### Snell’s law and geometry

Today’s post is somewhat inspired by this question on math.stackexchange. To begin with, recall Snell’s Law from geometric optics. It gives a rule for describing the propagation of light between two media of different indices of refraction: namely that across an interface where on one side the index of refraction is $n_1$ and on the other side $n_2$, the angles of incidence and refraction $\theta_1$ and $\theta_2$, as measured from the normal to the interface, satisfies the relation

$\displaystyle \frac{\sin \theta_1}{\sin \theta_2} = \frac{n_2}{n_1}$.

Here I pose two problems:

1. Given an observer situated above the interface, and a fish below the interface, find the “apparent position” of the fish according to the observer.
2. Given an observer situated above the interface, and the “apparent position” of a fish below the interface, find the actual position of the fish.

Without loss of generality we can assume that the interface is the $x$-axis, and the observer is situated at the point $(0,1)$ in the $x-y$ plane.

For the following we will denote by $r = n_2 / n_1$ the ratio of indices of refraction; and we will use the notation $S(a,b) = \sin a / \sin b$ and $T(a,b) = \tan a / \tan b$ as short-hands.

Question 1
Assume that the fish has coordinates $(F_x,F_y)$, where $F_x \geq 0$ and $F_y < 0$. Let the point $(X,0)$ be where the light-ray travelling from the fish to the observer intersection the interface. The angles of incidence and refraction obey the following relation

$\displaystyle X = \sqrt{1 + X^2} \sin \theta_1 \qquad F_x - X = \sqrt{ (F_x - X)^2 + F_y^2} \sin \theta_2$

applying Snell’s law we have

Eq (1)
$\displaystyle \frac{X^2}{1+X^2} = r^2 \frac{(F_x - X)^2}{(F_x-X)^2 + F_y^2}$.

The apparent position of the fish, however, is along the ray that connects the observer to the point $(X,0)$, but at distance

$\sqrt{1+X^2} + r\sqrt{(F_x-X)^2 + F_y^2}$

(using that $r$ is also the ratio of the velocities of light $v_1/v_2$ in the two media). So the apparent position of the fish will be at

$\displaystyle \left( X + r\sqrt{(F_x-X)^2 + F_y^2}\sin\theta_1, -r\sqrt{(F_x-X)^2 + F_y^2} \cos\theta_1 \right)$.

The $x$ coordinate can easily seen, by applying Snell’s law, to be $X + r^2(F_x - X)$. The $y$ coordinate, however, is

$\displaystyle \frac{\sin\theta_1\cos\theta_1}{\sin\theta_2\cos\theta_2} F_y = r^2 T(\theta_2,\theta_1) F_y$

and

$\displaystyle T(\theta_2,\theta_1) = -\frac{F_x - X}{X F_y}$

so we have that the apparent position of the fish will be

Eq (2)
$\displaystyle \left( F_x + (r^2-1)(F_x - X), - r^2 \frac{F_x - X}{X}\right)$.

Where we still need to solve (1), a quartic equation in $X$, before we can insert into (2). But in principle, the apparent position is an algebraic function of the real position and the ratio of refraction indices.

Notice that for the usual water-air interface, $r$ is bigger than 1, so the apparent position of the fish is horizontally displaced away from the observer.

Question 2
Suppose now the fish appears to be at $(A_x,A_y)$, with $A_x \geq 0$ and $A_y < 0$. Then the intersection point $(X,0)$ is simply given by

$\displaystyle X = \frac{A_x}{1 - A_y}$.

Reversing the computations above, we have that

$\displaystyle F_x = X+ \frac{1}{r^2}(A_x - X) = \frac{r^2 A_x - A_xA_y }{r^2 (1-A_y)}$.

Whereas

$\displaystyle F_y = \frac{A_y}{r}\sqrt{1 + \frac{A_x^2}{(1-A_y)^2}}$.

So somewhat interestingly: given the apparent position of the fish and the ratio $r$, the real position is a constructible number!