Compute the beam deflection $\delta$ (see figure) through a prism with angle $\epsilon$ as in the figure and refractive index $n_p$. When is the deflection $\delta$ minimal compared to the original irradiation direction.
We had the pleasure to visit a lab especially set up for new students, and some of us were allowed to take a test. Not really a test, since we could take it home with us at the end and have to give it back for grading during the next week. Anyway, I was stuck on this one and had no idea how to solve it. And since we were allowed to visit the lab at any time during the weekend I took up the opportunity to play around with a glass prism. I experimented with it some bit and noticed that when I managed to let light go through the prism in such a way that $\alpha_i=\alpha_t$ the deflection $\delta$ was smaller than all the other deflection I measured.
So I assumed that $\alpha_i$ must equal to $\alpha_t$ for the desired solution, but I don't know how to prove that.
Anyone got an idea how to do that?
This is a great problem, in that an explicit calculation would be somewhat tedious but the result can be explained without any calculation, just from symmetry. The required principle is that light trajectories are reversible; that is, if a light beam incident on the left under $\def\a#1{\alpha_{\text{#1}}}\a i$ exits on the right under $\a t$, then a light beam incident on the right under $\a t$ would exit on the left under $\a i$.
Now consider the situation in which you found $\delta$ to be minimal. If $\a i=\a t$, by the above reversibility principle, the entire trajectory must be symmetric, i.e. the path inside the prism must be horizontal (more precisely, perpendicular to the bisector of $\epsilon$, but that seems to be aligned with the vertical here). Now imagine changing $\a i$ and consider how $\a t$ changes. If it changes faster than $\a i$, reverse the roles (by virtue of the reversibility principle). In the reverse situation, changing $\a t$ would have to change $\a i$ slower than $\a t$ changes. But that contradicts the symmetry of the situation. By the same reasoning, $\a t$ also can't vary slower than $\a i$.
Thus, the geometrical symmetry and the abstract symmetry of the reversibility principle together require that $\a i$ and $\a t$ must vary at the same rate at this point. But if they vary at the same rate, the rate of change of their difference is zero. That is, the derivative of their difference with respect to either of them is zero. And since that difference differs from $\delta$ only by an added constant angle, the derivative of $\delta$ with respect to either of the angles is also zero at this point. And that derivative being zero is the necessary condition for $\delta$ to have an extremum.
Now, that doesn't quite show that $\delta$ has a minimum there – it has a stationary point, but that stationary point could be a minimum, a maximum or a saddle point. However, we know that as $\a i$ or $\a t$ approaches $90^\circ$, it changes very rapidly with respect to the other one; so we know that at the boundary $\delta$ decreases in the direction towards the symmetric situation. Thus there must be at least one minimum in between. Theoretically there could be two, with a maximum in the symmetric situation between them, but a) you measured one minimum and no maxima and b) intuitively it doesn't feel as if the deflection would have several extrema in this simple situation. If there is just one extremum, then it must be a minimum, and from the above it must be in the middle, in the symmetric situation.