LocusBlog

Snell's law, and the line you measure the angle from

Snell’s law relates the direction of a light ray on the two sides of a boundary between transparent materials:

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

$n_1$ and $n_2$ are the refractive indices of the two materials, $\theta_1$ is the angle of the incoming ray and $\theta_2$ the angle of the outgoing one. Both angles are measured from the normal, the line perpendicular to the surface, not from the surface itself. That convention causes more wrong answers than everything else in geometric optics put together, so it is worth stating before anything else: a ray that grazes along the glass has $\theta$ near $90^\circ$, and a ray that hits head-on has $\theta = 0$.

What the index actually is

The refractive index of a material is the ratio of the speed of light in vacuum to the speed of light in that material:

$$n = \frac{c}{v}$$

Glass with $n = 1.50$ carries light at $v = 3.00\times10^{8}/1.50 = 2.00\times10^{8}$ m/s. Since nothing carries light faster than $c$, every $n$ is at least 1, and a larger $n$ means a slower medium.

Slower, but not lower in pitch. When light crosses into a medium, its frequency does not change. The incoming wave drives the charges in the material at its own frequency and they radiate at that frequency; cycles cannot pile up or go missing at the boundary, because the fields on the two sides have to match at every instant. What gives instead is the wavelength:

$$\lambda = \frac{v}{f} = \frac{c}{nf} = \frac{\lambda_0}{n}$$

Take 600 nm red light in vacuum. Its frequency is $f = 3.00\times10^{8}/600\times10^{-9} = 5.00\times10^{14}$ Hz. Inside the glass the wavelength is $600/1.50 = 400$ nm and the speed is $2.00\times10^{8}$ m/s, so the frequency is $2.00\times10^{8}/400\times10^{-9} = 5.00\times10^{14}$ Hz. The same number, as required. Speed and wavelength drop by the factor $n$ together, which is exactly why their ratio survives.

Snell’s law comes out of that wavelength change. Picture a straight wavefront crossing the boundary at an angle, and follow it for a time $t$. The part still in medium 1 advances $v_1t$ while the part already in medium 2 advances $v_2t$, and both ends stay attached to the same stretch $L$ of the interface. Geometry gives $\sin\theta_1 = v_1t/L$ and $\sin\theta_2 = v_2t/L$, so:

$$\frac{\sin\theta_1}{\sin\theta_2} = \frac{v_1}{v_2} = \frac{n_2}{n_1}$$

Rearranged, that is Snell’s law. The bending is not a rule imposed on light; it is what a wavefront must do when one end of it is moving slower than the other.

A ray into glass, checked backwards

Send a ray from air onto a flat glass surface at $\theta_1 = 30^\circ$ from the normal, with $n_1 = 1.00$ and $n_2 = 1.50$. Since $\sin 30^\circ = 0.5$ exactly:

$$\sin\theta_2 = \frac{n_1\sin\theta_1}{n_2} = \frac{1.00 \times 0.5}{1.50} = \frac{1}{3}$$

$$\theta_2 = \arcsin(0.3333) = 19.47^\circ$$

The ray bent toward the normal, from $30^\circ$ to $19.47^\circ$, which is what entering a slower medium always does.

Check it by running the light backwards. A ray inside the glass heading out at $19.47^\circ$ from the normal gives $1.50 \times \sin 19.47^\circ = 1.50 \times 0.3333 = 0.5$, so $\sin\theta_2 = 0.5$ in the air and $\theta_2 = 30^\circ$. The original direction comes back. Snell’s law is symmetric in the two media, so a path that works one way works the other, and a worked refraction that fails to reverse has an arithmetic error in it somewhere.

When the equation has no solution

Run it the other way, glass to air, and the ray bends away from the normal. Push the incidence angle up and $\theta_2$ reaches $90^\circ$ before $\theta_1$ does. Set $\theta_2 = 90^\circ$, where $\sin\theta_2 = 1$, and Snell’s law gives the incidence angle at which that happens:

$$n_1\sin\theta_c = n_2 \quad\Longrightarrow\quad \sin\theta_c = \frac{n_2}{n_1}$$

This is the critical angle, and the ratio only makes sense when $n_1 > n_2$, so the effect exists only going from the denser medium to the thinner one. For glass to air:

$$\sin\theta_c = \frac{1.00}{1.50} = \frac{2}{3}, \qquad \theta_c = 41.81^\circ$$

Beyond that, the equation stops having an answer. A ray hitting the inside of the glass at $45^\circ$ would need $\sin\theta_2 = 1.50 \times \sin 45^\circ = 1.061$, and no angle has a sine above 1. Nothing is refracted; the whole beam reflects back into the glass, and this is total internal reflection. The absence of a solution is not a failure of the formula, it is the formula reporting that the refracted ray does not exist.

For water, with $n = 1.33$, the critical angle is $\arcsin(1/1.33) = 48.75^\circ$, which is why a swimmer looking up sees the whole sky compressed into a bright cone and mirrored water outside it. An optical fibre is the same physics used on purpose: keep every ray inside the core arriving beyond $\theta_c$ and none of the light escapes through the wall.

Measuring from the wrong line

Take the $30^\circ$ refraction again, but suppose the angle was read off from the glass surface rather than from the normal. The true angle of incidence is then $60^\circ$, not $30^\circ$, and:

$$\sin\theta_2 = \frac{\sin 60^\circ}{1.50} = \frac{0.8660}{1.50} = 0.5774, \qquad \theta_2 = 35.26^\circ$$

The refracted ray is $35.26^\circ$ from the normal, not $19.47^\circ$: an error of nearly 16 degrees from one convention slip, with no arithmetic mistake anywhere in the calculation.

The mix-up is self-detecting if you look for it. Suppose you measure from the surface, feed the number in as though it were from the normal, and then read the answer back out as an angle from the surface. Your $19.47^\circ$ result is $70.53^\circ$ from the normal, so the ray has bent away from the normal on its way into glass. Light entering a slower medium never does that. Any time a refraction into a denser material comes out with $\theta_2 > \theta_1$, the angles were taken from the wrong line. The reflection and Snell’s law skill varies which surface the ray meets so that the normal is somewhere different every time, which is the point.

One more thing the convention hides: a critical angle is a fair distance from grazing. $41.81^\circ$ from the normal is $48.19^\circ$ from the surface, and quoting the second number as though it were the first puts the whole total internal reflection condition on the wrong side of the boundary. The simulations browser draws the normal for you and traces the refracted ray, so you can drag the incidence angle up and watch the refracted ray flatten out and vanish exactly at $\theta_c$.

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