# reperiendi

## 0. Prelude

1. Derive Snell’s law: $\displaystyle \frac{\sin \theta_w}{\sin \theta_a} = \frac{c_w}{c_a}$ where $c_w$ and $c_a$ are the speed of light in water and air, respectively. Time from $A$ to $P$ is $\displaystyle \frac{\sqrt{x_a^2 + y_a^2}}{c_a}$.

Time from $P$ to $W$ is $\displaystyle \frac{\sqrt{(\ell-x_a)^2 + (h-y_a)^2}}{c_w}$.

Total time is minimized when the derivative is zero: $\displaystyle t(x_a) = \frac{\sqrt{x_a^2 + y_a^2}}{c_a} + \frac{\sqrt{(\ell-x_a)^2 + (h-y_a)^2}}{c_w}$ $\displaystyle \begin{array}{rl} \frac{dt(x_a)}{dx_a} & = \frac{x_a}{c_a\sqrt{x_a^2 + y_a^2}} - \frac{\ell-x_a}{c_w\sqrt{(\ell-x_a)^2 + (h-y_a)^2}} \\ &= \frac{\sin \theta_a}{c_a} - \frac{\sin \theta_w}{c_w} \\ & = 0 \end{array}$

so $\displaystyle \frac{\sin \theta_w}{\sin \theta_a} = \frac{c_w}{c_a}$.

2. Suppose the ant is outside a hemispherical bowl and the drop of honey is inside the bowl directly across from her. Find the shortest distance.

Imagine the bowl was formed by inverting the top half of an inflatable beach ball into itself. Then if we evert it after adding the ant and honey, we find the ant and honey at opposite poles, and the distance is half the circumference, $d = \pi r$. Of course, in order for the question to have an answer as asked, the result needs to be independent of the height, which was not given. We can therefore just place both of them on the equator and get the same answer that way.

3. What happens if the ant can crawl faster on the outside of the glass than on the inside? Snell’s law applies directly: following the text, “unroll” the surface of the glass and then “unfold” it. The rim of the glass becomes the boundary between the two media, where the outside plays the role of air and the inside that of water.

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