Geometric optics — rectilinear propagation, reflection, refraction: Visual walkthrough

Step 1 — What is a "ray", and what is "time along a path"?

WHAT. A ray is the thin straight arrow that shows which way light energy travels. We put down two points: a start point $A$ where the light leaves, and an end point $B$ where it must arrive. The light must get from $A$ to $B$ .

WHY. Before we can say "least time" we need to be able to measure the time of a trip. Time is just distance divided by speed. If a straight segment has length $ℓ$ and the light moves at speed $v$ , the time to cross it is

$t = \frac{ℓ}{v}$

$ℓ$ ::: the length of the straight piece of path (metres).
$v$ ::: the speed of light in that material (metres per second).
$t$ ::: the time the light spends on that piece (seconds).

PICTURE. One point $A$ , one point $B$ , one straight arrow between them, labelled with its length and the speed. That is the atom of everything that follows.

Step 2 — One medium: least time = straight line

WHAT. Suppose the whole space is one uniform material, so the speed $v$ is the same everywhere. Among all possible squiggly routes from $A$ to $B$ , which one takes least time?

WHY. Because $v$ is constant, $t = (total length) / v$ . Making $t$ small means making the length small. The shortest length between two points is the straight line — so light goes straight. That is rectilinear propagation, and we just derived it instead of assuming it.

PICTURE. Three candidate paths from $A$ to $B$ : two bent ones (longer, slower) and the straight one (shortest, fastest). The straight one wins.

Step 3 — Two mediums: set up the crossing

WHAT. Now split space into two halves by a horizontal boundary (the $x$ -axis). Above it the light moves at speed $v_{1}$ ; below it at a different speed $v_{2}$ . Point $A = (0, a)$ sits above; point $B = (d, - b)$ sits below. The light crosses the boundary at some point $P = (x, 0)$ that we get to choose.

WHY. This is the real question of refraction: the crossing point $P$ is the only freedom the light has. Move $P$ left or right and you trade a bit of the top journey for a bit of the bottom journey. We want the $P$ that makes the total time smallest.

$a$ ::: how high $A$ sits above the boundary (a fixed positive number).
$b$ ::: how deep $B$ sits below the boundary (a fixed positive number).
$d$ ::: the horizontal gap between $A$ and $B$ .
$x$ ::: the horizontal position of the crossing point $P$ — the one thing we vary.

PICTURE. The two-tone scene: top strip speed $v_{1}$ , bottom strip speed $v_{2}$ , the slanted top leg $A \to P$ , the slanted bottom leg $P \to B$ , and the movable dot $P$ sliding only along the allowed stretch $0 \leq x \leq d$ .

Step 4 — Write the total time as a formula

WHAT. Each leg is a straight line, so its length is the hypotenuse of a right triangle (Pythagoras). We ask why Pythagoras and not something else: because the legs are straight (Step 2), and the horizontal and vertical offsets of a straight segment are the two legs of a right triangle — its length is $(horizontal)^{2} + (vertical)^{2}$ .

Top leg $A \to P$ : horizontal offset $x$ , vertical offset $a$ , so length $x^{2} + a^{2}$ . Bottom leg $P \to B$ : horizontal offset $d - x$ , vertical offset $b$ , so length $(d - x)^{2} + b^{2}$ .

WHY. Divide each length by the speed of its medium to get the time on that leg, then add:

$T (x) = time in medium 1 \frac{x ^{2} + a ^{2}}{v _{1}} + time in medium 2 \frac{( d - x ) ^{2} + b ^{2}}{v _{2}} (0 \leq x \leq d)$

$x^{2} + a^{2}$ ::: length of the top leg (from $A$ down to $P$ ).
$(d - x)^{2} + b^{2}$ ::: length of the bottom leg (from $P$ down to $B$ ).
dividing by $v_{1}$ , $v_{2}$ ::: turns each length into a time, because $t = ℓ / v$ .
$T (x)$ ::: the total travel time as a function of where we put $P$ , valid on the allowed stretch $0 \leq x \leq d$ .

PICTURE. The two right triangles drawn explicitly, their legs labelled $x$ , $a$ and $d - x$ , $b$ , with the two hypotenuses highlighted.

Step 5 — "Least time" means the slope is zero

WHAT. $T (x)$ is a smooth curve: for small $x$ the top leg is short but the bottom leg is long, for large $x$ the reverse. Somewhere in between $T$ dips to a minimum. At the very bottom of a valley the curve is momentarily flat.

WHY this tool — the derivative. The derivative $\frac{d T}{d x}$ is a machine that reports the slope of $T$ at each $x$ : how fast the time changes if you nudge $P$ a tiny bit right. We use this tool and no other because the question "where is time smallest?" is exactly the question "where is the slope zero?" At the minimum, nudging $P$ either way doesn't change the time (to first order), so

$\frac{d T}{d x} = 0.$

PICTURE. The graph of $T$ against $x$ over the allowed stretch $0 \leq x \leq d$ (the two ends of the stretch are marked): a U-shaped valley, with a flat tangent line kissing its lowest point. Crucially the lowest point sits strictly between $x = 0$ and $x = d$ , so it is a genuine physical crossing, not an endpoint. To the left the slope is negative (downhill), to the right positive (uphill); it passes through zero exactly at the winning $P$ .

Step 6 — Do the differentiation

WHAT. Differentiate $T (x)$ term by term. The derivative of $x^{2} + a^{2}$ is $\frac{x}{x ^{2} + a ^{2}}$ (chain rule), and the derivative of $(d - x)^{2} + b^{2}$ is $\frac{- ( d - x )}{( d - x ) ^{2} + b ^{2}}$ (the extra minus comes from the $- x$ inside).

WHY. Setting the whole slope to zero:

$\frac{d T}{d x} = \frac{x}{v _{1} x ^{2} + a ^{2}} - \frac{d - x}{v _{2} ( d - x ) ^{2} + b ^{2}} = 0$

Move the second term across:

$\frac{x}{v _{1} x ^{2} + a ^{2}} = \frac{d - x}{v _{2} ( d - x ) ^{2} + b ^{2}}$

$\frac{x}{x ^{2} + a ^{2}}$ ::: horizontal offset ÷ hypotenuse of the top triangle.
$\frac{d - x}{( d - x ) ^{2} + b ^{2}}$ ::: horizontal offset ÷ hypotenuse of the bottom triangle.
$\frac{1}{v _{1}}$ , $\frac{1}{v _{2}}$ ::: the "time-cost per metre" of each medium — the factors that make this not a straight line.

PICTURE. The two triangles again, but now with the horizontal leg over the hypotenuse ratio highlighted on each — a big coloured wedge marking exactly the fractions in the equation.

Step 7 — Recognise the sines (the angles were hiding all along)

WHAT. Draw the normal — the vertical dashed line perpendicular to the boundary at $P$ . The angle between the top leg and the normal is $θ_{1}$ ; between the bottom leg and the normal is $θ_{2}$ .

WHY this tool — the sine. The sine of an angle in a right triangle is opposite over hypotenuse. Look at the top triangle: the side opposite $θ_{1}$ (the side across from that angle) is the horizontal offset $x$ , and the hypotenuse is $x^{2} + a^{2}$ . So

$sin θ_{1} = \frac{x}{x ^{2} + a ^{2}}, sin θ_{2} = \frac{d - x}{( d - x ) ^{2} + b ^{2}} .$

Those are exactly the two ugly fractions from Step 6! The sine is the perfect tool here because it is the one ratio that converts "how far $P$ slid sideways" into "how slanted the ray is from the normal." Substituting:

$\frac{sin θ _{1}}{v _{1}} = \frac{sin θ _{2}}{v _{2}}$

Finally use the definition of refractive index, $n = c / v$ . Here $c$ is the speed of light in vacuum — a fixed universal constant, about $3 \times 1 0^{8} m/s$ , the fastest speed anything can go. In any material light is slower, at speed $v < c$ , so $n = c / v$ is always $\geq 1$ ; the bigger $n$ is, the slower the light. From $n = c / v$ we get $\frac{1}{v} = \frac{n}{c}$ . Multiply both sides of the boxed relation by $c$ ; the $c$ cancels:

$n_{1} sin θ_{1} = n_{2} sin θ_{2}$

$θ_{1}$ ::: angle of the incoming ray from the normal, in medium 1.
$θ_{2}$ ::: angle of the outgoing ray from the normal, in medium 2.
$c$ ::: the speed of light in vacuum, a universal constant ( $\approx 3 \times 1 0^{8}$ m/s).
$n_{1}, n_{2}$ ::: refractive indices; $n = c / v$ , bigger $n$ = slower light.

PICTURE. The finished diagram: normal drawn, both angles marked at $P$ , the two sine-triangles shaded, and Snell's law written across the top.

Step 8 — Fast → slow: the ray bends TOWARD the normal

WHAT. Take the everyday case: light going from a faster medium into a slower one, i.e. $n_{1} < n_{2}$ (for example air → glass, or air → water). What does Snell's law predict for $θ_{2}$ ?

WHY. Rearrange $n_{1} sin θ_{1} = n_{2} sin θ_{2}$ to $sin θ_{2} = \frac{n _{1}}{n _{2}} sin θ_{1}$ . Since $n_{1} < n_{2}$ , the factor $\frac{n _{1}}{n _{2}} < 1$ , so $sin θ_{2} < sin θ_{1}$ , which means $θ_{2} < θ_{1}$ : the outgoing ray is less slanted — it bends toward the normal. This matches the marching-band picture: the row hits the slow medium and swings toward the perpendicular. It also matches least time — light "wants" fewer metres in the slow medium, so it dives closer to straight-down.

Worked number: air ( $n_{1} = 1$ ) → glass ( $n_{2} = 1.5$ ) at $θ_{1} = 30°$ . Then $sin θ_{2} = \frac{1}{1.5} sin 30° = \frac{0.5}{1.5} = 0.3333$ , giving $θ_{2} \approx 19.47°$ . Indeed $19.47° < 30°$ : bent toward the normal. ✔

PICTURE. The incoming ray at $θ_{1}$ in the fast (rarer) top medium and the refracted ray at the smaller $θ_{2}$ hugging the normal in the slow (denser) bottom medium.

Step 9 — The other cases (never leave a scenario unshown)

WHAT / WHY / PICTURE, all in one figure with three panels:

Panel A — same speed, $v_{1} = v_{2}$ (i.e. $n_{1} = n_{2}$ ). Snell's law gives $sin θ_{1} = sin θ_{2}$ , so $θ_{1} = θ_{2}$ : the ray goes straight through, no bend. This is Step 2 hiding inside the general law — a sanity check.
Panel B — head-on, $θ_{1} = 0$ . The ray arrives along the normal. Then $sin θ_{1} = 0 \Rightarrow sin θ_{2} = 0 \Rightarrow θ_{2} = 0$ : light entering perpendicular does not bend at all, whatever the two media. (It still slows down — speed changes, direction doesn't.)
Panel C — slow → fast ( $n_{1} > n_{2}$ ), the reverse of Step 8. Now $\frac{n _{1}}{n _{2}} > 1$ , so $sin θ_{2} = \frac{n _{1}}{n _{2}} sin θ_{1} > sin θ_{1}$ , giving $θ_{2} > θ_{1}$ : the ray bends away from the normal. Push $θ_{1}$ up until $θ_{2}$ would hit $90°$ — that is the critical angle $θ_{c}$ with $sin θ_{c} = n_{2} / n_{1}$ . For glass→air, $sin θ_{c} = 1/1.5 = 0.6667$ , so $θ_{c} \approx 41.81°$ . Beyond it, $sin θ_{2}$ would exceed 1, which is impossible, so the light cannot leave — it is totally internally reflected (this powers Total internal reflection and Optical fibres).

The one-picture summary

One figure compresses the whole journey: a valley-shaped time curve (with its minimum sitting between the endpoints $x = 0$ and $x = d$ ) whose lowest point selects the crossing point $P$ ; the bent ray at $P$ with its normal and two angles; and the boxed result $n_{1} sin θ_{1} = n_{2} sin θ_{2}$ tying them together — the geometry and the calculus in a single glance.

Recall Feynman: retell the whole walkthrough to a 12-year-old

Light is a runner who only cares about finishing fast, not about going straight. On one flat field (one medium) fastest is straight, so light goes straight — that's Step 1–2. Now put a swamp below the field: the runner sprints on grass ( $v_{1}$ ) and wades slowly in swamp ( $v_{2}$ ). If he aimed dead straight he'd wallow too long in the swamp; if he ran to the swamp's edge first he'd waste distance on grass. Somewhere between the start and the finish — never before one or past the other — is the sweet spot $P$ where the total time is smallest (Steps 3–4). To find it we ask "where does nudging $P$ stop helping?" — the flat bottom of the time-valley, slope zero, the derivative (Steps 5–6). Cleaning up the algebra, the two messy fractions turn out to be $sin θ_{1}$ and $sin θ_{2}$ — the slants from the up-down line — and dividing by the speeds (with $n = c / v$ , where $c$ is light's top speed in empty space) gives Snell's law (Step 7). The special cases: grass→swamp he angles toward straight-down (Step 8, bend toward normal); equal ground → straight; head-on → straight in; and sprinting out of the swamp too steeply → he can't escape and bounces back, total internal reflection (Step 9). Same runner, same rule — least time — every single time.

Connections

Fermat's principle — the least-time idea this whole page unpacks.
Total internal reflection, Optical fibres — Step 9, Panel C in action.
Wavefronts and Huygens' principle — the wave-level "why the wavefront pivots" behind the same bending.
Mirrors and Lenses, Dispersion — where Snell's law gets applied next.