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Home Unit 13 · Geometric Optics 13.1·13.2·13.3·13.4 Lesson
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Slower means bent toward the normal

Refraction is a change of speed showing up as a change of direction. The index of refraction is exactly that: $n = c/v$, so a larger $n$ means slower light. Snell's law, $n_1\sin\theta_1 = n_2\sin\theta_2$, then does the geometry, with one condition on how it is written: the index and the angle multiplied together must belong to the same medium.

§1

Tag each medium once, and keep the tag.

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

Crossing the pairs, writing $n_1\sin\theta_2 = n_2\sin\theta_1$, produces a refracted ray bending the wrong way, and the number it gives usually looks plausible, so nothing in the arithmetic catches it.

Light going from air into glass at $40^\circ$:

  1. Medium 1 is air: $n_1 = 1.0$, $\theta_1 = 40^\circ$.
  2. Medium 2 is glass: $n_2 = 1.5$, $\theta_2$ unknown.
  3. $(1.0)\sin 40^\circ = (1.5)\sin\theta_2$, so $\theta_2 \approx 25^\circ$.

Then check against the picture: entering the slower medium, the ray has to bend toward the normal and end at the smaller angle. It does. The crossed version gives about $74^\circ$, which fails that check immediately, so the picture is the audit.

§2

Which way it bends, and the case that does not bend at all.

One sentence covers it: higher index, slower light, smaller angle to the normal.

  1. Into a slower medium (higher $n$): bends toward the normal.
  2. Into a faster medium (lower $n$): bends away from the normal.
  3. Arriving along the normal ($\theta_1 = 0$): goes straight through, no bend at all.

Air into water at $30^\circ$ refracts to about $22^\circ$; reverse the trip and the angle grows back. The third case is the one that gets missed: at normal incidence there is nothing to bend, because the ray already lies along the normal, and the speed still changes even though the direction does not.

Reasoning that slowing down ought to spread a ray outward inverts the rule. The picture that gets it right is a car driving at an angle onto sand: the wheel that reaches the sand first slows first, and the car swings toward the perpendicular.

§3

Frequency is fixed by the source. Wavelength absorbs the change.

At a boundary, the surface passes along exactly as many cycles per second as it receives. So frequency does not change. What changes is the speed, and the wavelength follows:

$$v = \frac{c}{n}, \qquad \lambda = \frac{\lambda_0}{n}, \qquad f \ \text{unchanged}, \qquad v = f\lambda.$$

Light entering glass with $n = 1.5$ slows by a factor of $1.5$ and its wavelength shrinks by $1.5$, while the frequency sits exactly where the source put it.

Dragging the frequency down with the speed would recolour everything seen underwater, and it does not: an object keeps its colour through water. That observation is why colour is tied to frequency rather than to wavelength, and it is the reason the constancy of $f$ is worth stating as its own rule rather than being derived each time.

§4

Total internal reflection runs one way only, and dispersion runs through n.

Total internal reflection needs light travelling from the higher index medium toward the lower one, so that the refracted ray bends away from the normal and can be pushed all the way to $90^\circ$:

$$\sin\theta_c = \frac{n_2}{n_1}, \qquad n_1 > n_2.$$

Glass into air gives $\sin\theta_c = 1.0/1.5$, about $42^\circ$; past that, nothing is transmitted. Going the other way, from air into glass, the refracted angle is always the smaller one and never runs out of room, so no critical angle exists. Substituting anyway gives a sine above one, which is the equation saying exactly that rather than a number to round off.

Dispersion has a similar discipline: route it through the index. Violet meets a slightly larger $n$ in glass than red does, so violet travels slower there and Snell's law turns it through a larger angle at each surface. "Violet has more energy so it bends more" predicts the right order in glass by accident and then fails on a different medium, on a second surface, or on the question of why a vacuum disperses nothing. It also explains why a slab with parallel faces sends the colours back out parallel while a prism, whose faces are not parallel, keeps them spread.

§5

Skill Check.

Ten scenarios. Pick the chips that match your answer, then check. A scenario marks complete the first time every part is right. Progress saves on this device.

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