Thin film interference — reflected and transmitted: Visual walkthrough

Before line one, three words in plain language:

Everything below is about two rays born from one, and whether they come back in step.

Step 1 — One ray splits into two

WHAT. A single ray of light in air strikes the top of a transparent film. Part of it reflects straight back off the top surface (call it Ray 1, blue). The rest enters the film, travels down, reflects off the bottom, and climbs back out (call it Ray 2, orange).

WHY. These two rays started as one wave, so they are coherent — locked in a fixed rhythm relative to each other. That is the only reason a stable bright/dark pattern can exist. Two unrelated flashlights never do this.

PICTURE. Follow the split. Ray 2 clearly travels farther — it makes a round trip through the film that Ray 1 never takes. That extra distance is the seed of everything.

Step 2 — Measure the slant path inside the film

WHAT. Ray 2 goes down at angle $θ_{r}$ and comes back up. Each leg is a slanted line of length $t / cos θ_{r}$ , so both legs together are $2 t / cos θ_{r}$ .

WHY this length? Look at the right triangle the downward leg forms: the film thickness $t$ is the side along the normal (the "adjacent" side to $θ_{r}$ ), and the slant leg is the hypotenuse.

Two legs (down + up) ⇒ total slant $= \frac{2 t}{cos θ _{r}}$ .

PICTURE. The two shaded triangles below are mirror images. The vertical side is always $t$ ; the diagonal is the leg we just measured.

Step 3 — Compare fairly: drop a common wavefront

WHAT. A wavefront is a line perpendicular to the rays where every wave has the same phase — a "starting line" for a fair race. We drop the wavefront from the point where Ray 2 exits and mark where it crosses Ray 1. Only the path up to that line counts as "extra."

WHY. Phase is about how many wavelengths separate the two waves. If we measured Ray 2's full slant but let Ray 1 run free, we'd be timing runners from different starting lines — meaningless. The wavefront is the shared finish line.

PICTURE. The green segment is the piece of Ray 1's path (and the piece of Ray 2's slant) that gets subtracted off. What survives is the projection of the round trip onto the normal.

Recall The one-line algebra that makes it cosine

Question: why does $\frac{2 t}{c o s θ _{r}} - 2 t sin θ_{r} tan θ_{r}$ equal $2 t cos θ_{r}$ ? ::: Put over a common denominator: $\frac{2 t - 2 t s i n ^{2} θ _{r}}{c o s θ _{r}} = \frac{2 t ( 1 - s i n ^{2} θ _{r} )}{c o s θ _{r}} = \frac{2 t c o s ^{2} θ _{r}}{c o s θ _{r}} = 2 t cos θ_{r}$ , using $1 - sin^{2} θ_{r} = cos^{2} θ_{r}$ .

Step 4 — Turn distance into "optical distance"

WHAT. Multiply the geometric extra path by $n$ : $Δ_{opt} = 2 n t cos θ_{r} .$

WHY this tool — why multiply by $n$ , not add or square it? Inside the film light is slowed, so its wavelength shrinks to $λ_{film} = λ / n$ . What matters for interference is how many wavelengths fit into the path, and shorter wavelengths mean more of them fit. Counting wavelengths: $\frac{geometric path}{λ _{film}} = \frac{2 t c o s θ _{r}}{λ / n} = \frac{n \cdot 2 t c o s θ _{r}}{λ} .$ The factor $n$ appears naturally — it converts a real distance into an "as-if-in-vacuum" distance so we can keep using the ordinary air wavelength $λ$ on the other side.

PICTURE. Same geometric path, but the wave crammed into the film wiggles faster (shorter $λ$ ), so it accumulates more phase per millimetre. The $n$ counts that.

Step 5 — The half-wave hiccup on reflection

WHAT. Reflection itself can flip a wave upside down, adding half a wavelength ( $λ /2$ , a phase of $π$ ) — but only sometimes.

WHY / the rule. Bouncing off a denser medium (going from low $n$ to high $n$ , "rarer → denser") flips the wave. Bouncing off a rarer medium (high $n$ → low $n$ ) does not flip it. (This is Phase change on reflection.)

PICTURE. Ray 1 hits air→film (rarer→denser) ⇒ flip, drawn as a peak turning into a trough. Ray 2's bottom bounce depends on what's below the film — the picture shows both possibilities.

Step 6 — Case A: air–film–air (soap bubble), ONE flip

WHAT. Air on top, film in the middle, air below (a free soap bubble).

Top bounce: air→film = rarer→denser ⇒ flip.
Bottom bounce: film→air = denser→rarer ⇒ no flip.
Net = one flip ⇒ keep $+ \frac{λ}{2}$ .

WHY it matters. One flip swaps the usual bright/dark rules, so a formula memorised for "two flips" gives the exact opposite here.

PICTURE. The flip tags on each surface; only the top one fires.

Step 7 — Case B: anti-reflection coating, TWO flips

WHAT. Air ( $1.0$ ) → coating ( $n_{c} = 1.38$ ) → glass ( $1.52$ ). Now both bounces are rarer→denser ( $1 < 1.38$ and $1.38 < 1.52$ ), so two flips.

WHY. Two flips = a full $λ$ shift = no relative shift at all. The $λ /2$ term vanishes, and the standard rules return — flipped back relative to the soap bubble. This is the whole idea behind Anti-reflection coatings.

PICTURE. Both surfaces show a flip; together they cancel.

Step 8 — Degenerate case: the vanishing film ( $t \to 0$ )

WHAT. Let the soap bubble get so thin it is about to pop: $t \to 0$ , so $2 n t cos θ_{r} \to 0$ .

WHY it is the deepest test. With zero geometric path, naive thinking says "no path difference ⇒ everything in step ⇒ bright." But in the one-flip soap case the total is $Δ = 0 + \frac{λ}{2} = \frac{λ}{2}$ — exactly out of step ⇒ dark. This is why the top of a dying bubble goes black for every colour at once: the flip alone kills the reflection.

PICTURE. Ray 2's round trip collapses to nothing, yet the single flip still holds the two waves half a step apart.

Step 9 — Worked numbers to lock it in

The one-picture summary

Everything above, on one canvas: the split, the slant, the wavefront cut, the $n$ -stretch, the flip count, and the final boxed rule.

$Δ = 2 n t cos θ_{r} + (odd flips ? \frac{λ}{2} : 0)$

Recall Feynman retelling of the whole walkthrough

One beam of light hits a soap wall and splits into two runners. Runner 1 bounces off the front. Runner 2 dives in, sprints to the back wall, bounces, and climbs back out — so Runner 2 ran extra. To measure that extra fairly we draw a finish line (the wavefront) square across both runners; the extra distance that actually counts turns out to be $2 t cos θ_{r}$ , not the longer diagonal, because part of the dive happens before the finish line. Inside the film the runner takes shorter strides ( $λ / n$ ), so we multiply the distance by $n$ to count strides properly — that's $2 n t cos θ_{r}$ . Then the twist: bouncing off harder stuff (denser medium) gives a runner a half-stride hiccup. Count the hiccups. One (soap bubble) ⇒ add half a stride and the bright/dark rules swap. Two (coating on glass) ⇒ the hiccups cancel and the plain rules return. And if the wall is paper-thin, the extra distance is zero but the lone hiccup remains — so the runners meet exactly out of step and the spot goes black. That single hiccup is the whole game.

Recall Quick self-test

Why is the extra path a cosine, not a secant? ::: Because the two rays are compared at a shared wavefront; subtracting the pre-wavefront slice turns $2 t / cos θ_{r}$ into $2 t cos θ_{r}$ . Why multiply by $n$ ? ::: Inside the film the wavelength shrinks to $λ / n$ , so more wavelengths fit; the $n$ converts geometric path into optical path so we can keep the air $λ$ . A soap film about to pop looks black in reflection — why? ::: $t \to 0$ gives geometric path $0$ , but the single reflection flip leaves $λ /2$ ⇒ perfect cancellation.

Prerequisites revisited: Interference of light · Young's double slit · Refractive index · Snell's law · Phase change on reflection · Conservation of energy.