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What should I have told that student? The Fermat principle is interesting, but when in doubt use the basic laws of reflection and refraction. It is also good to realize the Fermat principle does not give one actual path which light ray will follow, it gives the possible paths. Which ones will be realized depends also on things like orientation of the ...


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Is this anything more than a mathematical curiosity? It is not a curiosity. The light travels all those paths. Does this come up as a problem/obstacle in higher physics? I'm not sure what the "this" is. Yes, people will sometimes wrongly think there is a unique solution when there isn't. There can be multiple geodesics between the same events ...


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Actually, the extra path is not irrelevant. If you put a light bulb at A and a $4\pi$ detector (this means $4\pi$ steradian coverage, i.e. it detects incoming light in any direction) at B, the detector will see light along both paths: direct, and bounced off the mirror, which is exactly the result you got from Fermat's Principle. If you want to exclude the ...


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Multiple classical solutions to Euler-Lagrange equations with pertinent/well-posed boundary conditions (such solutions are sometimes called instantons) are a common phenomenon in physics, cf. e.g. this related Phys.SE post and links therein. In optics, it is well-known that already e.g. two mirrors can create multiple classical paths.


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The reason we do this is that the Poynting vector $\vec{E}\times\vec{H}=\vec{E}\times\vec{B}/\mu$ is in the same direction as the wave vector $\vec{k}$ and shows the wave direction. So, to convert a forwards travelling plane wave to a backwards travelling version of itself, we need to change the sign of only one of the $\vec{E}$ or $\vec{B}$, If we changed ...


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I guess you wanted to say : $ q_w = h(T_\text{fluid} - T_\text{wall}) $ right? So this means that the heat flux density ($W.m^{-2}$) is proportional to the temperature difference between the fluid and the wall. It seems reasonable since when $T_\text{fluid} = T_\text{wall}$ , you don't have any heat flux anymore.


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Let wave function $\Psi$ be defined on domain $D \in \mathbb{R}^n$. The Neumann condition $\frac{\partial \Psi} {\partial {\bf n}} = 0$ on the boundary $\partial D$ has a simple interpretation in terms of the probability current of $\Psi$. For $\Delta \Psi = i \partial\Psi/\partial t$ (although it's usually taken as $i \partial\Psi/\partial t = - \Delta ...


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Suppose you want to analyze the stationary behavior of a particle in a potential well that is symmetric with respect to $x=0$ (picture below). In order to simplify your calculation, you could use as a boundary condition that $\frac{\partial \psi(x)}{\partial x}=0$ and solve the Schroedinger equation for x>0 only. This boundary condition then just reflects ...


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Once you admit the tunnelling the "topology" of the problem changes to "particle in a circle with an infinite barrier". In fact, to avoid arguments with energy when crossing the barrier, it can be described as "particle in a circle with an infinite barrier in one point". So physically it can be argued to be a different problem.


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Indeed the cosine function is valid for instance for other boundary conditions $$\psi'(0)=\psi'(L)=0$$ Your goal when you look for a set of solutions to the Schroedinger equation is to be able to decompose any general wavefunction as a sum over this set, and to do it consistently your boundary conditions must be the same for all solutions. So it is unlikely ...


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The particle in a box is subjected to the further restriction that the potential function rapidly tends to infinity at $ x=0 $ and $ x=L $ This means that the position of the particle is restricted to being within the box. As you said, the general solution to the schrodinger equation for particle in a box is $$ \psi (x)=A\cos kx + B\sin kx $$ Because the ...


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This problem is somewhat similar to the ammonia inversion. In that problem the probability densities $\psi_n(x)^2$ are similar for $n =1$ and $2$, for $n = 3$ and $4$ etc. As a result the Hamiltonian levels $E_1$ and $E_2$ are close together, as are $E_3$ and $E_4$, etcetera. See for example here and here. See also Energy levels for $NH_3$.



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