Why Does Snell's Law use Sine Instead of Angle Values? [closed]

I recognize that Snell's Law relates indices of refraction and the angles of incidence and refraction, but why does the equation use sine to describe those angles and not simply angle values? It would seem that angle values would make the most sense since it directly refers to the size of the angle entering or exiting mediums. What is it about the sine function that makes it essential to the correctness of Snell's Law?

closed as unclear what you're asking by Jon Custer, Bill N, Carl Witthoft, AccidentalFourierTransform, ZeroTheHeroMay 21 '17 at 13:54

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• Simply because that's how it is. – Pritt Balagopal May 19 '17 at 5:18
• Read section 26.2 of feynmanlectures.caltech.edu/I_26.html – Farcher May 19 '17 at 5:22
• Snell's law proof using Principle of Least Time – Yashas May 19 '17 at 12:40
• @Yashas I suspect that's what Farchar's link is about :-) – Carl Witthoft May 19 '17 at 17:41
• I am really disappointed at this question being closed, especially closed (unclear). It's obviously comprehensible from the title alone and we have two good answers too it. It is obviously not an unclear question. – Joshua Jun 9 at 18:38

Aside from the answer, "that's just the way Nature is, we aren't free to choose laws to be the way we want them to be", the most succinct answer is that Snell's law expresses the continuity of transverse component of the wave vector $\vec{k}$ across an interface; as such, it involves vector components, and thus linear functions of direction cosines rather than of angles. This proves Snell's law for any wave phenomenon described by the Helmholtz equation. Alternatively, it also follows from the continuity of transverse components of the electromagnetic field vectors across an interface.

Look up the derivation of Snell's law in either Section 1.5 of Born and Wolf, "Principles of Optics" or look at my sketch of a proof in my answer here.

The wavefront concept of Huygen's principle can be used to derive Snell's law. Perhaps your question will be answered if you take a look at the derivation of Snell's law.

In the following figure, wavefront $AB$ carries the incident rays and wavefront $A'B'$ carries the refracted rays. Note that each wavefront is perpendicular to the rays it carries. The ray (wavelet) at the A end of wavefront AB arrives at the interface first. The wavefront at the $B$ end arrives at the interface Δt seconds later because of distance $BB'$. Of course, $BB' = v_1\Delta t$ where $v_1$ is the wave speed in medium 1.

Meanwhile, the ray that arrived at A first travels a distance AA' in medium 2 that is shorter than BB' of medium 1. Of course, $AA' = v_2\Delta t$ where $v_2$ is the wave speed in medium 2. The assumption is that the speed of light in medium 2 is less than that of medium 1.

For this reason, light has to bend and get closer to the normal line if it moves slower in medium 2. Triangles $ABB'$ and $AA'B'$ both share the same hypotenuse $AB'$. The angle opposite to BB' is the same size as $\theta_1$, the angle of incidence. The angle opposite to the shorter length $AA'$ is the same size as $\theta_2$, the angle of refraction.

Proof :

$\sin{\theta_1}=\frac{BB'}{AB'}$ $\tag1$

$\sin{\theta_2}=\frac{AA'}{AB'}$ $\tag2$

From the above relations,

$\frac{\sin{\theta_1}}{\sin{\theta_2}}=\frac{v_1\Delta t}{v_2 \Delta t}$

$\frac{\sin{\theta_1}}{\sin{\theta_2}}=\frac{v_1}{v_2 }$ $\tag3$

But as we know, $n=\frac{c}{v}$, equation $(3)$ becomes,

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

$n_1\sin{\theta_1}=n_2\sin{\theta_2}$

This is how the $sine$ function introduces itself in the Snell's law.

• That's possibly the most confusing refraction diagram ever. – user1717828 May 19 '17 at 16:45
• Hehe, I agree with you but it is very elaborate though :P – Mitchell May 19 '17 at 16:47
• The lines in the middle serve no expository purpose. – Emre May 19 '17 at 20:27