fronts of waves. Whenever solutions of the D'Alembert / Helmholtz wave equation fulfill a slowly varying envelope approximation, the Eikonal equation follows and Snell's law for ray normals at interfaces is the inescapable conclusion of the Eikonal equation. In turn, all these concepts are equivalent to Fermat's "least time" principle.

The slowly varying envelope approximation is basically that, over regions of less than several wavelength's diameter, the wave can locally be thought of as a plane wave with a well defined phase front, i.e. that the solution $\psi(\mathbf{r})$ of the Helmoltz equation as a function of position $\mathbf{r}$ is of the form $\psi(\mathbf{r}) = \Psi(\mathbf{r})\,\exp(i\,\varphi(\mathbf{r}))$, where the amplitude $\Psi(\mathbf{r})$ is real-valued and varies significantly only over regions much larger than a wavelength. Over regions of a few wavelengths the phase is well approximated by $\varphi(\mathbf{r})\approx\mathbf{k}\cdot\mathbf{r}$.

A ray is then the integral curve of the vector field $\nabla\varphi$, and the more slowly the amplitude $\Psi(\mathbf{r})$ varies in comparison to a wavelength, the more accurately the Eikonal equation and Snell's law hold.

I show how to derive the Eikonal equation, Fermat's principle and the slowly varying envelope approximation from each other in this answer here and the answers that one links to.

But there is also an experimental answer that can be given to your question. Waves in wave tanks moving across interfaces between regions of different, constant depth can be experimentally shown to fulfill Snell's law. The Eikonal equation and Snell's law are also widely, successfully used in seismology and other, wholly wave-governed fields of physics.

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