Ray Optics: Snell's Law

Refraction is the bending of a light wave as it crosses the boundary between two different mediums. This is caused by the wave changing speed as it crosses into a different medium. The angle at which the light wave refracts away from the normal in the second medium can be found using Snell's Law, which states:

$n_{1}\sin(\theta_1) = n_{2}\sin(\theta_2)$

$n_1$ is the index of refraction of the incident medium and $n_2$ is the index of refraction of the refractive medium. Likewise, $\theta_1$ is the angle of incidence and $\theta_2$ is the angle of refraction.

Using simple logic, if $n_1$ is bigger than $n_2$ (for example, air and water have indices of refraction of 1 and 1.33), then $\theta_1$ has to be bigger than $\theta_2$. This means that $\theta_2$ will seemingly bend towards the normal (perpendicular axis to the border between the two mediums).

Snell's law (also known as the Snell–Descartes law and the law of refraction) is a formula used to describe the relationship between the angles of incidence and refraction, when referring to light or other waves passing through a boundary between two different isotropic media, such as water, glass, or air.

This is a derivation of Snell's Law. We will derive the law using Fermat's Principle, which states that, when light travels from one point to another, it takes the path that requires the least amount of time.

Suppose that we have a linear, horizontal boundary between two media (the fact that it's horizontal doesn't matter; we are just assigning an orientation for convenience of terminology). The first medium has refractive index $n_1$ and the second medium has refractive index $n_2.$ Point $A$ lies in the first medium at a distance $a$ from the boundary, and point $B$ lies in the second medium at a distance $b$ from the boundary. The horizontal separation between $A$ and $B$ is $d.$

Suppose that the light travels from $A$ to $B.$ It will travel in a straight line from $A$ to a point $X$ on the boundary and then travel in a straight line from $X$ to $B.$ The horizontal separation between $A$ and $X$ is $x,$ so the horizontal separation between $X$ and $B$ is $d-x.$ Let $\theta_1$ be the angle between segment $AX$ and the normal to the boundary at $X,$ and let $\theta_2$ be the angle between segment $BX$ and the normal to the boundary at $X.$

We want to find the point $X$ that minimizes the travel time $T.$ The Pythagorean Theorem allows us to easily calculate $T$: $$T = \frac{\sqrt{a^2+x^2}}{c/n_1} + \frac{\sqrt{b^2+(d-x)^2}}{c/n_2}$$ $$T = \frac{1}{c}\left(n_1\sqrt{a^2+x^2} + n_2\sqrt{b^2+(d-x)^2}\right)$$

For convenience, we will mininize $f(x) = cT.$ Computing the derivative, $$f'(x) = \frac{n_1x}{\sqrt{a^2+x^2}} - \frac{n_2(d-x)}{\sqrt{b^2+(d-x)^2}}$$ $$f'(x) = n_1\sin\theta_1 - n_2\sin\theta_2$$ Setting the derivative equal to $0,$ we get: $$n_1\sin\theta_1 = n_2\sin\theta_2.$$ To confirm that this is indeed a minimum, we can see that decreasing $x$ decreases $\sin\theta_1$ and increases $\sin\theta_2,$ and increasing $x$ does the opposite. Thus, $f'(x) < 0$ when $x$ is less than the desired value and $f'(x) > 0$ when $x$ is more than the desired value.

It follows that light takes the minimum time to travel from $A$ to $B$ if and only if: $$n_1\sin\theta_1 = n_2\sin\theta_2.$$ Thus, by Fermat's Principle, light will travel in a path that satisfies this equation. Thsi is called Snell's Law.